Current Electricity 1

Chapter
19
Current Electricity
Electric Current (5) For a given conductor current does not change with
(1) The time rate of flow of charge through any cross- change in cross-sectional area. In the following figure i1 = i2 = i3
ΔQ dQ
section is called current. i  Lim  . If flow is uniform
Δt 0 Δt dt i1 i2 i3
Q
then i  . Current is a scalar quantity. It's S.I. unit is ampere
t Fig. 19.2

(A) and C.G.S. unit is emu and is called biot (Bi), or ab ampere.

1A = (1/10) Bi (ab amp.) (6) Current due to translatory motion of charge : If n particle

(2) Ampere of current means the flow of 6.25  1018 each having a charge q, pass + +

+ +
electrons/sec through any cross-section of the conductor. nq
through a given area in time t then i  + +
t
(3) The conventional direction of current is taken to be the Fig. 19.3

direction of flow of positive charge, i.e. field and is opposite to

If n particles each having a charge q pass per second per
the direction of flow of negative charge as shown below.
i i unit area, the current associated with cross-sectional area A is

  i  nqA
E E
Fig. 19.1
If there are n particle per unit volume each having a charge

q and moving with velocity v, the current thorough, cross section

(4) The net charge in a current carrying conductor is zero. A is i  nqvA

Table : 19.1 Types of current

Alternating current (ac) Direct current (dc)

(i) (i) (Pulsating dc) (Constant dc)

i
i i
+

– t
t t
 2 Current Electricity
Magnitude and direction

both varies with time

ac  Rectifier  dc dc  Inverter  ac

(ii) Shows heating effect only (ii) Shows heating effect, chemical effect and magnetic effect of

current

(iii) It’s symbol is (iii) It’s symbol is + –

~

(7) Current due to rotatory motion of charge : If a point (1) Current density at point P is given by J 
di
n
dA
charge q is moving in a circle of radius r with speed v (frequency ˆ
dA
dA 
, angular speed  and time period T) then corresponding 
r q i P J i
J
q qv qω n
current i  q ν   
T 2 πr 2 π dA cos
Fig. 19.4 Fig. 19.5

(2) If the cross-sectional area is not normal to the current,

(8) Current carriers : The charged particles whose flow in a but makes an angle  with the direction of current then

definite direction constitutes the electric current are called di

J  di  JdA cos  J .dA  i   J  dA
dA cos
current carriers. In different situation current carriers are
(3) If current density J is uniform for a normal cross-
different.
i
section A then J 
(i) Solids : In solid conductors like metals current carriers A

are free electrons. (4) Current density J is a vector quantity. It's direction is
same as that of E . It's S.I. unit is amp/m2 and dimension [L–2A].
(ii) Liquids : In liquids current carriers are positive and
(5) In case of uniform flow of charge through a cross-
negative ions.
i
section normal to it as i  nqvA  J   nqv .
(iii) Gases : In gases current carriers are positive ions and A

free electrons. (6) Current density relates with electric field as

E
(iv) Semi conductor : In semi conductors current carriers J  E  ; where  = conductivity and  = resistivity or

are holes and free electrons. specific resistance of substance.

Current Density (J )
Drift Velocity
Current density at any point inside a conductor is defined
Drift velocity is the average uniform velocity acquired by
as a vector having magnitude equal to current per unit area free electrons inside a metal by the application of an electric
surrounding that point. Remember area is normal to the field which is responsible for current through it. Drift velocity is
very small it is of the order of 10–4 m/s as compared to thermal
direction of charge flow (or current passes) through that point. l
speed (~– 10 5 m / s) of electrons at room temperature.
A
vd
E

+ –
V
Fig. 19.6
 Current Electricity 1039

metallic lattice is defined as relaxation time

mean free path 
  . With rise in temperature
r.m.s. velocityof electrons v rms
vrms increases consequently  decreases.
(2) Mobility : Drift velocity per unit electric field is called
vd m2
mobility of electron i.e.   . It’s unit is .
E volt  sec
If suppose for a conductor Ohm's Law
n = Number of electron per unit volume of the conductor If the physical conditions of the conductor (length,
A = Area of cross-section temperature, mechanical strain etc.) remains some, then the
V = potential difference across the conductor current flowing through the conductor is directly proportional to
E = electric field inside the conductor
the potential difference across it’s two ends i.e. i  V 
i = current, J = current density,  = specific resistance,  =
V  iR where R is a proportionality constant, known as electric
 1
conductivity     then current relates with drift velocity as resistance.
 
i  neAv d we can also write (1) Ohm’s law is not a universal law, the substances, which
i J E E V
vd      . obey ohm’s law are known as ohmic substance.
neA ne ne ne  l n e
(1) The direction of drift velocity for electron in a metal is (2) Graph between V and i for a metallic conductor is a

opposite to that of applied electric field (i.e. current density J ). straight line as shown. At different temperatures V-i curves are
v d  E i.e., greater the electric field, larger will be the drift different.
V V
velocity. T1
1
(2) When a steady current flows through a conductor of T2
2
non-uniform cross-section drift velocity varies inversely with
 1 2
 1
area of cross-section  v d   vd 2 i i
 A (A) Slope of the line
i (B) Here tan1 > tan2
vd1 A1 < A2
V
so vd1  vd 2 = tan    R So R1 > R2
i i
A1 Fig. 19.9 i.e. T1 > T2
A2

Fig. 19.7

(3) If diameter (d) of a conductor is doubled, then drift

velocity of electrons inside it will not change. +
+ V – V – (3) The device or substances which don’t obey ohm’s law

e.g. gases, crystal rectifiers, thermoionic valve, transistors etc.

Less – d More – d
Same – vd Some – vd are known as non-ohmic or non-linear conductors. For these V-i
Crystal i
Fig. 19.8 rectifier
curve is not linear.

V 1
Static resistance R st  
i tan   
(1) Relaxation time () : The time interval between two
V
successive collisions of electrons with the positive ions in the Fig. 19.10
 1040 Current Electricity
V 1 gives R2 = R1 [1 +  (t2 – t1)]. This formula gives an approximate
Dynamic resistance R dyn  
I tan 
value.
Resistance Table 19.2 : Variation of resistance of some electrical material
(1) The property of substance by virtue of which it with temperature
opposes the flow of current through it, is known as the
Material Temp. coefficient of Variation of resistance
resistance.
resistance () with temperature rise
(2) Formula of resistance : For a conductor if l = length
of a conductor A = Area of cross-section of conductor, n = Metals Positive Increases
No. of free electrons per unit volume in conductor,  =
Solid non-metal Zero Independent
relaxation time then resistance of conductor
l m l Semi-conductor Negative Decreases
R  . ; where  = resistivity of the material of
A ne  A
2

conductor Electrolyte Negative Decreases

(3) Unit and dimension : It’s S.I. unit is Volt/Amp. or Ohm (). Ionised gases Negative Decreases
8
1volt 10 emu of potential
Also 1 ohm   = 109 emu of Alloys Small positive value Almost constant
1 Amp 10 1 emu of current
resistance. It’s dimension is [ML2 T 3 A 2 ] .
Resistivity (), Conductivity () and Conductance (C)
(4) Dependence of resistance : Resistance of a conductor
depends upon the following factors. (1) Resistivity : From R  
l
; If l = 1m, A = 1 m2 then
A
(i) Length of the conductor : Resistance of a conductor is
R   i.e. resistivity is numerically equal to the resistance of a
directly proportional to it’s length i.e. R  l and inversely
substance having unit area of cross-section and unit length.
1
proportional to it’s area of cross-section i.e. R 
A (i) Unit and dimension : It’s S.I. unit is ohm  m and

(ii) Temperature : For a conductor dimension is [ML3 T 3 A 2 ]

Resistance  temperatur e . m
(ii) It’s formula :  
ne 2
If R0 = resistance of conductor at 0oC
(iii) Resistivity is the intrinsic property of the substance. It
Rt = resistance of conductor at toC
is independent of shape and size of the body (i.e. l and A).
and ,  = temperature co-efficient of resistance
(iv) For different substances their resistivity is also
then Rt  R0 (1   t   t 2 ) for t > 300oC and
different e.g. silver = minimum = 1.6  10–8 -m and fused
Rt  R0
Rt  R0 (1  t ) for t  300oC or  
R0  t quartz = maximum  1016 -m

If R1 and R2 are the resistances at t1oC and t2oC insulator   alloy   semi - conductor   conductor
1   t1
(Maximum for fused quartz) (Minimum for silver )
R
respectively then 1  .
R2 1   t2
(v) Resistivity depends on the temperature. For metals
The value of  is different at different temperature. t  0 (1  t) i.e. resitivity increases with temperature.
Temperature coefficient of resistance
averaged over the
R 2  R1 (vi) Resistivity increases with impurity and mechanical
temperature range t1oC to t2oC is given by   which
R1 (t 2  t1 ) stress.
 Current Electricity 1041

(vii) Magnetic field increases the resistivity of all metals Ratio of resistances before and after stretching
except iron, cobalt and nickel. R1 l A l 
2
A
2
 r
4
 d 
4

 1  2  1   2   2   2 
R2 l 2 A1  l 2 

A
 1


r
 1


d
 1


(viii) Resistivity of certain substances like selenium,
cadmium, sulphides is inversely proportional to intensity of light
2
R1  l1 
(1) If length is given then R  l 2   
falling upon them. R 2  l 2 


(2) Conductivity : Reciprocal of resistivity is called

4
1 R r 
(2) If radius is given then R   1   2 
1 r4 R 2  r1 
conductivity () i.e.   with unit mho/m and dimensions

Electrical Conducting Materials For Specific Use
[M 1 L3 T 3 A 2 ] .
(1) Filament of electric bulb : Is made up of tungsten which
(3) Conductance : Reciprocal of resistance is known as
has high resistivity, high melting point.
1 1
conductance. C  It’s unit is or –1 or “Siemen”.
R  (2) Element of heating devices (such as heater, geyser or
i press) : Is made up of nichrome which has high resistivity and

high melting point.

 (3) Resistances of resistance boxes (standard resistances)

V
: Are made up of alloys (manganin, constantan or nichrome)
Fig. 19.11
these materials have moderate resistivity which is practically

independent of temperature so that the specified value of

Stretching of Wire
resistance does not alter with minor changes in temperature.
If a conducting wire stretches, it’s length increases, area of
(4) Fuse-wire : Is made up of tin-lead alloy (63% tin + 37%
cross-section decreases so resistance increases but volume
lead). It should have low melting point and high resistivity. It is
remain constant.
used in series as a safety device in an electric circuit and is
Suppose for a conducting wire before stretching it’s length =
designed so as to melt and thereby open the circuit if the current
l1, area of cross-section = A1, radius = r1, diameter = d1, and
l1 exceeds a predetermined value due to some fault. The function
resistance R1  
A1 of a fuse is independent of its length.

Before stretching After stretching Safe current of fuse wire relates with it’s radius as

l1 l2 i  r 3/2 .
 
(5) Thermistors : A thermistor is a heat sensitive resistor

Volume remains constant i.e. A1l1 = A2l2 usually prepared from oxides of various metals such as nickel,
Fig. 19.12 copper, cobalt, iron etc. These compounds are also semi-

conductor. For thermistors  is very high which may be positive

or negative. The resistance of thermistors changes very rapidly

After stretching length = l2, area of cross-section = A2,
l2 with change of temperature.
radius = r2, diameter = d2 and resistance  R2  
A2 i

V
Fig. 19.13
 1042 Current Electricity
O Orange 3 103

Y Yellow 4 104

G Green 5 105

B Blue 6 106

V Violet 7 107
Thermistors are used to detect small temperature change
G Grey 8 108
and to measure very low temperature.
W White 9 109
Colour Coding of Resistance
To know the value of resistance colour code is used. These To remember the sequence of colour code following

code are printed in form of set of rings or strips. By reading the sentence should kept in memory.

values of colour bands, we can estimate the value of resistance. B B R O Y Great Britain Very Good Wife.
Grouping of Resistance
The carbon resistance has normally four coloured rings or
(1) Series grouping
bands say A, B, C and D as shown in following figure.
A B C D
(i) Same current flows through each resistance but potential

difference distributes in the ratio of resistance i.e. V  R

R1 R2 R3

Fig. 19.14 V1 V2 V3
i

+ –
V
Colour band A and B : Indicate the first two significant Fig. 19.15

figures of resistance in ohm.

Band C : Indicates the decimal multiplier i.e. the number of

zeros that follows the two significant figures A and B. (ii) Req  R1  R 2  R 3 equivalent resistance is greater

Band D : Indicates the tolerance in percent about the than the maximum value of resistance in the combination.

indicated value or in other words it represents the percentage (iii) If n identical resistance are connected in series

accuracy of the indicated value. R eq  nR and potential difference across each resistance

V
The tolerance in the case of gold is  5% and in silver is  V' 
n
10%. If only three bands are marked on carbon resistance, then
(2) Parallel grouping
it indicate a tolerance of 20%. i1
R1
i2
(i) Same potential difference
Table 19.3 : Colour code for carbon resistance i3 R2
appeared across each resistance i
R3
Letters as an Colour Figure Multiplier
but current distributes in the
aid to memory (A, B) (C) reverse ratio of their resistance V

B Black 0 10o 1 Fig. 19.16

i.e. i 
R
B Brown 1 101

R Red 2 102
 Current Electricity 1043

1 1 1 1
(ii) Equivalent resistance is given by   
R eq R1 R 2 R 3
R1 R 2 R 3
or R eq  (R11  R 21  R 31 )1 or R eq 
R1 R 2  R 2 R 3  R 2 R1

Equivalent resistance is smaller than the minimum value of

resistance in the combination. (1) Emf of cell (E) : The potential difference across the
terminals of a cell when it is not supplying any current is called
(iv) If two resistance in parallel
R1 R 2 Multiplication it’s emf.
R eq  
R1  R 2 Addition (2) Potential difference (V) : The voltage across the
(v) Current through any resistance terminals of a cell when it is supplying current to external
 Resistance of oppositebranch  resistance is called potential difference or terminal voltage.
i'  i   
 Total resistance 
Potential difference is equal to the product of current and
Where i = required current (branch current),
resistance of that given part i.e. V = iR.
i1 R1
i = main current
(3) Internal resistance (r) : In case of a cell the opposition
 R2  i
i1  i   of electrolyte to the flow of current through it is called internal

 R1  R 2 
i2 R2 resistance of the cell. The internal resistance of a cell depends
 R1  on the distance between electrodes (r  d), area of electrodes [r
and i2  i  
 Fig. 19.17
 R1  R 2 
 (1/A)] and nature, concentration (r  C) and temperature of
(vi) In n identical resistance are connected in parallel electrolyte [r  (1/ temp.)].
R i A cell is said to be ideal, if it has zero internal resistance.
R eq  and current through each resistance i' 
n n
Cell in Various Positions
Cell
(1) Closed circuit : Cell supplies a constant current in the
R
circuit.

V = iR
i

E, r
Fig. 19.19

The device which converts chemical energy into electrical

energy is known as electric cell. Cell is a source of constant emf (i) Current given by the cell i 
E
Rr
but not constant current.
+ (ii) Potential difference across the resistance V  iR
A
Anode Cathode –
(iii) Potential drop inside the cell = ir

+ – + – (iv) Equation of cell E  V  ir (E > V)

– E 
+ (v) Internal resistance of the cell r    1   R
Symbol of cell V 
Electrolyte
(vi) Power dissipated in external resistance (load)
Fig. 19.18
 1044 Current Electricity
V2  E 
2
(i) Maximum current (called short circuit current) flows
P  Vi  i 2 R    .R
R R r E
momentarily isc 
Power delivered will be maximum when Rr so r
2
Pmax 
E
. (ii) Potential difference V = 0
4r
This statement in generalised from is called “maximum
Grouping of Cells

power transfer theorem”.

Pmax = E2/4r
P

R=r
R
Fig. 19.20

Group of cell is called a battery.

(vii) When the cell is being charged i.e. current is given to In series grouping of cell’s their emf’s are additive or
the cell then E = V – ir and E < V. subtractive while their internal resistances are always additive. If

(2) Open circuit : When no current is taken from the cell it dissimilar plates of cells are connected together their emf’s are
added to each other while if their similar plates are connected
is said to be in open circuit
R
together their emf’s are subtractive.
C D A B

E1 E2 E1 E2
Eeq = E1 + E2 Eeq = E1 – E2 (E1 > E2)
E, r req = r1 + r2 req = r1 + r2
Fig. 19.21 Fig. 19.23

(1) Series grouping : In series grouping anode of one cell is

(i) Current through the circuit i = 0
connected to cathode of other cell and so on. If n identical cells
(ii) Potential difference between A and B, VAB = E E, r E, r E, r E, r
are connected in series

(iii) Potential difference between C and D, VCD = 0

i
(3) Short circuit : If two terminals of cell are join together by
R
a thick conducting wire
R=0 Fig. 19.24

E, r
(i) Equivalent emf of the combination Eeq  nE
Fig. 19.22
(ii) Equivalent internal resistance req  nr
nE
(iii) Main current = Current from each cell  i 
R  nr
(iv) Potential difference across external resistance V  iR
 Current Electricity 1045

V
(v) Potential difference across each cell V ' 
n
2
 nE 
(vi) Power dissipated in the external circuit    .R
 R  nr 
(vii) Condition for maximum power R  nr and
 E2 
Pmax  n  

 4r 
(viii) This type of combination is used when nr << R.
(i) Equivalent emf of the combination Eeq  nE
(2) Parallel grouping : In parallel grouping all anodes are

connected at one point and all cathode are connected together (ii) Equivalent internal resistance of the combination
nr
at other point. If n identical cells areEconnected
,r in parallel req 
m
E, r (iii) Main current flowing through the load
nE mnE
E, r i 
nr mR  nr
R
i m
R
(iv) Potential difference across load V  iR
Fig. 19.25 V
(v) Potential difference across each cell V ' 
n
i
(vi) Current from each cell i ' 
n
nr
(vii) Condition for maximum power R
m
(i) Equivalent emf Eeq = E
E2
and Pmax  (mn )
(ii) Equivalent internal resistance Req  r / n 4r

(iii) Main current i 

E (viii) Total number of cell = mn
R r/n
Kirchoff's Laws
(iv) potential difference across external resistance = p.d.
across each cell = V = iR (1) Kirchoff’s first law : This law is also known as junction

i rule or current law (KCL). According to it the algebraic sum of

(v) Current from each cell i' 
n currents meeting at a junction is zero i.e. i = 0.
2
 E  i1
(vi) Power dissipated in the circuit P    .R
 R r/n  i4

(vii) Condition for max. power is R  r/n and i3

i2
E 2 
Pmax  n  
 Fig. 19.27
 4r 

(viii) This type of combination is used when nr >> R

In a circuit, at any junction the sum of the currents
(3) Mixed Grouping : If n identical cell’s are connected in
entering the junction must equal the sum of the currents leaving
a row and such m row’s are connected in parallel as shown.
the junction. i1  i3  i2  i4
E, r E, r E, r
1 (ii) This law is simply a statement of “conservation of
1 2 n

2 charge”.

i
m

R
Fig. 19.26
 1046 Current Electricity
(2) Kirchoff’s second law : This law is also known as loop (iv) The change in voltage in traversing an inductor in the

rule or voltage law (KVL) and according to it “the algebraic sum direction of current is  L
di
while in opposite direction it is
dt
of the changes in potential in complete traversal of a mesh L L
Adi i B A i B
L .
(closed loop) is zero”, i.e. V = 0 dt
di di
L L (B)
(i) This law represents “conservation of energy”. (A) dt dt

(ii) If there are n meshes in a circuit, the number of Fig. 19.31

independent equations in accordance with loop rule will be (n – 1).

(3) Sign convention for the application of Kirchoff’s law :

For the application of Kirchoff’s laws following sign convention

are to be considered Different Measuring Instruments

(i) The change in potential in traversing a resistance in

the direction of current is – iR while in the opposite direction +iR

A R B A R B
i i

– iR Fig. 19.28 + iR

(ii) The change in potential in traversing an emf source

from negative to positive terminal is +E while in the opposite

(1) Galvanometer : It is an instrument used to detect
direction – E irrespective of the direction of current in the circuit.
small current passing through it by showing deflection.

A E B A E B Galvanometers are of different types e.g. moving coil

galvanometer, moving magnet galvanometer, hot wire
–E +E
Fig. 19.29 galvanometer. In dc circuit usually moving coil galvanometer are

used.

(i) It’s symbol : G ; where G is the

(iii) The change in potential in traversing a capacitor from
q total internal resistance of the galvanometer.
the negative terminal to the positive terminal is  while in
C
(ii) Full scale deflection current : The current required for
q
opposite direction  .
C full scale deflection in a galvanometer is called full scale
C C
A B A B
– + – + deflection current and is represented by ig.
q q q
q
 
C C (iii) Shunt : The small resistance connected in parallel to
(A) (B)
Fig. 19.30 galvanometer coil, in order to control current flowing through the

galvanometer is known as shunt.

Table 19.4 : Merits and demerits of shunt

Merits of shunt Demerits of shunt

 Current Electricity 1047

i
To protect the galvanometer Shunt resistance decreases (c) To pass nth part of main current (i.e. ig  ) through
n
coil from burning the sensitivity of galvanometer. G
the galvanometer, required shunt S  .
It can be used to convert any
(n  1)

galvanometer into ammeter of (3) Voltmeter : It is a device used to measure potential

desired range. difference and is always put in parallel with the ‘circuit element’

across which potential difference

V
is to be measured.
(2) Ammeter : It is a device used to measure current and

is always connected in series with the ‘element’ through which

R
i
current is to be measured. R
+ –
V
i A Fig. 19.34

+ –
V
Fig. 19.32

(i) The reading of a voltmeter is always lesser than true

value.

(ii) Greater the resistance of voltmeter, more accurate will

(i) The reading of an ammeter is always lesser than
be its reading. A voltmeter is said to be ideal if its resistance is
actual current in the circuit.
infinite, i.e., it draws no current from the circuit element for its
(ii) Smaller the resistance of an ammeter more accurate
operation.
will be its reading. An ammeter is said to be ideal if its
(iii) Conversion of galvanometer into voltmeter : A
resistance r is zero.
galvanometer may be converted into a voltmeter by connecting
(iii) Conversion of galvanometer into ammeter : A R
a large resistance R in series with the galvanometer as shown
G
galvanometer may be converted into an ammeter by connecting in the figure. Vg = igG (V – Vg)
a low resistance (called shunt S) in parallel to the galvanometer ig
S
G as shown in figure. V
i – ig Fig. 19.35
i
G
ig

Ammeter
Fig. 19.33
(a) Equivalent resistance of the combination = G + R

(b) According to ohm’s law V = ig (G + R); which gives

V  V 
GS Required series resistance R  G  1G
(a) Equivalent resistance of the combination  ig  
GS  Vg 

(b) G and S are parallel to each other hence both will (c) If nth part of applied voltage appeared across
V
have equal potential difference i.e. ig G  (i  ig )S ; which gives galvanometer (i.e. Vg  ) then required series resistance
n

Required shunt S 
ig
G R  (n  1) G .
(i  i g )
 1048 Current Electricity
(4) Wheatstone bridge Potentiometer is a device mainly used to measure emf of a

: Wheatstone bridge is an B given cell and to compare emf’s of cells. It is also used to
P K1 Q
arrangement of four measure internal resistance of a given cell.
A C
resistance which can be G
(1) Circuit diagram : Potentiometer consists of a long
used to measure one of R S resistive wire AB of length L (about 6m to 10 m long) made up
them in terms of rest. Here + – D K2
of mangnine or constantan and a battery of known voltage e
arms AB and BC are called Fig. 19.36 and internal resistance r called supplier battery or driver cell.
ratio arm and arms AC and BD are called conjugate arms Connection of these two forms primary circuit.

One terminal of another cell (whose emf E is to be

(i) Balanced bridge : The bridge is said to be balanced measured) is connected at one end of the main circuit and the
when deflection in galvanometer is zero i.e. no current flows other terminal at any point on the resistive wire through a
through the galvanometer or in other words VB = VD. In the R
galvanometer G. This eforms
,r theKsecondary circuit.
h
Other details
P R
balanced condition  , on mutually changing the position Primary
Q S are as follows
circuit J
of cell and galvanometer this condition will not change. A B
Secondary
(ii) Unbalanced bridge : If the bridge is not balanced current E
circuit G
will flow from D to B if VD > VB i.e. (VA  VD )  (VA  VB ) which
Fig. 19.38
gives PS > RQ.

(iii) Applications of wheatstone bridge : Meter bridge, post

office box and Carey Foster bridge are instruments based on
the principle of wheatstone bridge and are used to measure
unknown resistance. J = Jockey
(5) Meter bridge : In case of meter bridge, the resistance
K = Key
wire AC is 100 cm long. Varying the position of tapping point B,
bridge is balanced. If in balanced position of bridge AB = l, BC R = Resistance of potentiometer wire,
Q (100  l) P R (100  l)
(100 – l) so that  . Also   S  R  = Specific resistance of potentiometer wire.
P l Q SS l
R R.B.
Rh = Variable resistance which controls the current

G through the wire AB

P B Q (i) The specific resistance () of potentiometer wire must

A C
l cm (100 – l) cm
be high but its temperature coefficient of resistance () must be
E K
low.
Fig. 19.37

(ii) All higher potential points (terminals) of primary and

secondary circuits must be connected together at point A and all

lower potential points must be connected to point B or jockey.

Potentiometer
 Current Electricity 1049

(iii) The value of known potential difference must be

greater than the value of unknown potential difference to be

measured.

(iv) The potential gradient must remain constant. For this

the current in the primary circuit must remain constant and the

jockey must not be slided in contact with the wire.

If V > E then current will flow in galvanometer circuit in
(v) The diameter of potentiometer wire must be uniform
one direction
everywhere.
If V < E then current will flow in galvanometer circuit in
(2) Potential gradient (x) : Potential difference (or fall in
opposite direction
potential) per unit length of wire is called potential gradient i.e.
  If V = E then no current will flow in galvanometer circuit
V volt e
x where V  iR   .R .

L m  R  R h  r  this condition to known as null deflection position, length l is

V iR iρ e R known as balancing length.

So x    .
L L A (R  Rh  r) L In balanced condition E  xl

(i) Potential gradient directly depends upon V iR  e  R

or E  xl  l l   .  l

L L  R  Rh  r  L
(a) The resistance per unit length (R/L) of potentiometer
x1 L l
wire. If V is constant then L  l   1  1
x 2 L2 l2
(b) The radius of potentiometer wire (i.e. Area of cross-
(6) Standardization of potentiometer : The process of
section) determining potential gradient experimentally is known as

(c) The specific resistance of the material of standardization of potentiometer.

potentiometer wire (i.e. ) e, r K Rh

(d) The current flowing through potentiometer wire ( i)

J
(ii) potential gradient indirectly depends upon A B
G
(a) The emf of battery in the primary circuit ( i.e. e) E0

(b) The resistance of rheostat in the primary circuit ( i.e.

E
Rh)
Fig. 19.40
(3) Working : Suppose jocky is made to touch a point J on

wire then potential difference between A and J will be V  xl

At this length (l) two potential difference are obtained

Let the balancing length for the standard emf E0 is l0 then
(i) V due to battery e and
E0
by the principle of potentiometer E0 = xl0  x 
(ii) E due to unknown cell l0

e, r K Rh

l
J1 J J2
A B

G G G
E
Fig. 19.39
 1050 Current Electricity
(7) Sensitivity of potentiometer : A potentiometer is said to

be more sensitive, if it measures a small potential difference (i) Initially in secondary circuit key K' remains open and
more accurately. balancing length (l1) is obtained. Since cell E is in open circuit

(i) The sensitivity of potentiometer is assessed by its so it’s emf balances on length l1 i.e. E = xl1 …. (i)

potential gradient. The sensitivity is inversely proportional to the (ii) Now key K is closed so cell E comes in closed circuit. If
potential gradient. the process of balancing repeated again then potential

(ii) In order to increase the sensitivity of potentiometer difference V balances on length l2 i.e. V = xl2

(a) The resistance in primary circuit will have to be …. (ii)

decreased. E 
(iii) By using formula internal resistance r    1  . R '
V 
(b) The length of potentiometer wire will have to be increased
l l 
so that the length may be measured more accuracy. r   1 2  . R'

 l2 
Table 19.5 : Difference between voltmeter and potentiometer
(2) Comparison of emf’s of two cell : Let l1 and l2 be the
Voltmeter Potentiometer balancing lengths with the cells E1 and E2 respectively then E1 =
It’s resistance is high but finite Its resistance is infinite E1 l1
xl1 and E2 = xl2  
E 2 l2 K Rh
It draws some current from It does not draw any current from e, r

source of emf the source of unknown emf

J
The potential difference The potential difference A B
measured by it is lesser than measured by it is equal to G
E1 1
the actual potential difference actual potential difference

Its sensitivity is low Its sensitivity is high

E2
2

It is a versatile instrument It measures only emf or Fig. 19.42

potential difference

It is based on deflection It is based on zero deflection

method method

Let E1 > E2 and both are connected in series. If balancing

Application of Potentiometer length is l1 when cell assist each other and it is l2 when they

(1) To determine the internal resistance of a primary cell oppose each other as shown then :
K Rh
e, r
E1 E2 E1 E2
+ – + – + – – +

J B
A
G (E1  E 2 )  xl 1 (E1  E 2 )  xl 2
E
E1  E 2 l E1 l l
R K   1 or  1 2
E1  E 2 l2 E 2 l1  l2
Fig. 19.41
 Current Electricity 1051

(3) Comparison of resistances : Let the balancing length for (iv) x  i 

iR
 e
iRl
where L = length of
L L
resistance R1 (when XY is connected) is l1 and let balancing
potentiometer wire,  = resistance per unit length, l = balancing
length for resistance R1 + R2 (when YZ is connected) is l2.
length for e
K Rh
(5) Calibration of ammeter : Checking the correctness of
ammeter readings with the help of potentiometer is called
J
A B calibration of ammeter.
G
X Y Z
(i) In the process of calibration of an ammeter the current

R1 flowing in a circuit is measured by an ammeter and the same

i R2
current is also measured with the help of potentiometer. By
Rh
K1 1 comparing both the values, the errors in the ammeter readings
e K1
Fig. 19.43
are determined. + –
+
A B
E1
+ –
1
2 G
R 2 l2  l1
Then iR1 = xl1 and i(R1 + R2) = xl2   1 3
R1 l1 A
+ –
+ –
K2
(4) To determine thermo emf
+ – K Rh Fig. 19.45

R A
A HRB
B
G
+ –
G
E0 1 2 3 (ii) For the calibration of an ammeter, 1  standard
Cold ice Hot sand
resistance coil is specifically used in the secondary circuit of the
Fig. 19.44
potentiometer, because the potential difference across 1  is

equal to the current flowing through it i.e. V = i.

(iii) If the balancing length for the emf E0 is l0 then E0 = xl0

(i) The value of thermo-emf in a thermocouple for ordinary x 
E0
(Process of standardisation)
l0
temperature difference is very low (10–6 volt). For this the
potential gradient x must be also very low (10–4 V/m). Hence a (iv) Let i ' current flows through 1 resistance giving
high resistance (R) is connected in series with the potentiometer potential difference as V '  i' (1)  xl1 where l1 is the balancing
wire in order to reduce current. E0
length. So error can be found as i  i  i'  i  xl1  i   l1
l0
(ii) The potential difference across R must be equal to the

emf of standard cell i.e. iR = E0  i 

E0 (6) Calibration of voltmeter
R
(i) Practical voltmeters are not ideal, because these do not
(iii) The small thermo emf produced in the thermocouple e =
have infinite resistance. The error of such practical voltmeter
xl
 1052 Current Electricity
can be found by comparing the voltmeter reading with conductor is zero.

calculated value of p.d. by potentiometer.

(ii) If l0 is balancing length for E0 the emf of standard cell by

connecting 1 and 2 of bi-directional key, then x = E0/l0.

(iii) The balancing length l1 for unknown potential difference

V is given by (by closing 2 and 3) V '  xl1  (E0 / l0 )l1 . 1

e K1 Rh  For a given conductor JA = i = constant so that J 
A
+ –
i.e., J1 A1 = J2 A2 ; this is called equation of continuity
+
A B
E0 1 C
+ –
2
+ – G
V 3
RB

+ –
K2
Rh
Fig. 19.46
 The drift velocity of electrons is small because of the
frequent collisions suffered by electrons.

 The small value of drift velocity produces a large amount

of electric current, due to the presence of extremely large
If the voltmeter reading is V then the error will be ( V – V)
number of free electrons in a conductor.
which may be + ve, – ve or zero.
The propagation of current is almost at the speed of light

and involves electromagnetic process. It is due to this reason

that the electric bulb glows immediately when switch is on.

 In the absence of electric field, the paths of electrons

between successive collisions are straight line while in
 Human body, though has a large resistance of the order of presence of electric field the paths are generally curved.
k (say 10 k), is very sensitive to minute currents even as
N Ax d
low as a few mA. Electrocution, excites and disorders the
 Free electron density in a metal is given by n 
A
nervous system of the body and hence one fails to control the where NA = Avogadro number, x = number of free electrons
activity of the body. per atom, d = density of metal and A = Atomic weight of

 dc flows uniformly throughout the cross-section of metal.

conductor while ac mainly flows through the outer surface  In the absence of radiation loss, the time in which a fuse will
area of the conductor. This is known as skin effect. melt does not depends on it’s length but varies with radius as

 It is worth noting that electric field inside a charged t  r4 .

conductor is zero, but it is non zero inside a current carrying  If length (l) and mass (m) of a conducting wire is given
V l2
conductor and is given by E  where V = potential then R  .
l m
difference across the conductor and l = length of the V
 Macroscopic form of Ohm’s law is R  , while it’s
conductor.
+ + + + + field out side the l current carrying
+Electric i
i
J1 + J2 –
Ein = 0 Ein = V/l
i
+ + + + + A+1 A2
 Current Electricity 1053

microscopic form is J =  E.

 After stretching if length increases by n times then

resistance will increase by n2 times i.e. R 2  n 2 R1 . Similarly 5
The longest diagonal (EC or AG)  R
1 6
if radius be reduced to times then area of cross-section
n
3
1 The diagonal of face (e.g. AC, ED, ....)  R
decreases times so the resistance becomes n times i.e.
4 4
n2
7
R 2  n 4 R1 . A side (e.g. AB, BC.....)  R
12
 After stretching if length of a conductor increases by x%  Resistance of a conducting body is not unique but
then resistance will increases by 2x % (valid only if x < 10%) depends on it’s length and area of cross-section i.e. how the

 Decoration of lightning in festivals is an example of potential difference is applied. See the following figures

series grouping whereas all household appliances connected c

c
in parallel grouping. 
 b
b
 Using n conductors of equal resistance, the number of
a
a
possible combinations is 2n – 1.

 If the resistance of n conductors are totally different, then

the number of possible combinations will be 2n.

 If n identical resistances are first connected in series and

then in parallel, the ratio of the equivalent resistance is given Length = a Length = b

Rp n2 Area of cross-section = b  c Area of cross-section = a  c

by  .
Rs 1  a   b 
Resistance R     Resistance R    
bc ac
 If a wire of resistance R, cut in n equal parts and then
 Some standard results for equivalent resistance
these parts are collected to form a bundle then equivalent
R1 R2
R
resistance of combination will be .
n2 A B
R5
 If equivalent resistance of R1 and R2 in series and
parallel be Rs and Rp respectively then
R3 R4
R1   Rs  Rs2  4 Rs R p  and R2   Rs  Rs2  4 Rs R p  .
1 1
2   2  

 If a skeleton cube is made with 12 equal resistance each R1 R2 (R3  R4 )  (R1  R2 )R3 R4  R5 (R1  R2 ) (R3  R4 )
R AB 
R5 (R1  R2  R3  R4 )  (R1  R3 )(R2  R4 )
having resistance R then the net resistance across
H G
R1 R2
E
F
A B
D R3
C

A B
R2 R1
 1054 Current Electricity
present in that branch. In practical situation it always happen

because we can never have an ideal cell or battery with zero

resistance.
2 R1 R2  R3 (R1  R2 )
R AB 
2 R3  R1  R2
 In series grouping of identical cells. If one cell is wrongly
connected then it will cancel out the effect of two cells e.g. If
R1 R1 R1 R1
A
in the combination of n identical cells (each having emf E and

R3 R3 R3 R3  internal resistance r) if x cell are wrongly connected then

B equivalent emf Eeq  (n  2 x ) E and equivalent internal

R2 R2 R2 R2
resistance req  nr .

 Graphical view of open circuit and closed circuit of a cell.

R AB 
1
2
1

(R1  R2 )  (R1  R2 )2  4 R3 (R1  R2 )
2
1/2
 V
Vmax =E; i = 0

R1 R1 R1 R1
A

imax =E/r ; V = 0 i
R2 R2 R2  

 If n identical cells are connected in a loop in order, then

1  R  emf between any two points is zero.
R AB  R1 1  1  4  2  E, r
2  
  R1 
E, r E, r
 It is a common misconception that “current in the circuit Close

will be maximum when power consumed by the load is E, r loop

E, r
maximum.”
n cell
Actually current i  E /(R  r) is maximum (= E/r) when R =

min = 0 with PL  (E / r)2  0  0 min . while power consumed

by the load E2R/(R + r)2 is maximum (= E2/4r) when R = r and

 In parallel grouping of two identical cell having no
i  (E / 2r)  max(  E / r). internal resistance
R R
 Emf is independent of the resistance of the circuit and
E E
depends upon the nature of electrolyte of the cell while
E E
potential difference depends upon the resistance between the

two points of the circuit and current flowing through the

circuit.
E eq  E E eq  0
 Whenever a cell or battery is present in a branch there
 When two cell’s of different emf and no internal
must be some resistance (internal or external or both)
resistance are connected in parallel then equivalent emf is

E1
 Current Electricity 1055

indeterminate, note that connecting a wire with a cell with no

resistance is equivalent to short circuiting. Therefore the total 1. Current of 4.8 amperes is flowing through a conductor.

current that will be flowing will be infinity. The number of electrons per second will be [CPMT 1986]

(a) 3  1019 (b) 7.68  10 21

(c) 7.68  10 20 (d) 3  10 20

2. When the current i is flowing through a conductor, the

drift velocity is v . If 2i current is flowed through the
same metal but having double the area of cross-section,

 In the parallel combination of non-identical cell's if they then the drift velocity will be

are connected with reversed polarity as shown then (a) v / 4 (b) v / 2

i1 E1,r1
equivalent emf (c) v (d) 4 v

3. When current flows through a conductor, then the order of

i i2 E2, r2
E1r2  E2r1 drift velocity of electrons will be [CPMT 1986]
Eeq 
r1  r2
R (a) 1010 m / sec (b) 10 2 cm / sec

(c) 10 4 cm / sec (d) 10 1 cm / sec

4. Every atom makes one free electron in copper. If 1.1

 Wheatstone bridge is most sensitive if all the arms of
ampere current is flowing in the wire of copper having 1
bridge have equal resistances i.e. P = Q = R = S
mm diameter, then the drift velocity (approx.) will be
 If the temperature of the conductor placed in the right (Density of copper  9  10 3 kg m 3 and atomic weight =
gap of metre bridge is increased, then the balancing length
63)
decreases and the jockey moves towards left.
[CPMT 1989]
 In Wheatstone bridge to avoid inductive effects the
(a) 0.3 mm / sec (b) 0.1 mm / sec
battery key should be pressed first and the galvanometer key
afterwards. (c) 0.2 mm / sec (d) 0.2 cm / sec

 The measurement of resistance by Wheatstone bridge is 5. Which one is not the correct statement [NCERT 1978]

not affected by the internal resistance of the cell. (a) 1 volt  1 coulomb  1 joule

 In case of zero deflection in the galvanometer current (b) 1 volt  1 ampere  1 joule / second
flows in the primary circuit of the potentiometer, not in the
(c) 1 volt  1 watt  1 H .P.
galvanometer circuit.
(d) Watt-hour can be expressed in eV
 A potentiometer can act as an ideal voltmeter.
6. If a 0.1 % increase in length due to stretching, the
percentage increase in its resistance will be

[MNR 1990; MP PMT 1996; UPSEAT 1999; MP PMT 2000]

(a) 0.2 % (b) 2 %

(c) 1 % (d) 0.1 %

Electric Conduction, Ohm's Law and Resistance
 1056 Current Electricity
7. The specific resistance of manganin is 50  10 8 ohm  m .
The resistance of a cube of length 50 cm will be

(a) 10 6 ohm (b) 2.5  10 5 ohm

(c) 10 8 ohm (d) 5  10 4 ohm

 Current Electricity 1051

8. The resistivity of iron is 1  10 7 ohm  m . The resistance of a iron 18. The specific resistance of a wire is  , its volume is 3 m 3 and its
wire of particular length and thickness is 1 ohm. If the length and resistance is 3 ohms, then its length will be
the diameter of wire both are doubled, then the resistivity in [CPMT 1984]
ohm  m will be [CPMT 1983; DPMT 1999]
1 3
(a) 1  10 7 (b) 2  10 7 (a) (b)
 
(c) 4  10 7 (d) 8  10 7
9. The temperature coefficient of resistance for a wire is 1 1
(c) 3 (d) 
0.00125 / C . At 300K its resistance is 1 ohm. The temperature at  3
which the resistance becomes 2 ohm is
[IIT 1980; MP PET 2002; KCET 2003; 19. 62.5  10 18 electrons per second are flowing through a wire of
MP PMT 2001; Orissa JEE 2002] area of cross-section 0.1 m 2 , the value of current flowing will be
(a) 1154 K (b) 1100 K (a) 1 A (b) 0.1 A
(c) 1400 K (d) 1127 K (c) 10 A (d) 0.11 A
10. When the length and area of cross-section both are doubled, then its 20. A piece of wire of resistance 4 ohms is bent through 180 at its
resistance [MP PET 1989] mid point and the two halves are twisted together, then the
resistance is [CPMT 1971]
(a) Will become half (b) Will be doubled
(a) 8 ohms (b) 1 ohm
(c) Will remain the same (d) Will become four times
(c) 2 ohms (d) 5 ohms
11. The resistance of a wire is 20 ohms. It is so stretched that the length
becomes three times, then the new resistance of the wire will be 21. When a[MPpiece of aluminium
PET 1989]
wire of finite length is drawn through a
series of dies to reduce its diameter to half its original value, its
(a) 6.67 ohms (b) 60.0 ohms resistance will become
(c) 120 ohms (d) 180.0 ohms [NCERT 1974; AIIMS 1997; MH CET 2000; UPSEAT 2001;
CBSE PMT 2002]
12. The resistivity of a wire [MP PMT 1984; DPMT 1982]
(a) Two times (b) Four times
(a) Increases with the length of the wire
(c) Eight times (d) Sixteen times
(b) Decreases with the area of cross-section 22. A wire 100 cm long and 2.0 mm diameter has a resistance of 0.7
(c) Decreases with the length and increases with the cross-section ohm, the electrical resistivity of the material is
of wire
(a) 4.4  10 6 ohm  m (b) 2.2  10 6 ohm  m
(d) None of the above statement is correct
13. Ohm's law is true (c) 1.1  10 6 ohm  m (d) 0.22  10 6 ohm  m
(a) For metallic conductors at low temperature 23. A certain wire has a resistance R . The resistance of another wire
identical with the first except having twice its diameter is
(b) For metallic conductors at high temperature
(a) 2R (b) 0.25 R
(c) For electrolytes when current passes through them
(d) For diode when current flows (c) 4R (d) 0 .5 R
14. The example for non-ohmic resistance is [MP PMT 1978] 24. In hydrogen atom, the electron makes 6.6  10 15 revolutions per
(a) Copper wire (b) Carbon resistance second around the nucleus in an orbit of radius 0.5  10 10 m . It
(c) Diode (d) Tungston wire is equivalent to a current nearly
(a) 1 A (b) 1 mA
15. Drift velocity v d varies with the intensity of electric field as per the
relation [CPMT 1981; BVP 2003] (c) 1 A (d) 1.6  10 19 A
1 25. A wire of length 5 m and radius 1 mm has a resistance of 1 ohm.
(a) vd  E (b) v d  What length of the wire of the same material at the same
E
temperature and of radius 2 mm will also have a resistance of 1
(c) v d  constant (d) v d  E 2 ohm
16. On increasing the temperature of a conductor, its resistance (a) 1.25 m (b) 2.5 m
increases because [CPMT 1982] (c) 10 m (d) 20 m
(a) Relaxation time decreases 26. When there is an electric current through a conducting wire along
its length, then an electric field must exist
(b) Mass of the electrons increases
(a) Outside the wire but normal to it
(c) Electron density decreases (b) Outside the wire but parallel to it
(d) None of the above (c) Inside the wire but parallel to it
17. In a conductor 4 coulombs of charge flows for 2 seconds. The value (d) Inside the wire but normal to it
of electric current will be [CPMT 1984] 27. Through a semiconductor, an electric current is due to drift of
(a) 4 volts (b) 4 amperes (a) Free electrons
(b) Free electrons and holes
(c) 2 amperes (d) 2 volts
(c) Positive and negative ions
(d) Protons
 1052 Current Electricity

28. In an electrolyte 3.2  10 18 bivalent positive ions drift to the right (a) 1.0 mm / sec (b) 1.0 m / sec
per second while 3.6  10 18 monovalent negative ions drift to the (c) 0.1 mm / sec (d) 0.01 mm / sec
left per second. Then the current is 39. It is easier to start a car engine on a hot day than on a cold day.
(a) 1.6 amp to the left (b) 1.6 amp to the right This is because the internal resistance of the car battery
(c) 0.45 amp to the right (d) 0.45 amp to the left (a) Decreases with rise in temperature
29. A metallic block has no potential difference applied across it, then (b) Increases with rise in temperature
the mean velocity of free electrons is T = absolute temperature of (c) Decreases with a fall in temperature
the block) (d) Does not change with a change in temperature
(a) Proportional to T 40. 5 amperes of current is passed through a metallic conductor. The
charge flowing in one minute in coulombs will be
(b) Proportional to T
[MP PET 1984]
(c) Zero (a) 5 (b) 12
(d) Finite but independent of temperature (c) 1/12 (d) 300
30. The specific resistance of all metals is most affected by 41. Two wires of the same material are given. The first wire is twice as
(a) Temperature (b) Pressure long as the second and has twice the diameter of the second. The
(c) Degree of illumination (d) Applied magnetic field resistance of the first will be
31. The positive temperature coefficient of resistance is for [MP PMT 1993]
(a) Carbon (b) Germanium (a) Twice of the second (b) Half of the second
(c) Copper (d) An electrolyte (c) Equal to the second (d) Four times of the second
32. The fact that the conductance of some metals rises to infinity at 42. An electric wire is connected across a cell of e.m.f. E. The current I
some temperature below a few Kelvin is called is measured by an ammeter of resistance R. According to ohm's law
(a) Thermal conductivity (b) Optical conductivity (a) E  I2R (b) E  IR
(c) Magnetic conductivity (d) Superconductivity
(c) E  R/I (d) E  I/R
33. Dimensions of a block are 1 cm  1 cm  100 cm . If specific
43. The resistances of a wire at temperatures tC and 0C are
resistance of its material is 3  10 7 ohm  m , then the resistance related by [MP PMT 1993]
between the opposite rectangular faces is (a) R t  R 0 (1   t) (b) R t  R 0 (1   t)
[MP PET 1993]
(a) 3  10 9 ohm (b) 3  10 7 ohm (c) R t  R 02 (1   t) (d) R t  R 02 (1   t)
44. An electric wire of length ‘I’ and area of cross-section a has a
(c) 3  10 5 ohm (d) 3  10 3 ohm
resistance R ohms. Another wire of the same material having same
34. In the above question, the resistance between the square faces is [MP area
length and PET 1993]
of cross-section 4a has a resistance of
(a) 3  10 9 ohm (b) 3  10 7 ohm (a) 4R (b) R/4
(c) 3  10 5 ohm (d) 3  10 3 ohm (c) R/16 (d) 16R
45. For which of the following the resistance decreases on increasing the
35. There is a current of 20 amperes in a copper wire of 10 6 square temperature [MP PET 1993]
metre area of cross-section. If the number of free electrons per cubic
(a) Copper (b) Tungsten
metre is 10 29 , then the drift velocity is
(c) Germanium (d) Aluminium
(a) 125  10 3 m / sec (b) 12.5  10 3 m / sec 46. If n, e,  and m respectively represent the density, charge relaxation
3 4 time and mass of the electron, then the resistance of a wire of
(c) 1.25  10 m / sec (d) 1.25  10 m / sec
length l and area of cross-section A will be
36. The electric intensity E , current density j and specific resistance
[CPMT 1992]
k are related to each other by the relation
[DPMT 2001] ml m 2 A
(a) (b)
(a) E  j/k (b) E  jk ne A
2
ne 2 l
(c) Ek/j (d) k  jE ne 2A ne 2 A
(c) (d)
37. The resistance of a wire of uniform diameter d and length L is 2ml 2m l
R . The resistance of another wire of the same material but 47. The relaxation time in conductors [DPMT 2003]
diameter 2d and length 4 L will be (a) Increases with the increase of temperature
[CPMT 1984; MP PET 2002] (b) Decreases with the increase of temperature
(a) 2R (b) R (c) It does not depend on temperature
(c) R / 2 (d) R / 4 (d) All of sudden changes at 400 K
38. There is a current of 1.344 amp in a copper wire whose area of 48. Which of the following statement is correct
cross-section normal to the length of the wire is 1 mm 2 . If the (a) Liquids obey fully the ohm's law
number of free electrons per cm 3 is 8.4  10 22 , then the drift (b) Liquids obey partially the ohm's law
velocity would be [CPMT 1990]
(c) There is no relation between current and p.d. for liquids
 Current Electricity 1053

(d) None of the above (c) Voltage (d) None of the above
49. A certain piece of silver of given mass is to be made like a wire. 58. A solenoid is at potential difference 60 V and current flows through
Which of the following combination of length (L) and the area of it is 15 ampere, then the resistance of coil will be
cross-sectional (A) will lead to the smallest resistance [MP PMT 1995; CBSE PMT 1997] [AFMC 1995]
(a) L and A (a) 4 (b) 8 
(b) 2L and A/2 (c) 0.25  (d) 2 
(c) L/2 and 2 A
59. All of the following statements are true except
(d) Any of the above, because volume of silver remains same [Manipal MEE 1995]
50. The resistance of a wire is 10  . Its length is increased by 10% by (a) Conductance is the reciprocal of resistance and is measured in
stretching. The new resistance will now be Siemens
[CPMT 2000; Pb PET 2004] (b) Ohm's law is not applicable at very low and very high
temperatures
(a) 12  (b) 1 .2 
(c) Ohm's law is applicable to semiconductors
(c) 13  (d) 11  (d) Ohm's law is not applicable to electron tubes, discharge tubes
and electrolytes
51. Resistance of tungsten wire at 150C is 133  . Its resistance
60. A potential difference of V is applied at the ends of a copper wire of
temperature coefficient is 0.0045 / C . The resistance of this wire length l and diameter d. On doubling only d, drift velocity
at 500C will be [DPMT 2004]
(a) Becomes two times (b) Becomes half
(a) 180  (b) 225  (c) Does not change (d) Becomes one fourth
(c) 258  (d) 317  61. If the resistance of a conductor is 5  at 50 C and 7  at 100 C then
o o

the mean temperature coefficient of resistance of the material is

52. A metal wire of specific resistance 64  10 6 ohm  cm and
(a) 0.008/ Co
(b) 0.006/ C o

length 198 cm has a resistance of 7 ohm, the radius of the wire will
be [MP PET 1994] (c) 0.004/ Co
(d) 0.001/ C o

(a) 2.4 cm (b) 0.24 cm 62. The resistance of a discharge tube is

(c) 0.024 cm (d) 24 cm [AFMC 1996; CBSE PMT 1999]
53. A copper wire of length 1 m and radius 1 mm is joined in series with (a) Ohmic (b) Non-ohmic
an iron wire of length 2 m and radius 3 mm and a current is passed
through the wires. The ratio of the current density in the copper (c) Both (a) and (b) (d) Zero
and iron wires is
63. We are able to obtain fairly large currents in a conductor because
[MP PMT 1994]
(a) 18 : 1 (b) 9 : 1 (a) The electron drift speed is usually very large
(c) 6 : 1 (d) 2 : 3 (b) The number density of free electrons is very high and this can
54. For a metallic wire, the ratio V / i (V  the applied potential compensate for the low values of the electron drift speed and
the very small magnitude of the electron charge
difference, i = current flowing) is [MP PMT 1994; BVP 2003]
(a) Independent of temperature (c) The number density of free electrons as well as the electron
drift speeds are very large and these compensate for the very
(b) Increases as the temperature rises small magnitude of the electron charge
(c) Decreases as the temperature rises
(d) The very small magnitude of the electron charge has to be
(d) Increases or decreases as temperature rises, depending upon divided by the still smaller product of the number density and
the metal drift speed to get the electric current
55. The resistance of a wire is R. If the length of the wire is doubled by
64. A platinum resistance thermometer has a resistance of 50  at
stretching, then the new resistance will be
[Roorkee 1992; AFMC 1995; KCET 1993; AMU (Med.) 1999; 20C . When dipped in a liquid the resistance becomes 76.8  .
CBSE PMT 1999; MP PET 2001; UPSEAT 2001] The temperature coefficient of resistance for platinum is
(a) 2R (b) 4R   3.92  10 3 / C . The temperature of the liquid is
R (a) 100C (b) 137C
(c) R (d)
4
(c) 167C (d) 200C
56. Which of the following has a negative temperature coefficient [AFMC 1995]
65. In a wire of circular cross-section with radius r, free electrons travel
(a) C (b) Fe with a drift velocity V when a current I flows through the wire.
(c) Mn (d) Ag What is the current in another wire of half the radius and of the
57. The reciprocal of resistance is [AFMC 1995] same material when the drift velocity is 2V
(a) Conductance (b) Resistivity [MP PET 1997]
 1054 Current Electricity
(a) 2I (b) I 74.  1 and  2 are the electrical conductivities of Ge and Na
(c) I/2 (d) I/4 respectively. If these substances are heated, then

66. The resistivity of a wire depends on its [MP PMT/PET 1998] (a) Both  1 and  2 increase

(a) Length (b) Area of cross-section (b)  1 increases and  2 decreases

(c) Shape (d) Material (c)  1 decreases and  2 increases
67. The conductivity of a superconductor is (d) Both  1 and  2 decrease
[Similar to KCET 1993; MP PMT/PET 1998]
75. 1.6 mA current is flowing in conducting wire then the number of
(a) Infinite (b) Very large electrons flowing per second is [RPMT 1999]
(c) Very small (d) Zero (a) 10 11
(b) 10 16

68. In a neon discharge tube 2.9  1018 Ne  ions move to the right (c) 10 19
(d) 10 15

76. A current I is passing through a wire having two sections P and Q

each second while 1.2  1018 electrons move to the left per second. of uniform diameters d and d/2 respectively. If the mean drift
Electron charge is 1.6  10 19 C . The current in the discharge tube velocity[MP
of PET 1999] in sections P and Q is denoted by v and v
electrons P Q

respectively, then [Roorkee 1999]

(a) 1 A towards right (b) 0.66 A towards right
1
(c) 0.66 A towards left (d) Zero (a) v = v
P Q
(b) v =
P
v Q

2
69. A steady current flows in a metallic conductor of non-uniform cross-
section. The quantity/ quantities constant along the length of the 1
(c) v =
P
v Q
(d) v = 2 v
P Q

conductor is/are 4
[KCET 1994, IIT 1997 Cancelled; CBSE PMT 2001] 77. If an electric current is passed through a nerve of a man, then man
(a) Current, electric field and drift speed (a) Begins to laugh
(b) Drift speed only (b) Begins to weep
(c) Current and drift speed (c) Is excited
(d) Current only (d) Becomes insensitive to pain
78. The resistance of a coil is 4.2  at 100 C and the temperature
The resistivity of alloys  Ralloy ; the resistivity of constituent metals
o

70. coefficient of resistance of its material is 0.004/ C. Its resistance at

o

R metal . Then, usually [KCET 1994] 0 C is

o
[KCET 1999]

(a) Ralloy  Rmetal (a) 6.5  (b) 5 

(c) 3  (d) 4 
(b) Ralloy  Rmetal
79. Masses of three wires of copper are in the ratio of 1 : 3 : 5 and their
(c) There is no simple relation between Ralloy and R metal lengths are in the ratio of 5 : 3 : 1. The ratio of their electrical
resistances are [AFMC 2000]
(d) Ralloy  Rmetal (a) 1 : 3 : 5 (b) 5 : 3 : 1
71. Two wires A and B of same material and same mass have radius (c) 1 : 15 : 125 (d) 125 : 15 : 1
2rand r. If resistance of wire A is 34  , then resistance of B will be 80. Conductivity increases
[RPET 1997] in the order of [AFMC 2000]
(a) Al, Ag, Cu (b) Al, Cu, Ag
(a) 544  (b) 272 
(c) Cu, Al, Ag (d) Ag, Cu, Al
(c) 68  (d) 17  81. A uniform wire of resistance R is uniformly compressed along its
length, until its radius becomes n times the original radius. Now
72. Two rods of same material and length have their electric resistance resistance of the wire becomes
in ratio 1 : 2 . When both rods are dipped in water, the correct [KCET 2000]
statement will be [RPMT 1997]
R R
(a) A has more loss of weight (a) (b)
n4 n2
(b) B has more loss of weight
R
(c) Both have same loss of weight (c) (d) nR
n
(d) Loss of weight will be in the ratio 1 : 2
82. The resistance of a conductor is 5 ohm at 50 C and 6 ohm at 100 C.
o o

73. 20 A current flows for 30 seconds in a wire, transfer of charge Its resistance at 0 C is
o
[KCET 2000]
will be [RPMT 1997] (a) 1 ohm (b) 2 ohm
(c) 3 ohm (d) 4 ohm
(a) 2  10 4 C (b) 4  10 4 C
83. If an electron revolves in the path of a circle of radius of 0.5 × 10 –1 0

(c) 6  10 4 C (d) 8  10 4 C m at frequency of 5 × 10 cycles/s the electric current in the circle is

15

(Charge of an electron = 1.6 × 10 C )

–19

[EAMCET 2000]
 Current Electricity 1055

(a) 0.4 mA (b) 0.8 mA 93. Calculate the amount of charge flowing in 2 minutes in a wire of
(c) 1.2 mA (d) 1.6 mA resistance 10  when a potential difference of 20 V is applied
between its ends [Kerala (Engg.) 2001]
84. Equal potentials are applied on an iron and copper wire of same
length. In order to have the same current flow in the two wires, the (a) 120 C (b) 240 C
ratio r (iron)/r (copper) of their radii must be (Given that specific (c) 20 C (d) 4 C
resistance of iron = 1.0  10 7 ohm–m and specific resistance of 94. If a wire of resistance R is melted and recasted to half of its length,
copper = 1.7  10 8 ohm-m) then the new resistance of the wire will be
[KCET (Med.) 2001]
[MP PMT 2000]
(a) R/4 (b) R/2
(a) About 1.2 (b) About 2.4
(c) R (d) 2R
(c) About 3.6 (d) About 4.8
95. The drift velocity does not depend upon [BHU 2001]
85. An electron (charge = 1.6 × 10 coulomb) is moving in a circle of
–19

radius 5.1 × 10 m at a frequency of 6.8 × 10 revolutions/sec. The

–11 15
(a) Cross-section of the wire (b) Length of the wire
equivalent current is approximately (c) Number of free electrons (d) Magnitude of the current
[MP PET 2000]
96. There is a current of 40 ampere in a wire of 10 6 m 2 area of
3 3
(a) 5.1  10 amp (b) 6.8  10 amp cross-section. If the number of free electron per m 3 is 10 29 ,
(c) 1.1  10 3 amp (d) 2.2  10 3 amp then the drift velocity will be [Pb. PMT 2001]

86. A rod of a certain metal is 1.0 m long and 0.6 cm in diameter. Its (a) 1.25 × 10 3 m/s (b) 2.50 × 10 3 m/s
3
resistance is 3.0 × 10 ohm. Another disc made of the same (c) 25.0 × 10 3 m/s (d) 250 × 10 3 m/s
metal is 2.0 cm in diameter and 1.0 mm thick. What is the
97. At room temperature, copper has free electron density of
resistance between the round faces of the disc [MP PET 2000]
8.4  10 per m 3 . The copper conductor has a cross-section of
28

(a) 1.35 × 10 8 ohm (b) 2.70 × 10 7 ohm 10 m and carries a current of 5.4 A. The electron drift velocity in
–6 2

copper is [UPSEAT 2002]

(c) 4.05 × 10 6 ohm (d) 8.10 × 10 5 ohm
(a) 400 m/s (b) 0.4 m/s
87. At what temperature will the resistance of a copper wire become
(c) 0.4 mm/s (d) 72 m/s
three times its value at 0 C (Temperature coefficient of resistance
o

for copper = 4 × 10 per C ) –3 o

98. The resistance of a 5 cm long wire is 10 . It is uniformly stretched
[MP PET 2000] so that its length becomes 20 cm. The resistance of the wire is
(a) 400 C o
(b) 450 C o
(a) 160  (b) 80 
(c) 500 C o
(d) 550 C o
(c) 40  (d) 20 
88. An electron revolves 6 × 10 times/sec in circular loop. The current
15
99. The resistance of an incandescent lamp is [KCET 2002]
in the loop is [MNR 1995; UPSEAT 2000]
(a) Greater when switched off
(a) 0.96 mA (b) 0.96  A
(b) Smaller when switched on
(c) 28.8 A (d) None of these
(c) Greater when switched on
89. The charge of an electron is 1.6 × 10 C. How many electrons strike
–19

the screen of a cathode ray tube each second when the beam (d) The same whether it is switched off or switched on
current is 16 mA [AMU (Med.) 2000]
100. In the figure a carbon resistor has bands of different colours on its
(a) 10 17
(b) 10 19

body as mentioned in the figure. The value of the resistance is

(c) 10 –19
(d) 10 –17
Silver
90. If potential V  100  0.5 Volt and current I  10  0.2 amp (a) 2.2 k 
are given to us. Then what will be the value of resistance [RPET 2001]
(b) 3.3 k 
(a) 10  0.7 ohm (b) 5  2 ohm
(c) 5.6 k 
Red
(c) 0.1  0.2ohm (d) None of these Brown
(d) 9.1 k  White
91. A nichrome wire 50 cm long and one square millimetre cross-
section carries a current of 4A when connected to a 2V battery. 101. By increasing the temperature, the specific resistance of a conductor
and a semiconductor [AIEEE 2002]
The resistivity of nichrome wire in ohm metre is
(a) Increases for both
[EAMCET 2001]
(b) Decreases for both
6 7
(a) 1 × 10 (b) 4 × 10 (c) Increases, decreases
(c) 3 × 10 7 (d) 2 × 10 7 (d) Decreases, increases
102. Which of the following is vector quantity [AFMC 2002]
92. If an observer is moving with respect to a stationary electron, then
he observes [DCE 2001] (a) Current density (b) Current
(a) Only magnetic field (b) Only electric field (c) Wattless current (d) Power

(c) Both (a) and (b) (d) None of the above

 1056 Current Electricity
103. Masses of 3 wires of same metal are in the ratio 1 : 2 : 3 and their 112. A strip of copper and another of germanium are cooled from room
lengths are in the ratio 3 : 2 : 1. The electrical resistances are in temperature to 80 K. The resistance of [AIEEE 2003]
ratio [CPMT 2002]
(a) Each of these increases
(a) 1 : 4 : 9 (b) 9 : 4 : 1
(b) Each of these decreases
(c) 1 : 2 : 3 (d) 27 : 6 : 1
(c) Copper strip increases and that of germanium decreases
104. A current of 1 mA is flowing through a copper wire. How many
electrons will pass a given point in one second (d) Copper strip decreases and that of germanium increases
[e = 1.6 × 10 Coulomb]
–19
[RPMT 2000; MP PMT 2002] 113. The length of a given cylindrical wire is increased by 100 %. Due to
19 15 the consequent decrease in diameter the change in the resistance of
(a) 6.25 × 10 (b) 6.25 × 10 the wire will be [AIEEE 2003]
(c) 6.25 × 10 31 (d) 6.25 × 10 8 (a) 300 % (b) 200 %
105. The drift velocity of free electrons in a conductor is ‘ v’ when a (c) 100 % (d) 50 %
current ‘i’ is flowing in it. If both the radius and current are
doubled, then drift velocity will be [BHU 2002] 114. Express which of the following setups can be used to verify Ohm’s
law [IIT-JEE (Screening) 2003]
v
(a) v (b) A
2
(a) (b)
v v
(c) (d) V V
4 8 A
106. A wire of radius r has resistance R. If it is stretched to a radius of
3r A V
, its resistance becomes [BHU 2002] (c) (d)
4
9R 16R
(a) (b)
16 9
V A
81R 256R 115. We have two wires A and B of same mass and same material. The
(c) (d)
256 81 diameter of the wire A is half of that B. If the resistance of wire A
107. The resistance of a conductor increases with is 24 ohm then the resistance of wire B will be
[CBSE PMT 2002] (a) 12 Ohm (b) 3.0 Ohm
(a) Increase in length (c) 1.5 Ohm (d) None of the above
(b) Increase in temperature 116. In a hydrogen discharge tube it is observed that through a given
(c) Decrease in cross–sectional area
cross-section 3.13  10 15 electrons are moving from right to left
(d) All of these
and 3.12  10 15 protons are moving from left to right. What is the
108. A copper wire has a square cross-section, 2.0 mm on a side. It
electric current in the discharge tube and what is its direction
carries a current of 8 A and the density of free electrons is
8  10 28 m 3 . The drift speed of electrons is equal to (a) 1 mA towards right (b) 1mA towards left
[AMU (Med.) 2002] (c) 2mA towards left (d) 2 mA towards right
3 2
(a) 0.156 × 10 m.s –1
(b) 0.156 × 10 m.s –1
117. A steady current i is flowing through a conductor of uniform cross-
3 2
section. Any segment of the conductor has
(c) 3.12 × 10 m.s –1
(d) 3.12 × 10 m.s –1

[MP PET 1996]

109. Two wires of same material have length L and 2L and cross– (a) Zero charge
sectional areas 4A and A respectively. The ratio of their specific (b) Only positive charge
resistance would be [MHCET 2002] (c) Only negative charge
(a) 1 : 2 (b) 8 : 1 (d) Charge proportional to current i
(c) 1 : 8 (d) 1 : 1 118. The length of the wire is doubled. Its conductance will be
[Kerala PMT 2004]
110. When a current flows through a conductor its temperature (a) Unchanged (b) Halved
[MHCET 2002] (c) Quadrupled (d) 1/4 of the original value
(a) May increase or decrease 119. A source of e.m.f. E = 15 V and having negligible internal resistance
(b) Remains same is connected to a variable resistance so that the current in the
circuit increases with time as i = 1.2 t + 3. Then, the total charge that
(c) Decreases will flow in first five second will be [
(d) Increases (a) 10 C (b) 20 C
111. What length of the wire of specific resistance 48  10 8  m is (c) 30 C (d) 40 C
needed to make a resistance of 4.2  (diameter of wire = 0.4 120. The new resistance of wire of R , whose radius is reduced half, is [J & K CET
mm) (a) 16 R (b) 3 R
[CBSE PMT 2000; Pb. PMT 2002] (c) 2R (d) R
121. A resistance R is stretched to four times its length. Its new
(a) 4.1 m (b) 3.1 m resistance will be [ISM Dhanbad 1994; UPSEAT 2003]
(c) 2.1 m (d) 1.1 m (a) 4 R (b) 64 R
 Current Electricity 1057

(c) R / 4 (d) 16 R (c) 21  10  5%

3
(d) 12  10  5%
3

122. What is the resistance of a carbon resistance which has bands of 130. A thick wire is stretched so that its length become two times.
colours brown, black and brown [DCE 1999]
Assuming that there is no change in its density, then what is the
(a) 100  (b) 1000  ratio of change in resistance of wire to the initial resistance of wire
(c) 10  (d) 1  (a) 2 : 1 (b) 4 : 1
123. The lead wires should have [Pb. PMT 2000]
(a) Larger diameter and low resistance (c) 3 : 1 (d) 1 : 4
(b) Smaller diameter and high resistance 131. The length of the resistance wire is increased by 10%. What is the
(c) Smaller diameter and low resistance corresponding change in the resistance of wire
(d) Larger diameter and high resistance [MH CET 2004]

124. The alloys constantan and manganin are used to make standard (a) 10% (b) 25%
resistance due to they have (c) 21% (d) 9%
[MH CET 2000; NCERT 1990]
(a) Low resistivity 132. The electric field E, current density J and conductivity  of a
(b) High resistivity conductor are related as [Kerala PMT 2005]
(c) Low temperature coefficient of resistance (a)   E / j (b)   j / E
(d) Both (b) and (c)
125. When a potential difference is applied across the ends of a linear (c)   jE (d)   1 / jE
metallic conductor [MP PET 1997]
(a) The free electrons are accelerated continuously from the lower 133. Two wires that are made up of two different materials whose
potential end to the higher potential end of the conductor specific resistance are in the ratio 2 : 3, length 3 : 4 and area 4 : 5.
(b) The free electrons are accelerated continuously from the higher The ratio of their resistances is [Kerala PMT 2005]
potential end to the lower potential end of the conductor (a) 6 : 5 (b) 6 : 8
(c) The free electrons acquire a constant drift velocity from the
lower potential end to the higher potential end of the (c) 5 : 8 (d) 1 : 2
conductor
(d) The free electrons are set in motion from their position of rest Grouping of Resistances
126. The electric resistance of a certain wire of iron is R. If its length and
radius are both doubled, then [CBSE PMT 2004] 1. The potential difference between points A and B of adjoining
(a) The resistance will be doubled and the specific resistance will figure is [CPMT 1991]
be halved 2
(a) V 5
(b) The resistance will be halved and the specific resistance will 3 5
A B
remain unchanged 8
(b) V
(c) The resistance will be halved and the specific resistance will be 9 2V
doubled 5 5
4
(c) V
(d) The resistance and the specific resistance, will both remain 3 5 5
unchanged D C
(d) 2 V
127. A wire of diameter 0.02 metre contains 10 free electrons per cubic
28

2. Two resistors of resistance R1 and R 2 having R1  R 2 are

metre. For an electrical current of 100 A, the drift velocity of the free
connected in parallel. For equivalent resistance R , the correct
electrons in the wire is nearly statement is [CPMT 1978; KCET (Med.) 2000]
[UPSEAT 2004] (a) R  R1  R 2 (b) R1  R  R 2
(a) 1  10 m/s
–19
(b) 5  10 m/s –10
(c) R2  R  (R1  R2 ) (d) R  R1
(c) 2  10 m/s
–4
(d) 8 10 m/s 3
3. A wire of resistance R is divided in 10 equal parts. These parts are
connected in parallel, the equivalent resistance of such connection
128. The following four wires are made of the same material and are at will be [CPMT 1973, 91]
the same temperature. Which one of them has highest electrical (a) 0.01 R (b) 0.1 R
resistance [UPSEAT 2004]
(c) 10 R (d) 100 R
(a) Length = 50 cm, diameter = 0.5 mm
4. The current in the adjoining circuit will be
(b) Length = 100 cm, diameter = 1 mm [IIT 1983; CPMT 1991, 92; MH CET 2002;
Pb. PMT 2001; Kerala PMT 2004]
(c) Length = 200 cm, diameter = 2 mm
1
(d) Length = 300 cm, diameter = 3 mm (a) ampere
45
129. The colour sequence in a carbon resistor is red, brown, orange and 1 i
silver. The resistance of the resistor is (b) ampere
15
30 30
[DCE 2004] 1 2V
(c) ampere
(a) 21  10  10% 3
(b) 23  10  10 1 10
30
 1058 Current Electricity

1 13. The effective resistance between the points A and B in the figure
(d) ampere is [MPDPET 1994]
5
5. There are 8 equal resistances R. Two are connected in parallel, such 3 3
four groups are connected in series, the total resistance of the (a) 5
system will be [MP PMT 1987]
6
(a) R / 2 (b) 2 R (b) 2  A C
(c) 4 R (d) 8 R
(c) 3
6. Three resistances of one ohm each are connected in parallel. Such 3 3
connection is again connected with 2 / 3  resistor in series. The (d) 4 
B
resultant resistance will be [MP PMT 1985] 14. Three resistances of magnitude 2, 3 and 5 ohm are connected in
5 3 parallel to a battery of 10 volts and of negligible resistance. The
(a)  (b) 
3 2 potential difference across 3  resistance will be

2 (a) 2 volts (b) 3 volts

(c) 1 (d) 
3 (c) 5 volts (d) 10 volts
7. The lowest resistance which can be obtained by connecting 10 15. A current of 2 A flows in a system of conductors as shown. The
resistors each of 1/10 ohm is potential difference (V A  VB ) will be [CPMT 1975, 76]
[MP PMT 1984; EAMCET 1994]
A
(a) 1 / 250  (b) 1 / 200 
(a) 2 V 2 3
(c) 1 / 100  (d) 1 / 10  2A
(b) 1 V
8. The reading of the ammeter as per figure shown is D C
1 (c) 1 V 3 2
(a) A
8 2
(d) 2 V B
3 2V
(b) A 2 16. Referring to the figure below, the effective resistance of the network
4 A
is [NCERT 1973, 75]
1 2
(c) A r r r
2 (a) 2 r
2
(d) 2 A r
(b) 4 r
9. Three resistors each of 2 ohm are connected together in a triangular (c) 10 r
shape. The resistance between any two vertices will be (d) 5r / 2 r r r
[CPMT 1983; MP PET 1990; MP PMT 1993; DCE 2004]
6
(a) 4/3 ohm (b) 3/4 ohm 17. Two resistances are joined in parallel whose resultant is ohm.
8
(c) 3 ohm (d) 6 ohm One of the resistance wire is broken and the effective resistance
10. There are n similar conductors each of resistance R . The resultant becomes 2  . Then the resistance in ohm of the wire that got
resistance comes out to be x when connected in parallel. If they are broken was
connected in series, the resistance comes out to be [DPMT 2004] [CPMT 1976; DPMT 1982]
(a) x /n 2
(b) n x 2 (a) 3/5 (b) 2
(c) 6/5 (d) 3
(c) x /n (d) nx
18. Given three equal resistors, how many different combination of all
11. Equivalent resistance between A and B will be [CPMT 1981] the three resistors can be made [NCERT 1970]
(a) Six (b) Five
3 3
(c) Four (d) Three
(a) 2 ohm
19. Lamps used for household lighting are connected in
(b) 18 ohm (a) Series (b) Parallel
3 3 (c) Mixed circuit (d) None of the above
3 3
(c) 6 ohm
20. The equivalent resistance of resistors connected in series is always [CPMT 1984;
(d) 3.6 ohm (a) Equal to the mean of component resistors
A B
3 3 (b) Less than the lowest of component resistors
12. A wire has a resistance of 12 ohm. It is bent in the form of
equilateral triangle. The effective resistance between any two corners (c) In between the lowest and the highest of component resistors
of the triangle is (d) Equal to sum of component resistors
21. A cell of negligible resistance and e.m.f. 2 volts is connected to series
(a) 9 ohms (b) 12 ohms combination of 2, 3 and 5 ohm. The potential difference in volts
(c) 6 ohms (d) 8/3 ohms between the terminals of 3 ohm resistance will be
(a) 0.6 (b) 2/3
 Current Electricity 1059

(c) 3 (d) 6 1
(c) One eight (d) th
22. Four wires of equal length and of resistances 10 ohms each are 16
connected in the form of a square. The equivalent resistance
between two opposite corners of the square is 29. Four resistances are connected in a circuit in the given figure. The
electric current flowing through 4 ohm and 6 ohm resistance is
[NCERT 1977] respectively [MP PET 1993]
(a) 10 ohm (b) 40 ohm 4 6
(c) 20 ohm (d) 10/4 ohm (a) 2 amp and 4 amp
23. Two resistors are connected (a) in series (b) in parallel. The (b) 1 amp and 2 amp 4 6
equivalent resistance in the two cases are 9 ohm and 2 ohm
(c) 1 amp and 1 amp
respectively. Then the resistances of the component resistors are [CPMT 1984]
(d) 2 amp and 2 amp
(a) 2 ohm and 7 ohm (b) 3 ohm and 6 ohm 20V
30. An infinite sequence of resistance is shown in the figure. The
(c) 3 ohm and 9 ohm (d) 5 ohm and 4 ohm
resultant resistance between A and B will be, when R1  1 ohm
24. Resistors of 1, 2, 3 ohm are connected in the form of a triangle. If a
and R 2  2 ohm [MP PET 1993]
1.5 volt cell of negligible internal resistance is connected across 3
ohm resistor, the current flowing through this resistance will be [CPMT 1984]
R1 R1 R1 R1 R1
(a) 0.25 amp (b) 0.5 amp A

(c) 1.0 amp (d) 1.5 amp R2 R2 R2 R2 R2

25. Resistances of 6 ohm each are connected in the manner shown in
adjoining figure. With the current 0.5 ampere as shown in figure, B
(a) Infinity (b) 1 
the potential difference VP  VQ is
(c) 2 (d) 1 .5 
6 6 6 [CPMT 1989]
31. In the figure, the value of resistors to be connected between C and
P 6 Q D so that the resistance of the entire circuit between A and B does
0.5 A not change with the number of elementary sets used is
6 6
(a) 3.6 V (b) 6.0 V R R R R C
(c) 3.0 V (d) 7.2 V A

26. The equivalent resistance of the arrangement of resistances shown in R R R R R

adjoining figure between the points A and B is
[CPMT 1990; BVP 2003] (a) R B R R (b) R( 3R  1) R D
8
(a) 6 ohm
16 20 (c) 3 R (d) R( 3  1)
(b) 8 ohm 16 32. In the figure shown, the total resistance between A and B is
(c) 16 ohm A B
9 2 C 1 1 1 1 1
(d) 24 ohm 6 A
18
8 8 4
27. In the network of resistors shown in the adjoining figure, the
equivalent resistance between A and B is
B 2 D 1 1 1 1 1
3 3 3 3 3 3
(a) 12  (b) 4 
A B
(c) 6 (d) 8 
3 3 3 3 3 3
33. The current from the battery in circuit diagram shown is
2 A 7 [IIT 1989]
(a) 54 ohm (b) 18 ohm
(c) 36 ohm (d) 9 ohm (a) 1 A 15V

28. A wire is broken in four equal parts. A packet is formed by keeping (b) 2 A
6 1
0.5
the four wires together. The resistance of the packet in comparison (c) 1.5 A
to the resistance of the wire will be
(d) 3 A
[MP PET 1985; AFMC 2005] 8 B
34. In the given figure, when key K is opened, 10
the reading of the
(a) Equal (b) One fourth ammeter A will be 10V
(a) 50 A + –

5
E A D

4 A
B C
K
 1060 Current Electricity
(b) 2 A resistance of the thicker wire is 10  . The total resistance of the
(c) 0.5 A combination will be [CBSE PMT 1995]

10 40
(d) A (a) 40  (b) 
9 3
35. In the given circuit, the potential of the point E is 5
(c)  (d) 100 
[MP PMT 2003] 2
(a) Zero + – E 1 42. The equivalent resistance of the following infinite network of
A D
8V
resistances is [AIIMS 1995]
(b)  8 V
2 2 2
(c) 4 /3V
C 2 2 2
(d) 4/3 V B
5
36. If a resistance R 2 is connected in parallel with the resistance R in
2 2 2
the circuit shown, then possible value of current through R and the (a) Less than 4 
possible value of R 2 will be
(b) 4 
I
(a) ,R R2 (c) More than 4  but less than 12 
3
(b) I, 2 R (d) 12 
R 43. In the figure given below, the current passing through 6  resistor
I
(c) , 2R I is [Manipal MEE 1995]
3
(a) 0.40 ampere 6
I A
(d) ,R + – (b) 0.48 ampere
2 1.2 A
37. Four wires AB, BC, CD, DA of resistance 4 ohm each and a fifth (c) 0.72 ampere
wire BD of resistance 8 ohm are joined to form a rectangle ABCD of (d) 0.80 ampere
which BD is a diagonal. The effective resistance between the points 4
44. Three equal resistances each of value R are joined as shown in the
A and B is [MP PMT 1994]
figure. The equivalent resistance between M and N is
(a) 24 ohm (b) 16 ohm
[MP PET 1995]
4 8 (a) R
(c) ohm (d) ohm
3 3 (b) 2R
38. A battery of e.m.f. 10 V is connected to resistance as shown in figure. M R
R
The potential difference VA  VB between the points A and B is (c) [MP PMT 1994] L R R N Z
2
R
(a) 2V 1 A 3 (d)
3
(b) 2V 3
45. The equivalent resistance between points A and B of an infinite
(c) 5V 3 B 1 network of resistances each of 1  connected as shown, is
20 10V
(d) V 1 1 1
11
A
39. Three resistances, each of 1 ohm, are joined in parallel. Three such
combinations are put in series, then the resultant resistance will be [MP PMT 1994]
1 1 1
(a) 9 ohm (b) 3 ohm
(a) Infinite (b) 2 
1 B
(c) 1 ohm (d) ohm
3 1 5
(c)  (d) Zero
40. A student has 10 resistors of resistance ‘r’. The minimum resistance 2
made by him from given resistors is 46. A copper wire of resistance R is cut into ten parts of equal length.
[AFMC 1995] Two pieces each are joined in series and then five such combinations
r are joined in parallel. The new combination will have a resistance [
(a) 10 r (b)
10 R
(a) R (b)
r r 4
(c) (d)
100 5 R R
(c) (d)
41. Two wires of same metal have the same length but their cross- 5 25
sections are in the ratio 3 : 1 . They are joined in series. The
 Current Electricity 1061

47. A wire has resistance 12  . It is bent in the form of a circle. The (c) 4.92 A
effective resistance between the two points on any diameter is equal (d) 2 A
to [JIPMER 1999] 54. What is the current (i) in the circuit as shown in figure
(a) 12  (b) 6  [AIIMS 1998]
i R2 = 2
(c) 3 (d) 24  (a) 2 A

R3 = 2
48. In the circuit shown, the point ‘B’ is earthed. The potential at the (b) 1.2 A
3V R1 = 2
point ‘A’ is (c) 1 A
5 7 B
(a) 14 V A
(d) 0.5 A
(b) 24 V 10 R4 = 2
55. n equal resistors are first connected in series and then connected in
(c) 26 V 50V C parallel. What is the ratio of the maximum to the minimum
3 resistance [KCET 1994]
(d) 50 V
E D 1
49. Three resistors each of 4  are connected together to form a (a) n (b)
n2
network. The equivalent resistance of the network cannot be
(a) 1.33  (b) 3 .0  1
(c) n2 (d)
n
(c) 6 .0  (d) 12.0 
56. A uniform wire of 16  is made into the form of a square. Two
50. In the circuit shown below, the cell has an e.m.f. of 10 V and internal opposite corners of the square are connected by a wire of resistance
resistance of 1 ohm. The other resistances are shown in the figure. 16  . The effective resistance between the other two opposite
The potential difference VA  VB is corners is [EAMCET (Med.) 1995]

[MP PMT 1997] (a) 32  (b) 20 

(a) 6 V E=10V
r=1 (c) 8 (d) 4 
(b) 4 V
4 A 2 1 57. For what value of R the net resistance of the circuit will be 18 ohms
(c) 2 V
R
(d) 2 V 2 B 4 (a) 8
51. A wire of resistance R is cut into ‘n’ equal parts. These parts are (b) 10  10 10
then connected in parallel. The equivalent resistance of the
(c) 16  10
combination will be [MP PMT/PET 1998; BHU 2005]
(d) 24  10 10 10
R A B
(a) nR (b) 58. In the figure, current through the 3  resistor is 0.8 ampere, then
n
potential drop through 4  resistor is
n R
(c) (d) [CBSE PMT 1993; AFMC 1999; MP PMT 2004]
R n2
3
52. The resistance between the terminal points A and B of the given (a) 9.6 V
infinitely long circuit will be [MP PMT/PET 1998]
(b) 2.6 V 4
1 1 1 (c) 4.8 V 6
A
(d) 1.2 V
+ –
1 1
Upto 59. Three resistances 4  each of are connected in the form of an
infinity
equilateral triangle. The effective resistance between two corners is
B (a) 8  (b) 12 
1 1 1
3 8
(a) ( 3  1) (b) (1  3 ) (c)  (d) 
8 3
(c) (1  3 ) (d) (2  3 ) 60. What will be the equivalent resistance between the two points A and
D [CBSE PMT 1996]
53. The current in the given circuit is [CBSE PMT 1999]
10 10 10
A B
(a) 8.31 A
(b) 6.82 A 10 10
RA = 3 RB = 6
4.8V
C D
RC = 6 10 10 10
 1062 Current Electricity
67. In the given figure, the equivalent resistance between the points A
and B is [AIIMS 1999]
(a) 10  (b) 20  R2 = 4 
(a) 8 
(c) 30  (d) 40 
61. What is the equivalent resistance between A and B in the figure (b) 6  R1 = 2 R4 = 2 
below if R  3  [SCRA 1996] (c) 4  A B
A B
R3 = 4 
(a) 9 (d) 2 
R R 68. An infinite ladder network is arranged with resistances R and 2 R as
(b) 12  R
shown. The effective resistance between terminals A and B is
(c) 15 
R R R
(d) None of these R R A
62. What is the equivalent resistance between A and B
[BHU 1997; MP PET 2001] 2R 2R 2R
2
(a) R
3 B
3 C (a)  (b) R
(b) R
2 A B (c) 2 R (d) 3 R
2R 2R D R
R 69. If all the resistors shown have the value 2 ohm each, the equivalent
(c) resistance over AB is [JIPMER 1999]
2
(d) 2 R (a) 2 ohm
A B
63. The current in the following circuit is [CBSE PMT 1997] (b) 4 ohm
1 2
(a) A (c) 1 ohm
8 3
2
(b) A 2
9 3 (d) 2 ohm
2V 3 3
2
(c) A
3 3
70. A battery of emf 10 V and internal resistance 3  is connected to a
(d) 1 A resistor as shown in the figure. If the current in the circuit is 0.5 A.
then the resistance of the resistor will be [
64. What is the equivalent resistance of the circuit [KCET 1998]

2 (a) 19 
4V, 1 
(a) 6 + – 2 (b) 17 
2 A
(b) 7  (c) 10  R
4 (d) 12 
(c) 8
71. The potential drop across the 3 resistor is [CPMT 2000]
(d) 9  V 3

65. 10 wires (same length, same area, same material) are connected in (a) 1 V 4

parallel and each has 1 resistance, then the equivalent resistance (b) 1.5 V 6
will be [RPMT 1999] (c) 2 V
(a) 10  (b) 1  (d) 3 V
72. V
In the given figure, potential difference3between A and B is
(c) 0.1  (d) 0.001 
[RPMT 2000]
66. The equivalent resistance of the circuit shown in the figure is (a) 0 10K D
[CPMT 1999] A
(b) 5 volt 30 V
2
(a) 8  (c) 10 volt 10K 10K
2 2
(b) 6  (d) 15 volt
2 B
(c) 5  73. If each resistance in the figure is of 9  then reading of ammeter is

(d) 4 
+
9V
–

A
 Current Electricity 1063

(d) 10 ohms
80. A uniform wire of resistance 9  is cut into 3 equal parts. They are
connected in the form of equilateral triangle ABC. A cell of e.m.f. 2
(a) 5 A (b) 8 A V and negligible internal resistance is connected across B and C.
Potential difference across AB is
(c) 2 A (d) 9 A
[Kerala (Engg.) 2001]
74. Four resistances 10 , 5 , 7  and 3  are connected so that they
(a) 1 V (b) 2 V
form the sides of a rectangle AB, BC, CD and DA respectively.
Another resistance of 10  is connected across the diagonal AC. The (c) 3 V (d) 0.5 V
equivalent resistance between A and B is 81. [EAMCET of resistances 2 , 4  and 8  are connected in
(Med.) 2000]
The resistors
parallel, then the equivalent resistance of the combination will be[KCET 2001]
(a) 2  (b) 5 
(c) 7  (d) 10  8 7
(a)  (b) 
7 8
75. Two wires of equal diameters, of resistivities 1 and  2 and
lengths l and l , respectively, are joined in series. The equivalent 7 4
1 2
(c)  (d) 
resistivity of the combination is 4 9
[EAMCET (Engg.) 2000]
82. Effective resistance between A and B is [UPSEAT 2001]
 1 l1   2 l 2  1 l 2   2 l1
(a) (b) (a) 15 
l1  l 2 l1  l 2 5
(b) 5 
 1 l 2   2 l1  1 l1   2 l 2
(c) (d) 5
l1  l 2 l1  l 2 5 A 5 5
(c)  B
76. Four resistances of 100  each are connected in the form of square. 2
Then, the effective resistance along the diagonal points is [MH CET 2000]
(d) 20  5
(a) 200  (b) 400 
(c) 100  (d) 150  83.
12
The effective resistance of two resistors in parallel is . If one
77. Equivalent resistance between the points A and B is (in ) 7
[AMU (Engg.) 2000] of the resistors is disconnected the resistance becomes 4 . The
resistance of the other resistor is [MH CET 2002]

A 1 1 1 1 1 B (a) 4  (b) 3 

12 7
(c)  (d) 
7 12
84. Two resistance wires on joining in parallel the resultant resistance is
1 1
(a) (b) 1 6
5 4 ohms . One of the wire breaks, the effective resistance is 2
5
1 1 ohms. The resistance of the broken wire is
(c) 2 (d) 3
3 2 [MP PET 2001, 2002]
78. Two wires of the same material and equal length are joined in
3
parallel combination. If one of them has half the thickness of the (a) ohm (b) 2 ohm
other and the thinner wire has a resistance of 8 ohms, the resistance 5
of the combination is equal to 6
[AMU (Engg.) 2000] (c) ohm (d) 3 ohm
5
5 8 85. In the circuit, the potential difference across PQ will be nearest to
(a) ohms (b) ohms
8 5
100 
3 8
(c) ohms (d) ohms (a) 9.6 V
8 3
(b) 6.6 V 48 V
79. In the circuit shown here, what is the value of the unknown resistor 80 
R so that the total resistance of the circuit between points P and Q (c) 4.8 V
100  Q
is also equal to R [MP PET 2001] (d) 3.2 V 20 
(a) 3 ohms
10 86. Three resistors are connected to form the sides of a triangleP ABC,
(b) 39 ohms the resistance of the sides AB, BC and CA are 40 ohms, 60 ohms
3 and 100 ohms respectively. The effective resistance between the
P Q
(c) 69 ohms 3 R points A and B in ohms will be
 1064 Current Electricity
[JIPMER 2002] 20
(a) V 8 B 6
(a) 32 (b) 64 7
(c) 50 (d) 200 40
(b) V
87. Find the equivalent resistance across AB [Orissa JEE 2002] 7 A
A 4 3
10
(a) 1  2 (c) V
2 7
(b) 2  2 10 V
(d) 0
(c) 3  2
2 95. In the circuit shown below, The reading of the voltmeter V is
(d) 4  B 4 16
88. The equivalent resistance between x and y in the circuit shown is V PMT 2002]
(a) 12 [MP
(a) 10  (b) 8 V
10  2A
(b) 40  (c) 20 V V

(c) 20  (d) 16 V 16 4

10  10  10 
x y 96. A wire has a resistance of 12 ohm. It is bent in the form of
5
(d)  equilateral triangle. The effective resistance between any two corners
2 10 
of the triangle is
89. The equivalent resistance between the points P and Q of the circuit
given is [Pb. PMT 2002] (a) 9 ohms (b) 12 ohms
R (c) 6 ohms (d) 8/3 ohms
(a)
4 97. A series combination of two resistors 1  each is connected to a 12
R R R
(b)
R P Q V battery of internal resistance 0.4 . The current flowing through
3 it will be [MH CET (Med.) 1999]
(c) 4 R (a) 3.5 A (b) 5 A
(d) 2 R (c) 6 A (d) 10 A
90. Two wires of the same dimensions but resistivities 1 and  2 are 98. In the circuit shown in the adjoining figure, the current between B
connected in series. The equivalent resistivity of the combination is [KCET
and2003]
D is zero, the unknown resistance is of
1   2 B [CPMT 1986]
(a) 1   2 (b)
2
4 X
(c) 1  2 (d) 2( 1   2 )
12
91. Three unequal resistors in parallel are equivalent to a resistance 1 A C
1
1
ohm. If two of them are in the ratio 1 : 2 and if no resistance value (a) 4
is fractional, the largest of the three resistances in ohms is [EAMCET 2003]
1
(a) 4 (b) 6 (b) 2  3
D
(c) 8 (d) 12 (c) 3
92. A 3volt battery with negligible internal resistance is connected in a
(d) em.f. of a cell is required to find the value of X
circuit as shown in the figure. The current I, in the circuit will be [AIEEE 2003]
99. In the circuit shown in the figure, the current flowing in 2 
I
(a) 1/3 A resistance [CPMT 1989; MP PMT 2004]
(b) 1 A 3
3V 3
(c) 1.5 A (a) 1.4 A 10 2
(d) 2 A 3 1.4A
(b) 1.2 A G
93. Find the equivalent resistance between the points a and b
[BHU 2003; CPMT 2004]
(c) 0.4 A
4 (d) 1.0 A 25 5
(a) 2 
100. Five resistors are connected as shown in the diagram. The equivalent
(b) 4  2 10 8
a b resistance between A and B is
(c) 8  [MP PMT 1996]
C
(d) 16  4 (a) 6 ohm 5 4
94. The potential difference between point A & B is (b) 9 ohm
[BHU 2003; CPMT 2004; MP PMT 2005] A 9
B
10 8
D
 Current Electricity 1065

(c) 12 ohm (c) 90  (d) 110 

(d) 15 ohm 106. Five resistances are connected as shown in the figure. The effective
resistance between the points A and B is
101. In the figure given the value of X resistance will be, when the p.d.
[MP PMT 1999; KCET 2001; BHU 2001, 05]
between B and D is zero [MP PET 1993]
10
B (a) 
X 3 2 3
6
20
(b)  7
8 3 3 A B
15 4
A C (c) 15  6
4
15 6 (d) 6 
4 107. In the given figure, when galvanometer shows no deflection, the
6 4 current (in ampere) flowing through 5  resistance will be
(a) 4 ohm D (b) 6 ohm
(c) 8 ohm (d) 9 ohm (a) 0.5
102. The effective resistance between points A and B is (b) 0.6 8 2
[NCERT 1974; MP PMT 2000] 2.1A
(c) 0.9
G
10 10 (d) 1.5
(a) 10 
A B 20 5
108. In the Wheatstone's bridge shown, P  2 , Q  3 , R  6 
(b) 20  10
and S  8  . In order to obtain balance, shunt resistance across 'S'
(c) 40  10 10 must be [SCRA 1998]
(d) None of the above three values
103. Five resistors of given values are connected together as shown in the P Q
figure. The current in the arm BD will be (a) 2
B [MP PMT 1995]
(b) 3 
R R
S R
C
(c) 6
4R
A
(d) 8 
R R
109. Five equal resistances each of value R are connected in a form
D shown alongside. The equivalent resistance of the network
(a) Half the current in the arm ABC (a) Between the points B and D is R B
(b) Zero R R R
(c) Twice the current in the arm ABC (b) Between the points B and D is
2
(d) Four times the current in the arm ABC A R C
(c) Between the points A and C is R
104. In the network shown in the figure, each of the resistance is equal to
R R
2  . The resistance between the points A and B is (d) Between the points A and C is
R
[CBSE PMT 1995] 2 D
(a) 1  110. In the circuit shown below the resistance of the galvanometer is 20
(b) 4 . In which case of the following alternatives are the currents
arranged strictly in the decreasing order
(c) 3 A i1
(d) 2 B (a) i, i i , i
10 ig 100
1, 2 g

105. In the arrangement of resistances shown below, the effective

(b) i, i i , i i2 G
resistance between points A and B is 2, 1 g

[MP PMT 1997; RPET 2001] (c) i, i i , i

2, g 1

2 20
5 10 15 (d) i, i i , i
1, g 2
i
P
111. Potential difference between the points P and Q in the electric
circuit shown is 2V 0 [KCET 1999]
A 10 10 B
P i = 1.5 A

Q (a) 4.5 V RA = 2 RB = 4
10 20 30 (b) 1.2 V
(a) 20  (b) 30  3
(c) 2.4 V
RD = 6 RC = 12
Q
 1066 Current Electricity
(d) 2.88 V (b) 40 
112. The current between B and D in the given figure is (c) 30 
B [RPET 2000; DCE 2001]
(d) 20 
30 30 118. If each of the resistance of the network shown in the figure is R, the
(a) 1 amp equivalent resistance between A and B is
A 60 C [KCET 2002]
(b) 2 amp
l
(a) 5 R
(c) Zero 30 30
(d) 0.5 amp D (b) 3 R R
R R
30V
113. In the given figure, equivalent resistance between A and B will be (c) R [CBSE PMT 2000]
R A
14 (d) R/2 B
(a)  R
3 119. The equivalent resistance of the following diagram A and B is
3 3 4 2
(b)  (a)  3 3
14 3
A 7 B (b) 9  A 3 B
9
(c)  (c) 6 
14 6 8 3 3
(d) None of these
14
(d)  120. Thirteen resistances each of resistance R ohm are connected in the
9 circuit as shown in the figure below. The effective resistance
114. In a typical Wheatstone network, the resistances in cyclic order are between A and B is [KCET 2003]
A = 10 , B = 5 , C = 4  and D = 4  for the bridge to be (a) 2R  R R
balanced [KCET 2000]
4R R
(b)  R R R
A = 10  B=5 3
A R B
2R
(c) 
3 R
R R R
D=4 C=4 (d) R 
121. In a Wheatstone’s bridge all the four R equal resistance R.
R arms have
(a) 10  should be connected in parallel with A
If the resistance of the galvanometer arm is also R, the equivalent
(b) 10  should be connected in series with A resistance of the combination as seen by the battery is
(c) 5  should be connected in series with B (a)
R
(b) R
2
(d) 5  should be connected in parallel with B
R
115. In the circuit shown in figure, the current drawn from the battery is (c) 2 R (d)
4A. If 10  resistor is replaced by 20  resistor, then current 4
drawn from the circuit will be 122. For what value of unknown resistance X, the potential difference
between B and D will be zero in the circuit shown in the figure
[KCET 2000; CBSE PMT 2001]
B
1 3 1
(a) 1 A (a) 4  12

(b) 2 A 10
10 1
(b) 6  A C
(c) 3 A 7 21 1
4A
(c) 2  X
(d) 0 A
+ – 1
116. Calculate the equivalent resistance between A and B (d) 5  6
D
[UPSEAT 2001] 123. Which arrangement of four identical resistances should be used to
9 draw maximum energy from a cell of voltage V
3 3 3
(a)  [MP PMT 2004]
2 (a)
B
(b) 3  A
3 3
(c) 6 
5 (b)
(d)  3 3 3
3
117. The equivalent resistance between P and Q in the given figure, is [MH CET (Med.) 2001]

20  20  (c)
(a) 50 
P 20  Q

20  20 
 Current Electricity 1067

132. A parallel combination of two resistors, of 1  each, is connected in

(d) series with a 1.5  resistor. The total combination is connected
across a 10 V battery. The current flowing in the circuit is
124. An unknown resistance R is connected in series with a resistance of
1

10 . This combinations is connected to one gap of a metre bridge (a) 5 A (b) 20 A

while a resistance R is connected in the other gap. The balance
2 (c) 0.2 A (d) 0.4 A
point is at 50 cm. Now, when the 10  resistance is removed the
balance point shifts to 40 cm. The value of R is (in ohm)
1
133. [KCET
If you are provided three resistances 2 , 3  and 6 . How will
2004]
(a) 60 (b) 40 you connect them so as to obtain the equivalent resistance of 4 
(c) 20 (d) 10 3 6
125. A wire has a resistance of 6 . It is cut into two parts and both half (a) (b) 3 2
values are connected in parallel. The new resistance is .... [KCET 2004]
2
(a) 12  (b) 1.5 
6
(c) 3  (d) 6  3
126. Six equal resistances are connected between points P, Q and R as (c) 2 (d) None of these
shown in the figure. Then the net resistance will be maximum
between [IIT-JEE (Screening) 2004] 6
P 134. The equivalent resistance and potential difference between A and B
for the circuit is respectively [Pb. PMT 2003]
(a) P and Q 6
(b) Q and R (a) 4 , 8 V
(c) P and R (b) 8 , 4 V 6
2A 2.5
(d) Any two points (c) 2 , 2 V C D
Q R A B
127. The total current supplied to the circuit by the battery is (d) 16 , 8 V 3
[AIEEE 2004] 135. Five equal resistances each of resistance R are connected as shown
(a) 1 A 2 in the figure. A battery of V volts is connected between A and B.
6V 6 The current flowing in AFCEB will be
(b) 2 A 3 [CBSE PMT 2004]
(c) 4 A
1.5 3V C
(d) 6 A (a)
R
128. An electric current is passed through a circuit containing two wires
of the same material, connected in parallel. If the lengths and radii V R
(b) R R
of the wires are in the ratio of 4/3 and 2/3, then the ratio of the R F
currents passing through the wire will be
V
[AIEEE 2004] (c)
2R R A
(a) 3 (b) 1/3 R B
(c) 8/9 (d) 2 2V D E
(d)
129. If a rod has resistance 4  and if rod is turned as half cycle then R
the resistance along diameter [BCECE 2004] 136. For the network shown in the figure the value of the current i is
(a) 1.56  (b) 2.44  9V
(a) 2
(c) 4  (d) 2  35
130. If three resistors of resistance 2, 4 and 5  are connected in 5V 4
parallel then the total resistance of the combination will be (b) 4
18 3
[Pb. PMT 2004]
20 19 5V 6
(c)
(a)  (b)  9
19 20 i
19 10 18V
(c)  (d)  (d) V
10 19 5
131. In circuit shown below, the resistances are given in ohms and the 137. When a wire of uniform cross-section a, length l and resistance R is
battery is assumed ideal with emf equal to 3 volt. The voltage across bent into a complete circle, resistance between any two of
the resistance R is diametrically opposite points will be
4

[CBSE PMT 2005]

[UPSEAT 2004; Kerala PMT 2004]
R R
50  (a) (b)
(a) 0.4 V 4 8
R1
+ R3 60  R4 30  R
(b) 0.6 V (c) 4R (d)
3V
2
(c) 1.2 V – 50  R2
R5 30  138. The current in a simple series circuit is 5.0 amp. When an additional
(d) 1.5 V resistance of 2.0 ohms is inserted, the current drops to 4.0 amp. The
original resistance of the circuit in ohms was
 1068 Current Electricity
(a) 1.25 (b) 8 4. By a cell a current of 0.9 A flows through 2 ohm resistor and 0.3 A
(c) 10 (d) 20 through 7 ohm resistor. The internal resistance of the cell is [KCET 2003]
(a) 0 .5  (b) 1 .0 
139. In the circuit given E = 6.0 V, R = 100 ohms, R = R = 50 ohms, R =
1 2 3 4

75 ohms. The equivalent resistance of the circuit, in ohms, is 2  2005]

(c) 1 .[KCET (d) 2.0 
(a) 11.875 R1 5. The e.m.f. of a cell is E volts and internal resistance is r ohm. The
(b) 26.31 resistance in external circuit is also r ohm. The p.d. across the cell
i will be [CPMT 1985; NCERT 1973]
(c) 118.75 R4 (a) E/2 (b) 2E
E R2 R3
(d) None of these (c) 4E (d) E/4
140. By using only two resistance coils-singly, in series, or in parallel one
6. A cell of e.m.f. E is connected with an external resistance R , then
should be able to obtain resistances of 3, 4, 12 and 16 ohms. The
separate resistances of the coil are p.d. across cell is V . The internal resistance of cell will be [MNR 1987; Kerala P
[KCET 2005] (E  V )R (E  V )R
(a) 3 and 4 (b) 4 and 12 (a) (b)
E V
(c) 12 and 16 (d) 16 and 3
141. In the given circuit, the voltmeter records 5 volts. The resistance of (V  E)R (V  E)R
(c) (d)
the voltmeter in ohms is [KCET 2005] V E
V 7. Two cells, e.m.f. of each is E and internal resistance r are
(a) 200 connected in parallel between the resistance R . The maximum
100 50 energy given to the resistor will be, only when
(b) 100
[MNR 1988; MP PET 2000; UPSEAT 2001]
(c) 10
10 V (a) R r/2 (b) Rr
(d) 50
(c) R  2r (d) R0
Kirchhoff's Law, Cells 8. Kirchhoff's first law i.e . i  0 at a junction is based on the law
of conservation of [CBSE PMT 1997; AIIMS 2000;
1. In the adjoining circuit, the battery E1 has an e.m. f . of 12 volt MP PMT 2002; RPMT 2001; DPMT 2005]
and zero internal resistance while the battery E has an e.m. f . of (a) Charge (b) Energy
2 volt . If the galvanometer G reads zero, then the value of the (c) Momentum (d) Angular momentum
resistance X in ohm is 9. Kirchhoff's second law is based on the law of conservation of
[NCERT 1990; AIEEE 2005] [RPET 2003; MH CET 2001]
500 
(a) 10 A G B (a) Charge (b) Energy
(b) 100 (c) Momentum (d) Sum of mass and energy
(c) 500 E1 X E 10. The figure below shows currents in a part of electric circuit. The
current i is [CPMT 1981; RPET 1999]
(d) 200
D C (a) 1.7 amp 1amp
2. The magnitude and direction of the current in the circuit shown will 2amp
be [CPMT 1986, 88] (b) 3.7 amp 1.3amp

1 e 2 (c) 1.3 amp

a b 2amp
10V 4V (d) 1 amp
i
11. The terminal potential difference of a cell is greater than its e.m.f.
3
when it is
7 d c (a) Being discharged
(a) A from a to b through e
3 (b) In open circuit
7 (c) Being charged
(b) A from b to a through e
3
(d) Being either charged or discharged
(c) 1A from b to a through e
(d) 1A from a to b through e 12. In the circuit shown, potential difference between X and Y will be
3. A cell of e.m. f . 1 .5 V having a finite internal resistance is (a) Zero 40 X Y
connected to a load resistance of 2  . For maximum power (b) 20 V
transfer the internal resistance of the cell should be (c) 60 V
[BIT 1988]
(a) 4 ohm (b) 0.5 ohm (d) 120 V 20
(c) 2 ohm (d) None of these
120V
 Current Electricity 1069

13. In the above question, potential difference across the 40  [MNR 1983]
resistance will be (a) The current decreases (b) The current increases
(a) Zero (b) 80 V (c) The e.m.f. increases (d) The e.m.f. decreases
(c) 40 V (d) 120 V 24. The internal resistance of a cell depends on
14. In the circuit shown, A and V are ideal ammeter and voltmeter (a) The distance between the plates
respectively. Reading of the voltmeter will be (b) The area of the plates immersed
2V (c) The concentration of the electrolyte
(a) 2 V (d) All the above
(b) 1 V 25. n identical cells each of e.m.f. E and internal resistance r are
A V connected in series. An external resistance R is connected in series
(c) 0.5 V to this combination. The current through R is
(d) Zero 1 1 [DPMT 2002]
15. When a resistance of 2ohm is connected across the terminals of a cell, nE nE
the current is 0.5 amperes. When the resistance is increased to 5 ohm, (a) (b)
R  nr nR  r
the current is 0.25 amperes. The internal resistance of the cell is [MP PMT 1996]
E nE
(a) 0.5 ohm (b) 1.0 ohm (c) (d)
R  nr R r
(c) 1.5 ohm (d) 2.0 ohm 26. A cell of internal resistance r is connected to an external resistance
R. The current will be maximum in R, if
16. The terminal potential difference of a cell when short-circuited is
[CPMT 1982]
( E = E.M.F. of the cell)
(a) R  r (b) R  r
(a) E (b) E / 2
(c) R  r (d) R  r / 2
(c) Zero (d) E / 3
27. To get the maximum current from a parallel combination of n
17. A primary cell has an e.m.f. of 1.5 volts, when short-circuited it gives identical cells each of internal resistance r in an external resistance
a current of 3 amperes. The internal resistance of the cell is R, when[CPMT 1976, 83] [DPMT 1999]
(a) 4.5 ohm (b) 2 ohm (a) R  r (b) R  r
(c) 0.5 ohm (d) 1/4.5 ohm (c) R  r (d) None of these
18. A 50V battery is connected across a 10 ohm resistor. The current is
28. Two identical cells send the same current in 2  resistance,
4.5 amperes. The internal resistance of the battery is [CPMT 1985; BHU 1997; Pb. PMT 2001]
whether connected in series or in parallel. The internal resistance of
(a) Zero (b) 0.5 ohm the cell should be
(c) 1.1 ohm (d) 5.0 ohm [NCERT 1982; Kerala PMT 2002]
19. The potential difference in open circuit for a cell is 2.2 volts. When a (a) 1  (b) 2 
4 ohm resistor is connected between its two electrodes the potential
difference becomes 2 volts. The internal resistance of the cell will be 1
(c)  (d) 2.5 
[MP PMT 1984; SCRA 1994; CBSE PMT 2002] 2
(a) 1 ohm (b) 0.2 ohm 29. The internal resistances of two cells shown are 0 .1  and 0 .3  .
(c) 2.5 ohm (d) 0.4 ohm If R  0.2  , the potential difference across the cell
20. A new flashlight cell of e.m.f. 1.5 volts gives a current of 15 amps,
2V, 0.1 2V, 0.3
when connected directly to an ammeter of resistance 0.04  . The
internal resistance of cell is [MP PET 1994] A B
(a) 0.04  (b) 0.06  (a) B will be zero
(c) 0.10  (d) 10  (b) A will be zero
(c) A and B will be 2V 0.2
21. A cell whose e.m.f. is 2 V and internal resistance is 0 .1  , is
(d) A will be  2V and B will be  2V
connected with a resistance of 3 .9  . The voltage across the cell
30. A torch battery consisting of two cells of 1.45 volts and an internal
terminal will be
[CPMT 1990; MP PET 1993; CBSE PMT 1999;
resistance 0.15  , each cell sending currents through the filament
AFMC 1999; Pb. PMT 2000; AIIMS 2001] of the lamps having resistance 1.5ohms. The value of current will be[MP PET 199
(a) 0.50 V (b) 1.90 V (a) 16.11 amp (b) 1.611 amp
(c) 0.1611 amp (d) 2.6 amp
(c) 1.95 V (d) 2.00 V 31. The electromotive force of a primary cell is 2 volts. When it is short-
22. The reading of a high resistance voltmeter when a cell is connected circuited it gives a current of 4 amperes. Its internal resistance in
across it is 2.2 V. When the terminals of the cell are also connected ohms is [MP PET 1995]
to a resistance of 5  the voltmeter reading drops to 1.8 V. Find (a) 0.5 (b) 5.0
(c) 2.0 (d) 8.0
the internal resistance of the cell [KCET 2003; MP PMT 2003]
32. The figure shows a network of currents. The magnitude of currents
(a) 1 .2  (b) 1 .3  is shown here. The current i will be
(c) 1 .1  (d) 1 .4  15A [MP PMT 1995]
3A
23. When cells are connected in parallel, then (a) 3 A

8A

i
5A
 1070 Current Electricity
(b) 13 A (d) i1  i2  i3
(c) 23 A 39. When a resistance of 2 ohm is connected across the terminals of a
cell, the current is 0.5 A. When the resistance is increased to 5 ohm,
(d) – 3 A the current is 0.25 A. The e.m.f. of the cell is
33. A battery of e.m.f. E and internal resistance r is connected to a [MP PET 1999, 2000; Pb. PMT 2002; MP PMT 2000]
variable resistor R as shown here. Which one of the following is true (a) 1.0[MPV PMT 1995] (b) 1.5 V
(c) 2.0 V (d) 2.5 V
E r 40. Two non-ideal identical batteries are connected in parallel. Consider
the following statements [MP PMT 1999]
(i) The equivalent e.m.f. is smaller than either of the two e.m.f.s
R
(ii) The equivalent internal resistance is smaller than either of the
two internal resistances
(a) Potential difference across the terminals of the battery is (a) Both (i) and (ii) are correct
maximum when R = r (b) (i) is correct but (ii) is wrong
(b) Power delivered to the resistor is maximum when R = r (c) (ii) is correct but (i) is wrong
(c) Current in the circuit is maximum when R = r (d) Both (i) and (ii) are wrong
(d) Current in the circuit is maximum when R  r 41. If six identical cells each having an e.m.f. of 6V are connected in
parallel, the e.m.f. of the combination is
34. A dry cell has an e.m.f. of 1.5 V and an internal resistance of
[EAMCET (Med.) 1995; Pb. PMT 1999; CPMT 2000]
0.05  . The maximum current obtainable from this cell for a very (a) 1 V (b) 36 V
short time interval is [Haryana CEE 1996]
1
(a) 30 A (b) 300 A (c) V (d) 6 V
6
(c) 3 A (d) 0.3 A
42. Consider the circuit shown in the figure. The current I3 is equal to
35. Consider the circuit given here with the following parameters
E.M.F. of the cell = 12 V. Internal resistance of the cell  2  . 28 54
(a) 5 amp
Resistance R  4 
E (b) 3 amp
(c) 3 amp 6V

(d) 5 / 6 amp I3
R
43. If VAB  4 V in the given8 figure,
V 12 V X will be
then resistance
Which one of the following statements in true
10 5V [RPET 1997]
(a) Rate of energy loss in the source is = 8 W
(a) 5
(b) Rate of energy conversion in the source is 16 W
(b) 10  A B
(c) Power output in is = 8 W
(d) Potential drop across R is = 16 V (c) 15 
2V X
36. A current of two amperes is flowing through a cell of e.m.f. 5 volts (d) 20 
and internal resistance 0.5 ohm from negative to positive electrode. 44. Two resistances R 1 and R 2 are joined as shown in the figure to
If the potential of negative electrode is 10V, the potential of positive
electrode will be two batteries of e.m.f. E1 and E 2 . If E 2 is short-circuited, the
[MP PMT 1997] current through R 1 is [NDA 1995]
(a) 5 V (b) 14 V R1
(a) E1 / R 1
(c) 15 V (d) 16 V
37. 100 cells each of e.m.f. 5 V and internal resistance 1 ohm are to be (b) E 2 / R1
arranged so as to produce maximum current in a 25 ohms E1
R2 E2
(c) E2 / R2
resistance. Each row is to contain equal number of cells. The
number of rows should be [MP PMT 1997] (d) E1 /(R 2  R1 )
(a) 2 (b) 4
45. A storage battery has e.m.f. 15 volts and internal resistance 0.05
(c) 5 (d) 10 ohm. Its terminal voltage when it is delivering 10 ampere is
38. The current in the arm CD of the circuit will be (a) 30 volts (b) 1.00 volts
[MP PMT/PET 1998; MP PMT 2000; DPMT 2000] (c) 14.5 volts (d) 15.5 volts
B
46. The number of dry cells, each of e.m.f. 1.5 volt and internal
resistance 0.5 ohm that must be joined in series with a resistance of
(a) i1  i2 20 ohm so as to send a current of 0.6 ampere through the circuit is
i2
i1
(b) i2  i3 A
(a) 2 (b) 8
O i3 (c) 10 (d) 12
(c) i1  i3 47. Emf is most closely related to [DCE 1999]
C (a) Mechanical force (b) Potential difference
D
 Current Electricity 1071

(c) Electric field (d) Magnetic field (a) 15 V (b) 12 V

48. For driving a current of 2 A for 6 minutes in a circuit, 1000 J of (c) 9 V (d) 6 V
work is to be done. The e.m.f. of the source in the circuit is 58. [CPMT
The current in the given 1999]is
circuit
(a) 1.38 V (b) 1.68 V [AIIMS 2000; MH CET 2003]
(c) 2.04 V (d) 3.10 V 10 
5V
49. Two batteries of e.m.f. 4V and 8 V with internal resistances 1  and (a) 0.1 A
2  are connected in a circuit with a resistance of 9  as shown in (b) 0.2 A
figure. The current and potential difference between the points P A B
and Q are [AFMC 1999] (c) 0.3 A
1 (d) 0.4 A 20 
(a) A and 3 V
3 2V
59. A current of 2.0 ampere passes through a cell of e.m.f. 1.5 volts
1 4V 8V 2
1 Q having internal resistance of 0.15 ohm. The potential difference
(b) A and 4 V P measured, in volts, across both the ends of the cell will be
6 r1 r2
1 (a) 1.35 (b) 1.50
(c) A and 9 V
9 (c) 1.00 (d) 1.20
9
1 60. A battery has e.m.f. 4 V and internal resistance r. When this battery
(d) A and 12V
2 is connected to an external resistance of 2 ohms, a current of 1 amp.
50. In the shown circuit, what is the potential difference across A and B flows in[AIIMS
the circuit.
1999] How much current will flow if the terminals of
20 V the battery are connected directly
(a) 50 V
[MP PET 2001]
(b) 45 V
(c) 30 V (a) 1 amp (b) 2 amp
(d) 20 V (c) 4 amp (d) Infinite
51. Four identical cells each having an electromotive
A B force (e.m.f.) of 61. Two batteries A and B each of e.m.f. 2 V are connected in series to
12V, are connected in parallel. The resultant electromotive force an external resistance R = 1 ohm. If the internal resistance of battery
(e.m.f.) of the combination is A is 1.9 ohms and that of B is 0.9 ohm, what is the potential
[CPMT 1999] difference between the terminals of battery A
A B
(a) 48 V (b) 12 V
(c) 4 V (d) 3 V (a) 2V
52. Electromotive force is the force, which is able to maintain a constant (b) V PMT 1999]
3.8[Pb.
(a) Current (b) Resistance (c) Zero
(d) None of the above R
(c) Power (d) Potential difference
53. A cell of emf 6 V and resistance 0.5 ohm is short circuited. The 62. When a resistor of 11  is connected in series with an electric cell,
current in the cell is [JIPMER 1999] the current flowing in it is 0.5 A. Instead, when a resistor of 5  is
(a) 3 amp (b) 12 amp connected to the same electric cell in series, the current increases by
0.4 A. The internal resistance of the cell is
(c) 24 amp (d) 6 amp
(a) 1.5  (b) 2 
54. A storage cell is charged by 5 amp D.C. for 18 hours. Its strength
after charging will be [JIPMER 1999] (c) 2.5  (d) 3.5 
(a) 18 AH (b) 5 AH 63. The internal resistance of a cell is the resistance of
(c) 90 AH (d) 15 AH [BHU 1999, 2000; AIIMS 2001]

55. A battery having e.m.f. 5 V and internal resistance 0.5  is (a) Electrodes of the cell
connected with a resistance of 4.5  then the voltage at the (b) Vessel of the cell
terminals of battery is [RPMT 2000] (c) Electrolyte used in the cell
(a) 4.5 V (b) 4 V (d) Material used in the cell
(c) 0 V (d) 2 V 64. How much work is required to carry a 6 C charge from the
56. In the given circuit the current I is
1
[DCE 2000] negative terminal to the positive terminal of a 9 V battery
30  [KCET (Med.) 2001]

(a) 0.4 A 3 6
I1 (a) 54 × 10 J (b) 54 × 10 J
(b) – 0.4 A 40 
(c) 54 × 10 9 J (d) 54 × 10 12 J
(c) 0.8 A I3
I2 40V 65. Consider four circuits shown in the figure below. In which circuit
40 
(d) – 0.8 A power dissipated is greatest (Neglect the internal resistance of the
80V power supply) [Orissa JEE 2002]
57. The internal resistance of a cell of e.m.f. 12 V is 5  10 2  . It is
connected across an unknown resistance. Voltage across the cell,
when a current of 60 A is drawn from it, is (a) (b) R
[CBSE PMT 2000]
E R R
E
R
 1072 Current Electricity
(c) Internal resistance is less than external resistance
(d) None of these
72. A battery is charged at a potential of 15 V for 8 hours when the
(c) (d) current flowing is 10 A. The battery on discharge supplies a current
R R R of 5 A for 15 hours. The mean terminal voltage during discharge is
R 14 V. The "Watt-hour" efficiency of the battery is
E
E (a) 82.5% (b) 80 %
R R
(c) 90% (d) 87.5%
66. The emf of a battery is 2 V and its internal resistance is 0.5 . 73. In the given current distribution what is the value of I
The maximum power which it can deliver to any external circuit will
be [AMU (Med.) 2002] [Orissa PMT 2004]

(a) 8 Watt (b) 4 Watt (a) 3A 4A

(c) 2 Watt (d) None of the above (b) 8 A I 2A
67. Kirchoff’s I law and II law of current, proves the (c) 2A 3A
[CBSE PMT 1993; BHU 2002; AFMC 2003] (d) 5A 5A
(a) Conservation of charge and energy
74. A capacitor is connected to a cell of emf E having some internal
(b) Conservation of current and energy resistance r. The potential difference across the
(c) Conservation of mass and charge
[CPMT 2004; MP PMT 2005]
(d) None of these
(a) Cell is < E (b) Cell is E
68. In the circuit, the reading of the ammeter is (assume internal
resistance of the battery be zero) (c) Capacitor is > E (d) Capacitor is < E
40 75. When the resistance of 9  is connected at the ends of a battery, its
(a) A
29 potential difference decreases from 40 volt to 30 volt. The internal
10 A resistance of the battery is [DPMT 2003]
(b) A
9 10V
4 (a) 6  (b) 3 
5 5
(c) A
3 (c) 9  (d) 15 
(d) 2 A
76. The maximum power drawn out of the cell from a source is given
69. In the above question, if the internal resistance of the battery is 1
ohm, then what is the reading of ammeter by (where r is internal resistance) [DCE 2002]

(a) 5/3 A (b) 40/29 A

(a) E 2 / 2r (b) E2 / 4r
(c) 10/9 A (d) 1 A
70. Eels are able to generate current with biological cells called (c) E2 / r (d) E 2 / 3r
electroplaques. The electroplaques in an eel are arranged in 100
rows, each row stretching horizontally along the body of the fish 77. Find out the value of current through 2 resistance for the given
containing 5000 electroplaques. The arrangement is suggestively circuit [IIT-JEE (Screening) 2005]
shown below. Each electroplaques has an emf of 0.15 V and internal
resistance of 0.25  [AIIMS 2004] (a) 5 A
0.15 V (b) 2 A
+ – + – + – 0.25  5 20V
10
(c) Zero 10V
+ – + – + –
(d) 4 A 2
5000 electroplaques per row
100 rows 78. Two batteries, one of emf 18 volts and internal resistance 2 and
the other of emf 12 volt and internal resistance 1 , are connected
as shown. The voltmeter V will record a reading of
+ – + – + –
(a) 15 volt
V
The water surrounding the eel completes a circuit between the head (b) 30 volt
500 
and its tail. If the water surrounding it has a resistance of 500 ,
the current an eel can produce in water is about (c) 14 volt 18V 2
(a) 1.5 A (b) 3.0 A (d) 18 volt
(c) 15 A (d) 30 A 12V
79. Two sources of equal emf are connected to1an external resistance R.
71. Current provided by a battery is maximum when
[AFMC 2004] The internal resistances of the two sources are R 1 and
(a) Internal resistance equal to external resistance
(b) Internal resistance is greater than external resistance
 Current Electricity 1073

R 2 (R 2  R1 ) . If the potential difference across the source having R1  R 2 R1  R2

(c) (d)
internal resistance R 2 is zero, then 2 2
[AIEEE 2005]
Different Measuring Instruments
(a) R  R1 R 2 /(R1  R 2 )
1. In meter bridge or Wheatstone bridge for measurement of
(b) R  R1 R 2 /(R 2  R1 ) resistance, the known and the unknown resistances are interchanged.
The error so removed is
(c) R  R 2  (R1  R 2 ) /(R 2  R1 )
[MNR 1988; MP PET 1995]
(d) R  R 2  R1 (a) End correction
(b) Index error
80. An energy source will supply a constant current into the load if its
internal resistance is [AIEEE 2005] (c) Due to temperature effect
(d) Random error
(a) Zero
2. A galvanometer can be converted into an ammeter by connecting
(b) Non-zero but less than the resistance of the load
[MP PMT 1987, 93; CPMT 1973, 75, 96, 2000;
(c) Equal to the resistance of the load MP PET 1994; AFMC 1993, 95; RPET 2000; DCE 2000]
(d) Very large as compared to the load resistance (a) Low resistance in series
81. The magnitude of i in ampere unit is [KCET 2005] (b) High resistance in parallel
(c) Low resistance in parallel
60
(a) 0.1 (d) High resistance in series
i 3. A cell of internal resistance 1 .5  and of e.m.f. 1.5 volt balances
(b) 0.3 15 5
500 cm on a potentiometer wire. If a wire of 15  is connected
(c) 0.6 1A 1A between the balance point and the cell, then the balance point will
shift [MP PMT 1985]
(d) None of these
(a) To zero (b) By 500 cm
10
82. To draw maximum current from a combination of cells, how should (c) By 750 cm (d) None of the above
the cells be grouped [AFMC 2005]
4. 10 3 amp is flowing through a resistance of 1000  . To measure
(a) Series the correct potential difference, the voltmeter is to be used of which
(b) Parallel the resistance should be [MP PMT 1985]

(c) Mixed (a) 0 (b) 500 

(d) Depends upon the relative values of external and internal (c) 1000  (d)  1000 
resistance 5. A galvanometer of 100  resistance gives full scale deflection when
83. The figure shows a network of currents. The magnitude of currents 10 mA of current is passed. To convert it into 10 A range ammeter,
is shown here. The current I will be [BCECE 2005] the resistance of the shunt required will be
1A (a) 10  (b) 1 
(a) 3 A (c) 0 .1  (d) 0.01 
(b) 9 A 10 A I 6. 50  and 100  resistors are connected in series. This
(c) 13 A connection is connected with a battery of 2.4 volts. When a
6A voltmeter of 100  resistance is connected across 100  resistor,
(d) 19 A then the reading of the voltmeter will be
2A [MP PMT 1985]
84. The n rows each containing m cells in series are joined in parallel.
Maximum current is taken from this combination across an external (a) 1.6 V (b) 1.0 V
(c) 1.2 V (d) 2.0 V
resistance of 3 resistance. If the total number of cells used are 24
and internal resistance of each cell is 0.5  then 7. A 2 volt[J &battery, a 15  resistor and a potentiometer of 100 cm
K CET 2005]
length, all are connected in series. If the resistance of potentiometer
(a) m  8, n  3 (b) m  6, n  4 wire is 5  , then the potential gradient of the potentiometer wire
(c) m  12, n  2 (d) m  2, n  12 is [AIIMS 1982]
(a) 0.005 V/cm (b) 0.05 V/cm
85. A cell of constant e.m.f. first connected to a resistance R1 and then
(c) 0.02 V/cm (d) 0.2 V/cm
connected to a resistance R 2 . If power delivered in both cases is
then the internal resistance of the cell is 8. An ammeter gives full scale deflection when current of 1.0 A is
[Orissa JEE 2005]
passed in it. To convert it into 10 A range ammeter, the ratio of its
resistance and the shunt resistance will be
R1 [MP PMT 1985]
(a) R1 R2 (b)
R2 (a) 1 : 9 (b) 1 : 10
 1074 Current Electricity
(c) 1 : 11 (d) 9 : 1 18. The tangent galvanometer, when connected in series with a standard
9. By ammeter, which of the following can be measured resistance can be used as [MP PET 1994]
[MP PET 1981; DPMT 2001] (a) An ammeter
(a) Electric potential (b) Potential difference (b) A voltmeter
(c) A wattmeter
(c) Current (d) Resistance
(d) Both an ammeter and a voltmeter
10. The resistance of 1 A ammeter is 0.018  . To convert it into 10 A
19. In Wheatstone's bridge P  9 ohm, Q  11 ohm, R  4 ohm and
ammeter, the shunt resistance required will be
S  6 ohm. How much resistance must be put in parallel to the
[MP PET 1982]
resistance S to balance the bridge
(a) 0.18  (b) 0.0018  [DPMT 1999]
(c) 0.002  (d) 0.12  44
(a) 24 ohm (b) ohm
11. For measurement of potential difference, potentiometer is preferred 9
in comparison to voltmeter because (c) 26.4 ohm (d) 18.7 ohm
[MP PET 1983] 20. A Daniel cell is balanced on 125 cm length of a potentiometer wire.
(a) Potentiometer is more sensitive than voltmeter Now the cell is short-circuited by a resistance 2 ohm and the
balance is obtained at 100 cm . The internal resistance of the Daniel
(b) The resistance of potentiometer is less than voltmeter cell is [UPSEAT 2002]
(c) Potentiometer is cheaper than voltmeter (a) 0.5 ohm (b) 1.5 ohm
(d) Potentiometer does not take current from the circuit (c) 1.25 ohm (d) 4/5 ohm
12. In order to pass 10% of main current through a moving coil 21. Sensitivity of potentiometer can be increased by
galvanometer of 99 ohm, the resistance of the required shunt is [MP PET 1990, 99; MP PMT 1994; [MP PET 1994]

RPET 2001; KCET 2003, 05]

(a) Increasing the e.m.f. of the cell
(b) Increasing the length of the potentiometer wire
(a) 9 .9  (b) 10 
(c) Decreasing the length of the potentiometer wire
(c) 11  (d) 9  (d) None of the above
13. An ammeter of 5 ohm resistance can read 5 mA. If it is to be used 22. A potentiometer is an ideal device of measuring potential difference
to read 100 volts, how much resistance is to be connected in series because
[MP PET 1991; MP PMT 1996; MP PMT 2000] (a) It uses a sensitive galvanometer
(a) 19.9995  (b) 199.995 (b) It does not disturb the potential difference it measures
(c) It is an elaborate arrangement
(c) 1999.95  (d) 19995  (d) It has a long wire hence heat developed is quickly radiated
14. The potential gradient along the length of a uniform wire is 23. A battery of 6 volts is connected to the terminals of a three metre
10 volt / metre . B and C are the two points at 30 cm and long wire of uniform thickness and resistance of the order of 100  .
60 cm point on a meter scale fitted along the wire. The potential The difference of potential between two points separated by 50 cm
on the wire will be
difference between B and C will be [CPMT 1986] [CPMT 1984; CBSE PMT 2004]
(a) 3 volt (b) 0.4 volt (a) 1 V (b) 1.5 V
(c) 7 volt (d) 4 volt (c) 2 V (d) 3 V
24. A galvanometer of 10 ohm resistance gives full scale deflection with
15. 100 mA current gives a full scale deflection in a galvanometer of 0.01 ampere of current. It is to be converted into an ammeter for
2  resistance. The resistance connected with the galvanometer to measuring 10 ampere current. The value of shunt resistance required
will be [MP PET 1984]
convert it into a voltmeter to measure 5 V is [MNR 1994; UPSEAT 2000]
10
(a) 98  (b) 52  (a) ohm (b) 0.1 ohm
999
(c) 50  (d) 48  (c) 0.5 ohm (d) 1.0 ohm
16. When a 12  resistor is connected with a moving coil 25. A potentiometer is used for the comparison of e.m.f. of two cells
E1 and E 2 . For cell E1 the no deflection point is obtained at
galvanometer then its deflection reduces from 50 divisions to 10
divisions. The resistance of the galvanometer is 20 cm and for E 2 the no deflection point is obtained at 30 cm .
[CPMT 2002; DPMT 2003] The ratio of their e.m.f.'s will be
[MP PET 1984]
(a) 24  (b) 36 
(a) 2/3 (b) 1/2
(c) 48  (d) 60  (c) 1 (d) 2
17. A galvanometer can be used as a voltmeter by connecting a 26. Potential gradient is defined as [MP PET 1994]
[AFMC 1993; MP PMT 1993, 95; CBSE PMT 2004] (a) Fall of potential per unit length of the wire
(a) High resistance in series (b) Low resistance in series (b) Fall of potential per unit area of the wire
(c) High resistance in parallel (d) Low resistance in parallel (c) Fall of potential between two ends of the wire
 Current Electricity 1075

(d) Potential at any one end of the wire 36. In the diagram shown, the reading of voltmeter is 20 V and that of
27. In an experiment of meter bridge, a null point is obtained at the ammeter is 4 A. The value of R should be (Consider given ammeter
centre of the bridge wire. When a resistance of 10 ohm is connected and voltmeter are not ideal) [RPMT 1997]
in one gap, the value of resistance in other gap is [MP PET 1994] V
(a) 10  (b) 5  (a) Equal to 5  20V
1 (b) Greater from 5 
(c)  (d) 500 
5 R
28. If the length of potentiometer wire is increased, then the length of (c) Less than 5  A
the previously obtained balance point will 4A
(a) Increase (b) Decrease (d) Greater or less than 5  depends on the material of R
(c) Remain unchanged (d) Become two times 37. A moving coil galvanometer has a resistance of 50  and gives full
29. In potentiometer a balance point is obtained, when scale deflection for 10 mA. How could it be converted into an
(a) The e.m.f. of the battery becomes equal to the e.m.f. of the ammeter with a full scale deflection for 1A
experimental cell
[MP PMT 1996]
(b) The p.d. of the wire between the +ve end to
jockey becomes equal to the e.m.f. of the experimental cell (a) 50 / 99  in series (b) 50 / 99  in parallel
(c) The p.d. of the wire between +ve point and jockey becomes
equal to the e.m.f. of the battery (c) 0.01  in series (d) 0.01  in parallel
(d) The p.d. across the potentiometer wire becomes equal to the 38. The current flowing through a coil of resistance 900 ohms is to be
e.m.f. of the battery reduced by 90%. What value of shunt should be connected across
30. In the experiment of potentiometer, at balance, there is no current the coil [Roorkee 1992]
in the
(a) 90  (b) 100 
(a) Main circuit
(b) Galvanometer circuit (c) 9 (d) 10 
(c) Potentiometer circuit 39. A galvanometer of resistance 25  gives full scale deflection for a
(d) Both main and galvanometer circuits
current of 10 milliampere, is to be changed into a voltmeter of range
31. If in the experiment of Wheatstone's bridge, the positions of cells 100 V by connecting a resistance of ‘R’ in series with galvanometer.
and galvanometer are interchanged, then balance points will The value of resistance R in  is
(a) Change [MP PET 1994]
(b) Remain unchanged (a) 10000 (b) 10025
(c) Depend on the internal resistance of cell and resistance of (c) 975 (d) 9975
galvanometer 40. In a potentiometer circuit there is a cell of e.m.f. 2 volt, a resistance
(d) None of these of 5 ohm and a wire of uniform thickness of length 1000 cm and
resistance 15 ohm. The potential gradient in the wire is
32. The resistance of a galvanometer is 90 ohms. If only 10 percent of
the main current may flow through the galvanometer, in which way 1 3
and of what value, a resistor is to be used (a) [MP VPET
/ cm
1996] (b) V / cm
500 2000
(a) 10 ohms in series (b) 10 ohms in parallel
3 1
(c) 810 ohms in series (d) 810 ohms in parallel (c) V / cm (d) V / cm
5000 1000
33. Two cells when connected in series are balanced on 8 m on a
41. The resistance of a galvanometer is 25 ohm and it requires 50 A
potentiometer. If the cells are connected with polarities of one of the
cell is reversed, they balance on 2m. The ratio of e.m.f.'s of the two for full deflection. The value of the shunt resistance required to
cells is convert it into an ammeter of 5 amp is
[MP PMT 1994; BHU 1997]
(a) 3:5 (b) 5 : 3
4
(a) 2.5  10 ohm (b) 1.25  10 3 ohm
(c) 3:4 (d) 4 : 3
(c) 0.05 ohm (d) 2.5 ohm
34. A voltmeter has a resistance of G ohms and range V volts. The value 42. Which is a wrong statement [MP PMT 1994]
of resistance used in series to convert it into a voltmeter of range
(a) The Wheatstone bridge is most sensitive when all the four
nV volts is resistances are of the same order
[MP PMT 1999; MP PET 2002; DPMT 2004; MH CET 2004] (b) In a balanced Wheatstone bridge, interchanging the positions of
(a) nG (b) (n  1)G galvanometer and cell affects the balance of the bridge
(c) Kirchhoff's first law (for currents meeting at a junction in an
G G electric circuit) expresses the conservation of charge
(c) (d) (d) The rheostat can be used as a potential divider
n (n  1)
43. A voltmeter having a resistance of 998 ohms is connected to a cell
35. Which of the following statement is wrong [MP PET 1994]
of e.m.f. 2 volt and internal resistance 2 ohm. The error in the
(a) Voltmeter should have high resistance measurement of e.m.f. will be [MP PMT 1994]
(b) Ammeter should have low resistance
(c) Ammeter is placed in parallel across the conductor in a circuit (a) 4  10 1 volt (b) 2  10 3 volt
(d) Voltmeter is placed in parallel across the conductor in a circuit
(c) 4  10 3 volt (d) 2  10 1 volt
 1076 Current Electricity
44. For comparing the e.m.f.'s of two cells with a potentiometer, a
standard cell is used to develop a potential gradient along the wires.
Which of the following possibilities would make the experiment
unsuccessful [MP PMT 1994]
(a) The e.m.f. of the standard cell is larger than the E e.m.f.'s of the (a) Decreases the resistance R
two cells (b) Increase the resistance R
(b) The diameter of the wires is the same and uniform throughout
(c) Reverse the terminals of battery V
(c) The number of wires is ten
(d) The e.m.f. of the standard cell is smaller than the e.m.f.'s of the (d) Reverse the terminals of cell E
two cells 51. In the Wheatstone's bridge (shown in figure) X  Y and A  B .
45. Which of the following is correct [BHU 1995] The direction of the current between ab will be
(a) Ammeter has low resistance and is connected in series a
(b) Ammeter has low resistance and is connected in parallel A B
(c) Voltmeter has low resistance and is connected in parallel (a) From a to b
(d) None of the above c d
(b) From b to a
46. An ammeter with internal resistance 90  reads 1.85 A when
(c) From b to a through c X Y
connected in a circuit containing a battery and two resistors 700 
and 410  in series. Actual current will be (d) From a to b through c b
[Roorkee 1995] 52. The figure shows a circuit diagram of a ‘Wheatstone Bridge’ to
measure the resistance G of the galvanometer. The relation
(a) 1.85 A (b) Greater than 1.85 A
P R
(c) Less than 1.85 A (d) None of these  will be satisfied only when
Q G
47. AB is a wire of uniform resistance. The galvanometer G shows no
current when the length AC = 20cm and CB = 80 cm. The resistance R Q
is equal to [MP PMT 1995; RPET 2001] P
(a) 2 R 80  S
G
R
(b) 8 
G
(c) 20 
A B
C (a) The galvanometer shows a deflection when switch S is closed
(d) 40 
(b) The galvanometer shows a deflection when switch S is open
48. The circuit shown here is used to compare the e.m.f. of two cells
(c) The galvanometer shows no change in deflection whether S is
E1 and E 2 (E1  E 2 ) . The null point is at C when the open or closed
galvanometer is connected to E1 . When the galvanometer is (d) The galvanometer shows no deflection
connected to E 2 , the null point will be [MP PMT 1995] 53. The resistance of a galvanometer is 50 ohms and the current
B required to give full scale deflection is 100 A . In order to convert
it into an ammeter, reading upto 10A, it is necessary to put a
(a) To the left of C resistance of [MP PMT 1997; AIIMS 1999]
(b) To the right of C C (a) 5  10 3  in parallel (b) 5  10 4  in parallel
(c) At C itself A B
E1 (c) 10 5  in series (d) 99,950  in series
(d) Nowhere on AB
G 54. A resistance of 4  and a wire of length 5 metres and resistance
49. E2
In an experiment to measure the internal resistance of a cell by
5  are joined in series and connected to a cell of e.m.f. 10 V and
potentiometer, it is found that the balance point is at a length of 2 m
when the cell is shunted by a 5  resistance; and is at a length of internal resistance 1  . A parallel combination of two identical cells
is balanced across 300 cm of the wire. The e.m.f. E of each cell is [MP PMT 199
3m when the cell is shunted by a 10  resistance. The internal
4 10V
resistance of the cell is, then
[Haryana CEE 1996] (a) 1.5 V 1

(a) 1 .5  (b) 10 
(b) 3.0 V 3m
(c) 15  (d) 1  5, 5m
(c) 0.67 V
50. A potentiometer circuit shown in the figure is set up to measure E
e.m.f. of a cell E. As the point P moves from X to Y the galvanometer (d) 1.33 V G
E
G shows deflection always in one direction, but the deflection
decreases continuously until Y is reached. In order to obtain balance 55. The resistivity of a potentiometer wire is 40  10 8 ohm  m and
point between X and Y it is necessary to its area of cross-section is 8  10 6 m 2 . If 0.2 amp current is
R flowing through the wire, the potential gradient will be
V

X P
Y

E
G
 Current Electricity 1077

(a) 10 2 volt / m (b) 10 1 volt / m

(a) IR  IG
(c) 3.2  10 2 volt / m (d) 1 volt / m
(b) IP  IG
56. If only 2% of the main current is to be passed through a
galvanometer of resistance G, then the resistance of shunt will be (c)  IPMT/PET
IQ[MP G 1998]
G G (d) IQ  IR
(a) (b)
50 49
65. In the following Wheatstone bridge P / Q  R / S . If key K is
(c) 50 G (d) 49 G
closed, then the galvanometer will show deflection
57. The resistance of an ideal voltmeter is [CPMT 1999]
[EAMCET (Med.) 1995; MP PMT/PET 1998; Pb. PMT 1999; P
CPMT 2000] (a) In left side Q
(a) Zero (b) Very low (b) In right side K
(c) Very large (d) Infinite (c) No deflection
R S
58. A 100 V voltmeter of internal resistance 20 k  in series with a
(d) In either side
high resistance R is connected to a 110 V line. The voltmeter reads 5
V, the value of R is [MP PET 1999] 66. A galvanometer having a resistance of 8 ohm is shunted by a wire of
(a) 210 k  (b) 315 k  resistance 2 ohm. If the total current is 1 amp, the part of it passing
through the shunt will be
(c) 420 k  (d) 440 k  [CBSE PMT 1998]
59. Constantan wire is used in making standard resistances because its [MPamp
(a) 0.25 PET 1999] (b) 0.8 amp
(a) Specific resistance is low (c) 0.2 amp (d) 0.5 amp
(b) Density is high 67. A potentiometer wire has length 10 m and resistance 20  . A 2. 5
(c) Temperature coefficient of resistance is negligible V battery of negligible internal resistance is connected across the
wire with an 80  series resistance. The potential gradient on the
(d) Melting point is high
wire will be [KCET 1994]
60. The net resistance of a voltmeter should be large to ensure that [MP PMT 1999]
(a) It does not get overheated (a) 5  10 5 V / mm (b) 2.5  10 4 V / cm
(b) It does not draw excessive current (c) 0.62  10 4 V / mm (d) 1  10 5 V / mm
(c) It can measure large potential difference 68. An ammeter whose resistance is 180  gives full scale deflection
(d) It does not appreciably change the potential difference to be when current is 2 mA. The shunt required to convert it into an
measured ammeter reading 20 mA (in ohms) is
61. A galvanometer has resistance of 7  and gives a full scale [EAMCET (Engg.) 1995]
(a) 18 (b) 20
deflection for a current of 1.0 A. How will you convert it into a
voltmeter of range 10 V [MP PMT 1999] (c) 0.1 (d) 10
69. A galvanometer whose resistance is 120  gives full scale
(a) 3  in series (b) 3  in parallel
deflection with a current of 0.05 A so that it can read a maximum
(c) 17  in series (d) 30  in series current of 10 A. A shunt resistance is added in parallel with it. The
resistance of the ammeter so formed is
62. A potentiometer consists of a wire of length 4 m and resistance [Bihar MEE 1995]
10  . It is connected to a cell of e.m.f. 2 V. The potential (a) 0.06  (b) 0.006 
difference per unit length of the wire will be
(c) 0 .6  (d) 6  s
[CBSE PMT 1999; AFMC 2001]
70. In a potentiometer experiment, the galvanometer shows no
(a) 0.5 V / m (b) 2V /m deflection when a cell is connected across 60 cm of the
(c) 5V /m (d) 10 V / m potentiometer wire. If the cell is shunted by a resistance of 6  ,
the balance is obtained across 50 cm of the wire. The internal
63. In a meter bridge, the balancing length from the left end (standard resistance of the cell is [SCRA 1994]
resistance of one ohm is in the right gap) is found to be 20 cm. The
(a) 0 .5  (b) 0 .6 
value of the unknown resistance is
[CBSE PMT 1999; Pb PMT 2004] (c) 1 .2  (d) 1 .5 
(a) 0 .8  (b) 0 .5  71. A voltmeter of resistance 1000  gives full scale deflection when a
current of 100 mA flow through it. The shunt resistance required
(c) 0 .4  (d) 0.25 
across it to enable it to be used as an ammeter reading 1 A at full
64. In the circuit shown P  R , the reading of the galvanometer is scale deflection is [SCRA 1994]
same with switch S open or closed. Then (a) 10000  (b) 9000 
[IIT-JEE (Screening) 1999]
(c) 222  (d) 111 
P Q

R G

V
 1078 Current Electricity
72. The resistance of 10 metre long potentiometer wire is 1ohm/meter. A (a) 20000  (b) 19989 
cell of e.m.f. 2.2 volts and a high resistance box are connected in
series to this wire. The value of resistance taken from resistance box (c) 10000  (d) 9989 
for getting potential gradient of 2.2 millivolt/metre will be[RPET 1997]
80. In a balanced Wheatstone’s network, the resistances in the arms Q
(a) 790  (b) 810  and S are interchanged. As a result of this
(c) 990  (d) 1000  [KCET 1999]
73. We have a galvanometer of resistance 25  . It is shunted by a (a) Network is not balanced
2.5  wire. The part of total current that flows through the (b) Network is still balanced
galvanometer is given as (c) Galvanometer shows zero deflection
[AFMC 1998; MH CET 1999; Pb. PMT 2002]
(d) Galvanometer and the cell must be interchanged to balance
I 1 I 1
(a)  (b)  81. The ammeter A reads 2 A and the voltmeter V reads 20 V. the
I0 11 I0 10
value of resistance R is (Assuming finite resistance's of ammeter and
I 3 I 4 voltmeter) [JIPMER 1999; MP PMT 2004]
(c)  (d) 
I0 11 I0 11 (a) Exactly 10 ohm R
74. In the adjoining circuit, the e.m.f. of the cell is 2 volt and the A
(b) Less than 10 ohm
internal resistance is negligible. The resistance of the voltmeter is 80
ohm. The reading of the voltmeter will be (c) More than 10 ohm
2V [CPMT 1991] (d) We cannot definitely say V
(a) 0.80 volt + –
82. The resistance of a galvanometer coil is R. What is the shunt
(b) 1.60 volt resistance required to convert it into an ammeter of range 4 times
80 
(c) 1.33 volt V R R
(a) (b)
(d) 2.00 volt 5 4
75. 20 wire be  80and
If the resistivity of a potentiometer  area of cross- R
(c) (d) 4 R
section be A, then what will be potential gradient along the wire 3[RPET 1996]
I I 83. If an ammeter is connected in parallel to a circuit, it is likely to be
(a) (b) damaged due to excess
A A [BHU 2000; BCECE 2004]
(a) Current (b) Voltage
IA
(c) (d) IA (c) Resistance (d) All of these

76. A voltmeter has resistance of 2000 ohms and it can measure upto 84. In the given figure, battery E is balanced on 55 cm length of
2V. If we want to increase its range to 10 V, then the required potentiometer wire but when a resistance of 10  is connected in
resistance in series will be parallel with the battery then it balances on 50 cm length of the
[CPMT 1997, SCRA 1994] potentiometer wire then internal resistance r of the battery is
(a) 2000  (b) 4000  (a) 1  2V

(c) 6000  (d) 8000  (b) 3 

1m
77. For a cell of e.m.f. 2V, a balance is obtained for 50 cm of the (c) 10 
potentiometer wire. If the cell is shunted by a 2  resistor and the B
(d) 5  A
balance is obtained across 40 cm of the wire, then the internal r
E
resistance of the cell is [SCRA 1998] 85. A galvanometer with a resistance of 12  gives full scale deflection
when a current of 3 mA is passed. It is required to convert it into a
(a) 0.25  (b) 0.50
voltmeter which can read up to 18 V. the resistance to be connected
(c) 0.80  (d) 1.00  is [Pb. PMT 2000]

78. The arrangement as shown in figure is called as (a) 6000  (b) 5988 
[CPMT 1999] (c) 5000  (d) 4988 
(a) Potential divider 86. The resistance of an ideal ammeter is [KCET 2000]
(b) Potential adder (a) Infinite (b) Very high
Total P.D.
(c) Potential substracter (c) Small (d) Zero

(d) Potential multiplier 87. A galvanometer of 25  resistance can read a maximum current of
Variable P.D. 6mA. It can be used as a voltmeter to measure a maximum of 6 V
79. A potentiometer wire of length 1 m and resistance 10  is by connecting a resistance to the galvanometer. Identify the correct
connected in series with a cell of emf 2V with internal resistance 1  choice in the given answers [EAMCET (Med.) 2000]
and a resistance box including a resistance R. If potential difference (a) 1025  in series (b) 1025  in parallel
between the ends of the wire is 1 mV, the value of R is [KCET 1999]
(c) 975  in series (d) 975  in parallel
 Current Electricity 1079

88. A galvanometer has a resistance of 25 ohm and a maximum of 0.01 G G

A current can be passed through it. In order to change it into an (c) (d)
20 19
ammeter of range 10 A, the shunt resistance required is [MP PET 2000]
96. A voltmeter having resistance of 50 × 10 ohm is used to measure the
3

(a) 5/999 ohm (b) 10/999 ohm voltage in a circuit. To increase the range of measurement 3 times
(c) 20/999 ohm (d) 25/999 ohm the additional series resistance required is
89. In the circuit shown, a meter bridge is in its balanced state. The (a) 10 ohm
5
(b) 150 k.ohm
meter bridge wire has a resistance 0.1 ohm/cm. The value of (c) 900 k.ohm (d) 9 × 10 ohm 6

unknown resistance X and the current drawn from the battery of 97. In a potentiometer experiment two cells of e.m.f. E and E are used
negligible resistance is [AMU (Engg.) 2000]
1

in series and in conjunction and the balancing length is found to be

2

X 6 58 cm of the wire. If the polarity of E is reversed, then the

2

(a) 6 , 5 amp E
balancing length becomes 29 cm. The ratio 1 of the e.m.f. of the
G E2
(b) 10 , 0.1 amp 40 cm 60 cm two cells is
A B [Kerala (Engg.) 2001]
(c) 4 , 1.0 amp (a) 1 : 1 (b) 2 : 1
(d) 12 , 0.5 amp (c) 3 : 1 (d) 4 : 1
5V
98. A milliammeter of range 10 mA has a coil of resistance 1 . To use
90. A galvanometer has 30 divisions and a sensitivity 16 A / div.It can it as voltmeter of range 10 volt, the resistance that must be
be converted into a voltmeter to read 3 V by connecting [Kerala
connected PMT 2005]
in series with it, will be [KCET 2001]
(a) Resistance nearly 6 k  in series
(a) 999  (b) 99 
(b) 6 k  in parallel (c) 1000  (d) None of these
(c) 500  in series 99. A voltmeter has a range 0-V with a series resistance R. With a
(d) It cannot be converted series resistance 2R, the range is 0-V. The correct relation between
91. Voltmeters V and V are connected in series across a D.C. line. V
1 2 1
V and V is [CPMT 2001]
reads 80 volts and has a per volt resistance of 200 ohms. V has a2
(a) V   2V (b) V   2V
total resistance of 32 kilo ohms. The line voltage is [UPSEAT 2000]
(a) 120 volts (b) 160 volts (c) V   2V (d) V '  2V
(c) 220 volts (d) 240 volts 100. The measurement of voltmeter in the following circuit is
92. A potentiometer having the potential gradient of 2 mV/cm is used to 6V [AFMC 2001]
measure the difference of potential across a resistance of 10 ohm. If + –
a length of 50 cm of the potentiometer wire is required to get the (a) 2.4 V
null point, the current passing through the 10 ohm resistor is (in
(b) 3.4 V 60
mA)
V
[AMU (Med.) 2000] (c) 4.0 V
(a) 1 (b) 2
(d) 6.0 V 40
(c) 5 (d) 10
93. AB is a potentiometer wire of length 100 cm and its resistance is 10 101. A 36  galvanometer is shunted by resistance of 4. The
ohms. It is connected in series with a resistance R = 40 ohms and a percentage of the total current, which passes through the
battery of e.m.f. 2 V and negligible internal resistance. If a source of galvanometer is [UPSEAT 2002]
unknown e.m.f. E is balanced by 40 cm length of the potentiometer (a) 8 % (b) 9 %
wire, the value of E is [MP PET 2001] (c) 10 % (d) 91 %
2V 102. An ammeter and a voltmeter of resistance R are connected in series
(a) 0.8 V R
to an electric cell of negligible internal resistance. Their readings are
(b) 1.6 V A and V respectively. If another resistance R is connected in parallel
with the voltmeter
40 cm
(c) 0.08 V [EAMCET 2000; KCET 2002]
A B
(d) 0.16 V (a) Both A and V will increase
94. An ammeter gives full deflection when a current of 2 amp. flows (b) Both A and V will decrease
E of ammeter is 12 ohms. If the same
through it. The resistance (c) A will decrease and V will increase
ammeter is to be used for measuring a maximum current of 5 amp.,
then the ammeter must be connected with a resistance of (d) A [MP
will PET 2001]and V will decrease
increase
(a) 8 ohms in series (b) 18 ohms in series 103. A wire of length 100 cm is connected to a cell of emf 2 V and
(c) 8 ohms in parallel (d) 18 ohms in parallel negligible internal resistance. The resistance of the wire is 3 .
The additional resistance required to produce a potential drop of 1
95. In a circuit 5 percent of total current passes through a milli volt per cm is [Kerala PET 2002]
galvanometer. If resistance of the galvanometer is G then value of
the shunt is [MP PET 2001] (a) 60  (b) 47 
(a) 19 G (b) 20 G (c) 57  (d) 35 
 1080 Current Electricity

104. A galvanometer of resistance 20  is to be converted into an 10 900 

ammeter of range 1 A. If a current of 1 mA produces full scale (a)
9 V
deflection, the shunt required for the purpose is
(b) 0.1 10 
[Kerala PET 2002]
(a) 0.01  (b) 0.05  (c) 1.0 100 
(c) 0.02  (d) 0.04  (d) 10.0
105. There are three voltmeters of the same range but of resistances 114. A cell of internal resistance 3 ohm and emf 10 volt is connected to a
10000  , 8000  and 4000  respectively. The best uniform wire of length 500 cm and resistance 3 ohm. The potential
gradient in the wire is [MP PET 2003]
voltmeter among these is the one whose resistance is [Kerala PET 2002]
(a) 30 mV/cm (b) 10 mV/cm
(a) 10000  (b) 8000 
(c) 20 mV/cm (d) 4 mV/cm
(c) 4000  (d) All are equally good
115. An ammeter of 100  resistance gives full deflection for the current
106. If an ammeter is to be used in place of a voltmeter then we must of 10 amp. Now the shunt resistance required to convert it into
–5

connect with the ammeter a ammeter of 1 amp. range, will be

[AIEEE 2002; AFMC 2002] [RPET 2003]
(a) Low resistance in parallel 4 5
(a) 10  (b) 10 
(b) High resistance in parallel
3 1
(c) High resistance in series (c) 10  (d) 10 
(d) Low resistance in series 116. A galvanometer of resistance 36  is changed into an ammeter by
107. A 10 m long wire of 20 resistance is connected with a battery of 3 using a shunt of 4 . The fraction f of total current passing
0

through the galvanometer is [BCECE 2003]

volt e.m.f. (negligible internal resistance) and a 10  resistance is
joined to it is series. Potential gradient along wire in volt per meter 1 1
(a) (b)
is [MP PMT 2003] 40 4
(a) 0.02 (b) 0.3 1 1
(c) (d)
(c) 0.2 (d) 1.3 140 10
108. A potentiometer has uniform potential gradient across it. Two cells 117. If the ammeter in the given circuit reads 2 A, the resistance R is
connected in series (i) to support each other and (ii) to oppose each 3
other are balanced over 6m and 2m respectively on the
(a) 1 ohm
potentiometer wire. The e.m.f.’s of the cells are in the ratio of [MP PMT 2002] R
(b) 2 ohm
(a) 1 : 2 (b) 1 : 1 6
(c) 3 ohm
(c) 3 : 1 (d) 2 : 1
109. The material of wire of potentiometer is (d) 4 ohm
[MP PMT 2002] 118. A 50 ohm galvanometer gets fullA scale deflection when a current of
6V
0.01 A passes through the coil. When it is converted to a 10 A
(a) Copper (b) Steel
ammeter, the shunt resistance is
(c) Manganin (d) Aluminium
[Orissa JEE 2003]
110. To convert a galvanometer into a voltmeter, one should connect a [CBSE PMT 2002]
(a) 0.01  (b) 0.05 
(a) High resistance in series with galvanometer
(c) 2000  (d) 5000 
(b) Low resistance in series with galvanometer
119. Resistance in the two gaps of a meter bridge are 10 ohm and 30
(c) High resistance in parallel with galvanometer
ohm respectively. If the resistances are interchanged the balance
(d) Low resistance in parallel with galvanometer point shifts by [Orissa JEE 2003]
111. To convert a 800 mV range milli voltmeter of resistance 40  into a (a) 33.3 cm (b) 66.67cm
galvanometer of 100 mA range, the resistance to be connected as (c) 25 cm (d) 50 cm
shunt is [CBSE PMT 2002]
120. A potentiometer has uniform potential gradient. The specific
(a) 10  (b) 20  resistance of the material of the potentiometer wire is 10 ohm– –7

meter and the current passing through it is 0.1 ampere; cross-section

(c) 30  (d) 40  of the wire is 10 m . The potential gradient along the potentiometer
–6 2

112. A 100 ohm galvanometer gives full scale deflection at 10 mA. How wire is [KCET 2003]
much shunt is required to read 100 mA
(a) 10 4 V/m (b) 10 6 V/m
[MP PET 2002]
(a) 11.11 ohm (b) 9.9 ohm (c) 10 2 V/m (d) 10 8 V/m
(c) 1.1 ohm (d) 4.4 ohm 121. Two resistances of 400  and 800  are connected in series with 6
volt battery of negligible internal resistance. A voltmeter of
113. The potential difference across the 100 resistance in the following resistance 10,000  is used to measure the potential difference
circuit is measured by a voltmeter of 900  resistance. The across 400 . The error in the measurement of potential difference
percentage error made in reading the potential difference is in volts[AMU (Med.) 2002]
approximately is [
(a) 0.01 (b) 0.02
(c) 0.03 (d) 0.05
 Current Electricity 1081

122. A galvanometer, having a resistance of 50  gives a full scale deflection for (a) 5040 (b) 4960
a current of 0.05 A. The length in meter of a resistance wire of area of (c) 2010 (d) 4050
cross-section 2.97× 10 cm that can be used to convert the galvanometer
–2 2

into an ammeter which can read a maximum of 5 A current is (Specific 130. For the post office box arrangement to determine the value of
resistance of the wire = 5 × 10 7 m) unknown
[EAMCET 2003]resistance the unknown resistance should be connected
between [IIT-JEE (Screening) 2004]
(a) 9 (b) 6
(c) 3 (d) 1.5 B C D
123. An ammeter reads upto 1 ampere. Its internal resistance is 0.81 (a) B and C
ohm. To increase the range to 10 A the value of the required shunt
is [AIEEE 2003] (b) C and D
(a) 0.09  (b) 0.03  (c) A and D
A
(c) 0.3  (d) 0.9  (d) B and C
1 1

124. The length of a wire of a potentiometer is 100 cm, and the emf of its B1 has 25 divisions.CA
131. A galvanometer of 50 ohm resistance 1 current of
standard cell is E volt. It is employed to measure the e.m.f of a
battery whose internal resistance is 0.5 . If the balance point is 4  10 ampere gives a deflection of one division. To convert this
–4

obtained at l = 30 cm from the positive end, the e.m.f. of the battery galvanometer into a voltmeter having a range of 25 volts, it should
is [AIEEE 2003] be connected with a resistance of
30 E [CBSE PMT 2004]
(a)
100
(a) 2500  as a shunt (b) 2450  as a shunt
30 E
(b) (c) 2550  in series (d) 2450  in series
100. 5
30 E 132. In a metre bridge experiment null point is obtained at 20 cm from
(c) one end of the wire when resistance X is balanced against another
(100  0 .5)
resistance Y. If X < Y, then where will be the new position of the null
30(E  0 .5 i) point from the same end, if one decides to balance a resistance of
(d) , where i is the current in the potentiometer
100 4X against Y
125. Resistance of 100 cm long potentiometer wire is 10, it is connected [AIEEE 2004]
to a battery (2 volt) and a resistance R in series. A source of 10 mV
gives null point at 40 cm length, then external resistance R is cm PMT 2003]
(a) 50[MP (b) 80 cm

(a) 490  (b) 790  (c) 40 cm (d) 70 cm

(c) 590  (d) 990  133. In the circuit given, the correct relation to a balanced Wheatstone
bridge is [Orissa PMT 2004]
126. The e.m.f. of a standard cell balances across 150 cm length of a wire
of potentiometer. When a resistance of 2  is connected as a shunt P R
(a)  P R
with the cell, the balance point is obtained at 100 cm . The internal Q S
resistance of the cell is G
P S
[MP PET 1993] (b) 
Q R S
(a) 0 .1  (b) 1  Q
P S
(c) 2 (d) 0 .5  (c) 
R Q
127. What is the reading of voltmeter in the following figure (d) None of these
10 V [MP PMT 2004]
134. A galvanometer coil of resistance 50 , show full deflection of
(a) 3 V 100 A . The shunt resistance to be added to the galvanometer, to
(b) 2 V 1000  work as an ammeter of range 10 mA is
V [Pb PET 2000]
(c) 5 V
(a) 5  in parallel (b) 0.5  in series
(d) 4 V
A 500  A 500  (c) 5  in series (d) 0.5  in parallel
128. The current flowing in a coil of resistance 90  is to be reduced by
90%. What value of resistance should be connected in parallel with 135. In given figure, the potentiometer wire AB has a resistance of 5 
it [MP PMT 2004] and length 10 m. The balancing length AM for the emf of 0.4 V is
(a) 9  (b) 90  R=45
(a) 0.4 m
(c) 1000  (d) 10  5V
(b) 4 m
129. The maximum current that can be measured by a galvanometer of M
resistance 40  is 10 mA. It is converted into a voltmeter that can (c) 0.8 m A B
read upto 50 V. The resistance to be connected in series with the (d) 8 m 0.4V
galvanometer is ... (in ohm)
[KCET 2004]
 1082 Current Electricity
136. A potentiometer consists of a wire of length 4 m and resistance 10 (a) 7 .5  (b) 45 
. It is connected to cell of emf 2 V. The potential difference per
unit length of the wire will be (c) 90  (d) 270 
[Pb. PET 2002]
2. Two uniform wires A and B are of the same metal and have
(a) 0.5 V/m (b) 10 V/m
(c) 2 V/m (d) 5 V/m equal masses. The radius of wire A is twice that of wire B . The
137. A voltmeter essentially consists of [UPSEAT 2004] total resistance of A and B when connected in parallel is
(a) A high resistance, in series with a galvanometer (a) 4  when the resistance of wire A is 4.25 
(b) A low resistance, in series with a galvanometer
(b) 5  when the resistance of wire A is 4.25 
(c) A high resistance in parallel with a galvanometer
(d) A low resistance in parallel with a galvanometer (c) 4  when the resistance of wire B is 4.25 
138. In a potentiometer experiment the balancing with a cell is at length (d) 4  when the resistance of wire B is 4.25 
240 cm. On shunting the cell with a resistance of 2 , the balancing
length becomes 120 cm. The internal resistance of the cell is 3. Twelve [DCE
wires2002;
of equal
AIEEElength
2005] and same cross-section are connected
(a) 4  (b) 2  in the form of a cube. If the resistance of each of the wires is R ,
then the effective resistance between the two diagonal ends would
(c) 1  (d) 0.5  be [J & K CET 2004]
139. With a potentiometer null point were obtained at 140 cm and 180
cm with cells of emf 1.1 V and one unknown X volts. Unknown emf is (a) 2 R
[DCE 2002] (b) 12 R
(a) 1.1 V (b) 1.8 V
5
(c) 2.4 V (d) 1.41 V (c) R
6
140. A moving coil galvanometer of resistance 100 is used as an
(d) 8 R
ammeter using a resistance 0.1 The maximum deflection current
in the galvanometer is 100A. Find the minimum current in the 4. You are given several identical resistances each of value R  10 
circuit so that the ammeter shows maximum deflection [IIT-JEE
and (Screening) 2005]of carrying maximum current of 1 ampere. It is
each capable
(a) 100.1 mA (b) 1000.1 mA required to make a suitable combination of these resistances to
(c) 10.01 mA (d) 1.01 mA produce a resistance of 5  which can carry a current of 4
amperes. The minimum number of resistances of the type R that
141. Two resistances are connected in two gaps of a metre bridge. The will be required for this job
balance point is 20 cm from the zero end. A resistance of 15 ohms is
connected in series with the smaller of the two. The null point shifts [CBSE PMT 1990]
to 40 cm. The value of the smaller resistance in ohms is (a) 4 [KCET 2005]
(b) 10
(a) 3 (b) 6 (c) 8 (d) 20
(c) 9 (d) 12
5. The resistance of a wire is 10 6  per metre. It is bend in the
142. If resistance of voltmeter is 10000 and resistance of ammeter is
form of a circle of diameter 2 m . A wire of the same material is
2 then find R when voltmeter reads 12V and ammeter reads 0.1 A [BCECE 2005]
connected across its diameter. The total resistance across its
(a) 118  (b) 120  diameter AB will be
(c) 124  (d) 114
143. Potentiometer wire of length 1 m is connected in series with 490 
resistance and 2V battery. If 0.2 mV/cm is the potential gradient, A B
then resistance of the potentiometer wire is [DCE 2005]

(a) 4.9  (b) 7.9 

4 2
(c) 5.9  (d) 6.9  (a)   10 6  (b)   10 6 
3 3
(c) 0.88  10 6  (d) 14  10 6 
6. In the figure shown, the capacity of the condenser C is 2 F . The
current in 2  resistor is [IIT 1982]
2

1. In an electrical cable there is a single wire of radius 9 mm of

copper. Its resistance is 5  . The cable is replaced by 6 different
3
insulated copper wires, the radius of each wire is 3 mm . Now the 2F
4
total resistance of the cable will be
[CPMT 1988] + –
6V 2.8
 Current Electricity 1083

(a) 9 A (b) 0.9 A 12. If in the circuit shown below, the internal resistance of the battery is
1.5  and V and V are the potentials at P and Q respectively, what
1 1 P Q

(c) A (d) A is the potential difference between the points P and Q

9 0 .9
7. When the key K is pressed at time t  0 , which of the following
(a) Zero
statements about the current I in the resistor AB of the given circuit 20 V 1.5 
is true [CBSE PMT 1995] (b) 4 volts (V > V )P Q
+ –
A B (c) 4 volts (V > V )Q P
3 P 2
2V K
1000 (d) 2.5 volts (V > V ) Q P

13. Two wires of resistance R and R have 2  temperature coefficient of

1000
1F C 1
Q 3 2

resistance  1 and  2 , respectively. These are joined in series. The

effective temperature coefficient of resistance is
(a) I = 2 mA at all t 1   2
(a) (b)  1 2
(b) I oscillates between 1 mA and 2mA 2
(c) I = 1 mA at all t
 1 R1   2 R 2 R1 R 2 1 2
(d) At t = 0 , I = 2 mA and with time it goes to 1 mA (c) (d)
R1  R 2 R12  R 22
8. A torch bulb rated as 4.5 W, 1.5 V is connected as shown in the
figure. The e.m.f. of the cell needed to make the bulb glow at full 14. Two cells of equal e.m.f. and of internal resistances r1 and
intensity is [MP PMT 1999]
r2 (r1  r2 ) are connected in series. On connecting this combination
4.5 W
1.5 V to an external resistance R, it is observed that the potential
(a) 4.5 V difference across the first cell becomes zero. The value of R will be
(b) 1.5 V [MP PET 1985; KCET 2005; Kerala PMT 2005]
1
(c) 2.67 V (a) r1  r2 (b) r1  r2
(d) 13.5 V E(r=2.67)
r1  r2 r1  r2
(c) (d)
9. In the circuit shown in the figure, the current through 2 2
3 2 2 [IIT 1998] 15. When connected across the terminals of a cell, a voltmeter measures
5V and a connected ammeter measures 10 A of current. A resistance
of 2 ohms is connected across the terminals of the cell. The current
9V 8 8 4 flowing through this resistance will be [

(a) 2.5 A (b) 2.0 A

2 2 2 (c) 5.0 A (d) 7.5 A
(a) The 3 resistor is 0.50A (b) The 3 resistor is 0.25 A
16. In the circuit shown here, E = E = E = 2 V and R = R = 4 ohms. The
1 2 3 1 2

(c) The 4 resistor is 0.50A (d) The 4 resistor is 0.25 A current flowing between points A and B through battery E is 2

10. There are three resistance coils of equal resistance. The maximum E1 R1
number of resistances you can obtain by connecting them in any (a) Zero
manner you choose, being free to use any number of the coils in any (b) 2 amp from A to B
way is E2
A B
(c) 2 amp from B to A
[ISM Dhanbad 1994]
(a) 3 (b) 4 (d) None of the above E3 R2

(c) 6 (d) 5 17. In the circuit shown below E = 4.0 V, R = 2 , E = 6.0 V, R = 4 

1 1 2 2

11. In the circuit shown, the value of each resistance is r, then and R = 2 . The current I is [MP PET 2003]
3 1

equivalent resistance of circuit between points A and B will be [Similar to CBSE PMT 1999; RPET 1999] R1 = 2 

r (a) 1.6 A E1 = 4 V
I1
(a) (4/3) r R3 = 2 
(b) 1.8 A
(b) 3r / 2
r r r (c) 1.25 A
r I2
(c) r / 3 R2 = 4 
(d) 1.0 A
r r
(d) 8r / 7 A B E2 = 6 V
C
 1084 Current Electricity

18. A microammeter has a resistance of 100  and full scale range of R5

50 A . It can be used as a voltmeter or as a higher range ammeter I

provided a resistance is added to it. Pick the correct range and R1 R3
R6
resistance combination
[SCRA 1996; AMU (Med.) 2001; Roorkee 2000] R2 R4
(a) 50 V range with 10 k  resistance in series (a) R1 R 2 R5  R 3 R 4 R 6

(b) 10 V range with 200 k  resistance in series 1 1 1 1

(b)   
R5 R 6 R1  R2 R3  R4
(c) 10 mA range with 1  resistance in parallel (c) R1 R 4  R 2 R 3
(d) 10 mA range with 0 .1  resistance in parallel (d) R1 R 3  R 2 R 4  R5 R6
19. The potential difference across 8 ohm resistance is 48 volt as shown 24. In the given circuit, with steady current, the potential drop across
the capacitor must be [IIT-JEE (Screening) 2001]
in the figure. The value of potential difference across X and Y points
will be [MP PET 1996] (a) V V R

X (b) V / 2
3 C
V
(a) 160 volt (c) V / 3
20 30 60
(b) 128 volt (d) 2V / 3 2V 2R
(c) 80 volt 25. A wire of length L and 3 identical cells of negligible internal
24 8 48V
resistances are connected in series. Due to current, the temperature
(d) 62 volt
1 of the wire is raised by  T in a time t. A number N of similar
20. Two resistances R1 Yand R are made of different materials. The cells is now connected in series with a wire of the same material and
2
cross–section but of length 2 L. The temperature of the wire is
temperature coefficient of the material of R1 is  and of the
raised by the same amount  T in the same time t. the value of N
material of R 2 is   . The resistance of the series combination of is
R1 and R 2 will not change with temperature, if R1 / R2 equals [MP PMT 1997] [IIT-JEE (Screening) 2001]
  (a) 4 (b) 6
(a) (b)
  (c) 8 (d) 9
26. What is the equivalent resistance between the points A and B of the
2  2 
(c) (d) network [AMU (Engg.) 2001]
 
57 2 3 2
21. An ionization chamber with parallel conducting plates as anode and (a)  A
7
cathode has 5  10 7 electrons and the same number of singly- 2
4 1
charged positive ions per cm 3 . The electrons are moving at 0.4 m/s. (b) 8 
10 1
The current density from anode to cathode is 4 A / m 2 . The
(c) 6 
velocity of positive ions moving towards cathode is [CBSE PMT 1992]
1.8 5
(a) 0.4 m/s (b) 16 m/s 57
(d)  2.2
(c) Zero (d) 0.1 m/s 5
B
22. A wire of resistance 10  is bent to form a circle. P and Q are 27. The effective resistance between points P and Q of the electrical
points on the circumference of the circle dividing it into a quadrant circuit shown in the figure is
and are connected to a Battery of 3 V and internal resistance 1  as
[IIT-JEE (Screening) 2002]
shown in the figure. The currents in the two parts of the circle are [Roorkee 1999]
2R 2R
6 18 (a) 2 Rr /(R  r)
(a) A and A
23 23 P (b) 8 R (R  r) /(3 R  r) 2R
r
5 15 r
(b) A and A
26 26 (c) 2r  4 R P Q
3V 2R
(c)
4
A and
12
A 1
Q (d) 5 R / 2  2r
25 25 2R 2R
28. In the circuit element given here, if the potential at point B, V = 0,
3 9 then the potentials of A and D are given as
B

(d) A and A
25 25 [AMU (Med.) 2002]
23. In the given circuit, it is observed that the current I is independent 1 amp 1.5  2.5  2V
of the value of the resistance R . Then the resistance values must
6
A B C D
satisfy [IIT-JEE (Screening) 2001]
 Current Electricity 1085

(a) VA  1.5 V, VD  2 V (b) VA  1.5 V, VD  2 V (a) 4/9

R R
(c) VA  1.5 V, VD  0.5 V (d) VA  1.5 V, VD  0.5 V (b) 8/9
R 6R R
29. The equivalent resistance between the points P and Q in the (c) 2 E
3
network given here is equal to (given r   ) (d) 18
2 R 4R
[AMU (Med.) 2002] 36. In the circuit shown here, the readings of the ammeter and
1 voltmeter are [Kerala PMT 2002]
(a)  r r
2 r 6 V, 1

(b) 1  r r (a) 6 A, 60 V
P Q
3 (b) 0.6 A, 6 V V
(c)  r 6 A
2 r r (c) 6/11 A, 60/11 V
(d) 2  (d) 11/6 A, 11/60 V 4

30. The current in a conductor varies with time t as I  2t  3 t 2 37. Length of a hollow tube is 5m, it’s outer diameter is 10 cm and
where I is in ampere and t in seconds. Electric charge flowing thickness of it’s wall is 5 mm. If resistivity of the material of the
through a section of the conductor during t = 2 sec to t = 3 sec is 1.7  10JEE
tube is [Orissa 2003]
m then resistance of tube will be
–8

(a) 10 C (b) 24 C
(a) 5.6  10  –5
(b) 2  10  –5

(c) 33 C (d) 44 C
31. A group of N cells whose emf varies directly with the internal (c) 4  10  –5
(d) None of these
resistance as per the equation E = 1.5 r are connected as shown in
38. A wire of resistor R is bent into a circular ring of radius r.
N N

the figure below. The current I in the circuit is

Equivalent resistance between two points X and Y on its
1 [KCET 2003]
circumference, when angle XOY is , can be given by
r1 2
(a) 0.51 amp r2 R
(a) (2   )
(b) 5.1 amp 4 2 X
N rN
r3 R
(c) 0.15 amp 3 (b) (2   )
r4 2 W  O Z
(d) 1.5 amp
(c) R (2 – )
32. In the shown arrangement of the experiment of 4the meter bridge if
Y
AC corresponding to null deflection of galvanometer is x, what 4
(d) (2   )
would be its value if the radius of the wire AB is doubled R
[IIT-JEE (Screening) 2003]
(a) x 39. Potential difference across the terminals of the battery shown in
figure is (r = internal resistance of battery)
(b) x/4
(a) 8 V
(c) 4x R1 R2 r =1
10 V
(b) 10 V
(d) 2x G (c) 6 V
33. The resistance of a wire of iron is 10 ohms and temp. coefficient of
(d) Zero 4
resistivity is 5  10 3 / C . AAt 20xC it carries
C B
30 milliamperes of
current. Keeping constant potential difference between its ends, the 40. As the switch S is closed in the circuit shown in figure, current
temperature of the wire is raised to 120C . The current in passed through it is
milliamperes that flows in the wire is [MP PMT 1994]
20 V 2 4 5V
(a) 20 (b) 15 (a) 4.5 A
(c) 10 (d) 40 A B
(b) 6.0 A 2
34. Seven resistances are connected as shown in the figure. The
equivalent resistance between A and B is [MP PET 2000] (c) 3.0 A
10 S
(a) 3  (d) Zero

(b) 4  A 10 3 B 41. In the following circuit a 10 m long potentiometer wire with
resistance 1.2 ohm/m, a resistance R and an accumulator of emf 2 V
(c) 4.5 
1

are connected in series. When the emf of thermocouple is 2.4 mV

5 8 6 6
(d) 5  then the deflection in galvanometer is zero. The current supplied by
the accumulator will be
35. A battery of internal resistance 4 is connected to the network of
resistances as shown. In order to give the maximum power to the + –
i R1
network, the value of R (in  ) should be (a) 4 
[IIT101995]
A
–4

5m
A B
G

Hot Cold
Junction Junction
 1086 Current Electricity

(b) 8  10 A –4 (a) 4 (b) 1

(c) 3 (d) 2
(c) 4  10 A –3

48. Following figure shows cross-sections through three long conductors

(d) 8  10 A –3
of the same length and material, with square cross-section of edge
lengths as shown. Conductor B will fit snugly within conductor A,
42. In the following circuit, bulb rated as 1.5 V, 0.45 W. If bulbs glows and conductor C will fit snugly within conductor B. Relationship
with full intensity then what will be the equivalent resistance between their end to end resistance is
between X and Y
3a
6V
(a) 0.45  2a
a
(b) 1  R
3 X
Y A B C
(c) 3 

(d) 5 
(a) R = R = R
B A B C

43. Consider the circuits shown in the figure. Both the circuits are (b) R > R > R
A B C

taking same current from battery but current through R in the (c) R < R < R
A B

1
second circuit is th of current through R in the first circuit. If R (d) Information is not sufficient
10
49. In the following star circuit diagram (figure), the equivalent
is 11 , the value of R 1 resistance between the points A and H will be
(a) 9.9  i i R1 A i
i/10 (a) 1.944 r r r 72°
(b) 11 
E R R2 R B r C D r E
(b) 0.973 r
(c) 8.8 
r r
(c) 0.486 r F J
(d) 7.7  (a) (b) H
(d) 0.243 r r r
44. In order to quadruple the resistance of a uniform wire, a part of its r r
50. I of 10 . The
In the adjoining circuit diagramG each resistance is
length was uniformly stretched till the final length of the entire wire
was 1.5 times the original length, the part of the wire was fraction current in the arm AD will be
E
equal to 2i
(a)
(a) 1 / 8 5
B F i
l 3i
(b) 1 / 6 (b)
5
(c) 1 / 10 i
0.5l
(d) 1 / 4 4i
(c) A C
5
45. In the circuit shown in figure reading of voltmeter is V when only S 1 1

is closed, reading of voltmeter is V when only S is closed and i

2 2 (d) D
reading of voltmeter is V when both S and S are closed. Then
3 1 2
5
51. In the circuit of adjoining figure the current through 12  resister
(a) V > V > V 3R
3 2 1

will be
(b) V > V > V R S1
2 1 3
(a) 1 A
6R
(c) V > V > V 5 5
3 1 2

S2 1
(b) A 10
(d) V > V > V V 5
1 2 3
5V 5V
46. Current through wire XY of circuit shown is A C
2
(c) A
1 XE 2 5 12
E F
(a) 1 A (d) 0 A
1 2
(b) 4 A 52. The reading of the ideal voltmeter in the adjoining diagram will be
Y
(c) 2 A 3 4 (a) 4 V A
(d) 3 A (b) 8 V
47. 12 cells each having same emf are connected
50V in series with some 10V 20
(c) 12 V
cells wrongly connected. The arrangement is connected in series V
with an ammeter and two cells which are in series. Current is 3 A (d) 14 V
10 4V
when cells and battery aid each other and is 2 A when cells and
battery oppose each other. The number of cells wrongly connected B N C
is
 Current Electricity 1087

53. The resistance of the series combination of two resistance is S. (a) AB

When they are joined in parallel the total resistance is P. If S = nP, (b) BC
then the minimum possible value of n is
(c) CD
[AIEEE 2004]
(d) DE
(a) 4 (b) 3
5. I-V characteristic of a copper wire of length L and area of cross-
(c) 2 (d) 1
section A is shown in figure. The slope of the curve becomes
54. A moving coil galvanometer has 150 equal divisions. Its current
sensitivity is 10 divisions per milliampere and voltage sensitivity is 2
divisions per millivolt. In order that each division reads 1 volt, the I
resistance in ohms needed to be connected in series with the coil
will be [AIEEE 2005]
(a) 99995 (b) 9995
(a) More if the experiment is performed at higher temperature
3 5 O
(c) 10 (d) 10 (b) More if a wire of steel of same dimensionVis used
(c) More if the length of the wire is increased
(d) Less if the length of the wire is increased
6. E denotes electric field in a uniform conductor, I corresponding
current through it, v d drift velocity of electrons and P denotes
thermal power produced in the conductor, then which of the
1. Which of the adjoining graphs represents ohmic resistance following graph is incorrect
[CPMT 1981; DPMT 2002] (a) vd (b) P

(a) V (b) V

E E
(c) P (d) P
I I
(c) (d)
V V

vd i
7. The two ends of a uniform conductor are joined to a cell of e.m.f. E
and some internal resistance. Starting from the midpoint P of the
I
2. Variation of current passing through a conductorI as the voltage conductor, we move in the direction of current and return to P. The
applied across its ends as varied is shown in the adjoining diagram. potential V at every point on the path is plotted against the distance
covered (x). Which of the following graphs best represents the
If the resistance (R) is determined at the points A, B, C and D, we resulting curve
will find that [CPMT 1988]
D V V
(a) R = R
C D V C (a) (b)
(b) R > R B E <E
B A

(c) R > R
C B
A
X X
(d) None of these V V
i (c) (d)
3. The voltage V and current I graph for a conductor at two different
temperatures T1 and T2 are shown in the figure. The relation E
between T1 and T2 is <E

[MP PET 1996; KCET 2002] X X

8. The resistance R t of a conductor varies with temperature t as
(a) T1  T2 T1 shown in the figure. If the variation is represented by
V Rt  R0 [1  t  t 2 ] , then
(b) T1  T2 [CPMT 1988]
T2
Rt
(c) T1  T2 (a)  and  are both negative

(d) T1  T2 I (b)  and  are both positive

4. From the graph between current I and voltage V shown below, (c)  is positive and  is negative
identify the portion corresponding to negative resistance [CBSE PMT 1997] t
(d)  is negative and  are positive

I E
C
B
D
V
A
 1088 Current Electricity
9. Variation of current and voltage in a conductor has been shown in (c) (d)
the diagram below. The resistance of the conductor is.

V 6
5
4
3
2 14. The V-i graph for a conductor at temperature T1 and T2 are as
1
1 2 3 4 5 6
shown in the figure. (T2  T1 ) is proportional to
i T2
(a) 4 ohm (b) 2 ohm (a) cos 2 V
(c) 3 ohm (d) 1 ohm (b) sin T1
10. Resistance as shown in figure is negative at [CPMT 1997]
(c) cot 2 
(d) tan  
I A C i

15. A cylindrical conductor has uniform cross-section. Resistivity of its

B
material increase linearly from left end to right end. If a constant
current is flowing through it and at a section distance x from left
(a) A V B
(b) end, magnitude of electric field intensity is E, which of the following
graphs is correct
(c) C (d) None of these
(a) E (b) E
11. For a cell, the graph between the potential difference (V) across the
terminals of the cell and the current (I) drawn from the cell is
shown in the figure. The e.m.f. and the internal resistance of the cell
are
O x O x
V(Volts)
(c) E (d)
2.0
E
1.5
1.0
0.5
O x O x
(a) 2V ,0.5  0
4 16. The V-i graph for a conductor makes an angle  with V-axis. Here
1 2 3(b) 4 2V
5 ,0I.(amperes)
V denotes the voltage and i denotes current. The resistance of
(c)  2V ,0.5  (d)  2V,0.4 
conductor is given by
12. The graph which represents the relation between the total resistance
R of a multi range moving coil voltmeter and its full scale deflection (a) sin (b) cos 
V is
(c) tan  (d) cot 
R R
17. A battery consists of a variable number 'n' of identical cells having
internal resistances connected in series. The terminals of battery are
short circuited and the current i is measured. Which of the graph
below shows the relation ship between i and n
V V
(ii) i i
(i)
(a) (b)
R R

O n O n
V (c) i (d) i
V
(a) (i) (iii) (b) (ii) (iv)
(c) (iii) (d) (iv)
13. When a current I is passed through a wire of constant resistance, it
produces a potential difference V across its ends. The graph drawn O n
O n
between log I and log V will be 18. In an experiment, a graph was plotted of the potential difference V
between the terminals of a cell against the circuit current i by
(a) (b) varying load rheostat. Internal conductance of the cell is given by
log I

log I

x A
log V log V V

y
log I

log I

log V log V
 Current Electricity 1089

4. Assertion : In the following circuit emf is 2V and internal

resistance of the cell is 1  and R = 1, then
reading of the voltmeter is 1V.
y
(a) xy (b) V
x
E=2V
x
(c) (d) (x – y)
y
r=1
19. V-i graphs for parallel and series combination of two identical A
resistors are as shown in figure. Which graph represents parallel R=1
combination 2
B Reason : V  E  ir where E = 2 V, i   1 A and R =
V 2
1 [AIIMS 1995]
A
5. Assertion : There is no current in the metals in the absence of
electric field.
(a) A (b)i B Reason : Motion of free electron are randomly.
[AIIMS 1994]
(c) A and B both (d) Neither A nor B
6. Assertion : Electric appliances with metallic body have three
20. The ammeter has range 1 ampere without shunt. the range can be connections, whereas an electric bulb has a two pin
varied by using different shunt resistances. The graph between connection.
shunt resistance and range will have the nature Reason : Three pin connections reduce heating of connecting
Q
wires.
R 7. Assertion : The drift velocity of electrons in a metallic wire will
decrease, if the temperature of the wire is
S P increased.
Shunt

S Reason : On increasing temperature, conductivity of metallic

R wire decreases.
Q 8. Assertion : The electric bulbs glows immediately when switch
is on.
O 1 2 3 4 Ampere Reason : The drift velocity of electrons in a metallic wire is
(a) P (b)
Range Q very high.
(c) R (d) S 9. Assertion : Bending a wire does not effect electrical resistance.
Reason : Resistance of wire is proportional to resistivity of
material.
10. Assertion : In meter bridge experiment, a high resistance is
always connected in series with a galvanometer.
Reason : As resistance increases current through the circuit
Read the assertion and reason carefully to mark the correct option out of increases.
the options given below : 11. Assertion : Voltameter measures current more accurately than
(a) If both assertion and reason are true and the reason is the correct ammeter.
explanation of the assertion. Reason : Relative error will be small if measured from
(b) If both assertion and reason are true but reason is not the correct voltameter.
explanation of the assertion. 12. Assertion : Electric field outside the conducting wire which
(c) If assertion is true but reason is false. carries a constant current is zero.
(d) If the assertion and reason both are false.
(e) If assertion is false but reason is true. Reason : Net charge on conducting wire is zero.
13. Assertion : The resistance of super-conductor is zero.
1. Assertion : The resistivity of a semiconductor increases with
temperature. Reason : The super-conductors are used for the transmission
of electric power.
Reason : The atoms of a semiconductor vibrate with larger
amplitude at higher temperatures thereby 14. Assertion : A potentiometer of longer length is used for
increasing its resistivity [AIIMS 2003] accurate measurement.
2. Assertion : In a simple battery circuit the point of lowest Reason : The potential gradient for a potentiometer of longer
potential is positive terminal of the battery length with a given source of e.m.f. becomes small.
Reason : The current flows towards the point of the higher 15. Assertion : The e.m.f. of the driver cell in the potentiometer
potential as it flows in such a circuit from the experiment should be greater than the e.m.f. of the
negative to the positive terminal.
cell to be determined.
[AIIMS 2002]
3. Assertion : The temperature coefficient of resistance is positive Reason : The fall of potential across the potentiometer wire
for metals and negative for p-type semiconductor. should not be less than the e.m.f. of the cell to be
Reason : The effective charge carriers in metals are determined.
negatively charged whereas in p-type semiconductor
they are positively charged. 16. Assertion : A person touching a high power line gets stuck
[AIIMS 1996] with the line.
 1090 Current Electricity
Reason : The current carrying wires attract the man towards 41 a 42 c 43 b 44 d 45 c
it. 46 d 47 c 48 b 49 b 50 d
17. Assertion : The connecting wires are made of copper. 51 d 52 c 53 d 54 a 55 c
Reason : The electrical conductivity of copper is high. 56 d 57 c 58 c 59 d 60 c
61 d 62 c 63 d 64 c 65 c
66 c 67 b 68 c 69 d 70 b
71 a 72 c 73 a 74 b 75 a
76 c 77 c 78 b 79 c 80 a
81 a 82 b 83 b 84 d 85 d
Electric Conduction, Ohm's Law and Resistance 86 a 87 a 88 a 89 b 90 b
91 b 92 c 93 b 94 d 95 a
1 a 2 c 3 b 4 b 5 c 96 d 97 b 98 b 99 d 100 a
6 a 7 a 8 a 9 d 10 c 101 c 102 a 103 b 104 d 105 a
11 d 12 d 13 a 14 c 15 a 106 a 107 b 108 d 109 bc 110 b

16 a 17 c 18 b 19 c 20 b 111 d 112 c 113 a 114 a 115 d

116 a 117 d 118 c 119 d 120 c
21 d 22 b 23 b 24 b 25 d
121 b 122 b 123 b 124 c 125 b
26 c 27 b 28 b 29 b 30 a
126 a 127 c 128 b 129 c 130 a
31 c 32 d 33 b 34 d 35 c
131 a 132 a 133 c 134 a 135 b
36 b 37 b 38 c 39 a 40 d 136 b 137 a 138 b 139 c 140 b
41 b 42 b 43 a 44 b 45 c 141 b
46 a 47 b 48 b 49 c 50 a
51 c 52 c 53 b 54 b 55 b Kirchhoff's Law, Cells
56 a 57 a 58 a 59 c 60 c
1 b 2 d 3 c 4 a 5 a
61 a 62 b 63 b 64 c 65 c
6 b 7 a 8 a 9 b 10 a
66 d 67 a 68 b 69 d 70 d
11 c 12 d 13 a 14 d 15 b
71 a 72 a 73 c 74 b 75 b 16 c 17 c 18 c 19 d 20 b
76 c 77 c 78 c 79 d 80 b 21 c 22 c 23 b 24 d 25 a
81 a 82 d 83 b 84 b 85 c 26 a 27 b 28 b 29 a 30 b
86 b 87 c 88 a 89 a 90 d 31 a 32 c 33 b 34 a 35 a
91 a 92 c 93 b 94 a 95 b 36 b 37 a 38 b 39 b 40 c
96 b 97 c 98 a 99 c 100 d 41 d 42 d 43 d 44 a 45 c
101 c 102 a 103 d 104 b 105 b 46 c 47 b 48 a 49 a 50 d

106 d 107 d 108 a 109 d 110 d 51 b 52 d 53 b 54 c 55 a

56 b 57 c 58 a 59 d 60 b
111 d 112 d 113 a 114 a 115 c
61 c 62 c 63 c 64 b 65 a
116 a 117 a 118 b 119 c 120 a
66 c 67 a 68 d 69 b 70 a
121 d 122 a 123 a 124 d 125 c
71 a 72 d 73 c 74 b 75 b
126 b 127 c 128 a 129 a 130 c
76 b 77 c 78 c 79 d 80 d
131 c 132 b 133 c
81 a 82 d 83 c 84 c 85 a

Grouping of Resistances Different Measuring Instruments

1 c 2 d 3 a 4 c 5 b 1 a 2 c 3 d 4 d 5 c
6 c 7 c 8 b 9 a 10 b 6 c 7 a 8 d 9 c 10 c
11 d 12 d 13 b 14 d 15 b 11 d 12 c 13 d 14 a 15 d
16 d 17 c 18 c 19 b 20 d 16 c 17 a 18 b 19 c 20 a
21 a 22 a 23 b 24 b 25 c 21 b 22 b 23 a 24 a 25 a
26 b 27 d 28 d 29 d 30 c 26 a 27 a 28 a 29 b 30 b
31 b 32 d 33 a 34 b 35 c 31 b 32 b 33 b 34 b 35 c
36 d 37 d 38 b 39 c 40 b 36 c 37 b 38 b 39 d 40 b
 Current Electricity 1091

41 a 42 b 43 c 44 d 45 a
46 b 47 c 48 a 49 b 50 a
51 b 52 c 53 b 54 b 55 a
56 b 57 d 58 c 59 c 60 d
61 a 62 a 63 d 64 a 65 d
66 b 67 a 68 b 69 c 70 c
71 d 72 c 73 a 74 c 75 a
76 d 77 b 78 a 79 b 80 a
81 c 82 c 83 a 84 a 85 b
86 d 87 c 88 d 89 c 90 a
91 d 92 d 93 d 94 c 95 d
96 a 97 c 98 a 99 d 100 d
101 c 102 d 103 c 104 c 105 a
106 c 107 c 108 d 109 c 110 a
111 a 112 a 113 c 114 b 115 c
116 d 117 a 118 b 119 d 120 c
121 d 122 c 123 a 124 a 125 b
126 b 127 d 128 d 129 b 130 c
131 d 132 a 133 c 134 d 135 d
136 a 137 a 138 b 139 d 140 a
141 c 142 a 143 a

Critical Thinking Questions

1 a 2 a 3 c 4 c 5 c
6 b 7 d 8 d 9 d 10 b
11 d 12 d 13 c 14 b 15 b
16 b 17 b 18 b 19 a 20 d
21 d 22 a 23 c 24 c 25 b
26 b 27 a 28 d 29 b 30 b
31 d 32 a 33 a 34 b 35 c
36 c 37 a 38 a 39 d 40 a
41 a 42 b 43 a 44 a 45 b
46 c 47 b 48 a 49 b 50 a
51 d 52 b 53 a 54 b

Graphical Questions
1 a 2 d 3 a 4 c 5 d
6 c 7 b 8 b 9 d 10 a
11 b 12 d 13 a 14 c 15 b
16 d 17 d 18 b 19 a 20 b
 1096 Current Electricity
 If length becomes 3 times so Resistance becomes 9 times
Assertion and Reason i.e. R'  9  20  180
12. (d) Resistivity is the property of the material. It does not depend
1 d 2 d 3 b 4 a 5 a upon size and shape.
6 c 7 b 8 c 9 a 10 c 13. (a) Because with rise in temperature resistance of conductor
11 a 12 a 13 b 14 a 15 a increase, so graph between V and i becomes non linear.
16 d 17 a
14. (c) Because V-i graph of diode is non-linear.
e V e El
15. (a) vd    or v d  .  (Since V  El)
m l m l
vd  E

1
16. (a) Resistance of conductor depends upon relation as R  .

With rise in temperature rms speed of free electron inside the
Electric Conduction, Ohm's Law and Resistance conductor increase, so relaxation time decrease and hence
resistance increases
1. (a) Number of electrons flowing per second q 4
17. (c) i   2 ampere
n i t 2
  4.8 / 1.6  10 19  3  1019
t e 3
18. (b) Volume  Al  3  A 
J l
2. (c) vd   vd  J (current density)
ne
l   l l 2 9 3
Now R   3   l2  
J1 
i
and J 2 
2i i
  J1 ;  (vd )1  (vd )2  v A 3 /l 3  
A 2A A
ne 62.5  10 18  1.6  10 19
3. (b) Order of drift velocity  10 4 m / sec  10 2 cm / sec 19. (c) i   10 ampere
t 1
4. (b) Density of Cu  9  10 3 kg / m 3 (mass of 1 m of Cu) 3

20. (b) In twisted wire, two halves each of resistance 2 are in

 6.0  10 atoms has a mass = 63  10 kg
23 –3

parallel, so equivalent resistance will be

2
 1 .
 Number of electrons per m are 3
2
1
6 .0  10 23 21. (d) In stretching of wire R 
  9  10 3  8 .5  10 28 r4
63  10  3
L  1
i 22. (b) R  0 .7 
Now drift velocity  v d  A 22
neA (1  10  3 ) 2
7
1.1
   2.2  10 6 ohm-m.
8.5  10 28
 1 .6  10 19    (0.5  10  3 )2
1 1 1
 0.1  10 3 m / sec 23. (b) R R 2  2 [d = diameter of wire]
A r d
5. (c) Because 1 H.P. = 746 J/s = 746 watt
24. (b) i = q  1.6  10 19  6.6  10 15  10.56  10 4 A  1mA
R 2 l R 0
6. (a) Rl  
2
 %  2  0 .1  0 .2% 2
R l R l R1 l r2 1 5 2
25. (d) R   1  22       l 2  20m
r2 R2 l 2 r1 1 l2  1 
l 8 50  10 2 6
7. (a) R  50  10   10  26. (c)
A (50  10  2 ) 2
27. (b) In semiconductors charge carries are free electrons and holes
8. (a) Resistivity of some material is its intrinsic property and is
constant at particular temperature. Resistivity does not depend 28. (b) Net current inet  i()  i()
upon shape.
+2e
 1 (1  t1 ) 1 (1  0 .00125  27) 
n()q()

n()q()
9. (d)    +
 2 (1  t 2 ) 2 (1  0 .00125  t) t t –
–e
 t  854C  T  1127 K n() n()
  2e  e inet
t t
l 2l l
10. (c) R1   R2  i.e. R 2 
A 2A A = 3.2  10  2  1.6  10 + 3.6  10  1.6  10
18 –19 18 –19

 R1  R 2 = 1.6 A (towards right)

11. (d) In case of stretching of wire R  l 2

 Current Electricity 1097

29. (b) In the absence of external electric field mean velocity of free 47. (b) Because as temperature increases, the resistivity increases and
3 KT  1
electron (V ) is given by Vrms   Vrms  T hence the relaxation time decreases for conductors   .
rms

m  
30. (a) With rise in temperature specific resistance increases 48. (b) In VI graph, we will not get a straight line in case of liquids.
31. (c) For metallic conductors, temperature co-efficient of resistance l
is positive. 49. (c) R
A
32. (d)
50. (a) Since R  l 2  If length is increased by 10%, resistance is
33. (b) Length l = 1 cm  10 2 m increases by almost 20%
Hence new resistance R'  10  20% of 10
20
 10   10  12 .
100
1 cm

100 cm R150 [1   (150)]

51. (c)  . Putting R150  133 and
R 500 [1   (500)]
1 cm
  0.0045 / C, we get R 500  258 
Area of cross-section A = 1 cm  100 cm
= 100 cm = 10 m 2 –2 2

l 64  10 6  198
52. (c) R 7   r  0.024 cm
10 2 A 22
 r2
Resistance R = 3  10  –7
= 3  10  –7

7
10  2
34. (d) In the above question for calculating equivalent resistance i i J i r2
53. (b) Current density J   2  1  1  22
between two opposite square faces. A r J 2 i2 r1
l = 100 cm = 1 m, A = 1 cm = 10 m , so resistance R = 3  10
2 –4 2 –7

But the wires are in series, so they have the same current,
1 J1 r22
  4 = 3  10  –3

hence i1  i2 . So   9 :1
10 J 2 r12
i 20
35. (c) vd   29  1.25  10 3 m / s 54. (b) As
V
 R and R  temperature
nAe 10  10  1.6  10 19
6
i

36. (b) Specific resistance k 

E 55. (b) R  l 2  If l doubled then R becomes 4 times.
j 56. (a) Temperature coefficient of a semiconductor is negative.
2 57. (a) The reciprocal of resistance is called conductance
l d 
2
l l R L  2d 
37. (b) R  2  1  1   2  
   1 Potential difference
A d R2 l2  d1  4L  d  58. (a) Resistance 
Current
 R = R = R.
2 1
59. (c) Ohm’s Law is not obeyed by semiconductors.
i 1.344
38. (c) vd   6 V
nAe 10  1.6  10 19  8.4  10 22 60. (c) Drift velocity vd  ; vd does not depend upon
 lne
1.344 diameter.
  0.01cm / s  0.1mm / s
10  1.6  8.4 61. (a) Using R T2  R T1 [1  (T2  T1 )]
1
39. (a) Internal resistance   R100  R 50 [1  (100  50)]
Temperatur e
(7  5)
40. (d) Charge = Current × Time =5 × 60 = 300 C  7  5 [1  (  50)]     0 .008 / o C
250
41. (b) By R  l / A
62. (b) This is because of secondary ionisation which is possible in the
42. (b) gas filled in it.
43. (a) 63. (b)
l l R R1 (1  t1 ) (1  3.92  10 3  20)
44. (b) R for first wire and R’=  for second wire.  
50

a 4a 4 64. (c)
R2 (1  t2 ) 76.8 (1  3 .92  10  3 t)
45. (c) For semiconductors, resistance decreases on increasing the
temperature.  t  167C
l n l i
46. (a) R  . 65. (c) From v d   i  vd A  i  vd r 2
A ne 2 A neA
66. (d) Resistivity depends only on the material of the conductor.
 1098 Current Electricity
67. (a) A particular temperature, the resistance of a superconductor is 77. (c) Human body, though has a large resistance of the order, of
1 1 K (say 10k  ), is very sensitive to minute currents even as
zero  G    
R 0 low as a few mA. Electrons, excites and disorders the nervous
system of the body and hence one fails to control the activity
(n ) (q  ) (n ) (q  ) of the body.
68. (b) Net current i  i  i  
t t
78. (c) R t  R 0 (1   t)
–
– e +  4.2 = R 0 (1  0.004  100)  1.4 R 0  R 0  3 .
Ne+
2 2 2
l2  l   l   l 
i 79. (d) R  R1 : R2 : R3   1  :  2  :  3 
(n ) (n )
m  m1   m 2   m 3 
 i   e   e
t t 25 9 1 1
 : :  25 : 3 :  125 : 15 : 1 .
19 19 1 3 5 5
 2.9  10  1.6  10
18
 1.2  10  1.6  10
18

80. (b)
 i  0.66 A
4
R1  r2 
4
(d) If E be electric field, then current density j = E R  nr  R
69. 81. (a)        R2  4 .
R 2  r1 
 R2  r  n
i
Also we know that current density j 
A R1 (1  t1 ) 5 (1    50) 1
82. (d)      per o C
Hence j is different for different area of cross-sections. When j R2 (1  t2 ) 6 (1    100) 200
is different, then E is also different. Thus E is not constant. The
j
Again by Rt  R0 (1  t)
drift velocity v d is given by v d  = different for different j
ne  1 
values. Hence only current i will be constant.  5  R0 1   50   R0  4 .
 200 
70. (d)
Q
71. (a) R
l
and mass m = volume (V)  density (d) = (A l) d 83. (b) i   Q  1.6  10 19  5  1015  0.8 mA .
A T

Since wires have same material so  and d is same for both. riron iron 1  10 7
84. (b)    2 .4 .
1 rCopper copper 1 .7  10  8
Also they have same mass  Al = constant  l 
A
85. (c) i  e  1.6  10 19  6.8  1015  1.1  10 3 amp.
2 4
R1 l1 A2  A2  r  86. (b) Resistivity of the material of the rod
      2
R2 l2 A1  A1  r 
 1 RA 3  10 3  (0 .3  10 2 )2
  = 27  10 9    m
4 l 1
34  r 
     R2  544  (Thickness)
R 2  2r  Resistance of disc R 
(Area of cross section)
l R A A R
72. (a) R  1  2 ( , L constant)  1  2  2
A R2 A1 A2 R1 (10 3 )
= 27  10 9    2.7  10 7 .
Now, when a body dipped in water, loss of weight   (1  10  2 )2
 V L g  AL L g 87. (c) By using R t  R 0 (1  t)
(Loss of weight)1 A 3  R 0  R 0 (1  4  10 3 t)  t  500 o C .
So,  1  2; so A has more loss of
(Loss of wight)2 A2
weight. 88. (a) i  6  10 15  1.6  10 19  0.96mA.
73. (c) Q = it = 20 × 10 × 30 = 6× 10 C n  1 .6  10 19
–6 –4

ne
89. (a) i  16  10 3   n  1017
74. (b) Ge is semiconductor and Na is a metal. The conductivity of t 1
semiconductor increases and that of the metals decreases with
the rise in temperature. V 100  0.5
90. (d) R   10  0.25  .
i 10  0.2
ne it 1 .6  10 3  1
75. (b) i   n   10 16 . V l 2 50  10 2
t e 1 .6  10 19 91. (a) R       1  10 6 m .
i A 4 (1  10  3 )2
i 1 1 92. (c)
76. (c) Drift velocity vd   vd  or vd  2
neA A d V Q Vt 20  2  60
93. (b) i   Q    240 C .
2 R t R 10
vP  dQ 
2
  
d /2 1 1
     v P  vQ .
v Q  d P   d  4 4
 Current Electricity 1099

2 2 115. (c) Same mass, same material i.e. volume is same or Al = constant
R 1  l1  R  l  R
94. (a) R  l2         4  R2  .
R 2  l 2  R2  l/2 
2 4
4 l R l A  A  d 
Also, R    1  1  2   2   2 
95. (b) A R 2 l2 A1  A1   d1 
i 40
96. (b) Vd   24  d 
4
neA 10 29  10 6  1.6  10 19     16  R2  1.5  .
R2  d /2 
= 2.5  10 3 m/sec .
116. (a) I  ne qe  np q p  1mA towards right
i 5 .4
97. (c) Vd   117. (a) As steady current is flowing through the conductor, hence the
nAe 8.4  10 28  10 6  1 .6  10 19
number of electrons entering from one end and outgoing from
= 0.4  10 3 m/sec  0.4 mm /sec . the other end of any segment is equal. Hence charge will be
2
zero.
R 1  l1 
2
10  5  1
98. (a)        R 2  160 . 1 A 1
R 2  l 2  R 2  20  16 118. (b) Conductance C  
R l
C 
l
1 dQ
99. (c) R ; where  = Relaxation time. 119. (c) i
t 5
 dQ  idt  Q   t 2 i dt  0 (1.2t  3) dt
 dt 1

When lamp is switched on, temperature of filament increase, 5

hence  decrease so R increases  1 .2 t 2 
  3 t   30 C
100. (d) R  91  10 2  9.1k .  2  0
101. (c) 4
R 2  r1 
4
R 2
102. (a) 120. (a) In stretching,     2     R2  16 R
R1  r2  R 1
l2 l2 l2 l2
103. (d) R  R1 : R 2 : R 3  1 : 2 : 3
m m1 m 2 m 3 121. (d) R'  n 2 R  R'  16 R
9 4 1 122. (a)
 R1 : R 2 : R 3  : :  27 : 6 : 1 .
1 2 3 Significant figures Multiplier
3
1  10 Brown Black Brown
104. (b) n   6 .25  10 15 .
1 .6  10 19 1 0 10 1

i i v i r 
2
v  R = 10  10 = 100 
1

105. (b) vd   vd  2   1   2   v '  .

ne r 2 r v ' i2  1
r 2 123. (a)
4
124. (d)
R 1  r2  4
R  3r/4  81 256 125. (c)
106. (d)         R2  R
R 2  r1 
 R2  r  256 81 2 2
l R2 l r 2 1 1
107. (d) 126. (b) R    2  12 =    
r2 R1 l1 r2 1 2 2
i 8
108. (a) v d   R1
nAe 8  10 28  (2  10 3 ) 2  1 .6  10 19  R2  , specific resistance doesn't depend upon length,
2
= 0.156  10 3 m/sec . and radius.
109. (d) Specific resistance doesn’t depend upon length and area. i 100
110. (d) Heating effect of current. 127. (c) By using vd  
neA 10 28  1 .6  10 19    (0 .02)2
R r 2 4 .2  3 .14  (0.2  10 3 )2 4
111. (d) l    1 .1m
 48  10  8
 2  10 4 m / sec
112. (d) For conductors, resistance  Temperature and for semi-
l l
conductor, resistance 
1 128. (a) R . For highest resistance 2 should be maximum,
Temperatur e r2 r
which is correct for option (a)
113. (a) If suppose initial length l1  100 then l2  100  100  200
129. (a) Red, brown, orange, silver red and brown represents the first
2 two significant figures.
R1  l1 
2
 100 
     R 2  4 R1 Significant figures Multiplier Tolerance
R2  l2   200 
Red Brown Orange Silver
R R  R1 4R  R
 100  2  100  1 1  100  300%. 2 1 10 3

 10%
R R1 R1
114. (a) Ammeter is always connected in series and Voltmeter is always  R = 21  10 3  10%
connected in parallel.
 1100 Current Electricity
2 2 R 1
R2 l R 2 6. (c) Resistance of 1 ohm group   
130. (c) In stretching R  l 2   22  2    n 3
R1 l1 R1  1 
2
 R 2  4 R1 . Change in resistance  R2  R1  3R1 This is in series with  resistor.
3
Change in resistance 3 R1 3 2 1 3
Now,    Total resistance      1
Original resistance R1 1 3 3 3
7. (c) Lowest resistance will be in the case when all the resistors are
2 connected in parallel.
R1  l1 
131. (c)    , If l1  100 then l = 110
R 2  l2  1

1

1
2

....... 10 times
R 0 .1 0 .1
2
R1  100  1
    R 2  1.21 R1  10  10....... 10 times
R 2  110  R
R2  R1 1
 100 i.e. R 
1

% change  100  21%
R1 R 100
2
132. (b) 8. (b) Resistance across XY  
3
l
133. (c) Resistance   Total resistance 2 2V
A A
2 8
R1  l A 2 3 5 5  2   2
  1  1  2     3 3 X Y
R 2  2 l 2 A1 3 4 4 8 2
Current through ammeter

Grouping of Resistances 2 6 3 2
   A
8/3 8 4
1. (c) The given circuit can be redrawn as follows
9. (a) Equivalent resistance of the combination
5 5 B 5 C
A (2  2)  2 8 4
2/3V 2/3V 2/3V    
222 6 3

2V 2 2

A D C
5 5 5 P Q
For identical resistances, potential difference distributes equally
R 2
among all. Hence potential difference across each resistance is 10. (b) In parallel, x  R  nx
2 4 n
V , and potential difference between A and B is V . In series, R + R + R .... n times = nR = n (nx) = n x 2

3 3
11. (d) The circuit reduces to
2. (d) Equivalent resistance of parallel resistors is always less than any 3
of the member of the resistance system.
3. (a) Each part will have a resistance r  R / 10
3 3
Let equivalent resistance be rR , then

1 1 1 1 A B
   .......... .10 times 6
rR r r r 9  6 9  6 18
R AB     3.6 
96 15 5
1 10 10 100 R
     rR   0.01R 12. (d) As resistance  Length
rR r R / 10 R 100
12
(30  30)30 60  30 Resistance of each arm   4
4. (c) R equivalent    20  3
(30  30)  30 90 4 8 8
 R effective   
V 2 1 4 8 3
i    ampere
R 20 10 13. (b) Given circuit is equivalent to
R 6 3 3
5. (b) Resistance of parallel group  A C
2
R
 Total equivalent resistance = 4   2R 6
2 A C  3

3
3 3
B B
 Current Electricity 1101

i1 3
So the equivalent resistance between points A and B is equal to   i1  i2  i2  0.5 A  i1
i2 3
63
R  2
63  6 12  6 
25. (c) V p  Vq     (0.5)  (2  4 ) (0.5)  3 V
14. (d) Potential difference across all resistors in parallel combination  3 12  6 
is same.
26. (b) 4
15. (b) Current through each arm DAC and DBC = 1A 24
8
VD  V A  2 and VD  VB  3  V A  VB  1V 20
16
3r 5 r 4 20
16. (d) R = r   A B A B
16
effecti ve

2 2 
R1 R 2 6
17. (c) If resistances are R 1 and R 2 then  …..(i) 6 6
R1  R 2 8 9
6 12
Suppose R 2 is broken then R1  2 …..(ii) 6 18 24  12
R AB   8
On solving equations (i) and (ii) we get R 2  6 / 5  (24  12)
18. (c) 27. (d) The network can be redrawn as follows
3 3 3
A B

 Req  9 
28. (d) Let the resistance of the wire be R, then we know that
19. (b) Because all the lamps have same voltage. resistance is proportional to the length of the wire. So each of
20. (d) R series  R1  R 2  R 3  ...... the four wires will have R/4 resistance and they are connected
in parallel. So the effective resistance will be
2 1 1 4
21. (a) Current supplied by cell i   A   4  R 1 
R
235 5 R1  R  16
2 3 5
4 4 66
29. (d) Equivalent resistance    5 ohm So the
4 4 66
i 20
current in the circuit   4 ampere Hence the current
5
flowing through each resistance = 2 ampere.
2V
3 1 30. (c) Let the resultant resistance be R. If we add one more branch,
So potential difference across 3 will be V 
 0.6 V then the resultant resistance would be the same because this is
5
an infinite sequence.
22. (a) According to the problem, we arrange four resistance as
follows A R1 = 1 X
A
10 10
D B R2 = 2 R

B
10 10 Y
RR 2
  R1  R  2 R  R  2  R 2  2 R
20 C20 R  R2
Equivalent resistance   10 
40  R 2  R  2  0  R  1 or R  2 ohm
R1 R2 31. (b) Cut the series from XY and let the resistance towards right of
23. (b) R1  R2  9 and  2  R1 R2  18
R1  R2 XY be R 0 whose value should be such that when connected
across AB does not change the entire resistance. The
R1  R2  (R1  R2 )2  4 R1 R2  81  72  3 combination is reduced to as shown below.
R1  6 , R2  3 E R X R R R C
A
1 .5
24. (b) i1  i2   1 amp
3/2 R R R R R

1 2 B
F R Y R R R D
i1 E R
A
i2 3
i R0
= R
1.5V
B
F
R
 1102 Current Electricity

Potential difference between C and A,

The resistance across EF,  R EF  (R 0  2 R) VC  VA  1  1  1V .......(i)
Potential difference between C and B,
(R 0  2 R)R R R  2R 2 VC  VB  1  3  3 V ......(ii)
Thus R AB   0  R0
R0  2R  R R0  3 R On solving (i) and (ii) VA  VB  2 volt

 R 02  2 RR 0  2 R 2  0  R 0  R( 3  1) i 2
Shot Trick : (VA  VB )  (R2  R1 )  (3  1)  2 V
2 2
32. (d) The last two resistance are out of circuit. Now 8  is in
1 1 1 1 3 1
parallel with (1  1  4  1  1) . 39. (c)      R  ohm
R 1 1 1 1 3
8 Now such three resistance are joined in series, hence total
 R  8  || 8    4   R AB  4  2  2  8 
2 1 1 1
R     1ohm
33. (a) The given circuit can be simplified as follows 3 3 3
40. (b) To obtain minimum resistance, all resistors must be connected
2 18 2 in parallel.
4.5
7 r
15V 15V Hence equivalent resistance of combination 
10
6 1  6 18 41. (a) For same material and same length
0.5 0.5
R2 A 3
10  1   R2  3R1
R1 A2 2
8 8
Resistance of thick wire R1  10
On further solving equivalent resistance R  15 
 Resistance of thin wire R2  30
15
Hence current from the battery i   1A Total resistance in series = 10 + 30 = 40 
15 42. (c) Similar to Q. No. 30
34. (b) The circuit will be as shown
10V 2R
R  22  2R  R 2  8  4 R  2R
2R
10 4  16  32
i  2A  R2  4 R  8  0  R   22 3
5 5 2
A

35. (c) The current in the circuit 

8

4 R cannot be negative, hence R  2  2 3  5.46
5 1 3 64
43. (b) P.d. across the circuit  1 .2   2.88 volt
4 4
Now VC  VE   1  VE   V 64
3 3 2 .88
Current through 6 ohm resistance   0 .48 A
36. (d) According to the figure, (I  I1 )R2  I1 R 6
I – I1 44. (d) Three resistances are in parallel.
R2 1 1 1 1 3
    
I1 R' R R R R
R
R The equivalent resistance R '  
I 3
R2  R
A 45. (c) Similar to Q. No. 30. By formula R  R1 
+ – I R2  R
and R
Only two values satisfying the above relation are
2 1 R
 R 1  R2  R  1  R  R
37. (d) Effective resistance between the points A and B is 1 R
32 8
R   1 14 1 5
12 3  R 2  R  1  0 or R  
2 2
10
38. (b) Req  5  , Current i   2A and current in each 1 5
5 Since R cannot be negative, hence R  
branch = 1A 2
i/2 1 3
46. (d) Rl
3 A
3 1
i
i/2 B
10V
 Current Electricity 1103

R V 4.8  15
Hence every new piece will have a resistance . If two  I  2A
10 R 36
pieces are connected in series, then their resistance 3
2R R 54. (a) Equivalent resistance of the circuit R  
  2
10 5
If 5 such combinations are joined in parallel, then net V 3
 Current through the circuit i    2A
R R R 3/2
resistance  
5  5 25 R max
55. (c) Rmax  nR and Rmin  R / n   n2
47. (c) 6 R min
56. (d) According to the principle of Wheatstone’s bridge, the effective
resistance between the given points is 4.
B

4 4
6
Req   3 6 A C
2 16
50
48. (b) Current in the given circuit i   2A 4 4
(5  7  10  3)
57. (c) D
Potential difference between A and B VA  VB  2  12
58. (c) Current through 6 resistance in parallel with 3 resistance =
 VA  0  24 V  VA  24 V 0.4 A
49. (b) If all are in series then R eq  12  So total current = 0.8 + 0.4 = 1.2 A

4
Potential drop across 4   1.2  4  4.8 V
If all are in parallel then R eq    1 .33 
3 59. (d) Two resistances in series are connected parallel with the third.
8 1 1 1 3 8
If two are in series then parallel with third, R eq   2.6  Hence     Rp  
3 Rp 4 8 8 3
If two are in parallel then series with third, R eq  6  60. (c) Resistances at C and B are not in the circuit. Use laws of
50. (d) Equivalent external resistance of the given circuit R eq  4  resistances in series and parallel excluding the two resistance.
61. (d) After simplifying the network, equivalent resistance obtained
E 10
Current given by the cell i   2A between A and B is 8.
R eq  r (4  1)
62. (c) The circuit consists of three resistances (2 R, 2R and R)
i 2 connected in parallel.
Hence, (VA  VB )   (R2  R1 )  (2  4 )  2 V .
2 2 63. (d) Resistance across the battery is
R 1 1 1 2 1 3 2
51. (d) Resistance of each part will be ; such n parts are joined in      R p  2  I   1 A
n Rp 3 6 6 6 2
R
parallel so Req  . 64. (c) The voltmeter is assumed to have infinite resistance. Hence (1 +
n2
52. (c) Let equivalent resistance between A and B be R, then 2 + 1) + 4 = 8.
equivalent resistance between C and D will also be R. R 1
65. (c) R'    0 .1
A 1 n 10
C
66. (c) The given circuit can be redrawn as follows

R R 2
1
2 2 2 1 2
1 
B D
R 2
R   2  R or R 2  2 R  2  0  R eq  5  .
R 1
2 4 8 R2 R3 44
 R  3 1 67. (b) R AB  R1   R4 = 2   2  6 .
2 R2  R3 44
53. (d) 6 and 6 are in series, so effective resistance is 12 which is 68. (c) Let equivalent resistance between A and B is R', so given circuit
in parallel with 3, so can be reduced as follows
1 1 1 15 36 R R R
    R A
R 3 12 36 15
R 2R 2R 2R 

B
R
A

 R 2R R

B
 1104 Current Electricity

 R AB  5  .
2R  R 
R'  R   R  2  RR   2 R 2  0 1l1 l
(2 R  R ) 75. (a) R1  and R2  2 2 . In series Req  R1  R2
A A
On solving the equation we get R'  2 R .
 eq. (l1  l 2 )  1 l1  2 l2  1 l1   2 l 2
R 2 8 2     eq  .
69. (d) R AB  R =  2 2 . A A A l1  l 2
3 3 3 3
76. (c) The figure can be drawn as follows
E 10 100 
70. (b) i   0.5   10 = 0.5R + 1.5  R = 17. D
C
Rr R3
36
71. (a) Equivalent resistance R  4   6  and main current 100 
36 100 
E 3
i    0.5 A
R 6 B
A 100 
Now potential difference across the combination of 3 and
 36  200  200
6, V  0 .5     1Volt R AC   100 .
36 200  200
The same potential difference, also develops across 3 77. (c)
resistance. 1 1 1 1 1
72. (c) A B
i

10k 1 1
i/2 A R AB  2   2 .
i/2 3 3
30V 10k
10k 1 r
78. (b)   same, l  same, A2  A1 (as r2  1 )
4 2
B
10
Equivalent resistance R  10   15 k  By using R  
l R A R 1
 1  2  1   R1  2
2 A R 2 A1 8 4
30
Current i =  2  10  3 A R1 R 2 28 8
15 Hence, Req    .
R1  R 2 (2  8 ) 5
Hence, potential difference between A and B
 2  10 3  79. (c) The given circuit can be simplified as follows
V     10  10 3  10 Volt.

2 10
 
3
9 Q
73. (a) Equivalent resistance R    3 R
9
1A 1A 1A 1A 1A 1A 1A 1A 1A 10
+
9V 9 9 9 9 9 9 9 9 9 3
–  P Q

9
A 10  (3  R(3) + R) 30  10 R
Current i 
 9A R3 3
1 10  3  R 13  R
Current passes through the ammeter = 5A. 39  3 R  30  10 R 69  13 R
R 
74. (b) The figure can be drawn as follows 13  R 13  R

D 7 C C 13 R  R 2  69  13 R  R  69  .
80. (a) The circuit can be drawn as follows
10 A
3  5
10 5
10
3 3
A B
A B 10
10 i1
B C
C i2 3
10 i
5
 5  2V

A B A B
10 10
 Current Electricity 1105

R
As, resistance is not fractional  2
R 1
 x  3, R  2, 2 x  6
3  (3  3)
Equivalent resistance R   2
3  (3  3) Hence, the value of largest resistance = 6.
2  3  1 (3  3)  3 3
Current i   1 A. So, i1  1    A. 92. (c) R  2  i   1 . 5 A .
2 36 3 (3  3)  3 2
1 93. (b) Given circuit is a balanced Wheatstone bridge circuit, hence it
Potential difference between A and B =  3  1volt. can be redrawn as follows
3
1 1 1 1 4  2 1 8 12 12
81. (a)      R eq   . 4 8
Req 2 4 8 8 7
a b a b
82. (b) The given figure is balance wheat stone bridge. 

7 1 1
83. (b)    R  3 2 4
12 4 R 6
6
84. (d) Suppose resistance of wires are R1 and R2 then 12  6
R AB   4 .
6 R1 R 2 (12  6)
 . If R 2 breaks then R1  2
5 R1  R 2 94. (d) The given circuit is a balanced wheatstone bridge circuit. Hence
potential difference between A and B is zero.
6 2  R2
Hence,   R 2  3 . 95. (a) In the following circuit potential difference between
5 2  R2
C and A is VC  V A  1  4  4 ……(i)
85. (d) Potential difference across PQ i.e. p.d. across the resistance of
20, which is V = i × 20 1A 4 A 16

48
and i   0 .16 A
(100  100  80  20) V
2A C
 V  0.16  20  3.2V .
86. (a) 1A 16 B 4
C Series
100+60 = 160 C and B is VC  VB  1  16  16 ……(ii)
60
On solving equations (i) and (ii) we get
100
160  40 VA  VB  12V .
R  32 .
160  40 96. (d) As resistance  Length
A B
40
87. (a) 12 8
A A  Resistance of each arm   4
3
2 4 4
2
2 
2 4 8 8
2  R effective   
2 2 4 8 3
4
B B 12
22 97. (b) i   5 A.
R AB   1 . (1  1)  0 .4
22
16 4
88. (a) Given circuit is a balance Wheatstone bridge circuit. 98. (b) By balanced Wheatstone bridge condition 
X 0 .5
R 8
89. (b) All of three resistance are in parallel So, R '  R/n  .  X  2
3 4

 eff.2l  (25  5) 
1 l  2l 1   2 99. (d) Current through 2  1 .4    1A
90. (b) R eq  R1  R 2     eff.  .  (10  2 )  (25  5 ) 
A A A 2
100. (a) Since the given bridge is balanced, hence there will be no
91. (b) Two resistance are in ratio 1 : 2 and third resistance is R current through 9  resistance. This resistance has no effect
and must be ignored in the calculations.
1 1 1 3 R 
So,   1 x    9
x 2x R 2  R 1
5 4

A B

10 8
14
 1106 Current Electricity
107. (b) Let current through 5  resistance be i. Then
10
9  18 i  25  (2.1  i)10  i   2 .1  0 .6 A
R AB  6 35
27
108. (d) Let the value of shunt be r. Hence the equivalent resistance of
101. (c) Potential difference between B and D is zero, it means Sr
branch containing S will be
Wheatstone bridge is in balanced condition S r
P Sr /(S  r)
B In balance condition,  . This gives r  8 
X Q R
6 8X
21 3
8 (8  X ) 109. (b, c) 2R
15 3 C
R R
A C B R D R
B D

15 4 6
18
R R 2R
6 4
A
6 4
1 1 1 1 R
So
P R
 
21 D 18
  X  8     R BD 
Q S 8X 6 R BD 2 R R 2 R 2
3
(8  X ) Between A and C circuit becomes equivalent to balanced
102. (a) This is a balanced Wheatstone bridge. Therefore no current will Wheatstone bridge so R AC  R .
flow from the diagonal resistance 10  1
110. (b) i 
(10  10)  (10  10) R
 Equivalent resistance   10  111. (d) Equivalent resistance between P and Q
(10  10)  (10  10)
1 1 1 1 48
103. (b) This is a balanced Wheatstone bridge circuit. So potential at B     RPQ  
and D will be same and no current flows through 4R R PQ (6  2) 3 (4  12) 25
resistance. Current between P and Q; i = 1.5A
104. (d) The equivalent circuits are as shown below So, potential difference between P and Q
C C 48
C VPQ  1 .5   2.88 V .
2 2 25
A A 112. (c) Given circuit is a balanced Wheatstone bridge i.e. potential
B B difference between B and D is zero. Hence, no current flows
A   between B and D.
2 2 113. (a) The given circuit is a balanced Wheatstone bridge, hence it can
D B be redrawn as follows
Clearly, the circuit is aD balanced Wheatstone Dbridge. So
effective resistance between A and B is 2  . 7
105. (a) By the concept of balanced Wheatstore bridge, the given circuit 3 4
can be redrawn as follows

30 A B
5 10 15
6 8
A B 14
7  14 14
 Req   .
(7  14 ) 3
114. (a) For a balance Wheatstone bridge.
10 20 30
60 A D 10 4
30  60    (Unbalanced)
 R AB   20  B C 5 4
(30  60)
A' D A' 4
106. (a) The given circuit is a balanced Wheatstone bridge type, hence     A'  5 
it can be simplified as follows B C 5 4
A' (5 ) is obtained by connecting a 10  resistance in
5 parallel with A.
2 3 115. (d) Given circuit is a balanced Wheatstone bridge circuit. So there

10 will be no change in equivalent resistance. Hence no further
A B R AB   current will be drawn.
3
4 6 116. (a) No current flow through vertical resistances
10
3 3 3 9
A B A B


3 3 3 9
 Current Electricity 1107

R 3R

R R  2  3 R
R PR  ||   R   3
3 2  R 3 R 11

3 2
9
R AB  . Hence it is clear that R PQ is maximum.
2
117. (d) The given circuit is a balanced Wheatstone bridge. 127. (c) Given circuit can be redrawn as follows
2
118. (c) The given circuit can be redrawn as follows 1.5
A
R 6
6V 3  6V 3
R R R
1.5 1.5

6
R B  6V 3 3  i 4A
1 .5
Equivalent resistance between A to B is R.
119. (d) Equivalent resistance of the given circuit is 3  . 2
r 
2
    
i1 R l 3 2 1
128. (b)  2  2   1 
120. (c) l1  r2 4 3
R R i2 R1  3
129. (c)
R R R 2R/3 2R/3 A A
R
A R B  A R B
R=4  R=4
R R R
R 2R/3 2R/3
R R
2R
Hence R eq  . B B
3 1 1 1 1 1 1 1 19 20
121. (b) 130. (a)         Req  
Req R1 R2 R3 2 4 5 20 19
P R
122. (b) For balanced Wheatstone bridge  131. (a) Equivalent resistance of the given network R eq  75 
Q S
12 x 6 i R1(50) i2 R1(50) i2
   x  6
(1 / 2) (1 / 2) R4
i1 i1
123. (b) For maximum energy equivalent resistance of combination 60 R3 (30)
should be minimum. 3V R2  R2
(50) (50) 50
10  R1 50 R5(30)
124. (c) For first balancing condition 
R2 50
3
 R 2  10  R1 . For second balancing condition  Total current through battery i 
75
R1 40 R1 2
    R1  20 
R2 60 10  R1 3 i1  i2 
3

3
75  2 150
125. (b) Given R  6  . When resistor is cut into two equal parts and
connected in parallel, then 3 60 3 60 2
Current through R4      A
R/2 R 6 150 (30  60) 150 90 150
Req     1.5 
2 4 4
2 2
126. (a) Resistance between P and Q V4  i4  R 4   30  V  0.4 V
150 5
5
R R 10 10
R R 5 132. (a) i   5A
1.5  1 || 1 1 .5  0 .5
R PQ  R ||     6  R
3 2 5
R  R 11
6 133. (c)
Resistance between Q and R Parallel
3
R 4R 2 2 2
 
R  R 3  4 R
R QR  ||  R    2 6
2  3 R 4R
 11
2 3 Req  4 
Resistance between P and R 134. (a) The equivalent resistance between C and D is
 1108 Current Electricity

1 1 1 1 2 3 141. (b) Since voltmeter records 5V, it means the equivalent. Resistance
    or R '   1 .5 
R' 6 6 3 3 2 of voltmeter and 100  must be 50, because in series grouping
if resistances are equal, they share equal potential difference. It
Now the equivalent resistance between A and B as R'  1.5 
conclude that resistance of voltmeter must be 100 .
and 2.5  are connected in series, so
R"  1. 5  2.5  4  Kirchoff's Law, Cells
Now by ohm's law, potential difference between A and B is
given by VA  VB  iR  2  4.0  8 volt 1. (b) For no current through galvanometer, we have

135. (b) The given circuit can be redrawn as follows  E1   12 

 X  E    X  2  X = 100 
A  500  X   500  X 
R
F C 2. (d) Since E1 (10 V )  E2 (4 V )
R R R So current in the circuit will be clockwise.

1 E1 e E2 2
D a b
E
R B
Equivalent resistance between A and B is R and 10V 4V
i
V
current i  3
R
136. (b) The given network is a balanced Wheatstone bridge. It's Applying Kirchoff's voltage law

equivalent resistance will be R 

18
  1  i  10  4  2  i  3i  0  i  1 A(a to b via e)
5
V 10  4
V V 5V  Current    1 .0 ampere
So current from the battery i    R 6
R 18 / 5 18
3. (c) For maximum power, external resistance = internal resistance.
R/2 R
137. (a)  R AB   R/2 4. (a) 0.9 (2 + r) = 0.3 (7 + r)  6 + 3r = 7 + r  r = 0.5 
2 4
5. (a) Since both the resistors are same, therefore potential difference
E
 V V  E  V 
2
A R/2 B V
6. (b) Let the current in the circuit  i 
1 i R 5 (R  2) R
138. (b) i   1  2    R = 8
R i2 R1 4 R EV EV EV
Across the cell, E  V  ir  r    R
139. (c) In given circuit three resistance R2 , R4 and R3 are parallel. i V/R  V 
7. (a) For maximum energy, we have
1 1 1 1 R1
   External resistance of the circuit
R R 2 R4 R3
i R4 r
1 1 1 = Equivalent internal resistance of the circuit i.e. R 
   R2 R3 2
50 50 75 E
8. (a) Kirchhoff's first law is based on the law of conservation of
75  75  50 charge.

50  75 9. (b) Kirchhoff's second law is based on the law of conservation of
50  75 50  75 75 energy.
R     18.75  10. (a) According to Kirchhoff's first law
75  75  50 200 4
At junction A, iAB  2  2  4 A
This resistance is in series with R1
At junction B, iAB  iBC  1  3 A

2A
Rresultant  R1  R  100  18.75  118.75 1A
1.3A

140. (b) When resistances 4  and 12 are connected in series A B

C
 4  12  16 2A
When these resistance are connected in parallel. t
At junction C, i  iBC  1.3  3  1.3  1.7 amp
1 1 1 4  12 4  12
   RP    3
RP 4 12 4  12 16 11. (c) In charging V > E.
12. (d) In open circuit of a cell V = E
 Current Electricity 1109

13. (a) Zero (Circuit open means no current and hence no potential Hence potential difference across A
difference across resistance). 20 4
14. (d) Zero (No potential difference across voltmeter).  2  0 .1   V (less than 2V)
3 3
15. (b) Let the e.m.f. of cell be E and internal resistance be r. Then 20
E E Potential difference across B  2  0 .3  0
0.5  and 0 .25  3
(r  2) (r  5) 30. (b) Here two cells are in series.
5 r Therefore total emf = 2E.
On dividing, 2   r  1 Total resistance = R + 2r
2r
2E 2  1.45 2.9 29
16. (c) In short circuiting R = 0, so V = 0  i     1.611 amp
R  2r 1.5  2  0.15 1.8 18
E 1 .5
17. (c) Short circuit current iSC   3  r  0.5  31. (a) E  V  ir
r r
E 2
50 50 5 After short-circuiting, V  0;  r    0.5 
18. (c) i  r  10   1.1 i 4
R r 4 .5 4 .5 32. (c) By Kirchhoff's current law.
19. (d) (4  r)i  2.2 ......(i) 33. (b) For power to be maximum
External resistance = Equivalent internal resistance of the
1
and 4 i  2  i  circuit
2
E 1.5
Putting the value of i in (i), we get r = 0.4 ohm. 34. (a) i    30 A
r 0.05
20. (b) Let the internal resistance of cell be r, then 12
35. (a) i   2A
E 1 .5 (4  2)
i  15   r  0.06
R r 0 .04  r Energy loss inside the source  i2r  (2)2  2  8W
21. (c) The voltage across cell terminal will be given by
36. (b) V2  V1  E  ir  5  2  0.5  4volt
E 2
 R   3.9  1.95 V  V2  4  V1  4  10  14 volt
R r (3 .9  0.1)
37. (a) If m = Number of rows
22. (c) E  2.2 volt, V  1.8 volt, R  5 R and n = Number of cells in a row
E   2 .2  Then m  n = 100 .....(i)
r    1 R    1   5  1 .1
V   1 .8  nr
Also condition of maximum current is R 
  m
 
(b) In parallel, equivalent resistance is low  i 
E  1n
23.  25   n = 25 m .....(ii)
 r  m
 R 
 n  On solving (i) and (ii) m = 2
24. (d) Internal resistance  distance 
1
 concentrat ion 38. (b) According to Kirchhoff's law iCD  i2  i3
Area
 E 
nE 39. (b) Since i   , we get
25. (a) Total e.m.f. = nE, Total resistance R + nr  i   R r
R  nr
E
26. (a) Current through R is maximum when total internal resistance 0 .5  ......(i)
of the circuit is equal to external resistance. 2r
27. (b) Cells are joined in parallel when internal resistance is higher E
0 .25  .....(ii)
then a external resistance. (R <<r) 5 r
E 5 r
i Dividing (i) by (ii), we get 2   r  1
r 2r
R
n E
 0 .5   E  1. 5 V
2E 2 1
28. (b) In series , i1 
2  2r r
40. (c) Because Eeq  E and req 
E 2E 2
In parallel, i2  
2
r 4 r 41. (d) In parallel combination Eeq  E  6 V
2 42. (d) Suppose current through different paths of the circuit is as
2E 2E follows.
Since i1  i2    r  2
4  r 2  2r 28 54
29. (a) Applying Kirchhoff law
20 6V
(2  2)  (0.1  0.3  0.2)i  i  A 1 2
3 i3

8V 12 V
 1110 Current Electricity
Applying Kirchoff’s voltage law for the loop ADEFA.
40i2  40i3  80  40  0
After applying KVL for loop (1) and loop (2)   40i2  40(i1  i2 )  120
1
We get 28i1  6  8  i1   A  i1  2i2  3 …….(iii)
2
1 On solving equation (ii) and (iii) i1  0.4 A .
and 54 i2  6  12  i2   A
3
57. (c) V  E  ir = 12  60  5  10 2 = 9V.
5
Hence i3  i1  i2   A 58. (a) Applying Kirchoff's voltage law in the loop
6 5V
10
5 X  2  10
43. (d) VAB  4   X  20 i
X  10
44. (a) After short circuiting, R becomes meaningless.
2 A B
i
45. (c) V  E  IR  15  10  0.05  14.5 V
nE n  1 .5 2V 20
46. (c) In series i   0 .6   n = 10 10i  5  20i  2  0  i  0.1 A
nr  R n  0 .5  20
47. (b) 59. (d) V  E  ir  1.5  2  0.15  1.20Volt .
W W 1000 E 4
48. (a) P  Vi  V    1.38 V 60. (b) i  1   r  2
t it 2  6  60 R r 2r
49. (a) Applying Kirchoff's voltage law in the given loop. Short circuit when terminals of battery connected directly then
4V i 8V 2 E 4
P 1 Q current flows which is iSC    2 A .
r 2
4V 22 4
i 61. (c) i  A
1  1.9  0.9 3.8
9
4
1 For cell A E  V  ir  V  2   1 .9  0 .
2i  8  4  1  i  9i  0  i  A 3.8
3
E
1 62. (c) By using i 
Potential difference across PQ   9  3V R r
3
E
50. (d) Because cell is in open circuit.  0.5   E  5.5  0. 5r ..…(i)
11  r
51. (b) In parallel combination Eeq  E  12V
E
52. (d) and 0 .9   E  4. 5  0.9r ..…(ii)
5r
53. (b) i 
E

6
 12 amp. On solving these equation, we have r  2.5 
r 0 .5 63. (c)
54. (c) Strength = 5 × 18 = 90AH. 64. (b) W  qV  6  10 6  9  54  10 6 J .
E 5 V2
55. (a) i   1A 65. (a) P ; for P to be maximum Req should be less. Hence
R  r 4.5  0.5 R eq
V  E  ir  5  1  0.5  4.5 Volt option (a) is correct.
56. (b) The circuit can be simplified as follows E2 (2)2
66. (c) Pmax    2W
B C 4 r 4  0.5
i1 30 67. (a)
i3 i3 68. (d) Applying Kirchhoff law in the first mesh
A D
40 10
i2 40V 10  5  i  i  2A
5
F E
69. (b) Applying Kirchhoff law in the first mesh
40 80V
Applying KCL at junction A 10  5i1  i .....(i)
i3  i1  i2 .….(i) i – i1 i

Applying Kirchoff’s voltage law for the loop ABCDA i1

10V
30i1  40i3  40  0 A
r = 1

4
  30i1  40(i1  i2 )  40  0 5

 7i1  4 i2  4 .….(ii)
 Current Electricity 1111
Applying in the second mesh 80. (d)
5i1  4 i  4 i1 ......(ii) 81. (a) Applying Kirchoff's law in following figure.
At junction A :
40 60
Solving equation (i) and (ii), we get i1  A i  i1  i2  1 .... (i)
29 I
70. (a) Given problem is the case of mixed grouping of cells For Loop (i) 15 1 5
A B
nE  60 i  (15  5)i1  0 1A I1 1A
So total current produced i  2
nr  i1  3i I2
R ...(ii)
m
For loop (2) 10
Here m  100, n  5000, R  500  – (15 + 5) i + 10 i = 0
1 2

E  0.15 V and r  0.25   i = i = (3 i) = 6i

2 1

5000  0 .15 750 On solving equation (i), (ii) and (iii) we get i = 0.1 A
 i   1.5 A
5000  0 .25 512.5 Short Trick : Branch current =
500 
100  Resistance of opposite branch 
main current  
71. (a)  Total resistance 
Discharging energy (Branch 60
72. (d) Watt hour efficiency  current)
Charging energy
 20  I
 
14  5  15  i  1  3 
  0.875  87.5% 20
15  8  10   60  1A 15 5
 3  10
73. (c) From Kirchoff's junction Law
= 0.1 A
20/3 
 4 2i5 3  0 i  2 A 82. (d) Maximum current will be drawn from the circuit if resultant
resistance of all internal resistances is equal to the value of
74. (b) In the given case cell is in open circuit (i = 0) so voltage across external resistance if the arrangement s mixed. In series,
the cell is equal to its e.m.f.
R  nr and in parallel, the external resistance is negligible.
75. (b) The internal resistance of battery is given by 83. (c) On applying Kirchoff's current law i = 13 A.
E   40  9  10 84. (c) Total cells = m  n = 24 .... (i)
r    1 R    1  9   3
V   30  30
For maximum current in the circuit R 
mr
n
E E2R
76. (b) i   P  i2 R  P  m
rR (r  R)2  3   (0 .5)  m = 6n ..... (ii)
n
Power is maximum when r = R  Pmax  E 2 / 4 r On solving equation (i) and (ii), we get m = 12, n =2
2
77. (c) Since the current coming out from the positive terminal is  E 
85. (a) Power dissipated  i2 R    R
equal to the current entering the negative terminal, therefore,  R r
current in the respective loop will remain confined in the loop
2 2
itself.  E   E 
   R1  


 R  r  R2
current through 2 resistor = 0  R1  r   2 
78. (c) Reading of voltmeter
 R1 (R22  r2  2 R2r)  R2 (R12  r 2  2 R1r)
E1 r2  E 2 r1 18  1  12  2
 E eq    14 V  R22 R1  R1r 2  2 R2r  R12 R2  R2r 2  2 R1 R2r
r1  r2 12
 (R1  R2 )r 2  (R1  R2 )r 2  (R1  R2 )R1 R2
2E
79. (d) i 
R  R1  R 2  r  R1 R2
From cell (2) E  V  iR2  0  iR2
R Different Measuring Instruments
1. (a) In meter bridge experiment, it is assumed that the resistance of
i the L shaped plate is negligible, but actually it is not so. The
error created due to this is called, end error. To remove this
the resistance box and the unknown resistance must be
interchanged and then the mean reading must be taken.
E, R2 2. (c) To convert a galvanometer into an ammeter a low value
E, R1
resistance is to be connected in parallel to it called shunt.
(1) (2)
2E 3. (d) Balance point has some fixed position on potentiometer wire. It
 E  R 2  R  R 2  R1 is not affect by the addition of resistance between balance
R  R1  R 2 point and cell.
 1112 Current Electricity
4. (d) Resistance of voltmeter should be greater than the external 21. (b) The sensitivity of potentiometer can be increased by decreasing
circuit resistance. An ideal voltmeter has infinite resistance. the potential gradient i.e. by increasing the length of
potentiometer wire.
iG 100  0 .01 1
5. (c) S  g    0.1 1
i  ig (10  0.01) 10 (Sensitivity   Length )
P.G.
6. (c) Equivalent resistance of the circuit Req  100
22. (b) In balance condition, potentiometer doesn't take the current
2 .4 from secondary circuit.
current through the circuit i  A 23. (a) Here same current is passing throughout the length of the
100
wire, hence V  R  l
P.D. across combination of voltmeter and 100  resistance
V1 l1 6 300

2 .4
 50  1 .2 V      V =1 V. 2

100 V2 l2 V2 50
Since the voltmeter and 100  resistance are in parallel, so the ig G 10  0.01 10
voltmeter reads the same value i.e. 1.2V. 24. (a) S   ohm
i  ig 10  0 .01 999
e R
7. (a) Potential gradient  . 25. (a) Ratio will be equal to the ratio of no deflection lengths i.e.
(R  Rh  r) L
E1 l1 2
2 5 V V  
   0 .5  0 .005 E2 l2 3
(15  5  0) 1 m cm
Potential difference
ig G G i  ig 10  1 9 26. (a) Potential gradient 
8. (d) S      Length
(i  ig ) S ig 1 1
27. (a) Wheatstone bridge is balanced, therefore
9. (c) Ammeter is used to measure the current through the circuit.
P R 10
iG 1  0.018 0.018  or 1   S  10 ohm
10. (c) S  g    0.002 Q S S
(i  ig ) 10  1 9
28. (a) When the length of potentiometer wire is increased, the
11. (d) Potentiometer works on null deflection method. In balance potential gradient decreases and the length of previous balance
condition no current flows in secondary circuit. point is increased.
ig G 10  99 29. (b)
12. (c) Shunt resistances S    11
(i  ig ) (100  10) 30. (b)
31. (b) The actual circuit is same.
V 100
13. (d) By using R  G  R   5  19,995
ig 5  10  3 32. (b)  ig  10% of i 
i
 S
G

90
 10 
14. (a) Potential gradient = Change in voltage per unit length 10 (n  1) (10  1)
V2  V1 E1 l1  l2 (8  2) 5
 10   V2  V1  3 volt 33. (b)   
30 / 100 E2 l1  l2 (8  2) 3
V 5 5000 34. (b) Suppose resistance R is connected in series with voltmeter as
15. (d) R  G  2   2  48 
ig 100 / 10 3
100 shown.
By Ohm's law ig ig R
iS 50  12 G
16. (c) ig   10   12  G  60  G  48  ig .R  (n  1)V
S G 12  G
V (n – 1)V
17. (a) To convert a galvanometer into a voltmeter, a high value V
 R  (n  1)G (where ig  ) nV
resistance is to be connected in series with it. G
18. (b) 35. (c) Ammeter is always connected in series with circuit.
P R 36. (c) If resistance of ammeter is r then
19. (c)  (For balancing bridge)
Q S 20  (R  r)4  R  r  5  R  5 
4  11 44 B
 S   ig  G 10  10 3  50 50
9 9 P = 9 Q = 11 37. (b) S     in parallel.
i  ig 1  10  3  10 99
1 1 1 A C
   i
S r 6 38. (b)  ig  (100  90)% of i 
6 10
9 1 1 R = 4
   r G 900
44 6 r  Required shunt S    100 
D (n  1) (10  1)
132 S'
 r  26.4  V 100
5 39. (d) R G   25  9975 
ig 10  10  3
l l 
 R  
25 
20. (a) r   1 2   2  0.5  V iR
 l2   100  40. (b) Potential gradient x  
L L
 Current Electricity 1113

2 15 3 V iR iL i
 x   volt / cm 55. (a) Potential gradient    
(15  5) 10 2000 L L AL A

S
G

25
 5
25 25
 5  2.5  10 4 
0 .2  40  10 8
41. (a)   10  2 V / m
i
1
5
1 10  1 10 8  10  6
ig 50  10  6 i G G G
42. (b) In balanced Wheatstone bridge, the arms of galvanometer and 56. (b) ig  2% of i   S  
50 (n  1) (50  1) 49
cell can be interchanged without affecting the balance of the
bridge. 57. (d) The resistance of an ideal voltmeter is considered as infinite.
43. (c) Error in measurement = Actual value – Measured value 58. (c)
R 20×103
998  V
V
5V
i i
Actual value = 2V 110V
110
i
2

1
A Here i 
998  2 500 + – 20  10 3  R
2V 2
 110 
1 998  V  iR  5     20  10 3
Since E  V  ir  V  E  ir  2  2  V  20  10  R 
3
500 500
10 5
 Measured value 
998
V  10 5  5 R  22  10 5  R  21   420 K
500 5
59. (c) Due to the negligible temperature co-efficient of resistance of
998 constantan wire, there is no change in it's resistance value with
 Error  2   4  10  3 volt
500 change in temperature.
44. (d) The emf of the standard cell must be greater than that of 60. (d) The resistance of voltmeter is too high, so that it draws
experimental cells, otherwise balance point is not obtained. negligible current from the circuit, hence potential drop in the
external circuit is also negligible.
45. (a)
61. (a) By connecting a series resistance
46. (b) In general, ammeter always reads less than the actual value
because of its resistance. V 10
R G  7  3 
R AC 20 ig 1
47. (c) By Wheatstone bridge,    R  20 
80 BC 80 62. (a) Since potential difference for full length of wire = 2 V
48. (a) E  l (balancing length)  P.D. per unit length of wire 
2
 0 .5
V
l l  4 m
l 2
49. (b) r   1 2   R '   1 5 ... (i) X 20 1
 2 
l  2  63. (d)   X    0 .25  .
1 80 4
l 3 64. (a) Reading of galvanometer remains same whether switch S is
and r   1   10 ... (ii)
 3  open or closed, hence no current will flow through the switch
i.e. R and G will be in series and same current will flow
On solving (i) and (ii) r = 10 
through them. IR  IG .
50. (a)
51. (b) In the part c b d, 65. (d) Pressing the key does not disturb current in all resistances as
the bridge is balanced. Therefore, deflection in the
Vc  Vd galvanometer in whatever direction it was, will stay.
Vc  Vb  Vb  Vd  Vb 
2 66. (b) ig S  (i  ig )G  ig (S  G)  iG
In the part c a d
Vc  Vd ig G 8
Vc  Va  Va  Vd   Va  Vb  Va     0 .8
2 i S G 28
52. (c) In balance condition, no current will flow through the branch e R
containing S. 67. (a) Potential gradient x  
(R  Rh  r) L
Gig 50  100  10 6
53. (b) Resistance in parallel S    x
2.5

20
 5  10 5
V
i  ig (10  100  10 6 ) (20  80  0) 10 mm
 S  5  10 4  68. (b) Given ig  2mA, i  20mA, G  180
V iR e R ig S 180
54. (b) Exl  l  E   l   180  S  10 S  S   20 
l L (R  Rh  r) L i GS 9
10 5 GS
E  3  3 V 69. (c) Resistance of shunted ammeter 
(5  4  1) 5 GS
 1114 Current Electricity

i G GS ig .G V 18
Also 1   85. (b) R G   12  5988 
ig S GS i ig 3  10  3
GS 0 .05  120 86. (d)
  = 0.6  V 6
GS 10 87. (c) R G   25  975 (In series).
(l  l )  60  50  ig 6  10  3
70. (c) r  1 2  R'     6  1.2 
l2  50  88. (d) ig  i
S
 0 .01  10
S
i G GS 25  S
71. (d) By using 1
ig S 25
 1000S  25  S  S  .
i 1000 1000 999
 1  S  111  89. (c)
100  10  3 S 9 X 6
V e R
72. (c) Potential gradient x   
L (R  Rh  r) L G

2.2 4 6
 2 .2  10  3   1  R  990  A
C
B
(10  Rh )
i GS ig S 2.5 1
73. (a)     
ig S i G  S 27.5 11
Resistance of the part AC 5V
80 R AC  0.1  40  4  and R CB  0.1  60  6 
74. (c) Total resistance of the circuit   20  60 
2
X 4
2 1 In balanced condition   X  4
 Main current i   A 6 6
60 30 Equivalent resistance R eq  5  so current drawn from
Combination of voltmeter and 80 resistance is connected in
5
series with 20, so current through 20 and this combination battery i   1A .
1 5
will be same  A.
30 V
90. (a) (R  G) ig  V  (R  G) 
Since the resistance of voltmeter is also 80, so this current is ig
ig R
equally distributed in 80 resistance and voltmeter G
3
1   6 .25 k 
(i.e.
60
A through each) 30  16  10  6

P.D. across 80 resistance 

1
 80  1 .33 V  Value of R is nearly equal to 6k 
60 This is connected in series in a voltmeter.
 L 
i  91. (d) V1 V2
A  i
 
V iR
75. (a) Potential gradient x   
L L L A
R1 = 16k R2 = 32k
10
76. (d) Here n  5 V1 = 80V
2
 R  (n  1)G  (5  1)2000  8000  V
R1  80  200  16000   16 k
l l  Current flowing through V1 = Current flowing through V2 =
77. (b) r   1 2  R  0 .5  .
 l1  80
 5  10  3 A .
78. (a) 16  10 3
e 2
79. (b) V  i.R.  . R  10  3   10 So, potential differences across V2 is
(R  Rh  r) (10  R  r)
V2  5  10 3  32  10 3  160 volt
 R  19,989 .
80. (a) Hence, line voltage V  V1  V2  80  160  240V .
81. (c) 2 R  20  R  10 . 92. (d) V  xl  iR  xl
i G 4 R R  2  10 3 
82. (c) 1   1  S  .  i  10     50  10  2  0 .1
ig S 1 S 3 2 
 10 
83. (a) When ammeter is connected in parallel to the circuit, net
resistance of the circuit decreases. Hence more current is  i  10  10 3 A  10 mA .
drawn from the battery, which damages the ammeter.
e R 2 10
l l   55  50  93. (d) E l    0.4  0.16 V .
84. (a) r   1 2   R '  r     10  1 . (R  Rh  r) L (10  40  0) 1
 2 
l  50 
 Current Electricity 1115

i G 5 12 S S
94. (c) 1   1  S  8 . (In parallel). 112. (a) ig  i  10  10 3   100  10 3
ig S 2 S GS 100  S
ig S 5 S G 1000
95. (d)     S 90 S  1000  S   11.11  .
i GS 100 G  S 19 90
96. (a) R  G(n  1)  50  10 3 (3  1)  10 5  . 113. (c) Before connecting the voltmeter, potential difference across
100 resistance
E1 l1  l2 58  29 3
97. (c)   
E2 l1  l2 58  29 1 100 10 10 100
Vi  V  V
V 10 (100  10) 11
98. (a) R G   1  999 . Vi
ig 10  10  3 Finally after connecting
voltmeter across 100
99. (d) For conversion of galvanometer (of resistances) into voltmeter, Equivalent resistance
a resistance R is connected in series. 100  900
 90  V
V1 V2 (100  900)
 ig  and ig 
RG 2R  G
Final potential difference
V1 V2 V 2 R  G 2(R  G)  G
   2   90 9 900
R  G 2R  G V1 RG (R  G) Vf  V  V
(90  10) 10
G VG Vi  Vf 100
 2  V2  2 V1  1  V2  2V1 10
(R  G) (R  G) % error =  100
Vi Vf
100. (d) If the voltmeter is ideal then given circuit is an open circuit, so
reading of voltmeter is equal to the e.m.f. of cell i.e., 6V. 10 9
V V V
 11 10  100  1 .0 .
ig S 4 1 10
101. (c)    i.e. 10%. V
i G  S 36  4 10 11
102. (d) After connecting a resistance R in parallel with voltmeter its e .R 10  3
effective resistance decreases. Hence less voltage appears across 114. (b) Potential gradient = = .
(R  r).L (3  3)  5
it i.e. V will decreases. Since overall resistance decreases so
more current will flow i.e. A will increase.  1V / m  10 mV / cm.
e R i G 1 100 100
103. (c) Potential gradient x  . 115. (c)  1   5  1   S  5  10 3  .
(R  R h  r) L ig S 10 S 10
ig
10 3 2 3 116. (d) 
S

4

4

1
 2
   R h  57  . i G  S 36  4 40 10
10 (3  R h  0) 1
V 6 6
i G 1 20 20 117. (a) i 2    R 1 .
104. (c) 1   1 S   0.02 . R 63 2  R
ig S 10  3 S 999 R
63
105. (a) Resistance of voltmeter should be high. S 0 .01 5 50
118. (b) ig  i    S  0 .05  .
106. (c) If ammeter is used in place of voltmeter (i.e. in parallel) it may GS 10 50  S 999
damage due to large current in circuit. Hence to control this
 100  l 
large amount of current a high resistance must be connected in 119. (d) S   .R
series.  l 

(c) Potential gradient x 

e R  100  l 
107. . Initially, 30     10  l  25 cm
(R  R h  r) L  l 
3 20  100  l 
   0 .2 Finally, 10     30  l  75 cm
(20  10  0) 10  l 
E1 l1  l2 (6  2) 2 So, shift = 50cm.
108. (d)   
E2 l1  l2 (6  2) 1 i 0 .1  10 7
120. (c) Potential gradient (x)    10  2 V/m
109. (c) Manganin or constantan are used for making the potentiometer A 10  6
wire. 121. (d) Before connecting voltmeter potential difference across 400
110. (a) resistance is
10,000
i G i.G G 100  10 3  40 40
111. (a) 1  1  1 V
ig S Vg S 800  10  3 S
 S  10 .
A 400 B 800

6V
 1116 Current Electricity

400 R1 l l1
Vi   6  2V 132. (a) In balancing condition,  1 
(400  800) R2 l 2 100  l1
X 20 1
   .....(i)
Y 80 4
After connecting voltmeter equivalent resistance between A and
4X l
400  10,000 and  .....(ii)
B   384.6  Y 100  l
(400  10,000)
4 l
Hence, potential difference measured by voltmeter    l  50 cm
384.6 4 100  l
Vf   6  1 .95 V
(384.6  800) 133. (c)
Error in measurement = Vi  Vf  2  1.95 = 0.05V.  i  100  10 6
134. (d) S   g   G   50  0 .5 
 i  ig  (10  10  3  100  10 6 )
i G 5 50  
122. (c) 1  1
ig S 0 .05 S (in parallel)
e R 5 5
50   l 50 2.97  10 2  10 4 135. (d) E .  l  0 .4   l
 S   l   3m . (R  Rh  r) L (5  45  0) 10
99 A 99 5  10 7
i G 10 0 .81 l=8m
123. (a) 1   1   S  0.09 .
ig S 1 S V 2
136. (a) Potential difference per unit length    0 .5 V / m
124. (a) From the principle of potentiometer V  l L 4
V l 137. (a)
  ; where V = emf of battery, E = emf of standard
E L l   240 
cell, L = Length of potentiometer wire 138. (b) r  R  1  1   2   1  2 
 2
l   120 
El 30 E
V  .
L 100 V
139. (d) E ; E is constant (volt. gradient).
e R l
125. (b) E . l
(R  R h  r) L V1 V 1 .1 V 180  1 .1
  2    V  1 .41 V
2 10 l1 l2 140 180 140
 10  10  3    0.4  R = 790
(10  R  0) 1  G
140. (a) IG  G  I  IG  S  I =  1   I G  I = 100.1 mA
l   150   S
126. (b) Using r  R  1  1   2   1   1
 l2   100  141. (c) Let S be larger and R be smaller resistance connected in two
gaps of meter bridge.
1000  500 1000
127. (d) Resistance between A and B  
(1500) 3  100  l  100  20
 S  R  R  4R .....(i)
 l  20
So, equivalent resistance of the circuit
10V
When 15  resistance is added to resistance R, then
1000 2500
R eq  500    100  40  6
3 3
1000 S  (R  15)  (R  15) .... (ii)
V  40  4
 Current drawn from the cell
From equations (i) and (ii) R  9 
10 3
i  A A 142. (a) According to following figure
(2500 / 3) 250 500 B 500 C
Reading of voltmeter i.e. Parallel 2
3 1000 A B
potential difference across AB   4V A
250 3 i=0.1A
i G 90
128. (d) ig   Required shunt S    10 
10 (n  1) (10  1)
V
50 Reading of voltmeter = Potential difference between A and B = i
129. (b) ig   40  4960 
10  10  3 (R + 2) 12 = 0.1 (R + 2)  R = 118 .
130. (c) Post office box is based on the principle of Wheatstone's bridge
e R
131. (d) Full deflection current ig  25  4  10 4  100  10 4 A 143. (a) Potential gradient x  .
(R  Rh  r) L
V 25
Using R  G   50  2450  in series. 0 .2  10 3 2 R
Ig 100  10  4     R = 4.9 .
10  2 (R  490  0) 1
 Current Electricity 1117

7. (d) At time t = 0 i.e. when capacitor is charging, current

Critical Thinking Questions i
2
 2mA
1. (a) Initially : Resistance of given cable 1000
When capacitor is full charged, no current will pass through it,
l
R ...
hence current through the circuit i 
2
  (9  10 3 ) 2  1mA
2000
(i)
P 4 .5
Finally : Resistance of each insulated copper wire is 8. (d) Current in the bulb    3A
l V 1.5
R'  . Hence equivalent resistance of 1 .5
  (3  10 3 ) 2 Current in 1  resistance   1 .5 A
1
R' 1  l  Hence total current from the cell i  3  1.5  4.5 A
cable R eq       ….(ii)

6 6    (3  10  3 ) 2  By using E  V  ir  E  1.5  4.5  (2.67)  13.5 V

l l 9. (d) Equivalent resistance of the circuit R  9 

V 9

 Main current i    1 A
 R 9
9 mm 3 2 2
On solving equation (i) and (ii) we get R = 7.5  0.25 A
eq
1A 0.5 A
4
R A  rB 
4
R 1 1
2. (a)    A     RB  16 R A
R B  rA  RB  2  16 9V 8 8 4

When R and R are connected in parallel then equivalent

A B 2 2 2
R A RB 16
resistance Req   RA After proper distribution, the current through 4 resistance is
(R A  RB ) 17 0.25 A.
If R A  4.25 then Req  4  i.e. option (a) is correct.
10. (b) Maximum number of resistance  2n 1  2 3 1  4
3. (c) The given circuit can be simplified as follows 11. (d) The given circuit can be simplified as follows.
r r r
r
A B C D r r r  r r
r
r r r r
5R
 R AD 
6 A B A  B
8r 2r/3
4. (c) Suppose n resistors are used for the required job. Suppose
equivalent resistance of the combination is R' and according to 3
 r r
energy conservation it's current rating is i'.
Energy consumed by the combination = n  (Energy consumed
by each resistance) B
A B A 2r
2 2 2r
 i'   R '   4   5 
 i'2 R'  n  i2 R  n              8 8r
 i   R   1   10  7
1 1 1 1 
5. (c) Resistance across AB     A B
R ' R R R1 5 20V, 1.5
12. (d) R eq  
6 R 2
R1  2  10 
i 3 i
20 2
and R    1  10 6  i  5A
R1 5
A B  1 .5 i/2 P
On solving, 2 X 2 3
R'  0.88  10 6  R
i/2 Q
6. (b) No current flows through the capacitor branch in steady state.
Total current supplied by the battery Potential difference between X and P,
6 3 5
i   . V X  VP     3  7 . 5 V ....(i)
2 .8  1 .2 2 2
3 3
Current through 2  resistor    0 .9 A
2 5
 Current Electricity 1119

5 On solving equation (i) and (ii) i1  1.8 A .

V X  VQ   2  5V …(ii)
2 18. (b) To convert a galvanometer into an ammeter, a shunt
On solving (i) and (ii) VP  VQ  2.5 volt; VQ  VP . I
S  g G is connected in parallel with it. To convert a
I  Ig
i 5
Short Trick : (VP  VQ )  (R2  R1 )  (2  3)   2 .5 V
2 2 galvanometer into a voltmeter, a resistance R   G is
Ig
 VQ  VP
connected in series with it.
13. (c) R t1  R1 (1   1 t) and Rt 2  R2 (1   2 t) 19. (a) The given circuit can be redrawn as follows
3
Also Req.  Rt1  Rt 2  Req  R1  R2 (R11  R2 2 )t X
i

  R   R2 2   
 Req  (R1  R2 )1   1 1 .t  10

  R1  R 2    20 30 60
R   R 2 2
So  eff  1 1
R1  R 2 A
i
14. (b) Let the voltage across any one cell is V, then
6
 2E 
V  E  ir  E  r1  

24 8 48 V
r 
 1 2 r  R  V
But V = 0 B
2 Er1 r1 r2 Y
 E 0 24  8i
r1  r2  R E E
Resistance between 1
A and B  6
32
 r1  r2  R  2r1 Current between A and B = Current between X and Y
 R  r1  r2 48
i  8A
5 6
15. (b) Emf E = 5V , Internal resistance r   0.5  Resistance between X and Y  (3  10  6  1)  20 
10
5  Potential difference between X and Y = 8  20 = 160 V
Current through the resistance i   2A
(2  0 .5) 20. (d) R1  R2  R1 (1  t)  R2 (1   t)
16. (b) The given circuit can be redrawn R1 
 R1  R2  R1  R2  R1t  R2 t  
E1 R1 R2 
Eeq Req
A B A B 21. (d) Current density of drifting electrons j = nev
n  5  107 cm 3  5  107  106 m 3 .
 i
E3 R2 v  0.4 ms 1 , e  1.6  10 19 C  j  3.2  10 6 Am 2
A
Current density of ions = (4 – 3.2)  10 = 0 .8  10  6
–6

E2 E2 m2
E1 R 2  E 2 R 1 2  4  2  4 This gives v for ions = 0.1 ms .
E eq    2 V and
–1

R1  R 2 44 22. (a) In the following figure M

i2
4 22 Resistance of part PNQ;
Req   2 . Current i   2 A from A to B i P
2 2 10
R1   2 .5  and
through E . 2 4 i1
17. (b) Applying Kirchhoff’s law for the loops (1) and (2) as shown in Resistance of part PMQ; N Q
figure 3 3V, 1
R1 = 2 i1 R2   10  7 .5 
For loop (1) 4
E1 = 4V
i1 R1 R2 2.5  7.5 15
1 (i1 – i2) Req   = .
R1  R2 (2.5  7.5) 8

R3 = 2 3 24
i2
Main Current i =  A
2 i2 15
R2 = 4  1 23
8
E2 = 6V  R2  24  7.5  18
2i1  2(i1  i2 )  4  0  2i1  i2  2 …(i) So, i = i   
   A
 R1  R2  23  2.5  7.5  23
1

For loop (2)

2(i1  i2 )  4 i2  6  0  i1  3i2  3 …(ii)
 1120 Current Electricity

24 18 6 In a circuit, any circuit element placed between points at the

and i2  i  i1    A. same potential can be removed, without affecting the rest of
23 23 23 the circuit. Here, by symmetry, points A, B and C are at same
23. (c) As I is independent of R 6 , no current flows through R 6 this potential, for any potential difference between P and Q.
requires that the junction of R 1 and R 2 is at the same potential The circuit can therefore be reduced as shown below
as the junction of R 3 and R 4 . This must satisfy the condition 4R

R1 R
 3 , as in the Wheatstone bridge.
R2 R4 2r
P Q
24. (c) Moving anticlockwise from A V R i
4R
 iR  V  2V  2iR  0
2 Rr
V V Effective resistance R eq  .
or 3iR  V or i  A Rr
3R B 28. (d) Potential difference between A and B
C
VA  VB  iR  V  V  iR 2V V A  VB  1  1 . 5
V 2R i  V A  0  1.5 V  V A  1.5 V
 Potential drop across C =
3 Potential difference between B and C
25. (b) Let R and m be the resistance and mass of the first wire, then VB  VC  1  2.5  2.5 V
the second wire has resistance 2R and mass 2m. Let E = emf of
each cell, S = specific heat capacity of the material of the wire.  0  VC  2.5 V  VC  2.5 V
3E Potential difference between C and D
For the first wire, current i1  and i12 Rt  mS T
R VC  VD  2V   2.5  VD  2  VD  0.5 V.
NE 29. (b) The given circuit can be simplifies as follows
For the second wire, i2  and i22 (2 R)t  2mS T .
2R r r
Thus, i1  i2 or N  6 . r
r r
26. (b) P Q
2 3 2
A r
2 r r
4 1


10 r r
1 r r
1.8 P Q
5 r
2.2 r

2  3 B


A 2r

1 2r
4 2 P Q
2r
2r 2 3
1.8 5
R'     1 .
3 3 2
Short circuited 2.2
B t 3  3 3 

2 3 4 30. (b) dQ = Idt  Q   t 2  2 
Idt   2 tdt  3 t 2 dt 
 
A 2

4 2
1    t 
= t2
3
2
3 3
2 = (9 – 4) + (27 – 8) = 5 + 19 = 24C.
E  E 2  E 3  .....  E n
31. (d) i  1
(r1  r2  r3  .......  rn )
5
 1 .5(r1  r2  r3  ......  rn )
  1.5 A .
2 1 B
5 (r1  r2  r3  .....  rn )
 A B
32. (a) Balancing length is independent of the cross sectional area of
R AB  8 . the wire.
27. (a) R1 (1  t1 ) 10 (1  5  10 3  20)
2R A 2R 33. (a)     R2  15 
R2 (1  t2 ) R2 (1  5  10  3  120)
2R i1 R 2 30 15
r r
Also     i2  20 mA
P B Q i2 R1 i2 10
2R
34. (b) The given circuit can be simplified as follows
2R C 2R 10

A 10 3
B
5 8 6 6
 Current Electricity 1121

R R
and R XZY   r(2   )  (2   )
2r 2
R R
 (2   )
R
 2 2
R XWY R XZY
5 3 Req   (2   )
A R XWY  R XZY R  R(2   ) 4 2
B 
2 2
 5 8 3
39. (d) Battery is short circuited so potential difference is zero.
Now it is a balance Wheatstone bridge. 40. (a) Let V be the potential of the junction as shown in figure.
Applying junction law, we have
So, 8
20 V 2 4 5V
A i1 i2 B
8
2
8  8 64
 R AB    4
8  8 16 i3

35. (c) The equivalent network is 0V

20  V 5  V V  0
R
2R or  
R R 2 4 2
R
or 40 – 2V + 5 – V = 2V or 5V = 45  V = 9V
6R V
 i3   4 .5 A
R 2
4R  2R 4R
R E E 2 .4  10 3
41. (a) E  x l  i l  i     4  10 4 A .
l l 1.2  5
E
4 4 E 42. (b) When bulb glows with full intensity, then voltage across it will
Clearly, the network of resistances is a balanced Wheatstone
bridge. So R AB is given by be 1.5 V and voltage across 3  resistance will be 4.5 V.
1 1 1 2 1 1
     R AB  2 R 6V
R AB 3 R 6 R 6R 2R
4.5 V 1.5 V
4
For maximum power transfer 2R  4   R   2 3 X R
2 Y
36. (c) The given circuit can be redrawn as follows
6V, 1
4 .5 B
Current through 3  resistance i   1 .5 A
3
Same current will flow between X and Y
A V B
So VXY  iRXY  1.5  1.5 R XY  R XY  1
i A
i 9i
43. (a) In figure (b) current through R2  i  
6 4 10 10
6 6
Current i   A Potential difference across R 2 = Potential difference across R
6  4  1 11
9 i R 11
P.D. between A and B, V 
6
 10 
60
V.  R2  i  R i.e. R2   
11 11 10 10 9 9
11 11
(a) By using R   .
l
; here A   (r22  r12 ) 
37. R2  R 1  11 
A Req   9
(R2  R) 11  11 10
Outer radius r = 5cm 2
9 1
Inner radius r = 5 – 0.5 = 4.5 cm 11
 R1  R  11  R1  9.9 
1

Total circuit resistance 

r2 5 mm 10
r1 44. (a) Let l be the original length of wire and x be its length stretched
10 cm uniformly such that final length is 1.5 l
l
8 5
So R  1 .7  10 
 {(5  10  2 )2  (4.5  10  2 )2 } (l – x) x 0.5l
 5.6  10 5  1.5l
R R  l (l  x ) (0.5 l  x ) x
38. (a) Here R XWY   (r )     Then 4 R    where A'  A
2r 2  r  A A' (0 .5 l  x )
 1122 Current Electricity

l lx (0 .5 l  x )2 Hence according to formula R  

l
 4   ; resistances of all the
A A xA A
conductors are equal i.e. R = R = R
1 l 2 x 2 lx
A B C

x 1
or 4 l  l  x    or  49. (b) Resistance of CD arm = 2r cos 72 = 0.62r o

4 x x x l 8
Resistance of CBFC branch
45. (b) In series : Potential difference  R
1 1 1 1  2 .62  A i
3
When only S is closed V1  E  0 .75 E     
1

4 R 2r 0 .62r r  2  0 .62  r r 72°

B r C D r E
6 1 2 .62 1 .24 r
When only S is closed V2  E  0 .86 E
2  R 
7 R 1 .24 r 2 .62 r r
and when both S and S are closed combined resistance of 6R
1 2
F H J
and 3R is 2R 1.24 r
Equivalent R'  2 R  r  2  r r r
2 2.62 r r
 V3    E  0 .67 E  V2  V1  V3 G I
3  2 .48 
 r  1   1 .946r
46. (c)  2 . 62 
1 X 2
Because the star circuit is symmetrical about the line AH
I1 Ixy  Equivalent resistance between A and H
1 2
1 1 1 R ' 1.946
Y    Req   r  0.973r
I2 3 4 Req R ' R ' 2 2
50. (a) E
A B (i – i1 – i2)
(i – i1 – i2)
 i1  0  ixy  3i2  0 i.e. i1  3i2 ......(i) F
B i
i2
Also  2(i1  ixy )  4(i2  ixy )  0 (i – i1)
i (i1 + i2)
i.e. 2i1  4 i2  6ixy ....(ii)
A C
Also VAB  1  i1  2(i1  ixy )  0  50  i1  2(i1  ixy )
i1
D
 3i1  2ixy .... (iii) Applying Kirchoff's law in mesh ABCDA
Solving (i), (ii) and (iii), ixy  2 A 10(i  i1 )  10i2  20i1  0  3i1  i2  i .......(i)
47. (b) Let n be the number of wrongly connected cells. and in mesh BEFCB
Number of cells helping one another  (12  n)  20 (i  i1  i2 )  10 (i1  i2 )  10i2  0
Total e.m.f. of such cells  (12  n)E
 3i1  4 i2  2i ......(ii)
Total e.m.f. of cells opposing = nE
Resultant e.m.f. of battery  (12  n)E  nE  (12  2n)E 2i i 2i
From equation (i) and (ii) i1  , i2   iAD 
5 5 5
Total resistance of cells = 12r
51. (d) Let the current in 12  resistance is i
( resistance remains same irrespective of connections of Applying loop theorem in closed mesh AEFCA
cells)
12i = – E + E = 0  i = 0
With additional cells
E 10  4 1
(a) Total e.m.f. of cells when additional cells help battery = (12 52. (b) Current flowing in the circuit i    A
R 20  10 5
– 2n) E + 2E
1
Total resistance = 12r + 2r = 14r P.D. across AC   20  4 V
5
(12  2n)E  2 E P.D. across AN = 4 + 4 = 8V
 3 ......(i)
14 r 53. (a) If two resistances are R 1 and R 2 then
(b) Similarly when additional cells oppose the battery R1 R 2
S  R1  R 2 and P 
(12  2n)E  2 E (R 1  R 2 )
 2 ......(ii)
14 r  RR 
From given condition S = nP i.e. (R1  R2 )  n  1 2 
Solving (i) and (ii), n = 1  R1  R2 
48. (a) All the conductors have equal lengths. Area of cross-section of  (R1  R2 )2  n R1 R2  (R1  R 2 )2  4 R1 R 2  nR1 R 2
A is {( 3 a)2  ( 2 a)2 }  a2 (R1  R2 )2
So n  4  . Hence minimum value of n is 4.
Similarly area of cross-section of B = Area of cross-section of C R1 R2
=a 2

Current sensitivity
54. (b) Voltage sensitivity 
Resistance of galvanomet er G
 Current Electricity 1123

10
 G 5.
2
150
Here ig  Full scale deflection current   15 mA . In this case, the value of R will not increase continuously.
10 Hence the correct option is (c).
V = Voltage to be measured = 150  1 = 150 V.
1
V 150 9. (d) Slope of V-i curve = resistance. Hence R   1
Hence R  G   5  9995  . 1
ig 15  10 3 10. (a) At point A the slope of the graph will be negative. Hence
resistance is negative.
11. (b) E.m.f. is the value of voltage, when no current is drawn from
Graphical Questions
2
the circuit so E = 2V. Also r = slope =  0 .4 
1. (a) For ohmic resistance V  i  V  Ri (here R is constant) 5
2. (d) From the curve it is clear that slopes at points A, B, C, D have 12. (d) For conversion of a galvanometer into a voltmeter
following order A > B > C > D. V V
 ig   ig ; where R = R + G = Total resistance
and also resistance at any point equals to slope of the V-i RG
V

RV
curve.
V
So order of resistance at three points will be  RV   RV  V
ig
R A  RB  RC  RD
3. (a) Slope of the V-i curve at any point equal to resistance at that 13. (a) According to ohm's law V  iR
point. From the curve slope for T > slope for T
1 2
 loge V  loge i  loge R  loge i  loge V  loge R
 RT1  RT2 . Also at higher temperature resistance will be The graph between loge I and loge V will be a straight line
higher so T > T
1 2
which cut loge V axis and it's gradient will be positive.
4. (c) For portion CD slope of the curve is negative i.e. resistance be 14. (c) As we know, for conductors resistance  Temperature.
negative.
From figure R  T  tan  T  tan = kT … (i)
 l 
1 1 1 1

5. (d) Slope of V-i curve  R    . But in given curve axis of i and R  T  tan (90 – )  T  cot = kT
2 2
o

2
….(ii)
2

 A From equation (i) and (ii) k (T2  T1 )  (cot   tan  )

1  A
and V are interchanged. So slope of given curve     cos  sin  (cos 2   sin2  )
R  l  (T2  T1 )      2 cot 2
 sin cos   sin cos 
i.e. with the increase in length of the wire. Slope of the curve
will decrease.  (T – T )  cot 2
2 1

iR i.  neAv d  15. (b) Let resistivity at a distance 'x' from left end be   (0  ax).
6. (c) E     vd  E (Straight line)
L A A Then electric field intensity at a distance 'x' from left end will
i i(0  ax)
 EA 
2
be equal to E   where i is the current
P  i2 R    R  P  E 2 (Symmetric parabola) A A
   flowing through the conductor. It means E   or E varies
linearly with distance 'x'. But at x = 0, E has non-zero value.
Also P  i2 (parabola) Hence (b) is correct.
Hence all graphs a, b, d are correct and c is incorrect. 16. (d) At an instant approach the student will choose tan will be the
7. (b) When we move in the direction of the current in a uniform right answer. But it is to be seen here the curve makes the
conductor, the potential difference decreases linearly. When we angle  with the V-axis. So it makes an angle (90 – ) with the
pass through the cell, from it's negative to it's positive terminal, i-axis.
the potential increases by an amount equal to it's potential So resistance = slope = tan (90 – ) = cot.
difference. This is less than it's emf, as there is some potential
drop across it's internal resistance when the cell is driving nE E
17. (d) Short circuited current i   i.e. i doesn't depend
current. nr r
upon n.
8. (b) Since the value of R continuously increases, both  and  must
be positive. x
18. (b) Here internal resistance is given by the slope of graph i.e. .
Actually the components of the given equation are as follows y
1 y
Rt t
2
But conductance  
t Resistance x
R 0
19. (a) RParallel  RSeries . From graph it is clear that slope of the line
A is lower than the slope of the line B. Also slope = resistance,
so line A represents the graph for parallel combination.
It  is positive,  is negative, the component
t will be shown in 20. (b) To make range n times, the galvanometer resistance should be
the following graph. G /n, where G is initial resistance.
Rt
t
Assertion and Reason
R 0

t 2
 1124 Current Electricity
1. (d) Resistivity of a semiconductor decreases with the temperature. 15. (a) If either the e.m.f. of the driver cell or potential difference
The atoms of a semiconductor vibrate with larger amplitudes across the whole potentiometer wire is lesser than the e.m.f. of
at higher temperatures thereby increasing it's conductivity not the experimental cell, then balance point will not obtained.
resistivity. 16. (d) Because there is no special attractive force that keeps a person
2. (d) It is quite clear that in a battery circuit, the point of lowest stuck with a high power line. The actual reason is that a
potential is the negative terminal of the battery and the current current of the order of 0.05 A or even less is enough to bring
flows from higher potential to lower potential. disorder in our nervous system. As a result of it, the affected
3. (b) The temperature co-efficient of resistance for metal is positive person may lose temporarily his ability to exercise his nervous
and that for semiconductor is negative. control to get himself free from the high power line.
In metals free electrons (negative charge) are charge carriers 17. (a) Due to high electrical conductivity of copper, it conducts the
while in P-type semiconductors, holes (positive charge) are current without offering much resistance. The copper being
majority charge carriers. diamagnetic material does not get magnetised due to current
through it and hence does not disturb the current in the
2 circuit.
4. (a) Here, E  2V , 1   1 A and r  1
2
Therefore, V  E  ir  2  1  1  1V
5. (a) It is clear that electrons move in all directions haphazardly in
metals. When an electric field is applied, each free electron
acquire a drift velocity. There is a net flow of charge, which
constitute current. In the absence of electric field this is
impossible and hence, there is no current.
6. (c) The metallic body of the electrical appliances is connected to
the third pin which is connected to the earth. This is a safety
precaution and avoids eventual electric shock. By doing this the
extra charge flowing through the metallic body is passed to
earth and avoid shocks. There is nothing such as reducing of
the heating of connecting wires by three pin connections.
7. (b) On increasing temperature of wire the kinetic energy of free
electrons increase and so they collide more rapidly with each
other and hence their drift velocity decreases. Also when
temperature increases, resistivity increase and resistivity is
inversely proportional to conductivity of material.
8. (c) In a conductor there are large number of free electrons. When
we close the circuit, the electric field is established instantly
with the speed of electromagnetic wave which cause electron
drift at every portion of the circuit. Due to which the current is
set up in the entire circuit instantly. The current which is set
up does not wait for the electrons flow from one end of the
conductor to the another end. It is due to this reason, the
electric bulb glows immediately when switch is on.
l
9. (a) Resistance wire R   . where  is resistivity of material
A
which does not depend on the geometry of wire. Since when
wire is banded, resistivity, length and area of cross-section do
not change, therefore resistance of wire also remain same.
10. (c) The resistance of the galvanometer is fixed. In meter bridge
experiments, to protect the galvanometer from a high current,
high resistance is connected to the galvanometer in order to
protect it from damage.
11. (a) Voltameter measures current indirectly in terms of mass of
ions deposited and electrochemical equivalent of the substance
 m
I  . Since value of m and Z are measured to 3rd
 Zt 
decimal place and 5th decimal place respectively. The relative
error in the measurement of current by voltmeter will be very
small as compared to that when measured by ammeter directly.
12. (a) When current flows through a conductor it always remains
uncharged, hence no electric field is produced outside it.
13. (b) Here assertion and reason both are correct but the reason is
not the correct explanation of assertion.
1
14. (a) Sensitivity   (Length of wire)
Potential gradient
 Current Electricity 1125

Current Electricity
1. Figure shows a simple potentiometer circuit for measuring a small (a) 1 V
e.m.f. produced by a thermocouple. The meter wire PQ has a (b) 2 V
resistance 5  and the driver cell has an e.m.f. of 2 V. If a balance (c) 6 V
point is obtained 0.600 m along PQ when measuring an e.m.f. of (d) 4 V
6.00 mV, what is the value of resistance R
9. A beam contains 2  10 doubly charged positive ions per cubic
8

(a) 995  R centimeter, all of which are moving with a speed of 10 m/s. The 5

2V current density is
(b) 1995  (a) 6.4 A/m 2
(b) 3.2 A/m 2

(c) 1.6 A/m (d) None of these

(c) 2995 
2

0.600m 10. In the circuit shown, the reading of ammeter when switch S is open
P Q
(d) None of these and when switch S is closed respectively are
Thermocouple
G (a) 3 A and 4 A 2
2. A car has a fresh battery of e.m.f. 12 V and internal resistance of S
6.00 mV
0.05 . If the starter motor draws a current of 90 A, the terminal (b) 4 A and 5 A 3
A
voltage when the starter is on will be 2
(c) 5 A and 6 A
(a) 12 V (b) 10.5 V (d) 6 A and 7 A 20V
(c) 8.5 V (d) 7.5 V
11. In the circuit as shown in figure the
3. If the balance point is obtained at the 35 cm in a metre bridge the
th

R
resistances in the left and right gaps are in the ratio of
(a) 7 : 13 (b) 13 : 7
0.5A
(c) 9 : 11 (d) 11 : 9 25V 10 10 20
4. Find the equivalent resistance across the terminals of source of e.m.f.
24 V for the circuit shown in figure (a) Resistance R = 46 
(a) 15  10 6 (b) Current through 20  resistance is 0.1 A
(c) Potential difference across the middle resistance is 2 V
(b) 10  15 (d) All option are correct
8 12. In figure shows a rectangular block with dimensions x, 2x and 4x.
(c) 5  8
E=24V Electrical contacts can be made to the block between opposite pairs
(d) 4  of faces (for example, between the faces labelled A-A, B-B and C-C).
4 Between which two faces would the maximum electrical resistance
5. In the circuit shown in figure, switch S is initially closed and S is
1 2
be obtained (A-A : Top and bottom faces, B-B : Left and right faces,
open. Find V – V
a b
C-C : Front and rear faces)
1 b 5
(a) 4 V
10F C
(b) 8 V 4x
S2
(c) 12 V B B
3 3
(d) 16 V a C
24V S1 (a) A-A (b) x B-B
2x
6. The figure here shows a portion of a circuit. What are the (c) C-C (d) Same for all three pairs
magnitude and direction of the current i in the lower right-hand
wire 13. A battery is connected to a uniform resistance wire AB and B is
1A earthed. Which one of the graphs below shows how the current
(a) 7 A density J varies along AB
2A
(b) 8 A 2A
– +
(c) 6 A 2A

(d) 2 A
3A 4A
i A B
7. A carbon resistor has colour strips as violet, yellow brown and
golden. The resistance is (a) J (b) J
(a) 641  (b) 741 
Zero at all
(c) 704  (d) 407  points
8. A voltmeter of resistance 1000  is connected across a resistance of 0 0
A B A B
500  in the given circuit. What will be the reading of voltmeter J J

10 V
V 0 0
A B A B
500 500
 1126 Current Electricity

(c) (d) 19. A moving coil galvanometer is converted into an ammeter reading
upto 0.03 A by connecting a shunt of resistance 4 r across it and
into an ammeter reading upto 0.06 A when a shunt of resistance
r is connected across it. What is the maximum current which can
14. A cylindrical metal wire of length l and cross sections area S, has be sent through this galvanometer if no shunt is used [
resistance R, conductance G, conductivity  and resistivity . Which
one of the following expressions for  is valid (a) 0.01 A (b) 0.02 A
GR R
(a) (b) (c) 0.03 A (d) 0.04 A
 G
GS Rl 20. Two conductors are made of the same material and have the same
(c) (d)
l S length. Conductor A is a solid wire of diameter 1.0 mm. Conductor B is a
15. A potential divider is used to give outputs of 4 V and 8 V from a 12 hollow tube of outside diameter 2.0 mm and inside diameter 1.0 mm.
V source. Which combination of resistances, (R , R , R ) gives the 1 2 3
The resistance ratio R /R will be
A B

correct voltages ? R : R : R1 2 3

(a) 1 (b) 2
+12V R3
(a) 2 : 1 : 2 (c) 3 (d) 4
+8V
(b) 1 : 1 : 1
R2 21. A wire has resistance of 24  is bent in the following shape. The
(c) 2 : 2 : 1 +4V effective resistance between A and B is
(d) 1 : 1 : 2 R1
0 Volt (a) 24  60°
16. Find equivalent resistance between A and B
(b) 10 
(a) R R R 60°
16 A
R (c)  B
3R 3 5 cm
(b) R R 10 cm
4 A R B
R R (d) None of these
R
R R
(c) 22. In the circuit shown in figure, find the current through the branch
2 R BD
R
(d) 2R R R 3
(a) 5 A A 6 B C
17. Following figure shows four situations in which positive and negative
charges moves horizontally through a region and gives the rate at (b) 0 A 15 V
which each charge moves. Rank the situations according to the (c) 3 A 3
30 V
effective current through the region greatest first
(d) 4 A
+ – D
7C/sec 3C/sec + 2C/sec
6C/sec 23. A battery of 24 cells, each of emf 1.5 V and internal resistance 2 is
4C/sec – – to be connected in order to send the maximum current through a 12
(i) (ii) + +  resistor. The correct arrangement of cells will be
(a) i = ii = iii = iv (b) i > ii5C/sec
> iii > iv 1C/sec
(a) 2 rows of 12 cells connected in parallel
(c) i = ii = iii > iv (d) i = ii(iii)= iii < iv (iv)

18. A and B are two square plates of same metal and same thickness but (b) 3 rows of 8 cells connected in parallel
length of B is twice that of A. Ratio of resistances of A and B is (c) 4 rows of 6 cells connected in parallel
(a) 4 : 1 (d) All of these
(b) 1 : 4
B
(c) 1 : 1
(d) 1 : 2
A

(SET - 19)
1. (a) The voltage per unit light of the metre wire PQ is 10 mV
current drawn from the driver cell is i   2 mA .
 6 .00 mV  5
  i.e. 10 mV/ m . Hence potential difference
 0 .600 m  (2 V  10mV ) 1990 mV
The resistance R    995  .
across the metre wire is 10 mV /m  1m  10 mV . The 2 mA 2 mA
 Current Electricity 1127

2. (d) V  E  i.r  12  90  0.05  12  4.5  7.5 V . 2 3

Now voltmeter reading  iv  R V    500  4 V .
P R R 3 250
3. (a) Using Wheatstone principle  
Q S 100  l 2  10 8  2  1.6  10 19  10 5
9. (a) J  nqv  n(ze)v  =6.4A/m 2


35

35

7 (10  2 )3
100  35 65 13 10. (b) When switch S is open total current through ammeter.
4. (c) Given circuit can be reduced to a simple circuit as shown in
figures below 20
i  4A .
Parallel (3  2)
10 6 6 6
20
When switch is closed i   5A .
15 3  (2 || 2)
 11. (d) R R
8
8 4 Parallel
0.5 A 0.5 A
Parallel 25 V
4 10 10 20 R'  4 R'
4 Series

1 1 1 1 20
    R'  4
R ' 10 10 20 5
i.e. Req  5  . 10 10
25 25
Now using ohm’s law i   0 .5 
5. (b) Switch S 2 is open so capacitor is not in circuit. R  R' R4
1 b 5 25
 R4   50  R  50  4  46 
O 0.5
3 3 0.5  5 2.5
Current through 20  resistor    0 .1 A
a 20  5 25
Potential difference across middle resistor
24 V 24 = Potential difference across 20   20  0.1  2V
Current through 3  resistor  4 A
33
12. (c) Let  is the resistivity of the material
Let potential of point ‘O’ shown in fig. is VO Resistance for contact A-A
then using ohm’s law x 
R AA   
VO  Va  3  4  12V ....(i) 2x  4 x 8 x
24 Similar for contacts B-B and C-C are respectively
Now current through 5  resistor   4A
5 1 2x  4
RBB  .  
So V0  Vb  4  1  4 V x  4 x 2x 8 x
.....(ii) 4x 2  16 
and RCC    
From equation (i) and (ii) Vb  Va  12  4  8 V. x  2x x 8x
6. (b) By using Kirchoff's junction law as shown below. It is clear maximum resistance will be for contact C-C.
13. (d) Wire AB is uniform so current through wire AB at every across
1A section will be same. Hence current density J  i / A  at every
2A point of the wire will be same.
2A 3A
1
5A
6A 2A 14. (a) Conductivity   .....(i)

3A
4A 1
i=8A and conductance G 
7. (b) Using standard colour codes R
Violet = 7, yellow = 4, brown = 1 and gold = 5 % (tolerance)  GR  1 .....(ii)

So R  74  10 1  5%  740  5% GR
From equation (i) and (ii)  

So its value will be nearest to 741  .
15. (b) Resistors are connected in series. So current through each
8. (d) Total current through the circuit resistor will be same
10 3 12  8 8  4 4  0 4 4 4
i  A i     
1000 250 R3 R2 R1 R3 R2 R1
 500
3 So, R1 : R2 : R3 :: 1 : 1 : 1 .
16. (c) Given circuit can be redrawn as follows
 1128 Current Electricity

Neglect
Series
Parallel R R R+R=2R
R/3
A B A R B

R Applying KVL along the loop ABDA, we get
R/3
– 6i – 3 i + 15 = 0 or 2i + i = 5 …..(i)
Parallel R R
1 2 1 2

Series Applying KVL along the loop BCDB, we get

Neglect  R+R=2R
2R
– 3(i – i ) – 30 + 3i = 0 or – i + 2i = 10 …..(ii)
1 2 2 1 2

3 Solving equation (i) and (ii) for i , we get i = 5 A.

R 2 2

R/3 23. (a) Suppose m rows are connected in parallel and each row
A B A B contains n identical cells (each cell having E = 15 V and r = 2)

For maximum current in the external resistance R, the
R/3
nr
R 2R necessary condition is R 
R m
 R eq  3
n2
2  12   n = 6m ..... (i)
m
17. (c) For figure (i) i1  7 A
Total cells = 24 = n  m ..... (ii)
For figure (ii) i2  4  3  7 A On solving equations (i) and (ii) n = 12 and m = 2
i.e. 2 rows of 12 cells are connected in parallel.
For figure (iii) i3  5  2  7 A
For figure (iv) i4  6  1  5 A
l    2l  R
18. (c) RA   and RB   i.e. A  1 : 1
lt t 2l  t t RB
ig S
19. (b)   ig G  (i  ig )S
i GS
 ig G  (0.03  ig )4 r .....(i) ***
and ig G  (0.06  ig )r .....(ii)
From (i) and (ii)
0.12  4 ig  0.06  ig  ig  0.02 A .
l
20. (c) For conductor A, R A  ,
r12
l
For conductor B, R B 
 (r22  r12 )
l

l r2
r1

2 mm


1 mm

1 mm

A
B
2 2
R A r22  r12  r2  d 
2
2
      1   2   1     1  3
RB r12  1
r  1
d 1
21. (b) Given resistance of each part will be

12 4
6 6


A 6 6 B A 6 6 B
R = 10  
eq 
B
22. (a) The current in the circuit are Aassumed
4as shown 6
in the fig.

6 i1 B 3 i1 – i2
A C

15 V 3 30 V
i2
i1 D