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 CURRENT
ELECTRICITY

CURRENT ELECTRICITY

1. ELECTRIC CURRENT 3. DRIFT VELOCITY

“The flow of charge in a definite direction constitutes the . . .

electric current and the time rate of flow of charge “If u1 , u2 , u3 , ...un are random thermal velocities of n free
through any cross-section of a conductor is the measure electrons in the metal conductor, then the average thermal
of current”. i.e., velocity of electrons is
. . . . .
u1 + u + u3 + ...+ un
net charge flown q dq n = 0
Electric current , I= =
dt As a result, there will be no net flow of electrons of
1. =
Though the “electric current represents the direction of flow
charge in one particular direction in a metal conductor,
of positive charge”.
hence no current”.
2. Yet it is treated as a scalar quantity.
“Drift velocity is defined as the average velocity with which
3. Current follows, the laws of scalar addition and not the laws
the free electrons get drifted towards the positive end of
of vector addition.
the conductor under the influence of an external electric
4. Because the angle between the wires carrying currents
field applied”.
does not affect the total current in the circuit.
1. The drift velocity of electons is of the order of 10–4 ms–1.
2. CURRENT CARRIERS 2. If V is the potential difference applied across the ends of
the conductor of length l, the magnitude of electric field set
(a) Current carriers in solid conductors :
up is
1. In solid conductors like metals, the valence electrons of
the atoms do not remain attached to individual atoms but Potential difference V
are free to move throughout the volume of the conductor. E= length =l

2. Under the effect of an external electric field, the valence

electrons move in a definite direction causing electric
current in the conductors.
3. Thus, valence electrons are the current carriers in solid
conductors.
(b) Current carriers in liquids :
1. In an electrolyte like CuSO4, NaCl etc., there are
3. Each free electrons in the conductor experience a force,
positively and negatively charged ions (like Cu++ , SO–– ,
4
Na+ , Cl– ). F = –e E.
2. These are forced to move in definite directions under the 4. The acceleration of each electron is
effect of an external electric field, causing electric current. eE
3. Thus, in liquids, the current carriers are positively and a .= .
m
negatively charged ions.
5. At any instant of time, the velocity acquired by electron
(c) Current carriers in gases : .
having thermal velocity uwill be
1. Ordinarily, the gases are insulators of electricity.
2. They can be ionized by applying a high potential v.1 = u.1 + aτ1
difference at low pressure
3. Thus, positive ions and electrons are the current carriers in where τ1 is the time elapsed since it has suffered its last
gases. collision with ion/atom of the conductor.
 CURRENT
ELECTRICITY
6. Similarly, the velocities acquired by other electrons in the
conductor will be ✦ The drift velocity of electrons is small because of the frequent
. collisions suffered by electrons.
v = . . .
2 u 2 + aτ 2 v 3 = u 3 + aτ3 , ....., v n = un + ✦ The small value of drift velocity produces a large amount of
, aτn .
7. The average velocity of all the free electrons in the electric current, due to the presence of extremely large
conductor under the effect of external electric field is the number of free electrons in a conductor. The propagation
drift velocity of current is almost at the speed of light and involves
. electromagnetic process. It is due to this reason that
vd of the free electrons.
the
. . . electric bulb glows immediately when switch is on.
. + v2 + ...+
Thus, = ✦ In the absence of electric field, the paths of electrons between
v vn
v successive collisions are straight line while in presence of
n
. . . . . . electric field the paths are generally curved.
u + + u2 + +... un +
aτ
=1 aτ2 aτn N xd
✦ Free electron density in a metal is given by n
n = A
. A
. . +...+un ⎞ . τ + τ +...+ τ .
( +
u1 . +a 2 n = 0 + aτ = where N = Avogrado number, x = number of free
=
u2 τ a
| | A
n n
⎝ ⎠ electrons per atom, d = density of metal and A = Atomic
weight of metal.
τ + τ2 +...
where,
+ τn τ = = average time that has
n 1. Mobility of charge carrier (µ), responsible for current is
elapsed
since each electron suffered its last collision with the ion/ defined as the magnitude of drift velocity of charge per unit
atom of conductor and is called average relaxation time. electic filed applied, i.e.,
8. Its value is the order of 10–14 second.
9. Putting the value of a in the above relation, we have
 CURRENT
ELECTRICITY
drift velocity vd q E τ / m q τ
µ= = =
= m
electric field E E
. –e e τe
Eτ 2. Mobility of electron, µ =
vd = e
me
m 3. The total current in the conducting material is the sum of
eE the currents due to positive current carriers and negative
Average drift speed, =
vd τ current carriers.
m vd = µe E
.
The negative sign show that vd is opposite to the direction
. 4. SI unit of mobility is m2S–1V–1 or ms–1 N–1 C
of E .
3.1 Relaxation time (τ ) 3.3 Relation between current and Drift Velocity
The time interval between two successive collisions of 1. Consider a conductor (say a copper wire) of length l and
electrons with the positive ions in the metallic lattice is defined of uniform area of cross-section
mean free path λ Volume of the conductor = Al.
as relaxation time τ = = . With
r.m.s. velocity of electrons vrms 2. If n is the number density of electrons, i.e., the number of
rise in temperature vrms increases consequently τ decreases. free electrons perunit volume of the conductor, then total
3.2 Mobility number of free electrons in the conducture = Aln.
3. Then total charge on all the free electrons in the conductor,
Drift velocity per unit electric field is called mobility of electron
i.e.
q=
v 4. The electric field set up across the conductor is given by
µ = d It's unit m2 E = V/l (in magnitude)
is E volt –
sec 5. Due to this field, the free electrons present in the conductor
✦ If cross-section is constant, I J i.e. for a given cross-sectional will begin to move with a drift velocity vd towards the left
area, greater the current density, larger will be current. hand side as shown in figure
 CURRENT
ELECTRICITY
V ml
or = = R = a constant for a given conductor for a
I A n e 2τ
given value of n, l and at a given temperature. It is known as
the electrical resistance of the conductor.
Thus, V = RI

6. Time taken by the free electrons to cross the conductors, this is Ohm's law.
t = l/vd (1) Ohm's law is not a universal law, the substances, which
q Alne obey ohm's law are known as ohmic substance.
Hence, current, dI = l
t= (2) Graph between V and i for a metallic conductor is a straight
= vd line as shown. At different temperatures V-i curves are
t different.

or I= Ane
( e Eτ ⎞
7. Putting the value of v = , we have
d| |
m
⎝ ⎠

I = Ane τE
2

4. OHM’S LAW (A) Slope of the line (B) Here tanθ > tanθ
Ohm's law states that the current (I) flowing through a V 1 2
= tan θ = = So, R > R i.e., T > T
conductor is directly proportional to the potential
difference R
(V) across the ends of the conductor”. i 1 2 1 2

(3) The device or substances which don't obey ohm's law

e.g. gases, crystal rectifiers, thermoionic valve,
transistors etc. are known as non-ohmic or non-linear
conductors. For these V-i curve is not linear.

Static resistance R
V 1
= =
st
i tan θ

i.e., I V or V I or V = RI ∆V 1
Dynamic resistance R = =
dyn
∆I tan ф
V
or = R = constant
4.1 Deduction of Ohm’s law
I

We know that
vd
eE
=
τ
m

But E = V/l vd eV 5. ELECTRICAL RESISTANCE

=
τ
Also, I = A n e v ml “The electrical resistance of a conductor is the obstruction
d
posed by the conductor to the flow of electric current
( eV ⎞ ( A n e2τ ⎞ through it”.
I=Ane | | | |V 1. i.e., R = V/I
⎝ ml τ⎠ ⎝ ml ⎠
 CURRENT
ELECTRICITY
volt
2. The SI unit of electrical resistance is ohm or . (vi) Resistivity increases with impurity and mechanical stress.
amp
(vii) Magnetic field increases the resistivity of all metals except
3. Dimensions of electric resistance iron, cobalt and nickel.
(viii) Resistivity of certain substances like selenium,
Pot. diff. work done/charge cadmium, sulphides is inversely proportional to
= current = current intensity of light
falling upon them.

ML2T–2 / AT V ml m l
= =「
1 2 –3 –2 3.
We have, R = = ×
A ne2τ A
= Ane2τ
M T A I l

5.1 Electrical, Resistivity or Specific Resistance comparing the above relation with the relation, R = p .
A
“The resistance of a conductor depends upon the 4.
We have, the resistivity of the material of a conductor,
following factors :
p m
(i) Length (l) : The resistance (R) of a conductor is directly
= ne τ
2
proportional to its length (l), i.e., R l
(ii) Area of cross-section (A) : The resistance (R) of a
5.3 Conductivity
conductor is inversely proportional to the area of
1
cross- section (A). of the conductor, i.e., R 1/A Reciprocal of resistivity is called conductivity (σ) i.e. σ = with
(iii) The resistance of conductor also depends upon the nature p
of material and temperature of the conductor. unit mho/m and dimensions [M –1 L–3 T 3 A 2 ] .
5.4 Conductance
l pl
From above ; R or R = .” 1
A A Reciprocal of resistance is known as conductance. C = It's
R
5.2 Resistivity (p)
unit is Ω1 or Ω or “Siemen”.
–1

1 Where p is constant of proportionality and is known as

specific resistance or electrical resistivity of the material
of the conductor
2. Specific resistance (or electrical resistivity) of the
material of a conductor is defined as the resistance of a
unit length with unit areas of cross section of the material
of the conductor.
(i) Unit and dimension : It's S.I. unit is ohm × m and
dimension is [ML3T–3A–2] 5.5 Stretching of Wire

pm= If a conducting wire stretches, it's length increases, area of cross-

(ii) It's formula : ne2τ section decreases so resistance increases but volume remain
constant.
(iii) Resistivity is the intrinsic property of the substance. It is
Suppose for a conducting wire before stretching it's length = l1,
independent of shape and size of the body (i.e. l and A). area of cross–section = A1, radius = r1, diameter = d1, and
(iv) For different substances their resistivity is also different l
e.g. p = minimum = 1.6 × 10–8 Ω-m and resistance = R = p 1
1 1
si
A
pfused quartz = maximum ≈ 1016 Ω-m
pconductor
pinsulato > palloy > psemi-conductor
r >
(Maximum for fused quartz) (Minimum for silver)

(v) Resistivity depends on the temperature. For metals

pt = p0(1 + α∆t) i.e. resitivity increases with temperature.
 CURRENT
ELECTRICITY
Volume remains constant i.e., A1l1 = A2l2
3. Semiconductors : These are those material whose
After stretching length = l2, area of cross-section = A2,
electrical conductivity lies inbetween that of insulators and
conductors.
Semiconductors can conduct charges but not so easily as is
radius = r2, diameter = d2 and resistance = R = p l2 in case of conductors. When a small potential difference is
2 A2 applied across the ends of a semiconductor, a weak current
Ratio of resistances before and after stretching flows through semiconductor due to motion of electrons and
holes.
2 4
R l A (l ⎞
(2 A (4 r (d ⎞ Examples of semiconductors are germanium, silicon etc.
⎞ ⎞
1
= 1× 2 =| 1| =| 2| =| 2
| =|
2
|
R2 l 2 A1 ⎝ l 2 ⎠ ⎝ A1 ⎠ ⎝ r1 ⎠ ⎝ d1 ⎠ The value of elecrical resistance R increases with rise of
temperature.
2
R(l ⎞
(1)
If length is given then R l2 ⇒ 1 = | 1 |
R2 ⎝ l 2 ⎠ R t – R0 increase in resistance
α
==
original resistance × rise of temp.
4 R0 × t
1 (r ⎞
R1 Thus, temperature coefficient of resistance is defined as the
(2) If radius is given then R ⇒ =| 2|
increase in resistance per unit original resistance per degree
r4 R r
2 ⎝ 1⎠ celsium or kelvin rise of temperature.
1. For metals like silver, copper, etc., the value of a is
6. CURRENT DENSITY, positive,
CONDUCTANCE therefore, resistance of a metal increases with rise in
temperature. The unit of α is K–1 or °C–1.
6.1 Relation between J, σ and E n Ae2τE
m 2. For insulators and semiconductors α is negative,
( eE ⎞ therefore, the resistance decreases with rise in temperature.
We know, I = n Aev = nAe τ
d
| 6.2 Non-Ohmic Devices
= m
⎝ ⎠ Those devices which do not obey Ohm's law are called non-
ne2τE or J = 1 ohmic devices. For example, vaccum tubes, semiconductor
or =
A m E diode, liquid electrolyte, transistor etc.
p For all non-ohmic devices (where there will be failure of
J=
Ohm's law), V–I graph has one or more of the following
1. Insulators : These are those materials whose electrical characteristics :
conducticity is either very very small or nil. (a) The relation between V and I is non-linear, figure
Insulators do not conduct charges. When a small potential
difference is applied across the two ends of an insulator,
the current through the insulator is zero.
Examples of insulators are glass, rubber, wood etc.
Variation of R, p with T
2. Conductors : These are those materials whose electrical
conductivity is very high
Conductor conduct charges very easily. When a small
potential difference is applied across the two ends of
conductor, a strong current flows through the conductor. (b) The relation between V and I depends on the sign of V. It
For super-conductor, the value of electrical conductivity means, if I is the current for a certain value of V, then
is infinite and electrical resistivity is zero. reversing the direction of V, keeping its magnitude fixed,
Examples of conductors are all metals like copper, silver, does not produce a current of same magnitude I, in the
aluminium, tungsten etc. opposite direction, figure.
 CURRENT
ELECTRICITY
To remember the value of colour coding used for carbon
resistor, the following sentences are found to be of great
help (where bold letters stand for colours).
B B ROY Green, Britain Very Good Wife Gold Silver.
Way of finding the resistance of carbon resistor from its
colour coding.
In the system of colour coding, Strips of different colours
are given on the body of the resistor, figure. The colours
on strips are noted from left to right.
(c) Therelationbetwe nV and I isnot unique, i.e., there
is more than one value of V for the same current I, figure.

(i) Colour of the first stip A from the end indicates the first
significant figure of resistance in ohm.
(ii) Colour of the second strip B indicate the second
significant figure of resistance in ohm.
(iii) The colour of the third strip C indicates the multiplier,
i.e., the number of zeros that will follow after the two
7. COLOUR CODE FOR CARBON significant figure.
(iv) The colour of fourth strip R indicates the tolerance limit
The colour code for carbon resistance is given in the
of the resistance value of percentage accuracy of resistance.
following table.

Colour code of carbon resistors

8. COMBINATION OF RESISTORS

8.1 Resistances in Series

Colour Letter as No. Mulitplier Colour Tolerance
anAid to Resistors are said to be connected in series, if the same
memory current is flowing through each resistor when some
poential difference is applied across the combination.
Black B 0 100 Gold 5%
1
Brown B 1 10 Silver 10%
2
Red R 2 10 No colour 20%
Orange O 3 103
Yellow Y 4 104
Green 5 105
Blue B 6 106 1. Let V be the potential difference applied across A and B
using the battery s. In series combination, the same
Violet V 7 107 current (say I) will be passing through each resistance.
Grey 8 108 2. Let V1, V2, V3 be the potential difference across R1, R2 and
White W 9 109 R3 respectively. According to Ohm's law
Gold –1 V1 = IR1, V2 = IR2, V3 = IR3
10
Silver –2 3. Here, V = V1 + V2 + V3 = IR1 + IR2 + IR3 = I (R1 + R2 + R3)
10
 CURRENT
ELECTRICITY
branches are different and I1, I2, I3 be the current through
the resistances R1, R2 and R3 respectively. Then,
I = I2 + I2 + I3
3. Here, potential difference across each resistor is V,
4. If Rs is the equivalent resistance of the given series therefore V = I1R1 = I2 R2 = I3R3
combination of resistances, figure, then the potential
difference across A and B is, V V V
or = , I3 =
= , I2 R R
R
I1
1 2 3
V = IRs.
We have Putting values, we get
IRs = I (R1 + R2 + R3) V V V
I= + +
or R1 R 2 R 3
Rs = R1 + R2 +

Memory 4. If Rp is the equivalent resistance of the given parallel

combination of resistance, figure, then
In a series resistance circuit, it should be noted that :
(i) the current is same in every resistor.
(ii) the current in the circuit is independent of the relative
positions of the various resistors in the series.
(iii) the voltage across any resistor is directly proportional to
the resistance of the resistor.
(iv) the total resistance of the circuit is equal to the sum of V = IRp or I = V/Rp
the individual resistances, plus the internal resistance of
a cell if any. we have
(v) The total resistance in the series circuit is obviously 1
V V V 1+ 1 + 1
more than the greatest resistance in the circuit. V R=
p R1
+R2 +R3 o =
Rp R1 R 2 R3

Thus, the reciprocal of equivalent resistance of a number of

8.2 Resistances in Parallel
resistor connected in parallel is equal to the sum of the
Any number of resistors are said to be connected in reciprocals of the individual resistances.
parallel if potential difference across each of them is the
same and is equal to the applied potential difference. Memory note

In a parallel resistance circuit, it should be noted that :

(i) the potential difference across each resistor is the same
and is equal to the applied potential difference.
(ii) the current through each resistor is inversely
proportional to the resistance of that resistor.
(iii) total current through the parallel combination is the sum
of the individual currents through the various resistors.
(iv) The reciprocal of the total resistance of the parallel
combination is equal to the sum of the reciprocals of the
individual resistances.
1. Let V be the potential difference applied across A and B (v) The total resistances are connected in series, the current
with the help of a battery s. through each resistance is same. When the resistance are
2. Let I be the main current in the circuit from battery. I divides in parallel, the pot-diff. accross each resistance is the
itself into three unequal parts because the resistances of same and not the current.
these
 CURRENT
ELECTRICITY
(iii) Potential drop inside the cell = ir
9. CELL (iv) Equation of cell E = V + (E > V)
The device which converts chemical energy into electrical ir
energy
is known as electric cell. Cell is a source of constant emf but (E ⎞
(v) Internal resistance of the cell r = –1 · R
not constant current. V
⎝| ⎠
(vi) Power dissipated in external resistance (load)
V2 ( E ⎞ 2
P = Vi = i2R = = | | .R
R ⎝R+r⎠

Power delivered will be maximum when R = so E2

Pmax
r = 4r .
This statement in generalised from is called “maximum
power transfer theorem”.

(1) Emf of cell (E) : The potential difference across the

terminals of a cell when it is not supplying any current
is called it's emf.
(2) Potential difference (V) : The voltage across the
terminals of a cell when it is supplying current to
external resistance is called potential difference or
terminal voltage. Potential difference is equal to the (vii) When the cell is being charged i.e. current is given to
product of current and resistance of that given part i.e. the cell then E = V – ir and E < V.
V = iR.
(2) Open circuit : When no current is taken from the cell it
(3) Internal resistance (r) : In case of a cell the opposition is said to be in open circuit.
of electrolyte to the flow of current through it is called
internal resistance of the cell. The internal resistance of
a
cell depends on the distance between electrodes (r d),
area of electrodes [r (1/A)] and nature, concentration
(r C) and temperature of electrolyte [r (1/ temp.)].
A cell is said to be ideal, if it has zero internal
resistance.
(i) Current through the circuit i = 0

9.1 Cell in Various Positions (ii) Potential difference between A and B, VAB = E
(iii) Potential difference between C and D, VCD = 0
(1) Closed circuit : Cell supplies a constant current in the
(3) Short circuit : If two terminals of cell are join together
circuit.
by a thick conducting wire

E
(i) Current given by the cell i
= R+r

(ii) Potential difference across the resistance V = iR

 CURRENT
ELECTRICITY
(i) Maximum current (called short circuit current) flows
plates of cells are connected together their emf’s are added to
momentarily isc = E
r each other while if their similar plates are connected together
their emf’s are subtractive.
(ii) Potential difference V = 0

Memory
1. It is important to note that during charging of a cell, the
positive electrode of the cell is connected to positive
terminal of battery charger and negative electrodes of
the cell is connected to negative terminal of battery
charger. In this process, current flows from positive (1) Series grouping : In series grouping anode of one cell is
electrode to negative electrode through the cell. Refer connected to cathode of other cell and so on. If n
figure identical cells are connected in series

V = s + Ir
Hence, the terminal potential difference becomes greater
than the emf of the cell.
2. The difference of emf and terminal voltage is called lost (i) Equivalent emf of the combination Eeq = nE
voltage as it is not indicated by a voltmeter. It is equal to Ir. (ii) Equivalent internal resistance req = nr
nE
(iii) Main current = Current from each cell = i =
9.2 Distinction between E.M.E. and Potential Difference
R + nr
E.M.F. of a Cell Potential Difference (iv) Potential difference across external resistance V = iR
V
1 The emf of a cells is the 1. The potential difference (v) Potential difference across each cell V ' =
maximum potential between the two points n
is
difference between the the difference of potential
two
cell electrodes of a between those two points ( nE ⎞2

when the cell is in (vi) Power dissipated in the external circuit = |⎝ | .R

the open circuit. in a closed circuit. R + nr
( E2 ⎞
2. It is independent of the
2. It depends upon the resis- (vii) Condition for maximum power R = nr and Pmax = n | |
resistance of the circuit ⎝ 4 ⎠
and depends upon the tance between the two
nature of electrodes points of the circuit and (viii) This type of combination is used when nr << R.
and the nature of current flowing through
(2) Parallel grouping : In parallel grouping all anodes are
electrolyte of the cell. the
connected at one point and all cathode are connected
circuit.
3. The term emf is used together at other point. If n identical cells are connected
for the source of electric in parallel
current. 3. The potential difference is
measured between any two E, r
4. It is a cause.
points of the electric
E, r
circuit.
4. It is an effect.
E, r
9.3 Grouping of Cells
i
R
In series grouping of cell’s their emf’s are additive or
subtractive while their internal resistances are always additive.
If dissimilar
 CURRENT
ELECTRICITY
(i) Equivalent emf Eeq = E
(iii) Main current flowing through the load
(ii) Equivalent internal resistance Req = r /
n nE mnE

E i= =
(iii)
Main current i = nr mR + nr
R+r/n R m
(iv)
Potential difference across external resistance = p.d. (iv) Potential difference across load V = iR
across each cell = V = iR V
(v) Potential difference across each cell V ' =
i
(v)
Current from each cell i ' = n
n i
2 (vi) Current from each cell i ' = n
( E ⎞
(vi) Power dissipated in the circuit P = | | .R
R+r/n (vii) Condition for maximum power R =
⎝ and
( E2 ⎞ nr
m
(vii)
Condition for max. power is R = r / n and Pmax = n | E 2
| ⎝ Pmax = (mn)
4r 4r
(viii)
This type of combination is used when nr >> R
(viii) Total number of cell = mn
Generalized Parallel Battery
Memory note
Note that (i) If the wo cells connected in parallel are of the
same emf s and same internal resistance r, then
sr + sr
seq = =s
r+r
r
1 1 1 or r =
2 = + = eq
req r r r 2

E1 E2
+ +... (ii) If n identical cells are connected in parallel, then the
En
equivalent emf of all the cells is equal to the emf of one
r1 r2 rn 1 1 1 1 cell.
= + +
Eeq 1 1 1 and r r r ... r .
=
+ + ... eq 1 2 n 1 1 1 n
or r = r/n
r1 r2 = + +...+ n terms =
rn

(3) Mixed Grouping : If n identical cell's are connected in a req r r r e

row and such m row's are connected in parallel as

shown. 10. ELECTRIC CURRENT

(1) The time rate of flow of charge through any cross-section

is called current.
∆Q dQ
i = Lim = . If flow is uniform
∆t→0 ∆t dt
Q
then i = . Current is a scalar quantity. It's S.I. unit
is t
ampere (A) and C.G.S. unit is emu and is called biot
(Bi), or ab ampere. 1A = (1/10) Bi (ab amp.)
(2) Ampere of current means the flow of 6.25 × 1018
(i) Equivalent emf of the combination Eeq = nE electrons/sec through any cross–section of the
conductor.
(ii) Equivalent internal resistance of the combination m
req = nr
 CURRENT
(3) The conventional direction of current is taken ELECTRICITY
to be the direction of flow of positive charge,
i.e. field and is
 CURRENT
ELECTRICITY
opposite to the direction of flow of negative charge as
shown below. (i) Solids : In solid conductors like metals current carriers
are free electrons.
(ii) Liquids : In liquids current carriers are positive and
negative ions.
(iii) Gases : In gases current carriers are positive ions and
free electrons.

(4) The net charge in a current carrying conductor is zero. (iv) Semi conductor : In semi conductors current carriers are
holes and free electrons.
(5) For a given conductor current does not change with
change in cross-sectional area. In the following figure (v) The amount of charge flowing through a crossection of
a conductor from t =i t to t =f t is given by :
i1 = i2 = i3
tf
q= I dt
∫
ti

From Graphs
(i) Slope of Q vs t graph gives instantaneous current.

(6) Current due to translatory motion of charge : If n

particle each having a charge q, pass through a given
area in time t then

If n particles each having a charge q pass per second per (ii) Area under the I vs t graph gives net charge flown.
unit area, the current associated with cross-sectional area
A is i = nqA
If there are n particle per unit volume each having a
charge q and moving with velocity v, the current thorough,
cross section A is i = nqvA
(7) Current due to rotatory motion of charge : If a point
charge q is moving in a circle of radius r with speed v
(frequency v, angular speed x and time period T)
then
q
corresponding current i = qv = = qv qω
= 11. KIRCHHOFF’S LAW
T 2πr 2π
11.1 Kirchhoff’s first law or Kirchhoff’s junction law
or Kirchhoff’s current law.

1. the algebraic sum of the currents meeting at a junction in

a closed electric circuit is zero, i.e., Σ I = 0
2. Consider a junction O in the electrical circuit at which
(8) Current carriers : The charged particles whose flow in the five conductors are meeting. Let I1, I2, I3, I4 and I5 be
a definite direction constitutes the electric current are the currents in these conductors in directions, shown in
called current carriers. In different situation current figure,
carriers are different.
 CURRENT
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3. Let us adopt the following sign convention : the current

flowing in a conductor towards the junction is taken as
positive and the current flowing away from the junction
We adopt the following sign convention :
is taken as negative.
Traverse a closed path of a circuit once completely in
4. According to Kirchhoff’s first law, at junction
clockwise or anticlockwise direction.
O (–I1) + (–I2) + I3 + (–I4) + I5 = 0
Difference between Kirchhoff’s I and II laws
or –I1 – I2 + I3 – I4 + I5 = 0
or Σ I= First Law Second Law
0
or I3 + I = I + I + I 1. This law supports the 1. This law supports the law
5 1 2 4

5. i.e., total current flowing towards the junction is equal to law of conservation of conservation of energy.
total current flowing out of the junction. of charge.
6. Current cannot be stored at a junction. It means, no point/ 2. According to this law 2. According to this law
junction in a circuit can act as a source or sink of charge. Σ I=0. Σ s = Σ IR

7. Kirchhoff’s first law supports law of conservation of 3. This law can be used in 3. This law can be used
open and closed in closed circuit only.
charge.
circuits.
11.2 Kirchhoff’s Second law or Kirchhoff’s loop
law or Kirchhoff’s voltage law.
12. EXPERIMENTS

The algebraic sum of changes in potential around any

closed path of electric circuit (or closed loop) involving 12.1 Galvanometer
resistors and cells in the loop is zero, i.e., Σ∆V = 0. It is an instrument used to detect small current passing through
In a closed loop, the algebraic sum of the emfs and it by showing deflection. Galvanometers are of different types
algebraic sum of the products of current and resistance in e.g. moving coil galvanometer, moving magnet galvanometer, hot
the various arms of the loop is zero, i.e., Σ s + Σ IR = wire galvanometer. In dc circuit usually moving coil
0. galvanometer are used.
Kirchhoff’s second law supports the law of conservation
of energy, i.e., the net change in the energy of a charge, (i) It’s symbol : ; where G is the total
after the charge completes a closed path must be zero. internal resistance of the galvanometer.
Kirchhoff’s second law follows from the fact that the (ii) Full scale deflection current : The current required for
electrostatic force is a conservative force and work done full scale deflection in a galvanometer is called full
by it in any closed path is zero. scale deflection current and is represented by ig.
Consider a closed electrical circuit as shown in figure. (iii) Shunt : The small resistance connected in parallel to
containing two cells of emfs. s1 and s2 and three resistors
of resistances R1, R2 and R3. galvanometer coil, in order to control current flowing
through the galvanometer is known as shunt.
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Table : Merits and demerits of shunt
i
(c) To pass nth part of main current (i.e. ig = ) through the
Merits of shunt Demerits of shunt n

To protect the galvano- galvanometer, required shunt S = G

Shunt resistance decreases the .
meter coil from burning . (n – 1)
sensitivity of galvanometer.
It can be used to convert 12.3 Voltmeter
any galvanometer into It is a device used to measure potential difference and is always
ammeter of desired put in parallel with the ‘circuit element’ across which potential
range. difference is to be measured.

12.2 Ammeter

It is a device used to measure current and is always connected

in series with the ‘element’ through which current is to be
measured.

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

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

(a) Equivalent resistance of the combination = G + R

(b) According to ohm’s law Maximum reading of V which
can be taken V = ig (G + R); which gives
V ( V ⎞
Required series resistance R = – G |=
i | V –1 G
|
|
g ⎝ g ⎠
(c) If nth part of applied voltage appeared across galvanometer
(a) Equivalent resistance of the combination GS
= G+ V
(i.e. V = ) then required series resistance R = (n – 1)
S
G.
g
n
(b) G and S are parallel to each other hence both will have
equal potential difference i.e. igG = (i – ig ) S ; 12.4 Wheatstone Bridge Principle
gives which
Wheatstone Bridge Principle states that if four resistances
ig
Required shunt S = G P, Q, R and S are arranged to form a bridge as shown in
(i – ig ) figure, if galvanometer shows no deflection, the bridge is
balanced.
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In that case
through the galvanometer or in other words VB = VD. In the
P P R
RQ=S balanced condition = , on mutually changing the
Q S
position of cell and galvanometer this condition will not
change.
(ii) Unbalanced bridge : If the bridge is not balanced current
will flow from D to B if VD > VB i.e. (VA – VD ) < (VA – VB)
which 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.

12.5 Slide Wire Bridge or Meter Bridge

Proof : A slide wire bridge is a practical form of Wheatstone bridge.
Let I be the total current given out by the cell. On It consists of a wire AC of constantan or manganin of 1
reaching the point A, it is divided into two parts :
metre length and of uniform area of cross-section.
1. I1 is flowing through P A meter scale is also fitted on the wooden board parallel to
2. (I – I1) through R. the length of the wire.
At B, the current I1 is divided into two parts, through the Copper strip fitted on the wooden board in order to provide
Ig two gaps in strips.
galvanometer G and (I1 – Ig) through Q.
A current (I – I1 + Ig) through S. Across one gap, a resistance box R and in another gap the
unknown resistance S are connected.
Applying Kirchhoff’s Second Law to the closed circuit
ABDA, we get The positive pole of the battery E is connected to terminal
A and the negative pole of the battery to terminal C through
I1P + Ig G – (I – I1) R = 0 ...(1)
one way key K.
where G is the resistance of galvanometer.
The circuit is now exactly the same as that of the
Again applying Kirchhoff’s Second Law to the closed Wheatstone bridge figure.
circuit BCDB, we get
(I1 – Ig) Q – (I – I1 + Ig) S – IgG = 0 ...(2)
The value of R is adjusted such that the galvanometer shows
no deflection, i.e., Ig = 0. Now, the bridge is balanced.
Putting Ig = 0 in (1) and (2) we have
I1P – (I – I1) R = 0 or I1P = (I – I1) R ...(3)
and I1Q – (I – I1) S = 0 or I1Q = (I – I1) S ...(4)

P R
Dividing (3) by (4), we get =
Q S
Note that in Wheatstone bridge circuit, arms AB and BC
having resistances P and Q form ratio arm. The arm AD,
having a resistance R, is a known variable resistance arm Adjust the position of jockey on the wire (say at B) where
on pressing, galvanometer shows no deflection.
and arm DC, having a resistance S is unknown resistance
arm. Note the length AB ( = l say) to the wire. Find the length
BC ( = 100 – l) of the wire.
(i) Balanced bridge : The bridge is said to be balanced when
deflection in galvanometer is zero i.e. no current flows
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ELECTRICITY
According to Wheatstone bridge principle
If I is the current flowing through the wire, then from Ohm’s
P R law; V = IR; As, R = pl/A
Q =S l (
where K =
Ip ⎞
V = Ip =
If r is the resistance per cm length of wire, Kl, ⎝| ⎠
then
P = resistance of the length l of the wire AB = lr or V l (if I and A are constant)
Q = resistance of the length (100– l) of the wire BC=(100 i.e., potential difference across any portion of potentiometer
– l) r. wire is directly proportional to length of the wire of that
protion.
lr R ( 100 – l ⎞
100 – l r = or S =| | ×R
S l
⎝ ⎠ Here, V/l = K = is called potential gradient, i.e., the fall
Knowing l and R, we can calculate S. of potential per unit length of wire.

12.6 Potentiometer and its principle of working 12.7 Determination of Potential Difference
using Potentiometer
Potentiometer is an apparatus used for measuring the emf A battery of emf s is connected between the end terminals
of a cells or potential difference between two points in an A and B of potentiometer wire with ammeter A 1,
electrical circuit accurately. resistance box R and key K in series. This circuit is called
an auxillary
A potentiometer consists of a long uniform wire generally
circuit. The ends of resistance R1 are connected to
made of manganin or constantan, stretched on a wooden terminals A and Jockey J through galvanometer G. A cell
s and key 1
board. K1 are connected across R1 as shown in figure.
Its ends are connected to the binding screws A and B. A
meter scale is fixed on the board parallel to the length of
the wire. The potentiometer is provided with a jockey J
with
the help of which, the contact can be made at any point
on the wire, figure. A battery s (called driving cell),
connected across A and B sends the current through the
wire which is
kept constant by using a rheostat Rh.

Working and Theory : Close key K and take out suitable

resistance R from resistance box so that the fall of
potential across the potentiometer wire is greater than the
potential difference to be measured.
It can be checked by pressing, firstly the jockey J on
potentiometer wire near end A and later on near end B, the
deflections in galvanometer are in opposite directions.

Principle : The working of a potentiometer is based on Close key K1. The current flows through R1. A potential
difference is developed across R1. Adjust the position of
the fact that the fall of potential across any portion of the
jockey on potentiometer wire where if pressed, the
wire is directly proportional to the length of that portion
galvanometer shows no deflection. Let it be when jockey
provided the wire is of uniform area of cross-section and
is at J. Note the length AJ (= l) of potentiometer wire.
a constant current is flowing through it.
This
Suppose A and p are respectively the area of cross-section
and specific resistance of the material of the wire. would happen when potential difference across R1 is
equal to the fall of potential across the potentiometer wire
Let V be the potential difference across the portion of the of length
wire of length l whose resistance is R. l. If K is the potential gradient of potentiometer wire, then
potential difference across R1, i.e.,
V = Kl
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If r is the resistance of potentiometer wire of length L, i.e., s1 = Kl1 ...(1)
then current through potentiometer wire is
where K is the potential gradient across the potentiometer
s
I= wire.
R+r Now remove the plug from the gap between 1 and 3 and
( s ⎞ insert in the gap between 2 and 3 of two way key so that
Potential drop across potentiometer wire = Ir = r cells of emf s comes into the circuit. Again find the
position
| | 2
R⎝ + r of jockey on potentiometer wire, where galvanometer shows
Potential gradient of potentiometer wire, i.e., fall of no deflection. Let it be at J2. Note the length of the wire AJ2
potential per unit length is ( = l2 say). Then
s2 = Kl2 ...(2)
( s ⎞r V=( s ⎞ rl
K= . |R+r|L s1 = l1
| | ⎝⎠
⎝ R + r L
Dividing (1) by (2), we get s2l2
Hence, V can be calculated.

12.8Comparison of emfs of two cells using Potentiometer 12.9 Precautions of experiment

A battery of emf s is connected between the end 1. The current in the potentiometer wire from driving cell
terminals A and B of potentiometer wire with rheostat Rh, must be kept constant during experiment.
ammeter A1 and key K in series.
2. While adjusting the position of jockey on potentiometer
The positive terminals of both the cells are connected to wire, the edge of jockey should not be rubbed on the
point A of the potentiometer. Their negative terminals are wire, otherwise area of cross-section of wire will not be
connected to two terminals 1 and 2 of two ways key, uniform and constant.
while its common terminal 3 is connected to jockey J
3. The current in the potentiometer wire from driving cell
through a galvanometer G.
should not be passed for long time as this would cause
Insert the plug in the gap between the terminals 1 and 3 heating effect, resulting the change in resistance of wire.
of two way key so that the cell of emf s1 is in the circuit.
Adjust the position of jockey on potentiometer wire, Memory note
where if pressed, the galvanometer shows no deflection. A balance point is obtained on the potentiometer wire if
Let it be when jockey be at J1. Note the length AJ1 (= l1 the fall of potential along the potentiometer wire, due to
say) of the wire. driving cell is greater than the e.m.f. of the cells to be
There is no current in arm As J . It means the potential balanced.
1
of
positive terminal of cell = potential of the point A, and the 12.10 Determination of Internal Resistance
potential of negative terminal of cell = potential of the point of a Cell by Potentiometer Method
J1.
To find the internal resistance r of a cell of emf s using
potentiometer, set up the circuit as shown in figure.

Therefore, the e.m.f. of the cell ( = s ) is equal to

potential difference between the points
1 A and J 1 of the
potentiometer wire.
CURRENT
ELECTRICITY
Close key K and maintain suitable constant current in the
potentiometer wire circuit with the help of rheostat and
potentiometer wire with the help of rheostat Rh. Adjust
using a single cell.
the position of jockey on the potentiometer wire where if
pressed, the galvanometer show no deflection. Let it be Difference between Potentiometer and Voltmeter
when
Potentiometer Voltmere
jockey is as J . Note the length AJ (= l ) of the
potentiometer
1 1 1
wire. Now emf of the cell, s = potential difference across
the length 1. It measures the emf 1. It measures the emf of
of the potentiometer wire.
l1 of a cell very a cell approximately.
...(1) accurately.
or s = Kl 2. While measuring emf, it
1
2. While measuring emf it
where K is the potential gradient across the wire. does not draw any drws some current from
current
Close key K1 and take out suitable resistance R from the from the source of the source of emf.
resistance box in the cell circuit. Again find the position known emf.
of the jockey on the potentiometer wire where 3. While measuring emf, 3. While measuring emf
galvanometer shows no deflection. Let it be at J2. Note the resistance of the resistance of
the length of the poten- voltmeter is high but
wire AJ2 ( = l2 say). As current is being drawn from the tiometer becomes finite.
cell, infinite.
its terminal potential difference V is balanced and not 4. Its sensitivity is 4. Its sensitivity is low.
emf high.
s. Therefore, potential difference between two poles of 5. It is based on deflection
the cell, V = potential difference across the length l 2 of 5. It is based on null method.
deflection 6. It can be used only to
the potentiometer wire
method.
6. It can be used for
i.e. V = Kl2 ...(2) various purposes. measure emf or potential
Dividing (1) by (2), we have difference.

s l1
= 13. HEATING EFFECT OF CURRENT
...(3)
V l2
We know that the internal resistance r of a cell of emf s,
when a resistance R is connected in its circuit is given by When some potential difference V is applied across a resistance
R then the work done by the electric field on charge q to flow
s–V ( s ⎞ through the circuit in time t will be
r= × R = | –1 | R ...(4)
V ⎝V ⎠ W = qV = Vit = i2R V2t Joule .
=
R
Putting the value (3) in (4), we get

( l1 ⎞ l –l
r= 1| = 1l 2 × R
| R
–
l
⎝ 2 ⎠ 2

Thus, knowing the values of l1, l2 This work appears as thermal energy in the resistor.
and R, the internal
resistance r of the cell can be determined. decreasing its potential gradient. The same can be achieved.
(i) By increasing the length of potentiometer wire.
12.11 Sensitiveness of Potentiometer
(ii) If the potentiometer wire is of fixed length, the potential
The sensitiveness of potentiometer means the smallest gradient can be decreased by reducing the current in the
potential difference that can be measured with its help.
The sensitiveness of a potentiometer can be increased by
 CURRENT
Heat produced by the resistance R is ELECTRICITY

W Vit i2Rt V2t

H= = = = Cal. This relation is called joules
J 4 · 2 4 · 2 4 · 2R
heating.
Some important relations for solving objective questions are as
follow :
 CURRENT
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of any electrical appliance can be calculated by rated
Condition Graph
V2
If R and t are constant power and rated voltage i.e. by using R = R e.g.
PR
H i2 and H V2
220×220
Resistance of 100W, 220 volt bulb is R = =
484Ω
If i and t are constant (series grouping) 100
H R (4) Power consumed (illumination) : An electrical appliance
(Bulb, heater, …. etc.) consume rated power (P R) only if
applied voltage (VA) is equal to rated voltage (VR) i.e. If
V2
A
If V and t are constant (Parallel grouping) VA = VR so Pconsumed = PR. If VA < VR then Pconsumed =
R
1
H 2
VR
R also we have R = so
PR

If V, i and R constant H t ( V2 ⎞
P (Brightness) = | A |.P
consumed
2 R
⎝V R
Pconsumed (Brightness)
e.g. If 100 W, 220 V bulb operates on 110 volt supply then
13.1 Electric Power
( 110 ⎞ 2
The rate at which electrical energy is dissipated into Pconsumed = | ×100 = 25 W
other forms of energy is called electrical power ⎝ 220
|
i.e.
If VA < VR then % drop in output power
W 2 V2 = (PR – Pconsumed ) ×100
P= = Vi = i R =
t R PR
For the series combination of bulbs, current through them will be
(1) Units : It's S.I. unit is Joule/sec or Watt consumed by them is in the reverse ratio of their
Bigger S.I. units are KW, MW and HP,
remember 1 HP = 746 Watt
(2) Rated values : On electrical appliances
(Bulbs, Heater … etc.)
resistance i.e. P 1
R

Wattage, voltage, ……. etc. are printed called rated (5) Thickness of filament of bulb : We know that resistance
values
R l
e.g. If suppose we have a bulb of 40 W, 220 V then of filament of bulb is given by R = , also R = p
V
rated power (P ) = 40 W while rated voltage (V ) = ,
220 V. It
R R PR A
means that on operating the bulb at 220 volt, the power
dissipated will be 40 W or in other words 40 J of 1 i.e. If rated
hence we can say that A PR
electrical energy will be converted into heat and light per Thickness

second. R
(3) Resistance of electrical appliance : If variation of power of a bulb is more, thickness of it's filament is also
resistance with temperature is neglected then resistance more and it's resistance will be less.
 CURRENT
ELECTRICITY

If applied voltage is constant then 1

P( consumed)
R
V2 If quantity of water is given n litre
(By P = A ). Hence if different bulbs (electrical
R
appliance) operated at same voltage supply then then 4180(4200) n ∆θ
t=
1 p
P P thickness
consumed R
R 13.2 Electric Energy

The total electric work done or energy supplied by the

source of emf in maintaining the current in an electric
Different bulbs
circuit for a given time is called electric energy consumed
in the circuit.
Electric energy, W = VIt = P.t
Electric energy = electric power × time
⇒ Resistance R251001000
> R> R
SI unit of electric energy is joule, wherre
⇒ Thickness of filament Brightness
t– > t> t
⇒ 100010040
1 joule = 1 volt × 1 ampere × 1 second = 1 watt × 1 second
B100010025
> B> B
The commercial unit of electric energy is called a
kilowatt- hour (kWh) or Board to Trade Unit (BOT) or
(6) Long distance power transmission : When power is
UNIT of Electricity, in brief, where
transmitted through a power line of resistance R,
power- 1 kWh = 1 kilo watt × 1 hour = 1000 watt × 1 hour
loss will be i R2
Thus 1 kilo watt hour is the total electric energy
Now if the power P is transmitted at voltage V consumed when an electrical appliance of power 1 kilo-
watt works for
P2 one hours.
P= i.e. i = (P/V) So, Power loss = ×R
Vi V 2
1 kWh = 1000 Wh = (1000 W) × (60 × 60 s) = 3.6 × 106 J.
Now as for a given power and line, P and R are constant
Note that the number of units of electricity consumed = No.
so Power loss (1/V2)
So if power is transmitted at high voltage, power loss watt × hour
will be small and vice-versa. e.g., power loss at 22 kV of kWh =
1000
is 10–4 times than at 220 V. This is why long distance
power transmission is carried out at high voltage. Electric energy = VI t = I2Rt = V2t /
13.3 Electricity Consumption
(7) Time taken by heater to boil the water : We know that
(1) The price of electricity consumed is calculated on the
heat required to raise the temperature ∆θ of any
basis of electrical energy and not on the basis of
substance of mass m and specific heat S is H = m.S.∆θ electrical power.
Here heat produced by the heater = Heat required to (2) The unit Joule for energy is very small hence a big
raise the temp. ∆θ of water. practical unit is considered known as kilowatt hour
J(m.S.∆θ) (KWH) or board of trade unit (B.T.U.) or simple unit.
i.e. p × t = J × m.S.∆θ ⇒ t =
p (3) 1 KWH or 1 unit is the quantity of electrical energy
which dissipates in one hour in an electrical circuit
{J = 4.18 or 4.2 J/cal)
when the electrical power in the circuit is 1 KWH thus
4180 ( or 4200) m ∆θ
for m kg water t = 1 KWH = 1000 W × 3600 sec = 3.6 × 106 J.
p
{S = 1000 cal/kgoC)
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(4) Important formulae to calculate the no. of consumed If they are connected If they are connected
units
in series in parallel
Total watt × Total hours
is n =
1000 1 1 1
= +
PS P1 P2 PP = P1 + P2
13.4 Combination of Bulbs (or Electrical Appliances)

1 1 1 HP H1 H2
⇒ HS / tS =H1 / +H2 / ⇒ tp = t1 + t2
Bulbs (Heater etc.) Bulbs (Heater etc.) t1 t2
are in series are in parallel
HS=H1= H2
Hp = H1 = H2
(1) Total power consumed (1) Total power consumed
1 1 1
so = +
1 1 1 so ts = t1+ t2
= + + .... tp t1 t2
Ptotal P1 P2 Ptotal = P1 + P2 + P3.........+ Pn i.e. time taken by i.e. time taken by parallel
combinationto boil the combination to boil the
same quantity of water same quantity of water

tp t1t2
ts = t1 + t2 t +t
=
1 2

(3) If three identical bulbs are connected in series as shown

in figure then on closing the switch S. Bulb C short
(2) In ‘n' bulbs are identical, (2) If ‘n' identical bulbs are circuited and hence illumination of bulbs A and B
P increases
P = in parallel. Ptotal = nP
total
N
Pconsumed Brightness
Pconsumed Brightness

1 1
V R P i
Prated R
R
i.e. in series combination i.e. in parallel combination Reason : Voltage on A and B increased.
bulb of lesser wattage will bulb of greater wattage will
(4) If three bulbs A, B and C are connected in mixed
give more bright light and give more bright light and
combination as shown, then illumination of bulb A
p.d. appeared across it will more current will pass
decreases if either B or C gets fused
be more. through it.
Some Standard Cases for Series and Parallel Combination

(1) If n identical bulbs first connected in series so PS = P

and

then connected in parallel. So PP

PP = nP hence = n2
PS
Reason : Voltage on A decreases.
(2) An electric kettle has two coils when one coil is switched
on it takes time to boil water and when the second (5) If two identical bulb A and B are connected in parallel
t1 coil
is switched on it takes time t2 to boil the same water. with ammeter A and key K as shown in figure.
 CURRENT
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It should be remembered that on pressing key reading of
ammeter becomes twice. 14. ELECTRICAL CONDUCTING
MATERIALS FOR SPECIFIC USE
(1) Filament of electric bulb : Is made up of tungsten which
has high resistivity, high melting point.
(2) Element of heating devices (such as heater, geyser or
press) : Is made up of nichrome which has high
resistivity and high melting point.

Reason : Total resistance becomes half. (3) Resistances of resistance boxes (standard resistances) :
Are made up of alloys (manganin, constantan or nichrome)
Concepts these materials have moderate resistivity which is
practically independent of temperature so that the
When a heavy current appliance such us motor,
specified value of resistance does not alter with minor
heater or geyser is switched on, it will draw a heavy
changes in temperature.
current from the source so that terminal voltage of
source decreases. Hence power consumed by the bulb (4) Fuse-wire : Is made up of tin-lead alloy (63% tin + 37%
decreases, so the light of bulb becomes less. lead). It should have low melting point and high
resistivity. It is used in series as a safety device in an
electric circuit and is designed so as to melt and thereby
open the circuit if the current exceeds a predetermined
value due to some fault. The function of a fuse is
independent of its length.

Safe current of fuse wire relates with it's radius as i r3/2

(5) Thermistors : A thermistor is a heat sensitive resistor
13.5 Some aspects of heating effects of current usually prepared from oxides of various metals such as
nickel, copper, cobalt, iron etc. These compounds are
1. The wire supplying current to an electric lamp are not also
practically heated while the filament of lamp becomes semi-conductor. For thermistors α is very high which
white hot. may be positive or negative. The resistance of thermistors
We know that in series connections the heat produced changes very rapidly with change of temperature.
due to a current in a conductor is proportional to its
resistance (i.e. H R). The filament of the lamp and the
supply wires are in series. The resistance of the wire
supplying the current to the lamp is very small as
compared to that of the filament of the lamp. Therefore,
there is more heating effect in the filament of the lamp
than that in the supply wires. Due to it, the filament of the
lamp becomes white hot whereas the wires remain Thermistors are used to detect small temperature change
practically unheated. and to measure very low temperature.
2. Electric Iron 15. SUPER CONDUCTIVITY
3. Electric Arc
Prof. K. Onnes, in 1911, discovered that certain metals and
4. Incandescent electric lamp alloys at very low temperature lose their resistance
considerably. This phenomenon is known as super-
conductivity. As the temperature decreases, the resistance of
the material also decreases, but when the temperature reaches a
certain critical value (called critical temperature or transition
temperature), the resistance of the material completely
disappears i.e., it becomes zero. Then the material behaves as if
5. Fuse wire it is a super-conductor and there will be flow of electrons
without any resistance whatsoever. The critical temperature is
different for different materials. It has been found
 CURRENT
ELECTRICITY
that mercury at critical temperature 4.2 K, lead at 7.25 K and
V = potential difference across the conductor and l =
niobium at critical temperature 9.2 K become super-conductors.
length of the conductor. Electric field out side the
A team of scientists discovered that an alloy of plutonium, cobalt current carrying conductor is zero.
and gallium exhibits super conductivity at temperatures below
18.5 K. Since 1987, many superconductors have been prepared
with critical temperature upto 125 K, as listed below
Bi2Ca2Sr2Cu3O10 at 105 K and Tl2Ca2Ba2Cu3O10 at 125 K.
The super-conductivity shown by materials can be verified by
simple experiment. If a current is once set up in a closed ring
of super-conducting material, it continues flowing for several 1
4. For a given conductor JA = i = constant so that J
weeks after the source of e.m.f. has been withdrawn. A
The cause of super-conductivity is that, the free electrons in super- i.e., J1 A1 = J2 A2 ; this is called equation of continuity
conductor are no longer independent but become mutually
dependent and coherent when the critical temperature is
reached. The ionic vibrations which could deflect free electrons
in metals are unable to deflect this coherent or co-operative
cloud of electrons in super-conductors. It means the coherent
cloud of electrons makes no collisions with ions of the super-
conductor and, as such, there is no resistance offered by the
super-conductor to the flow of electrons. 5. The drift velocity of electrons is small because of the
Super-conductivity is a very interesting field of research all frequent Collisions suffered by electrons.
over the world these days. The scientists have been working 6. The small value of drift velocity produces a large amount
actively to prepare super-conductor at room temperature and of electric current, due to the presence of extremely
they have met with some success only. large number of free electrons in a conductor. The
Application of super conductors propagation of current is almost at the speed of light and
1. Super conductors are used for making very strong involves electromagnetic process. It is due to this reason
electromagnets. that the electric bulb glows immediately when switch is
2. Super conductivity is playing an important role in on.
material science research and high energy partical 7. In the absence of electric field, the paths of electrons
physics. between successive collisions are straight line while in
3. Super conductivity is used to produce very high speed presence of electric field the paths are generally curved.
computers. NA x d
4. Super conductors are used for the transmission of 8. Free electron density in a metal is given by n =
A
electric power.
where NA = Avogadro number, x = number of free
electrons per atom, d = density of metal and A = Atomic
TIPS AND weight of metal.
1. Human body, though has a large resistance of the order 9. In the absence of radiation loss, the time in which a fuse
of kΩ (say 10 kΩ), is very sensitive to minute currents will melt does not depends on it's length but varies with
even
as low as a few mA. Electrocution, excites and radius
disorders the nervous system of the body and hence one as t r4
fails to control the activity of the body.
10. If length (l) and mass (m) of a conducting wire is given
2. dc flows uniformly throughout the cross-section of
conductor while ac mainly flows through the outer surface then R l2
m
area of the conductor. This is known as skin effect. V
11. Macroscopic form of Ohm's law is R = , while
it's i
3. It is worth noting that electric field inside a charged microscopic form is J = σ E.
conductor is zero, but it is non zero inside a current
12. After stretching if length increases by n times then
carrying conductor and is given by E = V where resistance will increase by n2 times i.e. = n2R
l 2
R
 CURRENT
ELECTRICITY
Similarly if radius be reduced to 1/n times then area of
cross-section decreases 1/n2 times so the resistance 24. Resistance of a conducting body is not unique but
depends on it's length and area of cross-section i.e. how
becomes n4 times i.e. = n4 R the potential difference is applied. See the following
R 2

13. After stretching if length of a conductor increases by x figures

% then resistance will increases by 2x % (valid only if x <
10%)
14. Decoration of lightning in festivals is an example of
series grouping whereas all household appliances
connected in parallel grouping.
15. Using n conductors of equal resistance, the number of
possible combinations is 2n – 1.
16. If the resistance of n conductors are totally different,
then the number of possible combinations will be 2n. Length = a Length = b
Area of cross-section = b × c Area of cross-section = a × c
17. If n identical resistances are first connected in series
and then in parallel, the ratio of the equivalent
resistance is
R ( a ⎞ ( b ⎞
n2 Resistance R=p Resistance R=p

given by | | | |
p
⎝ b×c ⎠ ⎝ a×c⎠
Rs = 1
18. If a wire of resistance R, cut in n equal parts and then 25. Some standard results for equivalent resistance
these parts are collected to form a bundle then
equivalent
R
resistance of combination will be 2 .
n
19. If equivalent resistance of R1 and R2 in series and parallel
be Rs and Rp respectively then
1
R = 「R + R 2 – 4R R and
2|
s p|
1 s s
]
1 2
「
R1R2 (R3 + R4 ) + (R1 + R2 )R3R4 + R5 (R1 + R2 )(R3 + R4

AB
)
R 2 = | Rs – Rs – 4Rs R =
Rp | 2 R5 (R1 + R2 + R3 + R4 ) + (R1 + R3 )(R2 + R4 )
]
20. If a skeleton cube is made with 12 equal resistance each
having resistance R then the net resistance across

2R1R 2 + R 3 (R1 + R 2 )
R AB = +R +R
2R
5 3 1 2
21. The longest diagonal (EC or AG) = R
6
3
22. The diagonal of face (e.g. AC, ED, ....) = R
4
 CURRENT
ELECTRICITY 7
23. A side (e.g. AB, BC.....) = R
12
 CURRENT
ELECTRICITY
1 1
R = (R + R ) + 32. If n identical cells are connected in a loop in order, then
「(R + R )2 + 4R (R + R )
1/ 2
emf between any two points is zero.
AB
2
1 2
2
1 2 3 1 2 ]

「 (
RAB 1 1+4
R1 |
= 1+ R

⎞
| ||
2

2 | ⎝ R1 ⎠ |] 33. In parallel grouping of two identical cell having no internal

resistance
26. It is a common misconception that “current in the
circuit will be maximum when power consumed by the
load is maximum.”
27. Actually current i = E/(R + r) is maximum (= E/r) when
R = min = 0 with P = (E/r)2 × 0 = 0 min. while power
L
consumed by the load E 2R/(R + r)2 is maximum (= E2/4r)
when R = r and i = (E / 2r) s max (= E / r).
28. Emf is independent of the resistance of the circuit and
34. When two cell's of different emf and no internal
depends upon the nature of electrolyte of the cell while
resistance are connected in parallel then equivalent
potential difference depends upon the resistance
emf is indeterminate, note that connecting a wire with a
between the two points of the circuit and current
cell with no resistance is equivalent to short circuiting.
flowing through the circuit.
Therefore the total current that will be flowing will be
29. Whenever a cell or battery is present in a branch there infinity.
must be some resistance (internal or external or both)
present in that branch. In practical situation it always
happen because we can never have an ideal cell or battery
with zero resistance.
30. In series grouping of identical cells. If one cell is
wrongly connected then it will cancel out the effect of
two cells 35. In the parallel combination of non-identical cell's if they
e.g. If in the combination of n identical cells (each are connected with reversed polarity as shown then
having emf E and internal resistance r) if x cell are equivalent emf
wrongly
connected then equivalent emf E = (n – 2 x) and E1rr2+– rE2 r1
eq Eeq =
E
r
equivalent internal resistance eq = nr 1 2

31. Graphical view of open circuit and closed circuit of a

cell.
CURRENT ELECTRICITY

36. Wheatstone bridge is most sensitive if all the arms of 39. The measurement of resistance by Wheatstone bridge is
bridge have equal resistances i.e. P = Q = R = S not affected by the internal resistance of the cell.
37. If the temperature of the conductor placed in the right 40. In case of zero deflection in the galvanometer current
gap of metre bridge is increased, then the balancing length flows in the primary circuit of the potentiometer, not in
decreases and the jockey moves towards left. the galvanometer circuit.
38. In Wheatstone bridge to avoid inductive effects the 41. A potentiometer can act as an ideal voltmeter.
battery key should be pressed first and the galvanometer
key afterwards.
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