Join our Telegram channel: @Jee_tricks

Current Electricity 343

16 Current Electricity
QUICK LOOK Where
ρ = Specific resistance
Electric Current: Current is a tensor quantity, while current
τ = Relaxation time,
density is a vector. Conventionally direction of current is taken
n = Electron density in meter–3
along the direction of flow of positive charges. In metals charge
carriers are only free electrons. In liquids charge carriers are Stretching of Wire: If a conducting wire stretches, it’s length
positive and negative ions. In gases charge carries are positive increases, area of cross-section decreases so resistance
ions and electrons. And in semi-conductors charge carriers are increases but volume remains constant.
electrons and holes.
Suppose for a conducting wire before stretching it’s length
Drift velocity of electrons in a metal is of the order of = l1 , area of cross-section = A1 , radius = r1 , diameter = d1 , and
10 −3 m / s and is directly proportional to electric field (or l1
resistance R1 = ρ
potential difference applied). The current flows with speed of A1
light. Mean velocity of electrons due to their thermal agitations After stretching length = l2 , area of cross-section = A2 , radius
(or random motion) is zero; while mean speed depends on
l2
temperature. = r2 , diameter = d 2 and resistance = R2 = ρ
d
A2
e Ratio of resistances before and after stretching
vd Area 2 2 4 4
A R1 l1 A2  l1   A2   r2   d 2 
= × =  =  =  = 
R2 l2 A1  l2   A1   r1   d1 
Current measuring point
2
Figure: 16.1 R l 

If length is given then R ∝ l ⇒ 1 =  1  2

q dq R2  l2 
Electric Current I = = (scalar quantity)
t dt 1 R r 
4

I
If radius is given then R ∝ 4 ⇒ 1 =  2 

Current Density J = where An = normal area r R2  r1 
An



Current I = j . A = JA cos θ = neAvd where vd is drift velocity.
Resistance of a conducting body is not unique but depends
on it’s length and area of cross-section i.e., how the
Ohm’s Law potential difference is applied. See the following figures
Under same physical conditions the voltage is directly
a
proportional to electric current in dc circuits. V = Rl (Under
same physical conditions)
b

The resistance of a conductor is directly proportional to a c b c
length and inversely proportional to cross-sectional area. i.e. Figure: 16.2

At a given temperature, the specific resistance of a
For length = a, area of cross-section = b × c
conductor is independent of dimensions but depends only
 a 
on material. Resistance R = ρ  

If a given mass of a material is stretched to decreases its  b×c 
cross-section, then its length also increase and then For length = b, Area of cross-section = a × c
l l  b 
Rα = or Rα 2 Resistance R = ρ  
a r  a×c 
ρ l 2m l 1
Resistance R = = .
Conductance K =
A ne 2τ A R

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

344 Quick Revision NCERT-PHYSICS

ne τ 2
If n identical resistances are first connected in series and

Specific resistance ρ = (for metals) then in parallel, the ratio of the equivalent resistance is
2m
1 2m Rs n 2

Conductivity σ = = 2 (for metals) given by = .
ρ ne τ Rp 1

Ohm’s Law J = σ E (alternative form) or V = Ri. For
If a wire of resistance R is cut in n equal parts and then
Ohmic conductors (like iron, silver), V − I graph is a straight these parts are collected to form a bundle, then equivalent
line. And for non-ohmic conductors (like junction diode, R
resistance of combination will be 2 .
torch bulb, thermistor) n
V − I graph is non-linear
If equivalent resistance of R1 and R2 in series and parallel
be Rs and Rp respectively then
A resistor at constant temperature A diode
1
Current Current Current R1 =  Rs + Rs2 − 4 Rs R p 
2 
1
R2 =  Rs − Rs2 − 4 Rs Rp  .
Potential Potential Potential 2 
difference difference difference
Internal Resistance r: Potential difference across the terinals
Figure: 16.3 of a cell V = E − ir where r = internal resistance, E = emf of

Effect of temperature on resistance: Generally metals offer E 

cell here V = iR. Internal resistance r =  − 1 R Where R =
more electrical resistance if temperature is increased. On the V 
other hand the resistance offered by a non – metallic external resistance.
substance normally decreases with increase of Current flow = 1

temperature. Rt = R0 (1 + α t + β t 2 ); α > β Electron drift

For linear variation or if t is not too large

Load = R
p.d = IR = V
e.m.f. = E
Source

Rt = R0 (1 + α t ) For metals α is positive and for semi-

conductor α is negative.
Combination of Resistances

Resistance in series:
Figure: 16.4
Net resistance R = R1 + R2 + R3
Kirchoff’s Laws
Net potential difference, V = V1 + V2 + V3
Kirchoff’s first law (or current law) is based on conservation of
Current i = i1 + i2 + i3 (same in all resistance) charge. Junction Law: Σi = 0 at any junction: Current law: the

Resistances in parallel: sum of the currents into any junction is equal to the sum of the
1 1 1 1 current out.
Net resistance R is given by = + +
R R1 R2 R3 I = I1 + I2

Net current i = i1 + i2 + i3 I2 R2 I2

Potential difference V = V1 + V2 + V3 (same across all

resistances) I1 R1 I1

Note I I
V 0
Decoration of lights in festivals is an example of series
grouping whereas all household appliances are connected in V
Figure: 16.5
parallel grouping.
For any branch of the circuit, the current out of the branch must

Using n conductors of equal resistance, the number of
be equal to the current into the branch. This is required by the
possible combinations is 2 n −1. conservation of electric charge. Any cross-section of the circuit

If the resistances of n conductors are totally different, then must carry the total current. For a series circuit, the current is
the number of possible combinations will be 2 n. the same at any point in the circuit.

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

Current Electricity 345

I I I Wheatstone’s Bridge: When Wheatstone’s bridge is balanced,
R1 R2 the resistance in arm BD may be ignored while calculating the
I I equivalent resistance of bridge between A and C.
P R
I I
Condition of balance is =
VB 0 Q S
VB
Equivalent resistance between terminals connected to
Figure: 16.6
1 1 1
battery at balance = +
Req P + Q R + S
Kirchoff’s second law (or voltage law) is based on conservation
of energy. Loop law ΣV = 0 or ΣiR = ΣE for a closed circuit:
When battery and galvanometer arms of a Wheatstone’s
bridge are interchanged, the balance position remains
Voltage Law: The net voltage. Drop around any closed undisturbed while sensitivity of bridge charges.
looppath must be zero.
A Wheatstone’s bridge is most sensitive if its all resistance
P, Q, R, S are equal.
V 0
R2 100 − 1
Metre Bridge: Unknown resistance S = × R , where l =
l
V R1 0 balancing length in cm
B
V 0 0 P Q S
R R.B
V i1 i1
Figure: 16.7 G C
R S G
For any path you follow around the circuit, the sum of the i2 i2 Q
A P B C
voltages rises (like batteries) must equal the sum of the voltage
i = i1 + i2 D i 1 cm (100 – 1) cm
drops. Voltage represents energy per unit charge, and
E
conservation of energy demands that energy is neither created E K
nor destroyed. Figure: 16.10
V 0
R1 R2
Potentiometer: If L is length of potentiometer wire AB,
V
V 0 0 Potential gradient k = AB = i ρ , where ρ is resistance per unit
L
VB length of potentiometer wire
Figure: 16.8

EMF of a cell, E = kl
Equivalent Resistance for Cube: If a skeleton cube is made E1 l1

For same potential gradient =
with 12 equal resistances each having resistance R then the net E2 l2
resistance across
5 Power supply

The longest diagonal ( AG or EC or BH or DF) = R
6
C
3 A B

The diagonal of face (e.g. AC, ED… etc.) = R K1
4
7 ε0

A side (e.g. AB, BC etc.) = R G
12 R1
Standard cell
H
G Figure: 16.11
E F
Combinations of Cells
nE
D C
In series: If n identical cells are in series i =
R + nr
A B
Where
Figure: 16.9

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

346 Quick Revision NCERT-PHYSICS

R = external resistance; R1 R2

r = internal resistance of a cell and

E = emf of a cell A R5 B
n R3 R4
E r E r E r
Figure: 16.15
I I
R
2 R1 R2 + R3 ( R1 + R2 )
Case (ii): RAB =
V 2 R3 + R1 + R2
Figure: 16.12
R1 R2
E

In parallel: n cells in parallel i =
r
R+  A R5 B
n
R1 R2

E r Figure: 16.16
n
1 1 1/ 2
Case (iii): RAB = ( R1 + R2 ) +  ( R1 + R2 )2 + 4 R3 ( R1 + R2 ) 
2 2
E r R1 R1 R1 R1
I I A
R
RAB R3 R3 R3 tends to infinity
V
Figure: 16.13
B
R2 R2 R2 R2

Mixed grouping: n cells in a now, m such rows in
Figure: 16.17
mnE
parallel i =
mR + nr  
n Case (iv): R AB = 1 R1 1 + 1 + 4  R 2  
2   R1  
E r E r E r
m R1 R1 R1
A
n
RAB R2 R2 R2 tends to infinity
E r E r E r
I I
R B

V Figure: 16.18
Figure: 16.14
Transformation between Y or Star and delta connection
nr
For maximum current Rext = Rint or R =
m N N

If two cells of different emfs are correctly connected in R
E + E2
series i = 1 R R
R + r1 + r2 R R

If two cells of different emfs are wrongly connected in R
N N N N
series i.e., (positive terminals connected together) Figure: 16.19
E − E2
i= 1
The transformation from ∆ -load to Y-load. To compute the
R + r1 + r2
impedance Ry at a terminal node of the Y circuit with

Some Standard Results for Equivalent Resistance R ' R"

impedances R ', R " to adjacent node i Ry = n the ∆
Case (i): ΣR∆
R R ( R + R4 ) + ( R1 + R2 ) R3 R4 + R5 ( R1 + R2 )( R3 + R4 ) R 'R"
RAB = 1 2 3 circuit by Ry = where R∆ are all impedances in the
R5 ( R1 + R2 + R3 + R4 ) + ( R1 + R3 )( R2 + R4 ) ΣR∆
∆ circuit. This yields the specific formulae

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

Current Electricity 347

Rb Rc Ra Rc sensitive. Induced e.m.f. a thermo-couple E = at + bt 2
R1 = , R2 = and
Ra + Rb + Rc Ra + Rb + Rc G
Cu Cu
Ra Rc
R3 = Fe
Ra + Rb + Rc
A B

Equations for the transformation from Y-load to ∆ -load
Hot junction Ice
The general idea is to compute an impedance R∆ in the ∆
Rp
circuit by R∆ =
Ropposite
Figure: 16.20 Variation of thermo e.m.f. with temperature
Where R p = R1 R2 + R2 R3 + R3 R1 is the sum of the products of all
 a 
pairs of impedances in the Y circuit and Ropposite is the Neutral temperature, t n = −   °C Neutral temperature is
 2b 
impedance of the node in the Y circuit which is opposite the independent of temperature of cold junction. At neutral
edge with R∆ . The formula for the individual edges are thus temperature, the thermo e.m.f. is maximum; but thermoelectric
R1 R2 + R2 R3 + R3 R1 R R + R2 R3 + R3 R1 power is zero.
Ra = , Rb = 1 2 and
R1 R2
R1 R2 + R2 R3 + R3 R1
Rc =
R3
Ti
Temp Diff.
Z12 Z13  1 1 1  O Tn
Z1 = ; Z12 = Z1Z 2  + + 
Z12 + Z13 + Z 23 Z
 1 Z 2 Z 3 
Figure: 16.21

Temperature of inversion depends on temperature of cold

Faraday’s Laws junction (t0 ) tn − t0 = ti − tn
Mass of element deposited at electrode (i) m = Zq = Zit ,
a
m W Inversion temperature when t0 is 0° then ti is = 2tn = −   °C
(ii) 1 = 1 b
m2 W2
I
Z = Electrochemical equivalent, W = Chemical equivalent
atomic weight
Where W = I
valency T1 T2
W Figure: 16.22
Faraday number F = = 69500Coul/g-equivalent
Z dE

Thermoelectric power of a thermocouple, P = = a + bt
dt
Thermo-electric Effects
dE
A thermocouple is a temperature-measuring device consisting
Peltier coefficient, π = T
of two dissimilar conductors that contact each other at one or dt
more spots, where a temperature differential is experienced by Π12 I ≡ Power evolved at junction
the different conductors (or semiconductors). It produces a dP

Thomson coefficient, σ = −T
voltage when the temperature of one of the spots differs from dt
the reference temperature at other parts of the circuit. τ I ∇T ≡ Power evolved per unit volume
Thermocouples are a widely used type of temperature sensor
Thomson coefficient of lead is zero
for measurement and control, and can also convert a B
temperature gradient into electricity. Commercial A
thermocouples are inexpensive, interchangeable, are supplied B B
with standard connectors, and can measure a wide range of
temperatures. Seebeck effect is reversible. Seebeck effect is A third dissimilar metal is introduced
Figure: 16.23
reversible. The direction of current in Cu-Fe thermocouple is
for Cu to Fe through hot junction and in Bi- Sb couple it is
Law of Intermediate metals EAB + EBC = E AC
form Bi to Sb through hot junction. Bi-Sb couple is most

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

348 Quick Revision NCERT-PHYSICS

MULTIPLE CHOICE QUESTIONS Red Silver

Current Density (J)

1. The potential difference applied to an X-ray tube is 5 KV
White Brown
and the current through it is 3.2 mA. Then the number of
electrons striking the target per second is: a. 2.2 kΩ b. 3.3 kΩ
a. 2 × 1016 b. 5 × 106 c. 5.6 kΩ d. 9.1 kΩ
c. 1 × 1017 d. 4 × 1015 Ohm’s Law and Resistance
2. A beam of electrons moving at a speed of 106 m/s along a 9. Two wires of resistance R1 and R2 have temperature co-
line produces a current of 1.6 × 10–6 A. The number of efficient of resistance α1 and α2 respectively. These are
electrons in the 1 metre of the beam is: joined in series. The effective temperature co-efficient of
a. 106 b. 107 resistance is:
13
c. 10 d. 1019 α1 + α 2
a. b. α1α 2
3. An electron is moving in a circular path of radius 5.1 × 2
10–11 m at a frequency of 6.8 × 1015 revolution/sec. The α1 R1 + α 2 R2 R1 R2α1α 2
c. d.
equivalent current is approximately: R1 + R2 R12 + R22
a. 5.1 × 10–3 A b. 6.8 × 10–3 A
c. 1.1 × 10 A
–3
d. 2.2 × 10–3 A 10. A wire of length L and resistance R is stretched to get the
radius of cross-section halved. What is new resistance?
4. A copper wire of length 1m and radius 1mm is joined in a. 5 R b. 8 R
series with an iron wire of length 2m and radius 3mm and c. 4 R d. 16 R
a current is passed through the wire. The ratio of current
densities in the copper and iron wire is: 11. The V-i graph for a conductor at temperature T1 and T2 are
a. 18 : 1 b. 9 : 1 as shown in the figure. (T2 – T1) is proportional to:
V
c. 6 : 1 d. 2 : 3 T2

5. A conducting wire of cross-sectional area 1 cm2 has 3 ×

T1
1023 m–3 charge carriers. If wire carries a current of 24 mA, θ
the drift speed of the carrier is: θ
i
a. 5 × 10–6 m/s b. 5 × 10–3 m/s
c. 0.5 m/s d. 5 × 10–2 m/s a. cos 2θ b. sinθ
c. cot 2θ d. tanθ
6. In a wire of circular cross-section with radius r, free
electrons travel with a drift velocity v, when a current i 12. The resistance of a wire at 20°C is 20 Ω and at 500°C is
flows through the wire. What is the current in another 60Ω. At which temperature resistance will be 25Ω:
wire of half the radius and of the some material when the a. 50°C b. 60°C
drift velocity is 2v? c. 70°C d. 80°C
a. 2i b. i
13. The specific resistance of manganin is 50 ×10−8 Ω m. The
i i
c. d. resistance of a manganin cube having length 50 cm is:
2 4
7. Two wires A and B of the same material, having radii in a. 10–6 Ω b. 2.5 × 10–5 Ω
the ratio 1 : 2 and carry currents in the ratio 4 : 1. The c. 10–8 Ω d. 5 × 10–4 Ω
ratio of drift speeds of electrons in A and B is: 14. A rod of certain metal is 1 m long and 0.6 cm in diameter.
a. 16 : 1 b. 1 : 16 It’s resistance is 3 × 10–3Ω. A disc of the same metal is 1
c. 1 : 4 d. 4 : 1 mm thick and 2 cm in diameter, what is the resistance
Electronics Colour Code between it’s circular faces.
8. In the figure a carbon resistor has band of different colours a. 1.35 × 10 −6 Ω b. 2.7 × 10 −7 Ω
on its body. The resistance of the following body is: c. 4.05 × 10−6 Ω d. 8.1× 10−6 Ω

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

Current Electricity 349

15. An aluminum rod of length 3.14 m is of square cross- Kirchoff’s Laws
section 3.14 × 3.14 mm2. What should be the radius of 1 m 23. Two resistance R1 and R2 provides series to parallel
long another rod of same material to have equal equivalents as n/1 then the correct relationship is:
resistance: 2 2 3/ 2 3/ 2
R  R  R  R 
a. 2 mm b. 4 mm a.  1  +  2  = n 2 b.  1  + 2  = n3 / 2
c. 1 mm d. 6 mm  R2   R1   R2   R1 
1/ 2 1/ 2
16. If a copper wire is stretched to make it 0.1% longer, the R   R2  R  R 
c.  1 +  = n d.  1  + 2  = n1/ 2
percentage increase in resistance will be:  R2   R1   R2   R1 
a. 0.2 b. 2
24. Five resistances are combined according to the figure. The
c. 1 d. 0.1
equivalent resistance between the point X and Y will be:
17. The temperature co-efficient of resistance of a wire is 10 Ω
0.00125/oC. At 300K. It’s resistance is 1Ω. The resistance
of the wire will be 2 Ω at: X Y
a. 1154 K b. 1127 K 10 Ω 20 Ω 10 Ω

c. 600 K d. 1400 K
10 Ω
18. Masses of three wires are in the ratio 1 : 3 : 5 and their a. 10 Ω b. 22 Ω
lengths are in the ratio 5 : 3 : 1. The ratio of their electrical c. 20 Ω d. 50 Ω
resistance is:
25. What will be the equivalent resistance of circuit shown in
a. 1 : 3 : 5 b. 5 : 3 : 1
figure between points A and D:
c. 1 : 15 : 125 d. 125 : 15 : 1 10 Ω 10 Ω 10 Ω
A B
19. Dimensions of a block are 1 cm × 1 cm × 100 cm. If
specific resistance of its material is 3 × 10–7 ohm-m, then 10 Ω 10 Ω
the resistance between it’s opposite rectangular faces is:
a. 3 × 10–9 ohm b. 3 × 10–7 ohm C D
c. 3 × 10–5 ohm d. 3 × 10–3 ohm 10 Ω 10 Ω 10 Ω
a. 10 Ω b. 20 Ω
20. A new flashlight cell of emf 1.5 volts gives a current of 15
c. 30 Ω d. 40 Ω
amps, when connected directly to an ammeter of
resistance 0.04 Ω. The internal resistance of cell is: 26. In the network shown in the figure each of resistance is
a. 0.04 Ω b. 0.06 Ω equal to 2Ω. The resistance between A and B is:
C
c. 0.10 Ω d. 10 Ω
21. For a cell, the terminal potential difference is 2.2 V when
the circuit is open and reduces to 1.8 V, when the cell is
O
connected across a resistance, R = 5Ω. The internal A
B
resistance of the cell is: D E
10 9 a. 1 Ω b. 2 Ω c. 3 Ω d. 4 Ω
a. Ω b. Ω
9 10
27. The equivalent resistance between points A and B of an
11 5
c. Ω d. Ω infinite network of resistance, each of 1Ω, connected as
9 9
shown is:
22. The internal resistance of a cell of emf 2V is 0.1Ω. It’s 1Ω 1Ω 1Ω
A
connected to a resistance of 3.9Ω. The voltage across the
cell will be:
1Ω 1Ω 1Ω ⇒∞
a. 0.5 volt b. 1.9 volt
c. 1.95 volt d. 2 volt
B

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

350 Quick Revision NCERT-PHYSICS

a. Infinite b. 2 Ω 33. To get maximum current in a resistance of 3 Ω one can
1+ 5 use n rows of m cells connected in parallel. If the total no.
c. Ω d. Zero of cells is 24 and the internal resistance of a cell is 0.5
2
then:
28. The equivalent resistance between A and B in the circuit
a. m = 12, n = 2 b. m = 8, n = 4
shown will be:
r
c. m = 2, n = 12 d. m = 6, n = 4

34. 100 cells each of emf 5V and internal resistance 1Ω are to

r r r r be arranged so as to produce maximum current in a 25Ω
resistance. Each row contains equal number of cells. The
A B number of rows should be:
r C r
5 6 7 8 a. 2 b. 4 c. 5 d. 100
a. r b. r c. r d. r
4 5 6 7 35. n identical cells, each of emf E and internal resistance r,
29. In the given figure, equivalent resistance between A and B are joined in series to form a closed circuit. The potential
will be: difference across any one cell is:
a. Zero b. E
3Ω 4Ω E  n −1 
c. d.  E
A 7Ω B n  n 

6Ω 8Ω Network Simplification
36. In the adjoining circuit, the battery E1 has as emf of 12
14 3 volt and zero internal resistance, while the battery E has
a. Ω b. Ω
3 14 an emf of 2 volt. If the galvanometer reads zero, then the
9 14 value of resistance X ohm is:
c. Ω d. Ω
14 9 500 Ω O
A G B
30. If each resistance in the figure is of 9Ω then reading of
ammeter is: E1 XΩ E

D C
9V P
a. 10 b. 100 c. 500 d. 200
A 37. In the circuit shown here E1 = E2 = E3 = 2V and R1 = R2 =
a. 5 A b. 8 A 4 Ω. The current flowing between point A and B through
c. 2 A d. 9 A battery E2 is:
31. A wire has resistance 12Ω. It is bent in the form of a E1 R1

circle. The effective resistance between the two points on

E2
any diameter is equal to: A B
a. 12 Ω b. 6 Ω E3 R2
c. 3 Ω d. 24 Ω
a. Zero b. 2 A from A to B
Combinations of Cells
c. 2 A from B to A d. None of these
32. In a mixed grouping of identical cells 5 rows are connected
38. The magnitude and direction of the current in the circuit
in parallel by each row contains 10 cell. This combination
shown will be:
send a current i through an external resistance of 20Ω. If 1Ω 2Ω
e
the emf and internal resistance of each cell is 1.5 volt and a b
10 V 4V
1 Ω respectively then the value of i is:
a. 0.14 b. 0.25 3Ω
c. 0.75 d. 0.68 d c

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

Current Electricity 351

7 Different Measuring Instruments
a. A from a to b through e
3 44. The scale of a galvanometer of resistance 100Ω contains
7 25 divisions. It gives a deflection of one division on
b. A from b and a through e
3 passing a current of 4 × 10−4 A. The resistance in ohms to
c. 1.0 A from b to a through e be added to it, so that it may become a voltmeter of range
d. 1.0 A from a to b through e 2.5 volt is:
39. Figure represents a part of the closed circuit. The potential a. 100 b. 150 c. 250 d. 300
difference between points A and B (VA – VB) is: 45. A galvanometer, having a resistance of 50 Ω gives a full
2A 2Ω 3V 1Ω
A B scale deflection for a current of 0.05 A. the length in meter
a. + 9 V b. – 9 V of a resistance wire of area of cross-section 2.97 × 10–2
c. + 3 V d. + 6 V cm2 that can be used to convert the galvanometer into an
ammeter which can read a maximum of 5A current is:
40. In the following circuit E1 = 4V, R1 = 2Ω, E2 = 6V, R2 = (Specific resistance of the wire = 5 × 10–7 Ωm)
2Ω and R3 = 4Ω. The current i1 is: a. 9 b. 6 c. 3 d. 1.5
E1 R1
46. 100 mA current gives a full scale deflection in a
i2 R2 galvanometer of resistance 2Ω. The resistance connected
with the galvanometer to convert it into a voltmeter of 5 V
i2 E2 R3 range is:
a. 98 Ω b. 52 Ω
a. 1.6 A b. 1.8 A
c. 2.25 A d. 1 A c. 80 Ω d. 48 Ω

41. Determine the current in the following circuit: 47. A ammeter of range 10 mA has a coil of resistance 1Ω. To
10V 2Ω use it as voltmeter of range 10 volt, the resistance that
must be connected in series with it will be:
a. 999 Ω b. 99 Ω
c. 1000 Ω d. None of these
5V 3Ω
48. In the following figure ammeter and voltmeter reads 2
a. 1 A b. 2.5 A amp and 120 volt respectively. Resistance of voltmeter is:
c. 0.4 A d. 3 A 75 Ω
X Y
A
42. The figure shows a network of currents. The magnitude of
current is shown here. The current i will be:
15 A V
3A
a. 100 Ω b. 200 Ω
c. 300 Ω d. 400 Ω
i Potentiometer
5A
a. 3 A b. 13 A 49. A battery with negligible internal resistance is connected
c. 23 A d. – 3 A with 10m long wire. A standard cell gets balanced on 600
cm length of this wire. On increasing the length of
43. Consider the circuit shown in the figure. The current i3 is
potentiometer wire by 2m then the null point will be
equal to:
displaced by:
28 Ω 54 Ω
a. 200 cm b. 120 cm
c. 720 cm d. 600 cm
6V
i3 50. In the following circuit a 10 m long potentiometer wire
8V 12V
with resistance 1.2 ohm/m, a resistance R1 and an
a. 5 amp b. 3 amp accumulator of emf 2 V are connected in series. When the
c. – 3 amp d. – 5/6 amp emf of thermocouple is 2.4 mV then the deflection in

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

352 Quick Revision NCERT-PHYSICS

galvanometer is zero. The current supplied by the NCERT EXEMPLAR PROBLEMS
accumulator will be:
More than One Answer
R1 56. Heater of an electric kettle is made of a wire of length L
i
and diameter d. It takes 4 minutes to raise the temperature
5m
A B of 0.5 kg water by 40 K. This heater is replaced by a new
G heater having two wires of the same material, each of
length L and diameter 2d. The way these wires are
connected is given in the options. How much time in
Hot Junction Cold Junction
minutes will it take to raise the temperature of the same
a. 4 × 10 A
–4
b. 8 × 10–4 A amount of water by 40 K?
c. 4 × 10–3 A d. 8 × 10–3 A a. 4 if wires are in parallel b. 2 if wires are in series
c. 1 if wires are in series d. 0.5 if wires are in parallel
51. The resistivity of a potentiometer wire is 40 × 10–8 Ωm
and its area of cross section is 8 × 10–6 m2. If 0.2 amp. 57. Capacitor C1 of capacitance 1µF and capacitor C2 of
Current is flowing through the wire, the potential gradient capacitance 2µF are separately charged fully by a
will be: common battery. The two capacitors are then separately
a. 10–2 volt/m b. 10–1 volt/m allowed to discharge through equal resistors at time t = 0:
c. 3.2 × 10–2 volt/m d. 1 volt/m a. The current in each of the two discharging circuits is
zero at t = 0
52. A deniel cell is balanced on 125 cm length of a
b. The currents in the two discharging circuits at t = 0 are
potentiometer wire. When the cell is short circuited with a
equal but not zero
2 Ω resistance the balancing length obtained is 100 cm.
c. The currents in the two discharging circuits at t = 0 are
Internal resistance of the cell will be:
unequal
a. 1.5 Ω b. 0.5 Ω
d. Capacitor C1, loses 50% of its initial charge sooner than
c. 1.25 Ω d. 4/5 Ω
C2 loses 50% of its initial charge
53. A potentiometer wire of length 10 m and a resistance 30 Ω
58. A micro ammeter has a resistance of 100 Ω and full scale
is connected in series with a battery of emf 2.5 V and
range of 50µA. It can be used as a voltmeter or as a higher
internal resistance 5 Ω and an external resistance R. If the
range ammeter provided a resistance is added to it. Pick
fall of potential along the potentiometer wire is 50 µV/mm,
the correct range and resistance combination(s):
the value of R is: (in Ω)
a. 50 V range with 10 kΩ resistance in series
a. 115 b. 80
b. 10 V range with 200 kΩ resistance in series
c. 50 d. 100
c. 5 mA range with 1 Ω resistance in parallel
54. A 2 volt battery, a 15 Ω resistor and a potentiometer of d. 10 mA range with 1 Ω resistance in parallel
100 cm length, all are connected in series. If the resistance
59. When a potential difference is applied across, the current
of potentiometer wire is 5 Ω, then the potential gradient of
passing through:
the potentiometer wire is:
a. an insulator at 0 K is zero
a. 0.005 V/cm b. 0.05 V/cm
b. a semiconductor at 0 K is zero
c. 0.02 V/cm d. 0.2 V/cm
c. a metal at 0 K is finite
55. In an experiment to measure the internal resistance of a d. a p-n diode at 300 K is finite, if it is reverse biased
cell by potentiometer, it is found that the balance point is
60. For the circuit shown in the figure:
at a length of 2 m when the cell is shunted by a 5 Ω
resistance; and is at a length of 3 m when the cell is
1 2 kΩ R1
shunted by a 10 Ω resistance. The internal resistance of
the cell is, then:
a. 1.5 Ω b. 10 Ω 24V 6 kΩ R2 RL 1.5 kΩ
c. 15 Ω d. 1 Ω

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

Current Electricity 353

a. the current I through the battery is 7.5 mA 65. Two resistors having equal resistances are joined in series
b. the potential difference across RL is 18 V and a current is passed through the combination. Neglect
c. ratio of powers dissipated in R1and R2 is 3 any variation in resistance as the temperature changes. In
d. if R1 are interchanged, magnitude of the power a given time interval:
dissipated in RL will decrease by a factor of 9 a. equal amounts of thermal energy must be produced in
the resistors
61. For the resistance network shown in the figure, choose the
b. unequal amounts of thermal energy may be produced
correct options(s).
c. the temperature must rise equally in the resistors
P I2 2Ω S
2Ω 2Ω d. the temperature may rise equally in the resistors
1Ω 1Ω 66. Which of the following statements is/are correct?
4Ω 4Ω a. Resistance of the filament of a bulb is inversely
Q 4Ω T proportional to the power of the bulb
I1
12 V b. Resistance of the filament of a bulb is directly
proportional to the power of the bulb
a. The current through PQ is zero c. Higher is the wattage of bulb, higher is the current that
b. I1 = 3 A can be allowed to pass through a bulb
c. The potential at S is less than that at Q d. Higher is the wattage of bulb, lower is the current that
can be allowed to pass through a bulb
d. I 2 = 2 A
67. Which of the following statements, concerning Kirchoff’s
62. In the given circuit, the AC source has ω = 100 rad/s. laws, is/are correct?
Considering the inductor and capacitor to be ideal, the a. In a network of conductors, the algebraic sum of
correct choice(s) is:(are) currents meeting at any junction in the circuit is always
100 µ F 100 Ω equal to zero
b. Kirchoff’s first law is an alternative form of the law of
conservation of charge
0.5 H 50 Ω
c. In any closed mesh, the algebraic sum of the products
I 20V of the currents and resistances of the different parts of the
~ mesh is always equal to the algebraic sum of different
a. The current through the circuit, I is 0.3 A e.m.f.’s acting in the mesh.
b. The current through the circuit, I is 0.3 2 A d. none of the above statement is correct

c. The voltage across 100Ω resistor = 10 2 V 68. If three resistances R1 , R2 and R3 are joined in series

d. The voltage across 50Ω resistor = 10 V through a battery, then:

a. equivalent resistance > highest individual resistance
63. When same quantity of electricity is passes through b. current supplied by source = current in each resistance
different electrolytes simultaneously, the masses of c. applied potential difference is shared among the three
different element liberated are found to be proportional to resistance directly in their ratio
their: d. applied potential difference is shared among the three
a. electrochemical equivalent b. chemical equivalent resistance inversely in their ratio
c. atomic masses d. atomic numbers
69. If three resistances R1, R2 and R3 are joined in parallel
64. When a potential difference is applied between the ends of across some potential source, then:
a conductor: a. equivalent resistance < lowest individual resistance
a. the free electrons start moving b. applied potential difference = potential difference
b. the free electrons are accelerated continuously from the across each resistance
lower potential end to higher potential end of the conductor c. current supplied by the source is shared among three
c. the free electrons acquire a constant drift velocity from resistance inverselyin their ration
the lower potential end to the higher potential end d. current supplied by the source is shared among three
d. ions starts vibrating with a larger amplitude resistances directly in their ratio

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

354 Quick Revision NCERT-PHYSICS

70. In potentiometer experiments: Reason: The resistance of a metallic conductor increases
a. the wire must be uniform and must be very much long with increase in temperature.
b. e.m.f. of the battery in main circuit must be greater than
75. Assertion: A 60 W bulb is connected in parallel with a
e.m.f. to be measured
room heater and this combination is connected across the
c. +ve terminals of the battery and cell must be connected
mains. If the 60 W bulb is replaced by a 100 W bulb, the
at the same point
heat produced by the heater will remain the same.
d. none of above
Reason: When resistance are connected in parallel across
71. A wire of resistance R is cut into ‘n’ equal parts. These the mains, the heat produced is inversely proportional to
parts are then connected in parallel. The equivalent the resistance.
resistance of the combination will be:
76. Assertion: If an electric field is applied to a metallic
R n R conductor, the free electrons experience a force but do not
a. nR b. c. d.
n R n2 accelerate; they only drift at a constant speed.
72. In the adjoining circuit, the e.m.f. of the cell is 2 volt and Reason: The force exerted by the electric field is
the internal resistance is negligible. The resistance of the completely balanced by the Coulomb force between
voltmeter is 80 ohm. The reading of the voltmeter will be: electrons and protons.

2V
77. Assertion: In the metre bridge experiment shown in Fig.,
80 Ω the balance length AC corresponding to null deflection of
V the galvanometer is x. If the radius of the wire AB is
doubled, the balanced length becomes 4x.
20 Ω 80 Ω

a. 0.80 volt b. 1.60 volt

R1 R2
c. 1.33 volt d. 2.00 volt
73. A steady current flows in a metallic conductor of non- G
uniform cross-section. The quantity/ quantities constant A B
x C
along the length of the conductor is/are:
Reason: The resistance of a wire is inversely proportional
a. Current, electric field and drift speed
to the square of its radius.
b. Drift speed only
c. Current and drift speed d. Current only 78. Assertion: In the potentiometer circuit shown in Fig. E1
and E2 are the emfs of cells C1 and C2 respectively with
Assertion and Reason
E1 > E2. Cell C1 and C2 respectively with E1 > E2. Cell C1
Note: Read the Assertion (A) and Reason (R) carefully to mark has negligible internal resistance. For a given resistor R,
the correct option out of the options given below: the balance length is x. If the diameter of the
a. If both assertion and reason are true and the reason is the potentiometer wire AB is increased, the balance length x
correct explanation of the assertion. will decrease.
b. If both assertion and reason are true but reason is not the
correct explanation of the assertion. C1 R
c. If assertion is true but reason is false.
x D
d. If the assertion and reason both are false. A B
e. If assertion is false but reason is true. C2 G

74. Assertion: Figure shows the current-voltage (I–V) graphs

Reason: At the balance point, the potential difference
for a given metallic wire at two different tempera-tures T1
and T2. It follows from the graphs that T2 is greater than between AD due to cell C1= E2, the emf of cell C2.
T1 . 79. Assertion: In the potentiometer circuit shown in above
I T1 question, the wire AB is not changed but the value of resistor
R is decreased. Then the balance length x will decrease.
T2
Reason: At the balance point, the potential difference
between A and D due to cell C1 = emf E2 of cell C2.
V

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

Current Electricity 355

80. Assertion: Electric appliances with metallic body have resistance R and a valve in the balloon is opened.
three connections, whereas an electric bulb has a two pin The gas inside the balloon escapes from the valve
connection. and the size of the balloon decreases. The rate of
Reason: Three pin connections reduce heating of decrease in radius of the balloon is controlled in
connecting wires. such a manner that potential of the balloon remains R
constant. Assume the electric permittivity of the
81. Assertion: Bending a wire does not effect electrical
surrounding air equals to that of free space (ε0) and
resistance.
charge cannot leak to the surrounding air.
Reason: Resistance of wire is proportional to resistivity of
material. 88. Rate at which radius r of the balloon changes with time is
best represented by the equation:
82. Assertion: Electric field outside the conducting wire
dr 1 dr 1
which carries a constant current is zero. a. = b. =−
Reason: Net charge on conducting wire is zero. dt 4πε 0 R dt 4πε 0 R
dr r dr r
83. Assertion: The e.m.f. of the driver cell in the c. = d. =−
dt 4πε 0 aR dt 4πε 0 aR
potentiometer experiment should be greater than the e.m.f.
of the cell to be determined. 89. How much heat is dissipated in the resistance R till the
Reason: The fall of potential across the potentiometer wire moment radius of the balloon becomes half?
should not be less than the e.m.f. of the cell to be a. 0.5πε 0 aV02 b. πε 0 aV02
determined.
c. 2πε 0 aV02 d. 4πε 0 aV02
84. Assertion: The temperature coefficient of resistance is
Paragraph -II
positive for metals and negative for p-type semiconductor.
A thermal power plant produces electric power of 600 kW at
Reason: The effective charge carriers in metals are
4000 V, which is to be transported to a place 20 km away from
negatively charged whereas in p-type semiconductor they
the power plant for consumers’ usage. It can be transported
are positively charged.
either directly with a cable of large current carrying capacity or
85. Assertion: In a simple battery circuit the point of lowest by using a combination of step-up and step-down transformers
potential is positive terminal of the battery at the two ends. The drawback of the direct transmission is the
Reason: The current flows towards the point of the higher large energy dissipation. In the method using transformers, the
potential as it flows in such a circuit from the negative to dissipation is much smaller. In this method, a step-up
the positive terminal. transformer is used at the plant side so that the current is
86. Assertion: The resistivity of a semiconductor increases reduced to a smaller value. At the consumers’ end, a step-down
with temperature. transformer is used to supply power to the consumers at the
Reason: The atoms of a semiconductor vibrate with larger specified lower voltage. It is reasonable to assume that the
amplitude at higher temperatures thereby increasing its power cable is purely resistive and the transformers are ideal
resistivity with the power factor unity. All the currents and voltage
mentioned are rms values.
87. Assertion: The drift velocity of electrons in a metallic
90. If the direct transmission method with a cable of
wire will decrease, if the temperature of the wire is
resistance 0.4 Ω km–1 is used, the power dissipation (in %)
increased.
during transmission is:
Reason: On increasing temperature, conductivity of
a. 20 b. 30 c. 40 d. 50
metallic wire decreases.
91. In the method using the transformers, assume that the ratio
Comprehension Based of the number of turns in the primary to that in the
Paragraph –I secondary in the step-up transformer is 1 : 10. If the power
A conducting balloon of radius ‘a’ is charged to a potential V0 to the consumers has to be supplied at 200 V, the ratio of
and held at a large height above the earth surface. The large the number of turns in the primary to that in the secondary
height of the balloon from the earth ensures that charge in the step-down transformer is:
distribution on the surface of the balloon remains unaffected by a. 200 : 1 b. 150 : 1
the presence of the earth. It is connected to the earth through a c. 100 : 1 d. 50 : 1

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

356 Quick Revision NCERT-PHYSICS

Match the Column Integer
92. Match the statement of Column with those in Column II: 95. The length of a potentiometer wire is 10 cm. A cell of emf
Column I Column II 10
(A) When the length and 1. For metallic E is balanced at a length cm from the positive end of
3
area of cross-section conductors at low
the wire. If the length of the wire is increased by 5cm, at
both are doubled, then temperature
what distance (in cm) from positive end will the same cell
its resistance
give a balance point?
(B) The resistivity of a wire 2. does not depend on
the length and the 96. Fig., for what value of R (in ohm) will the current in
cross-section of wire galvanometer G be zero?
(C) Ohm's law is true 3. Will remain the same 5Ω

(D) The example for non- 4. Diode 7V

ohmic resistance is R 12V
a. A→3, B→2, C→1, D→4
b. A→4, B→1, C→3, D→2
c. A→1, B→2, C→3, D→4
3V, 0.5W
d. A→2, B→1, C→3, D→4
93. Match the statement of Column with those in Column II: 4Ω 6V

Column I Column II 0.5Ω

(A) Through a semiconductor, an 1. Temperature R
electric current is due to drift of
(B) The specific resistance of all 2. Free electrons 97. A battery of emf E and internal resistance r is connected
metals is most affected by and holes to an external resistance R. The maximum power in the
(C) The positive temperature 3. Copper external circuit is 9W. The current flowing in the circuit
coefficient of resistance is for under the condition is 3A. What is the value of E in volt?
(D) The fact that the conductance of 4. 98. In the circuit, Fig., what should be the value of r in ohm
some metals rises to infinity at Superconductivity
so that power developed in the resistor r will be
some temperature below a few
maximum?
Kelvin is called r
a. A→4, B→2, C→3, D→1
b. A→3, B→1, C→4, D→2
2Ω 4Ω
c. A→2, B→1, C→3, D→4
d. A→2, B→1, C→4, D→3
3Ω V
94. Match the statement of Column I with those in Column II.
Column I Column II
(A) Dielectric ring uniformly 1. Time independent 99. When two identical batteries of internal resistance 1Ω
charged electrostatic field out each are connected in series across a resistor R, the rate of
of system heat produced in R is J1. When the same batteries are
(B) Dielectric ring uniformly 2. Magnetic field connected in parallel across R, the rate is J2. If J1 = 2.25
charged rotating with J2, then the value of R in ohm is:
angular velocity ω 100. A wire of length L and 3 identical cells of negligible
(C) Constant current in ring i0 3. Induced electric field internal resistances are connected in series. Due to this
(D) i = i0 cos ω t 4. Magnetic moment current, the temperature of the wire is raised by ∆T in time
a. A→ 1; B → 2,4; C → 2,4; D → 2,3,4 t. A number N of similar cells is now connected in series
b. A→ 2; B → 3,4; C → 1,3; D → 1,2,3 with a wire of the same material and cross-section but of
c. A→ 3; B → 1,3; C → 1,2; D → 2,3,4 length 2L. The temperature of wire is raised by same
d. A→ 4; B → 3,4; C → 2,3; D → 1,2,3 amount ∆T in the same time t. The value of N is:

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

Current Electricity 357

ANSWER r
2
neπ r 2 v i
⇒ i ' = neπ   ⋅ 2v = =
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 2 2 2
a b c b b c a d c d
7. (a) As i = neAvd
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c d a b c a b d b b i1 A1 vd1 r12 vd1 vd 16
⇒ = × = ⋅ ⇒ 1 =
21. 22. 23. 24. 25. 26. 27. 28. 29. 30. i2 A2 vd2 r22 vd2 vd 2 1
a c d a c b c d a a
31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 8. (d) R = 91×102 ± 10% ≈ 9.1 k Ω
c d a a a b b d a b
9. (c) Suppose at toC resistances of the two wires becomes
41. 42. 43. 44. 45. 46. 47. 48. 49. 50.
R1t and R2t respectively and equivalent resistance
a c d b c d a c b a
51. 52. 53. 54. 55. 56. 57. 58. 59. 60. becomes Rt. In series grouping Rt = R1t + R2t, also R1t = R1
a b a a b b,d b,d b,c a,b,d a,d (1 + α1t) and R2t = R2 (1 + α2t)
61. 62. 63. 64. 65. 66. 67. 68. 69. 70. Rt = R1(1 + α1t) + R2(1 + α2t)
a,b,d a,c a,b c,d a,d a,c a,b,c a,b,c a,b,c a,b,c = (R1 + R2) + (R1α1 + R2α2)t
71. 72. 73. 74. 75. 76. 77. 78. 79. 80.  R α + R2α 2 
c c d a a c d d a e = ( R1 + R2 ) 1 + 1 1 t
 R1 + R2 
81. 82. 83. 84. 85. 86. 87. 88. 89. 90.
b a c b d d b b c b R1α1 + R2α 2
Hence effective temperature co-efficient is .
91. 92. 93. 94. 95. 96. 97. 98. 99. 100. R1 + R2
a a c a 5 7 6 2 4 6 4
R1  r2 
10. (d) By using = 
SOLUTION R2  r1 
Multiple Choice Questions 4
R  r/2
⇒ =  ⇒ R ' = 16 R
q ne R'  r 
1. (a) i = =
t t 11. (c) As we know, for conductors resistance ∝ Temperature.
−3
it 3.2 × 10 × 1 From figure R1 ∝ T1
⇒ n= = = 2 × 1016
e 1.6 × 10 −19 ⇒ tanθ ∝ T1
q q qv nev ⇒ tanθ = kT1 . . . (i)
2. (b) i = = = =
t ( x / v) x x (k = constant) and R2 ∝ T2
ix 1.6 × 10 −6 × 1 ⇒ tan (90o – θ) ∝ T2
⇒ n= = = 107
ev 1.6 × 10−19 × 106 ⇒ cotθ = kT2 . . . (ii)
From equation (i) and (ii) k (T2 − T1 ) = (cot θ − tan θ )
3. (c) ν = 6.8 × 1015
 cos θ sin θ  (cos θ − sin θ )
2 2
1 (T2 − T1 ) =  − =
⇒ T= sec  sin θ cos θ  sin θ cos θ
6.8 ×1015
cos 2θ
Q = = 2 cot 2θ ⇒ (T2 – T1) ∝ cot 2θ
⇒ i= = 1.6 × 10 −19 × 6.8 × 1015 = 1.1 × 10 −3 A sin θ cos θ
T
R1 (1 + α t1 )
i 1 12. (d) By using =
4. (b) We know J = when i = constant J ∝ R2 (1 + α t2 )
A A
20 1 + 20α 1
J c Ai  ri   3 
2 2 ⇒ = ⇒α =
9 60 1 + 500α 220
⇒ = =  =  =
J i Ac  rc   1  1 Again by using the same formula for 20Ω and 25Ω
 1 
i 24 × 10−3 1+ × 20 
5. (b) vd = = = 5 × 10 −3 m / s 20  220  ⇒ t = 80°C.
neA 3 × 1023 × 1.6 ×10−19 ×10−4 ⇒ =
25  1 
(c) i = neAvd = neπ r 2v 1 + ×t 
6.  220 

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

358 Quick Revision NCERT-PHYSICS

−8 −2
l 50 ×10 × 50 ×10 E 
13. (a) R = ρ = = 10−6 Ω 22. (c) By using r =  − 1 R
A (50 ×10−2 )2 V 
2 
l R l A ⇒ 0.1 =  − 1 × 3.9 ⇒ V = 1 . 95 volt
14. (b) By using R = ρ ⋅ ; disc = disc × rod V 
A Rrod lrod Adisc
Rdisc 10−3 π (0.3 × 10−2 )2 23. (d) Series resistance RS = R1 + R2 and parallel resistance
⇒ = ×
3 × 10−3 1 π (10−2 )2 R1 R2
RP =
⇒ −7
Rdisc = 2.7 × 10 Ω R1 + R2
RS ( R1 + R2 ) 2 R1 + R2
l ⇒ = =n ⇒ = n
15. (c) By using R = ρ ⋅ ⇒ l ∝ A RP R1 R2 R1 R2
A
3.14 3.14 × 3.14 ×10−6 R12 R22 R1 R2
⇒ = ⇒ + = n ⇒ + = n
1 π × r2 R1 R2 R1 R2 R2 R1
⇒ r = 10–3 , m = 1 mm
24. (a) The equivalent circuit of above can be drawn as.
16. (a) In case of streching R ∝ l2 Which is a balanced wheatstone bridge. So current
∆R ∆l through AB is zero.
So, = 2 = 2 × 0.1 = 0.2
R l A 10 Ω
Y
17. (b) By using Rt = Ro (1 + α∆t) 10 Ω
20 Ω
R1 1 + α t1 10 Ω
⇒ =
R2 1 + α t2 X
B
10 Ω
1 1 + (300 − 273)α
So, = ⇒ t2 = 854°C = 1127 K 1 1 1 1
2 1 + α t2 So, = + =
R 20 20 10
l l2 l2  m ⇒ R = 10 Ω
18. (d) R = ρ = ρ = ρ σ ∵σ = 
A V m  V 25. (c) The equivalent circuit of above fig between A and D
l2
l2
l 2
9 1 can be drawn as
R1 : R2 : R3 = 1
: 2
: = 25 : : = 125 :15 :1
3
Balanced
m1 m2 m3 3 5 10 10
Wheatstone bridge
A
−2
19. (b) Length l = 1 cm = 10 m
10 10
Area of cross-section A = 1 cm × 100 cm
= 100 cm2 = 10–2 m2 D
−2
10 10
10
Resistance R = 3 ×10−7 × = 3 ×10−7 Ω
10−2 10
A Series
E ⇒
20. (b) By using i = 10
R+r
1.5 D
⇒ 15 = ⇒ r = 0.06 Ω
0.04 + r 10

So, R eq = 10 + 10 + 10 = 30 Ω
21. (a) In open circuit, E = V = 2.2 V
In close circuit, V = 1.8 V, R = 5Ω 26. (b) Taking the portion COD is figure to outside the
E   2.2  triangle (left), the above circuit will be now as resistance
So, Internal resistance, r =  − 1 R =  − 1 × 5
V   1.8  of each is 2Ω the circuit will behaves as a balanced
10 wheatstone bridge and no current flows through CD.
⇒ r= Ω Hence RAB = 2Ω
9

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

Current Electricity 359

27. (c) Suppose the effective resistance between A and B is 31. (c) Equivalent resistance of the following circuit will be
Req. Since the network consists of infinite cell. If we 6
R eq = = 3 Ω
exclude one cell from the chain, remaining network have 2
infinite cells i.e. effective resistance between C and D will 6Ω
also Req
R R
A C 6Ω

R Req 32. (d) No. of cells in a row n = 10; No. of such rows m = 5
2V R = (R1 || R2) = 2Ω
B D
R Req
So, Now Req = Ro + ( R || Req ) = R + 2V
R + Req A B
1 nE 10 × 1.5 15
⇒ Req = [1 + 5] ⇒ i= = = = 0.68 amp
2  nr   10 × 1  22
 R +   20 + 
28. (d) In the circuit, by means of symmetry the point C is at  m  5 
zero potential. The equivalent circuit can be drawn as
33. (a) In this question R = 3Ω, mn = 24, r = 0.5Ω and
r Series (r S r) = 2r r Parallel
mr
R= . On putting the values we get n = 2 and m = 12.
n
r r r r ⇒ r 2r r
34. (a) Total no. of cells, = mn = 100 . . . .(i)
A B A B nr n ×1
r r r r Series Current will be maximum when R = ; 25 =
⇓ m m
2 8
r+r+ r = r
Series ⇒ n = 25 m . . . (ii)
3 3 2/3 r
From equation (i) and (ii) n = 50 and m = 2
⇐ r r
nE E
35. (a) Current in the circuit i = =
A B A B nr r
2r 2r The equivalent circuit of one cell is shown in the figure.
 8r  8
⇒ Req =  || 2r  = r Potential difference across the cell
 3  7
E
P R = V A − VB = − E + ir = − E + .r = 0
29. (a) Given Wheatstone bridge is balanced because = . r
Q S
36. (b) For zero deflection in galvanometer the potential
Hence the circuit can be redrawn as follows
different across X should be E = 2V
3Ω 4Ω Series 3 + 4 = 7 Ω 12 X
In this condition =2
500 + X
A B
∴ X = 100 Ω
6Ω 8Ω Series 6 + 8 = 14 Ω
37. (b) The equivalent circuit can be drawn as since E1 & E3
14 are connected in parallel,
7Ω Parallel Req = Ω
3
2+2
So, Current i = = 2 Amp from A to B.
A B 2
14Ω
10 − 4
38. (d) Current i = = 1 A from a to b via e
9 3 + 2 +1
30. (a) Main current through the battery i = = 9 A . Current
1
39. (a) The given part of a closed circuit can be redrawn as
through each resistance will be 1A and only 5 resistances
follows. It should be remember that product of current and
on the right side of ammeter contributes for passing
resistance can be treated as an imaginary cell having emf
current through the ammeter. So reading of ammeter will
= iR.
be 5A.

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

360 Quick Revision NCERT-PHYSICS

4V 3V 2V 45. (c) Given G = 50 Ω, ig = 0.05 Amp., i = 5A,
A B
9V
A = 2.97 × 10–2 cm2 and ρ = 5 × 10–7Ω-m
⇒ A B i G
By using = 1+
⇒ 7 Hence VA – VB = +9 V ig S

40. (b) For loop (1) −2i1 − 2(i1 − i2 ) + 4 = 0 G.ig ρl Gig Gig A
⇒ S = ⇒ = ⇒ l=
4V 2Ω (i − ig ) A (i − ig ) (i − ig )ρ

i1 i1
On putting values l = 3 m.
1 2Ω (i2 – i2)
V 5
46. (d) R = −G = − 2 = 50 − 2 = 48 Ω
i2 2 i2 Ig 100 ×10−3

4Ω V
6V 47. (a) By using R = −G
ig
⇒ 2 i1 − i2 = 2 . . . (i)
10
For loop (2) −4 i2 + 2(i1 − i2 ) + 6 = 0 ⇒ R= − 1 = 999Ω
10 × 10 −3
⇒ 3 i2 − i1 = 3 . . . (ii)
48. (c) Let resistance of voltmeter be RV. Equivalent
After solving equation (i) and (ii) we get i1 = 1 . 8 A and
75 RV
i2 = 1 . 6 A resistance between X and Y is RXY =
75 + RV
41. (a) Similar plates of the two batteries are connected Reading of voltmeter = potential difference across X and Y
together, so the net emf = 10 – 5 = 5V = 120 = i × RXY
Total resistance in the circuit = 2 + 3 = 5Ω 75 RV
ΣV 5 = 2× ⇒ RV = 300Ω
∴ i= = = 1A 75 + RV
ΣR 5
L1 l
42. (c) i = 15 + 3 + 5 = 23 A 49. (b) By using = 1
L2 l2
43. (d) Suppose current through different paths of the circuit 10 600
⇒ = ⇒ l2 = 720 cm.
is as follows. 12 l2
28 Ω 54 Ω
Hence displacement = 720 – 600 = 120 cm

50. (a) E = x l = iρ l
1 6V 2
E E 2.4 × 10−3
i3 ∴ i= = = = 4 × 10−4 A.
ρl ρl 1.2 × 5
8V 12V
V iR i ρ L i ρ
After applying KVL for loop (1) and loop (2) 51. (a) Potential gradient = = = =
L L AL A
1
We get 28i1 = −6 − 8 ⇒ i1 = − A 0.2 × 40 ×10−8
2 = = 10−2 V / m
8 ×10−6
1
and 54i2 = −6 − 12 ⇒ i2 = − A
3 l1 − l2
52. (b) By using r = ×R'
5 l2
Hence i3 = i1 + i2 = − A
6 125 − 100 1
⇒ r= × 2 = = 0.5 Ω
44. (b) Current sensitivity of galvanometer = 4 ×10−4 Amp/div. 100 2
So full scale deflection current (ig) = Current sensitivity ×
e R
Total number of division = 4 × 10–4 × 25 = 10–2 A 53. (a) By using x = ⋅
( R + Rh + r ) L
To convert galvanometer in to voltmeter, resistance to be
V 2.5 50 × 10−6 2.5 30
put in series is R = − G = −2 − 100 = 150 Ω ⇒ −3
= × ⇒ R = 115
ig 10 10 (30 + R + 5) 10

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

Current Electricity 361

e R 6 × 1.5
54. (a) By using x = ⋅ 59. (a, b, d) Rtotal = 2 + = 3.2 k Ω
( R + Rh + r ) L 6 + 1.5
M 2Ω P 2Ω S 2Ω
2 5
⇒ x= × = 0.5V /m = 0.005V /cm I2
(5 + 15 + 0) 1
N Q T
I3 4Ω 4Ω 4Ω
l −l  I1
55. (b) By using r =  1 2 R'
 l2 
l − 2 12 V
⇒ r =  1  × 5 . . . (i) 24V
 2  (a) I = = 7.5mA = I R1
3.2k Ω
l − 3
r =  1  × 10 . . . (ii)  RL 
 3  ⇒ I R2 =  I ⇒ I =
1.5
× 7.5 = 1.5 mA
 RL + R2
7.5

On solving (i) and (ii) r = 10 Ω
⇒ I RL = 6mA
NCERT Exemplar Problems (b) VRL = ( I RL )( RL ) = 9V
More than One Answer
PR1 ( I R2 ) R1 (7.5) 2 (2) 25
V2 V2 V2 (c) = = =
56. (b, d) H = 4= t1 t2 PR2 ( I R2 ) R2 (1.5) 2 (6) 3
R R / 2 R /8
⇒ t1 = 2 min. ⇒ t2 = 0.5 min . (d) When R1 and R2 are inter changed, then
R2 RL 2 × 1.5 6
= = k Ω . Now potential difference
57. (b, d) The discharging current in the circuit is, i = i0 e −t / CR R2 + RL 3.5 7
V  6/7 
Here, i0 = initial current = across RL will be VL = 24 
R  = 3V
6 + 6/7
Here, V is the potential with which capacitor was charged.
Earlier it was 9 V
+ –
V2
Since, P = or P ∝ V 2
R
In new situation potential difference has been decreased
R three times. Therefore, power dissipated will decrease by
a factor of 9.
Since, V and R for both the capacitors are same, initial
discharging current will be same but non-zero. 60. (a, d) the current I through the battery is 7.5 mA. if R1 are
∴ Correct option is (b). interchanged, magnitude of the power dissipated in RL will
Further, τ C = CR C1 < C2 or τ C1 < τ C2 decrease by a factor of 9.

Or C1 loses its 50% of initial charge sooner than C2. 61. (a, b, d) Due to symmetry on upper side and lower side,
points P and Q are at same potentials. Similarly, points S
58. (b, c) To increase the range of ammeter a parallel
and T are at same potentials. Therefore, the simple circuit
resistance (called shunt) is required which is given by
can be drawn as shown below
 i  12 12
S = g  G I2 = = 2 A or I3 = = 1A
 i − ig
  2+2+2 4+4+4
 50 × 10−6  ∴ I1 = I 2 + I 3 = 3 A ⇒ I PQ = 0
For option (c) S =  −3 −6 
(100) ≈ 1Ω
 5 × 10 − 50 ×10  Because VP = VQ
To change it in voltmeter, a high resistance R is put in Potential drop (from left to right) across each resistance is
V 12
series, where R is given by R = − G = 4V
ig 3
10 ∴ VMS = 2 × 4 = 8V ⇒ VNQ = 1× 4 = 4V or VS < VQ
For option (b) R = − 100 ≈ 200k Ω
50 × 10 −6

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

362 Quick Revision NCERT-PHYSICS

1 wattage of bulb, higher is the current that can be allowed
62. (a, c) Circuit 1- X C = = 100Ω
ωC to pass through a bulb.
100 µ F 100 Ω 67. (a,b,c) In a network of conductors, the algebraic sum of
I1 Z1 currents meeting at any junction in the circuit is always
Z2 equal to zero. Kirchoff’s first law is an alternative form of
I2 0.5 H 50 Ω
the law of conservation of charge. In any closed mesh, the
I 20V algebraic sum of the products of the currents and
~ resistances of the different parts of the mesh is always equal
2 2
to the algebraic sum of different e.m.f.’s acting in the mesh.
∴ Z1 = (100) + (100) = 100 2Ω
68. (a,b,c) equivalent resistance > highest individual
−1 
R  resistance. current supplied by source = current in each
φ1 = cos  1  = 45°
 Z1  resistance. applied potential difference is shared among
In this circuit current leads the voltage. the three resistance directly in their ratio.
V 20 1 69. (a,b,c) equivalent resistance < lowest individual resistance.
I1 = = = A
Z1 100 2 5 2 applied potential difference = potential difference across
1 each resistance. current supplied by the source is shared
V100Ω = (100) I1 = (100) V = 10 2V
5 2 among three resistance inverselyin their ration.
Circuit 2- X L = ω L = (100)(0.5) = 50Ω 70. (a, b, c) the wire must be uniform and must be very much
long. e.m.f. of the battery in main circuit must be greater
R 
Z 2 = (50) 2 + (50)2 = 50 2Ω ⇒ φ2 = cos −1  2  = 45° than e.m.f. to be measured. +ve terminals of the battery
 Z2  and cell must be connected at the same point.
In this circuit voltage leads the current. 71. (c) Let equivalent resistance between A and B be R, then
V 20 2 equivalent resistance between C and D will also be R.
I2 = = = A 1Ω
Z 2 50 2 5 A C

 2
V50 Ω = (50) I 2 = 50  = 10 2V 1Ω
 5 
R R
  1Ω
B D
Further, I1 and I2 have a mutual phase difference of 90°
R
1 4 1 ⇒ R′ = + 2 = R or R 2 − 2R − 2 = 0
∴ I = I12 + I 22 = + = A ≈ 0.3 A R +1
50 50 10
2± 4+8
∴ R= = 3 +1
m W Z 2
63. (a, b) According to faraday’s laws: 1 = 1 = 1
m2 W2 Z 2 80
72. (c) Total resistance of the circuit = + 20 = 60 Ω
64. (c, d) When a potential difference is applied between the 2
two ends of a conductor, the electrons at every point of 2 1
⇒ Main current i = = A
conductor are accelerated but for a very short interval of 60 30
time (= τ ) . After that electrons always collide with some Combination of voltmeter and 80Ω resistance is
+ ve ion and lose their velocity i.e. electrons move with an connected in series with 20Ω,
average velocity called as drift velocity and as a result of So, current through 20Ω and this combination will be same
their collisions with + ve ions amplitude of vibrations of + 1
ve ions is increased. = A . Since the resistance of voltmeter is also 80Ω, so
30
65. (a, d) equal amounts of thermal energy must be produced this current is equally distributed in 80Ω resistance and
in the resistors. the temperature may rise equally in the 1
voltmeter (i.e. A through each)
resistors. 60
66. (a,c) Resistance of the filament of a bulb is inversely 1
P.D. across 80 Ω resistance = × 80 = 1.33V
proportional to the power of the bulb. Higher is the 60

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

Current Electricity 363

73. (d) If E be electric field, then current density j = σ E 81. (b) Neutral temperature is the temperature of hot junction,
i at which the thermo e.m.f. produced in the thermocouple
Also we know that current density j = Hence j is becomes maximum. It is independent of cold junction and
A
depends on the nature of materials of two metals used to
different for different area of cross-sections. When j is
form thermocouple.
different, then E is also different. Thus E is not constant.
j 82. (a) When lamp B or C gets fused equivalent resistance of
The drift velocity vd is given by vd = = different for
ne B and C increases. In series voltage distributes in the ratio
different j values. Hence only current i will be constant. of resistance, so voltage appears across B increases or in
other words voltage across A decreases.
Assertion and Reason
83. (c) The electrical appliances with metallic body like
74. (a) From Ohm’s law, the slope of the I – V gives the
heater, press etc. have three pin connections. Two pins are
reciprocal of the resistance of the wire.
for supply line and third pin is for earth connection for
Since the slope of the graph is smalle at temperature T2,
safety purposes.
the resistance of the wire is greater at temperature T2 than
at temperature T1. Hence T2 is greater than T1. 84. (b) The temperature co-efficient of resistance for metal is
positive and that for semiconductor is negative. In metals
75. (a) When a bulb and a heater a connected in parallel and
free electrons (negative charge) are charge carriers while
this combination is connected across the main, potential
in P-type semiconductors, holes (positive charge) are
difference across each is the same equal to the voltage V
majority charge carriers.
of the mains, irrespective of the resistance of the bulb. If R
is the resistance of the heater, the rate at which heat is 85. (d) It is quite clear that in a battery circuit, the point of
produced will be V2/R in both cases. lowest potential is the negative terminal of the battery and
the current flows from higher potential to lower potential.
76. (c) The electrons suffer a large number of collisions with
the positive ions of the conductor. Although the electric 86. (d) Resistivity of a semiconductor decreases with the
field accelerates an electron between two collisions, it is temperature. The atoms of a semiconductor vibrate with
decelerated by collision. The net acceleration averages out larger amplitudes at higher temperatures thereby
to zero and the electron acquires a constant average speed. increasing it's conductivity not resistivity.
The gain in speed between collisions is lost in the next 87. (b) On increasing temperature of wire the kinetic energy of
collision free electrons increase and so they collide more rapidly with
77. (d) The condition for no deflection of the galvanometer is each other and hence their drift velocity decreases. Also
R1 RAC when temperature increases, resistivity increase and resistivity
= where RAC and RCB are the resistances of the
R2 RCB is inversely proportional to conductivity of material.
bridge wire of length AC and CB respectively.If the radius Comprehension Based
of the wire AB is doubled, the ratio RAC/RCB will remain dr 1
unchanged. Hence the balance length will remain the 88. (b) =−
dt 4πε 0 R
same.
89. (c) 2πε 0 aV02
78. (d) If the diameter of wire AB is increased, its resistance
will decrease. Hence, the potential difference between A 90. (b) For direction transmission P = i 2 R = (150) 2 (0.4 × 20)
and B due to cell C1 will decrease. Therefore, the null
= 1.8 ×105 w
point will be obtained at a higher value of x.
1.8 × 105
79. (a) If the value of R is decreased, the potential difference Fraction (in %) = × 100 = 30%
6 × 105
between A and B due to cell C1 will increase. Hence the
91. (a) 40000 / 200 = 200
balance length will be smaller than x.
Match the Column
80. (e) Voltage of dc source is constant but in ac, peak value
92. (a) A→3, B→2, C→1, D→4
of voltage is 2 times the rms. voltage. Hence bulb will
glow with more brightness when connected to an ac 93. (c) A→2, B→1, C→3, D→4
source of the same voltage. 94. (a) A→ 1; B → 2,4; C → 2,4; D → 2,3,4

WWW.EXAMBAT.IN
 Join our Telegram channel: @Jee_tricks

364 Quick Revision NCERT-PHYSICS

Integer Using Kirchhoff’s second law, in closed circuit ABEFA,
95. (5) Let E0 be the potential difference applied across the we have, 6 I1 + 3 ( I1 + I 2 ) − V = 0
total length l (= 10 cm) of potentiometer wire. or V = 9 I1 + 3I 2 . . .(i)
E Using Kirchhoff’s second law, in closed circuit BCDEB,
Potential gradient in the first case = 0
l I2
we have rI 2 − 6 I1 = 0 or I1 = r
lE  E 6
As per question, E =  0  = 0 . . .(i)
3 l  3 I2 3 3I
From (i), V = 9 × r + 3I 2 = I 2 r + 3I 2 = 2 ( r + 2)
E0 2E 6 2 2
Potential gradient in second case = = 0 .
3l / 2 3l 2V
or I2 = Power developed in resistor r,
If x is the desired length of potentiometer to balance the 3( r + 2)
2 E0
emf E of the cell, then E = x × . . .(ii) 4V 2
3l P = I 22 r = ×r
9 ( r + 2)
2

E0 2E l 10
From (i) and (ii), we have = x × 0 or = = 5cm. Power developed is maximum, when ( r + 2) 2 is minimum
3 3l 2 2
( r + 2)
2
96. (7) When the current through galvanometer is zero, then or =0
Potential difference across R Ω = 7 V
or r 2 + 4r + 4 = 0
Potential difference across 5 Ω = 12 – 7 = 5 V
or r 2 − 4r + 4 + 8r = 0
5 7
Current in 5 Ω = = 1A , Resistance, R = = 7 Ω. or ( r − 2)
2
+ 8r = 0 or r = 2Ω.
5 1
97. (6) Refer Fig., there will be maximum power in the J1
external circuit if R = r. 99. (4) Given, = 2.25
J2
2 2
E, r  2E   2E 
⇒   R = J1 and   R = J2
I = 3A  2r + R   r + 2R 
( r + 2R )
2
R J1
∴ = = 2.25
( 2r + R )
2
Maximum power consumed, J2
2
 E  E2 2 + 2R 3
Pmax =   ×r = =9 …(i) Or = 1.5 =
r+r 4r 2r + R 2
E On solving, we get, R = 4r = 4 × 1 = 2Ω.
Current, I = = 3, E = 6 r …(ii)
r+r
Solving (i) and (ii), we get E = 6 V. 100. (6) Heat = mS∆T = i2Rt
Length (L) ⇒ Resistance = R and mass = m
98. (2) The equivalent circuit with the distribution of current
Length (2L) ⇒ Resistance = 2R and mass = 2m
is shown in Fig.
m1 S1∆T1 i12 R1t1
r So =
D C m2 S 2 ∆T2 i22 r2 t2
I2 I2
mS ∆T i 2 Rt
6Ω ⇒ = 21 ⇒ i1 = i2
E B 2mS ∆T i2 2 Rt
(I1 + I2) (I1 + I2) (3 E ) 2 ( NE ) 2
3Ω ⇒ = ⇒N=6
F A 12 2R
V

***

WWW.EXAMBAT.IN