2

ELECTRIC CIRCUITS
Introduction

This chapter deals with the movement of charges in conductors due to batteries.

2.1 Current
The time rate of flow of charge through any cross-section is called current.
q
i.e., Iavg =
t

q dq
I = lim 
t 0 t dt
Unit of current is ampere
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Currents may be classified, for ease of analysis into two types :

 If magnitude and direction of current does not vary with time, it is said to be direct current.
 If current is periodic (with constant magnitude) with half cycle positive and half negative, it is said
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to be alternating current.

Current density
an

Current at any cross section is defined as

i = J ·d S

wher J is current density and d S is cross-section area.

ka

I dI
or J  lim n i.e., J  n
S S dS
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where n is a unit vector in the direction of flow of current.

If cross-section is not perpendicular to the current, the cross-section area normal to current in
accordance to given figure will be dS = dS0 cos .

dI dS0
 J= dS 
d S0 cos  J
I
Also, J  E

where  is conductivity of the material and E is electric field across the ends of the conductor..

Drift velocity
It is defined as average velocity with which charge (free electron for conductor) flows when potential
difference is applied across conductor.

-2. 1 -
 Electric Circuits

eE
vd = 
m
Where  is time constant, E is electric field across conductor and m is mass of electron.
Drift velocity and current are related as
I = neAvd
where n = no. of free electron per unit volume
A = cross section area of conductor
vd = drift velocity of electrons

Resistivity ()
It is given by
m
=
ne 2 
 unit of resistivity is -m
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 for conductor  increases with temperature

 for semiconductor and insulator  decreases with temperature
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drift velocity vd
Mobility of electron (µe) = electric field  E , where vd = eE/m,  = relaxation time.

Conductivity
an

Inverse of resistivity is defined as conductivity of material.

2.2 Electric Circuit

ka

An electric circuit consists of active and passive elements. An active element like a cell, battery or
power supply provides current and electrical energy to an electric circuit. The passive elements like
resistor (R), capacitor (C) and inductor (L) consume or store the electrical energy. A resistor opposes
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flow of current. A capacitor C offers a low resistance to flow of alternating current and does not allow
1 Q2
direct current to pass through it at steady state. Energy stored in a capacitor, U = CV 2 = . An
2 2C
inductor L opposes the variations of current and it does not oppose steady or direct current. The
1 2
energy stored in an inductor, U = LI .
2

Ohm’s Law
Ohm’s law states that the ratio of potential difference V maintained across a conductor to the current
flowing through the conductor is a constant which is equal to the resistance of the conductor provided
the physical conditions remain unchanged. If V is the potential difference in volts and current I is in
amperes, the electrical resistance is measured in ohms in the S.I. units.

V
i.e., R
I

-2. 2 -
 Electric Circuits

Resistance
Resistance R of material is given by
l
R 
A
Where l is the length and A is the area of cross-section of the conducting body of resistance R.
The resistance of most of the conductors increases with increase in temperature whereas the resistance
of few of the materials decreases with temperature. In the case of some of the conductors, the
resistivity changes linearly with the increase of temperature. If 0 is the resistivity at lower temperature
T0 and  is the resistivity at higher temperature T, then
 = 0[1 + (T – T0)]
where  is called the temperature coefficient of resistivity.
For semi-conductors and insulators resistance decreases with increase in temperature.

Resistances in series
If a number of resistances R1, R2, R3, ....., Rn are connected in series to a battery of e.m.f. E, the
same current will pass through all the resistances. The equivalent or effective resistance Rseries for
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such resistance is equal to the sum of all resistances. R1 R2 R3

Rseries = R1 + R2 +  + Rn I
I v1 v2 vn
v1 = v2 = vn
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E
Resistances in parallel
A parallel combination of resistances can be made by connecting one end of all resistances to the
positive terminal of the battery and the second end of all resistances is connected to the negative
an

terminal of the battery. The potential difference across all the resistances is the same whereas the

current I in the main circuit divides itself across the respective resistances R1, R2, R3, ....., Rn as
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I1 R1
I1, I2, I3, ....., In so that I = I1 + I2 + I3 + ..... + In
I2 R2
The equivalent or effective resistance in parallel R11 is given
by : I R3
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1 1 1 1 1 I3
   
R eq R1 R 2 R 3 Rn E

Electromotive Force and Potential Difference

EMF of a battery is the work done by the battery in transfering one unit R

charge from its negative terminal to its positive terminal.

I
If a battery of e.m.f. E volts and internal resistance r ohm is connected
E, r
to an external resistance R ohm, the current I flows throughout the
circuit.
E = IR + Ir
The potential difference across the external resistance R is less than the e.m.f. of the cell. This
potential difference across R when the current is flowing through it is called as the closed circuit
voltage. It is less than the e.m.f. of the cell when no current flows through the battery. Terminal
potential difference across R = IR = E – Ir.

-2. 3 -
 Electric Circuits

Grouping of Cells

Series Combination
If n cells are connected in series, as shown in figure, then they
can be replaced by a single cell of emf E and resistance r, I r r r

given by E E E
E = E1 + E2 +  + En
and r = r1 + r2 +  + rn
R
Parallel Combination
If m cells are connected in parallel as shown in figure, then their equivalent cell of emf E and internal
resistance r is given by E1
E1 E Em r1
 2  E2
r1 r2 rm
r2
E = 1 1 1 E3
 
r2 r2 rm r3
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1 1 1 1
and    
r r2 r2 rm R

Series and Parallel Grouping of Cells

ES

Suppose there are N identical cells each of emf E and internal resistance r. And, they are grouped
together to form an arrangement as shown in figure. In each branch there are n cells in series and
there are m such branches in parallel. The group of cells is joined across an external resistor R.
an

Since the total number of cells N, therefore

n
N = mn
The equivalent cell of the group has the emf
ka

E0 = nE
R
n
and internal resistance r0 = r
m
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Thus, the total current flowing through the external resistor R is

E nE mn E
I=  or I =
r0  R nr nr  m R
R
m
Condition for Maximum Current through the External Resistance R.

 For maximum current through R, we must have

nr = mR
m r
or 
n R
 The maximum current is given by

nE mE E mn
Imax =  
2R 2r 2 rR

-2. 4 -
 Electric Circuits

Wheatstone’s bridge
If a net work of four resistances R1, R2, R3 and R4 are connected along arms AB, BC, AD and CD and
ends A and C are connected across the terminals of a battery so that the galvonometer connected
across B and D does not show any deflection i.e. potential at point B and D is equal. Such an arrangement
is called a balanced Wheatstone’s Bridge for which current I1, flows across ABC and I2 flows across
ADC. B

 I1R1 = I2R3 R1 R2
G
I1R2 = I2R4 A I1 C
I2
R1 R R1 R R3 R4
or  3 or  2
R2 R4 R3 R4 D
+ –
Kirchhoff’s Law
If an electrical circuit has more than one path of electrical closed circuits, it is called a network. The
electric currents in different portions of such electric circuits can be found by the application of the
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following Kirchhoff’s Laws :

Kirchhoff’s First Law of Current : The algebraic sum

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of currents at any junction in a circuit is zero. It implies that

the sum of the currents entering at any junction is equal to
the sum of currents leaving that junction. At junction O,
I1 + I2 + I3 – I4 – I5 – I6 = 0.
an

Kirchhoff’s Voltage Law (KVL): Around any closed circuit loop the sum of voltage changes (drops
or gains) across all the circuit elements while traversing in one direction (clockwise or anticlockwise)
must be zero.
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 While moving from higher potential to lower potential while crossing element to apply. Kirchhoff’s
IInd law, P.D. is taken as negative.
Sample Problem 2.1:
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A wire of uniform area of cross-section has a resistance of 10 . It is bent into a circle and points A
and B situated at a distance of a quarter of the circumference apart are connected to a battery of
e.m.f. 6.0 V and of internal resistance 1 . Find the current in two parts AB and ACB of the circuit.
Sol.: Total resistance of the wire = 10 
10 6.0 V, 1 
Resistance of wire AB = r1 = = 2.5  ( resistance  length)
4
10  3 A I
Resistance of wire ACB = r2 = = 7.5  I1
4
Let equivalent resistance of r1 & r2 is R I2 B

1 1 1 1 1 2 2 8
       
R r1 r2 2.5 7.5 5 15 15 C

15
R= 
8

-2. 5 -
 Electric Circuits

E 6 .0 48
I=   A
r  R 1  15 23
8
By current division
r2 48 7.5 36
I1 = I    A
r1  r2 23 2.5  7.5 23
48 36 12
I2 = I  I1 =   A
23 23 23
Sample Problem 2.2:
A uniform copper wire of mass 2.23 gm carries a current of 1 A when an e.m.f. of 1.7 volt is applied
across it. Find the length and area of cross-section. If the wire is uniformly stretched to double its
length, f ind the new resistance. Density of copper = 8.92  103 kg/m3 and its resistivity is
1.7  10–8 m.
Sol.: If A is area of cross-section and L is length of wire,
mass 2.23  10 3
10 6
Volume = L  A = density   ..... (i)
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8.92  103 4

V 1 .7
Resistance of wire, R = = = 1.7 
I 1
1. 7
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L L R
R= or =  = 1.7  108 = 108 .....(ii)
A A

L 10 6 10 2
Multiplying (i) and (ii) L  A  = × 108 = = 25
A 4 4
 L = 5m
an

10 6
A = 4  5 = 5  10–8 m2
When the wire is stretched to double its length, its volume remains constant. Its new area A is given
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as
A
A  2L = A  L or A = = 2.5  10–8 m2
2
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  2L 4 L
New resistance = = = 4  1.7 = 6.8 
 A A
2
 
Sample Problem 2.3:
Calculate the equivalent resistance between points P and Q of the network of resistances shown in
figures (a), (b) and (c).

7 2 2 2
O P 
 
8

8

10 
4

3 5   

 
Q
P Q 2 2 2
10  P  Q
(a) (b) (c)

-2. 6 -
 Electric Circuits

Sol.: The network of resistances in these circuits are arranged in various branches such that these can be
grouped into series or parallel combinations of resistances. The resistances should be grouped from
the sides farthest from points P and Q.
(a) Resistors 3 and 7 in series have a total resistance of 10 which is parallel to 10 resistance.
This resistance of 10 is in parallel to a resistance of 10 along OP which gives an equivalent
10  10
resistance of 10  10 = 5. This resistance of 5 along OP is in series with 5 resistance along

OQ yielding a total resistance of 10. This resistance of 10 is parallel to 10 along PQ which
gives an equivalent resistance of 5 between the points P and Q.
(b) Resistances 2, 4 and 2 on extreme right side are in series and gives a total resistance of 8
88
which in parallel with 8 resistance and gives 8  8 = 4. Again 2, 4 and 2 in series combine

to give a resistance of 8 which is parallel to 8 resistance and it gives total resistance of 4.
The resistances of 2 and 2 are in series with this 4 resistance on extreme left combine to yield
an equivalent resistance of 8 between points P and Q.
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(c) Consider the 3 resistance in series with 3 along extreme left side of the network which
combines to give 6 which inturn is parallel to 6 and yields a total resistance 3 and so on. At
last, there will be a resistance of 3 along PO in series with 3 along OQ which is equal to 6.
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63
The resistance of 3 along PQ alongwith 6 in parallel will yield an equivalent of 6  3 = 2

resistance between points P and Q.

Sample Problem 2.4:
an

Calculate the equivalent resistances between points P and Q of the network of resistances shown in
figures (a), (b) and (c).
ka
B B
A R B R
4 6 P
R
10  C 2R R
P Q R 3R R A
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A
8 P
12  Q R Q
R D 4R C D C
D R
(b) (c)
(a)

Sol.: All these network of resistances represent balanced Wheatstone’s Bridge in which no current flows
through the side BD.
(a) The resistance 10 through BD is ineffective in balanced Wheatstone’s Bridge. The resistance
across P and Q is equivalent to 4 and 6 in series which is parallel to 8 and 12 in series.
Equivalent resistance R eq is
1 1 1 1 1 2 1 3
     
Req 4  6 812 10 20 20 20

20
R eq  = 6.67 .
3

-2. 7 -
 Electric Circuits

(b) In balanced Wheatstone’s Bridge ABCD, 3R across BD is ineffective. Arm AD and BC have
respective resistances 2R and 2R. R eq across PQ is sum of resistances R and 2R in series which
is parallel to resistance, 2R on side AD in series with 4R along CD.
1 1 1 1 1 2 1
    
Req R  2R 2 R  4 R 3R 6 R 6R
R eq = 2R
(c) The circuit ABCD is equivalent to balanced Wheatstone’s Bridge with each arm having equal
resistance R and resistance 2R along BD is ineffective. The equivalent resistance R eq is given by:
1 1 1 2R  2R
  or Req = 2 R  2 R = R
Req RR R R
Sample Problem 2.5:
A battery of 6.0 V is connected to an infinite network of resistances of 1 connected in series and 2
in parallel as shown in the circuit. Find
(a) effective resistance between A and B
(b) the current through 2 resistance nearest to the battery.
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A
1 1 1 1 1
+
2

2
2

2
2
6.0 V – 
ES

B
Sol.: (a) Let R be equivalent resistance between A and B. Assume that one more set of resistances 1
and 2 is connected between them. As there are infinite resistances having a total resistance R
between A and B, the addition of resistances 1 and 2 will not affect the resistance R. Let R
an

be total resistance between A and B. As addition of one resistance does not make any difference
for infinite resistances.
R = R 1  A
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A
If R1 is resistance between A and B.
2

1 1 1 2 R 2R R
   or R1 =
R1 R 2 2R 2 R
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B
2R 2  3R B
Resistance between AB = R = 1   =R
2 R 2 R
 2R + R2 – 3R – 2 = 0 or R2 – R – 2 = 0
(R + 1) (R – 2) = 0 or R = +2, –1
R = –1 is not possible
 R = 2 ohm.
22
(b) Resistance between A and B = 2  2 = 1

Resistance across AB = 1 + 1 = 2
Let a current I flows through 1 and current I1 flows from A to B.
6 1 I A I – I1
Current I = = 3 ampere A
2
P.D. across AB = RAB  I = 1  3 = 3V 6V I1 2
2
3
current across AB = = 1.5 ampere.
2 B B

-2. 8 -
 Electric Circuits

Sample Problem 2.6:

Twelve wires, each of resistance 6 are connected to form a network of skelton cube. A current
enters at one corner and leaves from the diagonally opposite corner. Find the effective resistance
between the opposite corners.

Sol.: Let i be the current flowing towards junction A which will divide itself across different 12 wires as
shown. This is due to symmetry. AB, AE and AD are symmetrically placed so that current equal to i/3
flows through each of them, similarly i/3 current flows through each of the resistance CG, FG & HG
such that the current i flows out of junction G. Let R be equivalent resistance between A and G.
E i/6
Consider path ABCG, F
i/3
i/6
VAG = VAB + VBC + VCG A B
i/3
i
i/3 i/3
i i  i  5i  i/6
iReq    r    r    r    r H G
3 6 3  6  i/3
i/3
i/6 i
i/3
5 5
Req    r   6 = 5 ohm. D i/6 C
6 6
JE
 
1.5 V
2
A D
Sample Problem 2.7:
Find the potential difference between points B and C in the 2 3
ES
8
circuit shown in the figure where the battery of 1.5 V has B C

an internal resistance of 2 and the battery of 2V has an 2 3

internal resistance of 1.
E F
1
an
2V
Sol.: Figure between show current distribution according to Kirchhoff’s Ist Law
Applying Kirchhoff’s second law for loops ADCBA and BCFEB.
–1.5 + 2i + 3i + 8i1 + 2i = 0
ka
1.5 V
i 2
or 7i + 8i1 = 1.5 ...(i) A D

and –8i1 + 3(i – i1) + 1(i – i1) + 2 + 2(i – i1) = 0

2 3
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or 3i – 7i1 = –1 ...(ii) i1 8
B C
Solving (i) and (ii) i – i1
23 3 2 3
i1 = and i =
146 146 1
E F
Potential difference between B and C,
2V
23 92
VB – VC = 8 × = V
146 73 2V 2
A B
Sample Problem 2.8:
1V
In the circuit shown, find :
1 1
(a) potential difference between B and D 2
(b) potential difference across cell of 1V between B and 1V
C and cell of 3V between C and D D C
3 3V

-2. 9 -
 Electric Circuits

Sol.: Figure shows distribution of current according to Kirchhoff’s first law

Applying Kirchhoff’s second law to loop ABDA and BCDB
2V 2 i
–2 – 2i – 2i1 – 1 × i + 1 = 0 A B
or 3i + 2i1 = –1 ...(i) i – i1
i1
and –1(i – i1) + 1 – 3 – 3(i – i1) + 2i1 = 0 1 1
or 4i – 6i1 = –2 = 0 ...(ii) 2
1V
Solving i and i1
D C
3 3V
5 1
i=  A , i1 = A
13 13

2
(a) P.D across B and = VB – VD =  V
13
5 1 6
(b)  i – i1 =   
13 13 13
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6 19
VBC = 1 + 1 × = V
13 13

6 21
ES

VDC = 3 – ×3= V
13 13
Sample Problem 2.9:
Two capacitors C1 = 10 F and C2 = 20 F are connected 10  12 V I1
in the circuit diagram showing three cells of E.M.F. 12V,
an

C1 20 F
6V and 3V connected alongwith various resistors. In the 5 5 I2
steady state, find : 3V
I3
(a) the currents I1, I2 and I3
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6V 10 
(b) the energy stored in capacitors C1 and C2
Sol.: (a) In steady state, capacitor C2 acts as an open circuit.
 I1 = 0.
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Applying Kirchhoff’s First Law at junction D,

I1 – I2 – I3 = 0.
or I3 = –I2 ...(i)
Applying Kirchhoff’s Second Law to mesh GHDEFG, IR = E.
–3 – 5I3 + 5I2 – 10I3 –6 = 0
or 15I3 – 5(–I3) = –9 A
10  B 12 V I1
C
9 9 C1 10 F C2 20 F
 I3 = – , I2 =
20 20 H I1
G D
9 9 3 V I3 5  5 I2
I3
 I1 = 0, I2 = A, I3 = – A. I3
20 20
F E
It shows that the direction of current I3 is 6 V I3 10 
clockwise and is opposite to what is shown
in the figure.

-2. 1 0 -
 Electric Circuits

(b) Applying Kirchhoff’s Second Law to mesh ABHGA, where the potential difference across
condenser C1 is V1,
5I3 + 3 + V1 = 0  V1 = –5I3 – 3
59 3
or V1 = –3=– V (Negative sign shows that polarity of charge on capacitor plates
20 4
are opposite to what shown in figure)
2
1 1  3
Energy stored in capacitor C1 = C1V12 =  10  10–6    = 2.8  10–6 J
2 2  4 

Applying Kirchhoff’s Second Law to mesh BCDHB, where the potential difference across C2 is
V2,
12 + V2 – I2  5 = 0  V2 = 5I2 – 12
9
or V2 = 5 × – 12
20
JE

9 39
 V2 = – 12 = V i.e. C is at a higher potential than D.
4 4
Energy stored in capacitor C2
2
1 1   39 
ES

= C2V22 =  20  10–6   4 
2 2  
= 9.506  10–4 J.

OBJECTIVE QUESTIONS: 2 – I
an

1. Calculate the drift velocity of electrons in silver wire with cross-sectional area 3.14 × 10–6 m2 carrying
a current of 20 A. Given atomic weight of Ag = 108, density of silver = 10.5 × 103 kg/m3.
(a) 2.798 × 10–4 m/sec. (b) 67.98 × 10–4 m/sec.
ka

(c) 0.67 × 10–4 m/sec. (d) 6.798 × 10–4 m/sec.

Sol. [d]
6.023  1026
Number of electrons per kg of silver =
lp

108
Number of electrons per unit volume of silver
6.023  1026
n  10.5  103
108
 20
vd  =  108 = 6.798 × 10–4 m/sec.
neA 6.023  10  10.5  10  1.6  10 19  3.14  10 6
26 3

2. Calculate the relaxation time and mean free path in Cu at room temperature 300 K, if number density
of free electrons is 8.5 × 1028 /m³ and resistivity  = 1.7 × 10–8 mho-m. Given k = 1.38 × 10–23 J/K.
(a) 25 Å (b) 20 Å (c) 5 Å (d) 30 Å
Sol. [a]
m m
Relaxation time   
2
ne ne2 
9.1  1031
= = 2.5×10–14 sec
8.5  1028  (1.6  1019 ) 2  1.7  10 8

-2. 1 1 -
 Electric Circuits

3kT 3  1.38  1023  300

vrms    31
 105 m/s
m 9.1 10
So mean free path  = vrms ×  = 105 × 2.5 × 10–14 = 25 Å

3. Eight resistances each of resistance 5 are connected in the circuit shown in figure. The equivalent
resistance between A and B is

B
(a) (8/3)  (b) (16/3)  (c) (15/7)  (d) (19/2) 
Sol. [a]
The given circuit can be redrawn as
8
JE

 RAB  
3
ES
an

4. The equivalent resistance between points A and B is

R R
A B
ka

R R

(a) 2R (b) (3/4) R (c) (4/3) R (d) (3/5) R

lp

Sol. [d]
The circuit can be redrawn as
3
and finally RAB  R
R 5 R
A B A
3R/2
R
R R R 2

A R B
B

5. In the circuit shown in figure potential difference between points A and B is 16 V. The current passing
through 2 resistance will be
9V 3V
A B
4 1 4

2
(a) 2.5 A (b) 3.5 A (c) 4.0 A (d) zero

-2. 1 2 -
 Electric Circuits

Sol. [b]
 4i1 + 2(i1 + i2) – 3 + 4i1 = 16 V …(i)
Using Kirchhoff’s second law in the closed loop we have
9 – i2 – 2(i1 + i2) = 0 …(ii) 3V
i1 i2 9V i1
Solving equations (i) and (ii), we get A B
4 1 4
i1 = 1.5 A and i2 = 2 A i1+i2
 Current through 2W resistor = 2 + 1.5 = 3.5 A. 2
6. In the given circuit, it is observed that the current I is independent of the value of the resistance R6.
Then the resistance values must satisfy
R5

I
R1 R3
R6

R2 R4
JE

1 1 1 1
(a) R1R2R5 = R3R4R6 (b)   
R5 R6 R1  R2 R3  R4
ES

(c) R1R4 = R2R3 (d) R1R3 = R2R4 = R5R6

Sol. [c]
Since current I is independent of R6, it follows that the resistances R1, R2, R3 and R4 must form a
balanced Wheatstone bridge.
an

7. Current passing through 3 resistance is

A
10V 2 4V
B
ka

3

C
(a) (14/3) A (b) 3 A (c) 2 A (d) (12/5) A
lp

A
Sol. [c]
10V 2 4V
VA – VB = 4V …(i)
VA – VC = 10 V …(ii) B
3
 VB – VC = 6 V
C
8. The effective resistance between points P and Q of the electrical circuit shown in the figure is
2R 2R

2R

P Q

r
2R

2R 2R
2Rr 8R ( R  r ) 5R
(a) (b) (c) 2r + 4R (d)
Rr 3R  r 2  2R

-2. 1 3 -
 Electric Circuits

Sol. [a]
From symmetry the equivalent circuit reduces to as shown in the figure.
2R 2R
2 Rr
Req 
Rr

P Q

2R 2R
9. The two ends of a uniform conductor are joined to a cell of emf e and some internal resistance. Starting
from the midpoint P of the conductor, we move in the direction of the current and return to P. The
potential V at every point on the path is plotted against the distance covered (x). Which of the following
best represents the resulting curve ?



JE
V V V V
(a) (b)  (c)  (d)
x x x x
Sol. [b]
ES

When we move in the direction of the current in a uniform conductor, the potential decreases linearly.
When we pass through the cell, from its negative to its positive terminal, the potential increases by an
amount equal to its potential difference. This is less than its emf, as there is some potential drop across
its internal resistance when the cell is driving current.
an

10. Calculate the effective resistance between A and B in following network.

5 10 15
ka

A 10 10 B

10 20 30

lp

(a) 5  (b) 10  (c) 20  (d) 30 

Sol. [c]
Ratio of upper resistances 5 : 10 : 15 = 1 : 2 : 3
Ratio of lower resistances 10 : 20 : 30 = 1 : 2 : 3
The ratio is same so resistance in middle are nonuseful.
Equivalent resistance = (5 + 10 + 15) | | (10 + 20 + 30)
30  60
So, R eq   20 
30  60

2.3 Potentiometer
1. Potentiometer consists of a uniform wire of large length of 4 m, 6 m, 8 m, etc.
2. The potential drop across the potentiometer wire is directly proportional to its length (this is
because the wire is of uniform area of cross-section).

-2. 1 4 -
 Electric Circuits

E
()

A D
B

V (i)

E
()

D2 D1 B
A
E1 (ii)
E2 G

E K1
()

D2 D1
A B
JE
G (iii)
()

K2
R.B

3. Through the path of the voltmeter [fig. (i)], current from the battery of e.m.f. E can be varied i.e.
ES
if point D is closer to point A, less current flows through the path of galvanometer; and if point D
is farther from point A, then more current flows through the path of galvanometer.
4. For comparing e.m.f.s of two cells [fig. (ii)], the positive terminals of both the cells and the
positive terminal of the battery (of e.m.f. E) are connected to the same end A of the potentiometer
an

wire. Now, we locate point D on the potentiometer wire so that no current flows through the path
of galvanometer. Then [see figure (ii)]
E.M.F. of cell = P.D. across A and D on potentiometer wire ...(i)
ka

1 E 1 AD1 l
For two cells, E  AD  l .....(ii)
2 2 2
lp

5. For finding internal resistance (r) of a cell :

(i) We connect a cell [see figure (iii)] and locate a point D1 on potentiometer wire so that no
current flows through the path of galvanometer, when key K2 is not used.
(ii) Now, we take out some resistance R from the resistance-box (R.B.) and use key K2 also,
and locate a point D2 on the potentiometer wire so that no current flows through the path of
galvanometer.
 l1 
Then, r = R  l  1 , where l1 = length AD1 of potentiometer wire
2 
l2 = length AD2 of potentiometer wire

Conversion of Galvanometer into ammeter and voltmeter

In a galvanometer, a moving coil is placed in a uniform magnetic field. When the current to be measured
is passed through the coil of a galvanometer, the coil is deflected through a certain angle depending

-2. 1 5 -
 Electric Circuits

upon the current. The current I can be measured by converting a galvanometer into an ammeter by
connecting a small resistance Rs in parallel to the galvanometer of resistance Rg through which the
I Rg
current Ig is flowing.
IG G
Ig
Rs   Rg
I I g
RS
The potential difference V between any two points A and B is measured by connecting a voltmeter
across these points. A galvanometer is converted into a voltmeter by connecting a high resistance R in
series with a galvanometer of resistance Rg. Rg R

Vg V Vg G

Rg R IG
Vg V–Vg
V B
A
Heating Effects of Current
If a current I ampere passes through a heating element of resistance R for time t when a potential
JE

difference V volt is applied across it, the power produced is P.

V2
P = VI = I 2R =
R
ES

Electrical energy or work = VIt J

Work VIt
Heat produced = H = = cal
J J
where Joule’s constant, J = 4.18 Joule/calorie
an

VIt 0.24 V 2 t
H= = 0.24 VIt = calorie
4.2 R
1 kilowatt hour = 1000  60  60 = 3.6  106 Joule.
Fuse Wire
ka

It is used in series with electrical installations to protect from high electric current, this fuse melts
causing breakage in the circuit when high electric current flows. Fuse wire has high resistance and
lp

low melting point, it is generally prepared from tin lead alloy (63% tin + 37 lead).
Maximum power transfer
If a cell of e.m.f. E and internal resistance r is connected to a load resistance R,
E
current I 
Rr
E2R
Electric Power = P = I 2R = ( R  r ) 2 E, r

If maximum power is transferred, R

dP d  E2R   1 2R 
 E2   0
dR
=0 or dR  ( R  r ) 2   (R  r)
2 ( R  r )3 
or R=r

d2p
For R = r, <0  P is maximum
d R2
i.e., The power transferred is maximum when load resistance is equal to internal resistance of a cell.

-2. 1 6 -
 Electric Circuits

2.4 Charging and Discharging

1. Charging: When a capacitor C is connected to a battery through a resistance R, the plates of a
capacitor will acquire equal and opposite charge and the P.D. across it becomes equal to the emf
of the battery. The process (called charging) takes sometime and during this time there is an
electric current through the resistance. If at any time t, I is the current through the resistance R
and q is the charge on capacitor C, the equation of emf for the circuit will be

VC + VR = E, i.e., V + IR = E

But I = (dq/dt) and q = CV

dq q
So, R  E C
dt C R

q dq q dq t dt t
or 0 (CE  q)
 
0 q0  q
  CR
0
or loge (q0 – q) = –
CR
+k
JE

which on solving for q gives

q = q0 (1 – e–t/CR) with q0 = CE (for t = ) .....(8)

ES

This is the required result and from this it is clear that:

(a) During charging, charge on the capacitor increases from 0 to q0(= CE) non-linearly.
an

(b) The density CR is called capacitive time constant  of the circuit [as it has dimensions of time]
and physically represents the time in which charge on the capacitor reaches 0.632 times of its
maximum value during charging.
ka

(c) During charging current at any time t in the circuit will be

dq d E
 [q0(1 – e–t/CR) = I0e–t/CR with I0 =
lp

I= (at t = 0)
dt dt R
i.e., initially it acts as short circuit or as a simple conducting wire If t , I  0, i.e., it acts as
open circuit or as a broken wire.
q0 or E I0
or Voltage

I
Charge

O t O t
2. Discharging: If a charged capacitor C having charge q0 is discharged through a resistance R
then at any time t,
V = IR
But as I = (–dq/dt) and q = CV

-2. 1 7 -
 Electric Circuits

dq q q dq t dt
R   0 i,e., q 
dt C 0 dt 0 CR

or q = q0 e–t/CR
This is the required result and from this it is clear that
(a) During discharging, the charge on capacitor decreases exponentially from q0 to 0
(b) The capacitive time constant  = CR is the time in which charge becomes (1/e), i.e., 0.368
times of its initial value (q0)
(c) During discharging current at any time t in the circuit:
dq d E
I=–  (q0e–t/CR) = I0e–t/CR with I0 =
dt dt R
and its direction is opposite to that of charging.
(d) As in discharging of a capacitor through a resistance
t
q = q0e–t/CR i.e., R =
C log e (q0 / q)
JE

So resistance R can be determined from the value of t and (q0/q), i.e., (V0/V). Using this concept
in laboratory we determine the value of high resistances (~ M).
Sample Problem 2.10:
ES

A cell of steady emf. 2.0 V is put across a potentiometer wire. For finding internal resistance of a
cell of emf 1.5 V, it is connected as shown under. The balance point for this cell in open circuit is
76.3 cm. When a resistance of 9.5  is put across this cell, the balance point shifts to 64.8 cm. Find
the internal resistance of the cell.
an

2V

A B
1.5 V
G
ka

9.5 
Sol.: R = 9.5  l1 = 76.3  l2 = 64.8  r=?
lp

l   76.3 
r = R  1  1  9.5    1 = 9.5 × [1.1775 – 1] = 9.5 × 0.1775 = 1.686 
 l2   64.8 

Sample Problem 2.11:

An accumulator is connected first to an external resistance R1 and then to another external resis-
tance R2 for the same time. At what value of the internal resistance of the accumulator will the
amount of heat dissipated in the external resistances be the same in the two cases ?
Sol.: When an accumulator of emf E and internal resistance r is connected across a load resistance R, the

E 2 Rt  E 
heat dissipated in the external circuit H  I 2 Rt   as I  
(R  r )2  (R  r) 
According to given problem :
E 2 R1t E 2 R2t
 i.e., (R2 – R1) (r2 – R1R2) = 0
( R1  r ) 2 ( R2  r ) 2
And as R2  R1 (given) r2 – R1R2 = 0

-2. 1 8 -
 Electric Circuits

Sample Problem 2.12:

A uniform potential gradient is established across a potentiometer wire. Two cells of emf E1 and E2
connected to support and oppose each other are balanced over 1 = 6m and 2 = 2m. Find E1/E2.
Sol.: E1 + E2 = x1 = 6x and E1 – E2 = 2x
E1  E2 6 E1 2
 or 
E1  E2 2 E2 1
Sample Problem 2.13:
Figure shows use of potentiometer for comparison of two resistances. The balance point with standard
resistance R = 10 is at 58.3 cm, while after exchanging with unknown resistance X is 68.5 cm. Find
X.
JE

Sol.: Let E1 and E2 be potential drops across R and X

E2 IX X E
so E  IR  R or X 2R
1 E1
ES

E2  2
But E  
1 1

2 68.5
an

so X R  10  11.75
1 58.3
Sample Problem 2.14:
A galvanometer of coil resistance 20 ohm, gives a full scale deflection with a current of 5 mA. What
ka

arrangements should be made in order to measure currents upto 1.0 A ?

Sol.: The upper limiting value of current to be measured is to be increased by a factor.
lp

1.0 A
n  200
5A

 The resistance of the shunt required will be

G 20
S   0.1 
n  1 200  1
Hence, a shunt of resistance 0.1 should be connected in parallel across the galvanometer coil.
Sample Problem 2.15:
A voltmeter having 100 resistance can measure a potential difference of 25V. What resistance R
is required to be connected in series, to make it read voltages upto 250 V ?
Sol.: The upper limiting value of voltage is to be increased by a factor

250V
n  10
25V

R = (n – 1) G = (10 – 1) 100 = 900 

-2. 1 9 -
 Electric Circuits

Sample Problem 2.16:

The current in a potentiometer wire of 100 cm. length is adjusted to yield a null point at 40 cm with
a cadmium cell of emf 1.018 V. Calculate the potential gradient of the wire. Also, find the balancing
length corresponding to an emf of 1.450 V.
Sol.: From the principle of potentiometer wire, the fall in potential along a 40 cm. length is 1.018 V (the
emf of the balancing length).
 Potential gradient = fall in potential per unit length of the wire

1.018V
=  24.45 V / m
40  10 2 m

If  be the balancing length corresponding to the cell of emf 1.450V, then

1.450 
since, E   or = 57 cm.
1.018 40 cm
JE

Sample Problem 2.17:

A 10F condenser C is charged through a resistance R of 0.1 M from a battery of 1.5 V. Find the
ES

time required for the capacitor to get charged upto 0.75 V for the circuits shown in figure in case (a)
and (b) when switch S is switched on.
R S
an

E C C R
S
ka
(a) (b)
Sol.: (a) In the case of charging of a capacitor C through the resistance R,
q = q0 [1 – e–t/RC]
lp

q
e–t/RC = 1  q
0

For a capacitor, q = CV
q V 0.75 1
  
q0 V0 1.5 2

1 1 t
 e–t/RC = 1 – = or  –log e 2
2 2 RC

t = RC log e 2 = (0.1  106)  (10  10–6) log e 2

= log e 2 = 0.693 second.
(b) In the case of the capacitor C being connected directly to the battery initially, it acts like short
circuit. The capacitor will get charged instantaneously.
 t = 0 second.

-2. 2 0 -
 Electric Circuits

OBJECTIVE QUESTIONS: 2 – II

1. The correct circuit for the determination of internal resistance of a primary cell by using potentiometer
is :
–
+ .
( )
+ ( )

+ – E – +
(a) G (b) G
E

R .
( )
R ( )

+ –
.
( )
+ –
.
( )

(c) R (d) + –E

+ – E .
( ) G
R .
( ) G
Sol. [d]
JE

2. A milliammeter of range 10 mA and resistance 9 is joined in a circuit as shown. The metre gives full-
scale deflection for current I when A and B are used as its terminals, i.e., current entres at A and leaves
at B (C is left isolated). The value of I is
9, 10 mA
ES

0.1 0.9
A B C
an

(a) 100 mA (b) 900 mA (c) 1 A (d) 1.1 A

Sol. [c]
9
ig = 10 mA = 0.01 A ig 0.9

VA – VB = (I – ig)0.1 = ig × 9.9
ka

or I × 0.1 = 10ig 0.1

(I-i ) I I
10  0.01 g

or I 1A
0.1 A B
lp

3. The charge flowing through a resistance R varies with time t as Q = at – bt2. The total heat produced
in R by the time current ceases is
a3 R a3 R
(a) a3R/6b (b) a3R/3b (c) (d)
2b b
Sol. [a]
Q = at – bt2
dQ
i  a  2bt
dt
i = 0 for t = t0 = a/2b, i.e., current flow from t = 0 to t = t0.
t0
a3 R
The heat produced =  i 2 R dt. Putting the value of i we get heat produced = .
0
6b
4. Three voltmeters A, B and C having resistances R, 1.5R and 3R, respectively, are connected as shown.
When come potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC
respectively. Then

-2. 2 1 -
 Electric Circuits

X A Y

(a) VA  VB = VC (b) VA = VB  VC (c) VA  VB = VC (d) VA = VB = VC

Sol. [d]
The division of current I into the two parallel branches will be as shown, VA = IR

 2I 
VB    1.5 R  IR
 3 

I
VC    3R  IR  VA = VB = VC
3
5. A galvanometer of resistance 12 shows full scale deflection for a current of 2.5 mA. If you convert
JE

it to a voltmeter of range 10 volt, resistance of the voltmeter will be

(a) 2000  (b) 5000  (c) 400  (d) 4000 
Sol. [d]
ES

V 10
Series resistance required for voltmeter R  G   12  3988 
g 2.5  103

Resistance of voltmeter Rv = R + G = 3988 + 12 = 4000 .

6. In the circuit shown in figure, the power which is dissipated as heat in the 6 resistor is 6W. What is
an

the value of resistance R in the circuit ? R

(a) 6 (b) 10 6

(c) 13  (d) 24 

ka

Sol. [d] 8

8R
6  12
8 R
lp

12V
8R = 6(8 + R)
2R = 48 R = 24 
7. An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V respectively. If a
resistance is now joined in parallel with the voltmeter.
(a) both A and V will increase (b) both A and V will decrease
(c) A will decrease, V will increase (d) A will increase, V will decrease.
Sol. [d]
When a resistance is joined in parallel with the voltmeter, the total resistance of the circuit decreases.
Current will increase and ammeter reading will increase.

V A

Potential difference across the ammeter increases thus potential difference across voltmeter decreases.

-2. 2 2 -
 Electric Circuits

8. A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to
the current, the temperature of the wire is raised by T in a time t. A number N of similar cells is now
connected in series with a wire of the same material and cross-section but of length 2L. the temperature
of the wire is raised by the same amount T in the same time t. The value of N is
(a) 4 (b) 3 (c) 8 (d) 9
Sol. [b]

[(3)2 / (L / A)] ( LA) s T

2

[( N ) / (  2 L / A)] (2 L A) s T

(using V2/R = msT)

 = resistivity; s = specific heat capacity of material of the wire. A = area of cross section
9 1
2
  N 2  9  N  3.
N 2 2
9. 50 V battery is supplying current of 10 amp when connected to a resistor. If the efficiency of battery at
this current is 25%. Then internal resistance of battery is
JE

(a) 2.5 (b) 3.75  (c) 1.25  (d) 5

Sol. [b]
50 = 10 [R + r] R + r = 5
R R
ES
  0.25 
Rr Rr
R + r = 4R  r = 3R
5
then R   1.25  and r = 3.75 .
4
an

10. In the circuit shown in figure the heat produced in the 5 ohm resistor due to the current flowing through
it is 10 calories per second. The heat generated in the 4 ohms resistor is
(a) 1 calorie/sec
(b) 2 calories/sec
ka

(c) 3 calories/sec
(d) 4 calories/sec
Sol. [b]
lp

Let I1 be the current flowing in 5 resistance and

(I – I1) in 4 and 6 resistance.
The heat generated in 5 resistor is 10 cal/s = 4.2 × 10 J/s
 4.2 × 10 = I12R
4.2  10
 I1   8.4  2.9 amp ...(i)
5
Since AB and CD are in parallel.
 The potential difference remains the same between C and D;
and between A and B.
 (I – I1) (4 + 6) = I1 × 5 on solving using I1 from (i) we get
(I – 2.9) 10 = 2.9 × 5
 I – 2.9 = 1.45  I = 4.35
Heat released/sec in 4  resistance will be = (4.35 – 2.9)2 × 4 = 8.4 J/s = 2 cal/s

-2. 2 3 -