9. Current Electricity

UNIT--Kirchhoff’s Laws of electrical network

9.1 Introduction:
The study of electrical charges in motion is called current electricity. The motion of
electric charges in a conductor produces electric current. The electric current is defined as
the rate flow of charge. If ‘q’ charge flows in time ‘t’ through a conductor, then current ‘I’ is
given by I = q / t. When potential difference is applied across the conductor, the electric
current flows through a conductor. The relation between current ( I ) and potential
difference ( V ) is given by ohm’s law which is given by V = R I . Where R is resistance of
conductor.
In XI th science we study how to apply ohm’s law to simple circuit containing resistance
only or how to determine the effective resistance in series and parallel combination of
resistances.
Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 3

However this ohm’s law cannot be applied to complicated electrical circuits containing
large number of electrical components as shown in figure. The more complicated circuit
can be analysed by using Kirchhoff’s laws. The scientist Kirchhoff formulated two laws for
analysing complicated circuit . In this chapter we will discuss these laws and their
applications.

9.2 Kirchhoff’s laws for electrical network :
Before to study these laws we will define some terms used for electrical circuits
1. Junction : Any point in an electrical circuit where two or more conductors joined
together is a junction
2. Loop : Any closed conducting path in an electrical network is called loop or mesh
3. Branch : A branch in any part of network that lies between two junctions.
In Fig 9.1 there are two junctions
labelled as ‘a’ and ‘b’. There are three
branches- These are three paths 1, 2, 3
from a to b
Fig 9.1

9.2.1 Kirchhoff’s First Law (Current Law OR Junction Law ) :
Statement: ”The algebraic sum of currents at a junction is zero in an electrical
network”.
where ‘𝐼𝑖’ is the current in the i th conductor at a junction
having ‘n ‘ conductors
Explanation :
Sign convention :
1) Currents arriving at junction are considered as positive.
2) Currents leaving the junction are considered as negative.
σ𝑖 0
𝑛
(𝐼𝑖) = 0

Appling sign convention we can write,
I1- I2 + I3 + I4 – I5 – I6 = 0
Arriving currents I1, I3, I4 are taken as positive and
leaving currents I2, I5, and I6 are taken as negative.
Therefore, I1+ I3 + I4 = I2 + I5 + I6
Thus the total currents flowing towards the junction is equal to the total
current flowing away from the junction. This is Kirchhoff’s first law.
Consider a junction point P in circuit where six conductors meet as shown in fig. 9.2 .

9.2.2 Kirchhoff’s Second Law ( Voltage Law ) :
Statement : “ The algebraic the sum of potential differences ( product of current and
resistance) and the electromotive forces (emf’s) in a closed loop is zero.“
i.e. σ(𝐼 𝑅) + σ (𝐸) = 0
Explanation:
Sign convention :
1. While tracing a loop through a resistor ,if we are travelling along the
direction of conventional current ,the potential across that resistor is
considered negative . If loop is traced against the direction of
conventional current, the that resistor is considered positive.
2. The emf of electrical source is positive while tracing the loop within
the source from negative terminal of source to its positive terminal. It
is taken as negative while tracing within the source from positive
terminal to negative terminal.

Consider the loop ABFGH in clockwise
direction. Appling Sign convention to this
loop we get,
-I1R1 - I3R5 - I1R3 + E1 = 0
E1 = I1R1 + I3R5 + I1R3
Now consider the loop BFDCB in
anticlockwise direction, applying Sign
convention we get,
- I2R2 – I3R5 – I2R4 + E2 = 0
E2 = I2R2 + I3R5 + I2R4
Consider an electrical network as shown in fig. 9.3
E1
E2

Steps usually followed while solving a problem using Kirchhoff’s laws :
1. Choose some direction of the currents.
2. Reduce the number of variables using Kirchhoff’s first law.
3. Determine number of independent loops
4. Apply voltage law to all the independent loops
5. Solve the equations obtained simultaneously
6. In case the answer of current is negative, the conventional current is flowing in the
direction opposite to that chosen by us.

Problem on Kirchhoff’s First law :

Problem on Kirchhoff’s Second law ( voltage law) :

UNIT– Wheatstone Bridge

9.3 Wheatstone Bridge
The Wheatstone bridge was originally developed by Charles Wheatstone
(1802-1875),
To measure the values of unknown resistance, range from tens of ohms to
hundreds of ohms.
It calibrates measuring instruments , voltmeter, ammeters , etc.

P, Q, R & S – Resistance
G – Galvanometer
E – emf (Battery)
k– key
Circuit Diagram:
Fig. 9.4

From Kirchhoff”s current law - I=I1+I2 …………(1)
Apply Kirchhoff's voltage law for loop ABDA,
- I1P – IgG +I2 S = 0 ( ‫؞‬ I g = 0)
- I1 P + I2 S = 0
‫؞‬ I1 P = I2 S………………(2)
Apply Kirchhoff”s voltage law for loop BCDB,
- (I1-Ig) Q + (I2+Ig) R+ Ig G = 0 ( ‫؞‬ I g = 0)
- I1 Q + I2 R = 0
‫؞‬ I1 Q = I2 R…………………(3)
At Balance condition
VB = VD, I g = 0

This is the condition for balancing the Wheatstone Bridge.
If any three resistances in the bridge are known, the fourth resistance can be
determined by using eq.(4).
Uses of Wheatstone Bridge:
1. To measure the value of very low resistance precisely.
2. To measure the quantities such as galvanometer resistance, capacitance,
inductance and impedance.
Dividing Eq. 2 by Eq. 3 , we get
P/Q = S/R……….4

Applications of Wheatstone Bridge:-
9.3.1 Meter Bridge
Meter bridge is the modification of Whetstone's bridge used to
determine unknown resistance.
It is an instrument which works on the principle of Whetstone's
bridge.
So it is also called as Whetstone's Meter bridge.
The length of wire used is One meter , so it is called Meter bridge

Circuit Diagram:
AB - Metal wire(1m)
X- Unknown Resistance
G- Galvanometer
E- Battery
J – Jockey
Rh- Rheostat
D- Null point
lx- Length of wire bet. A&D
lR- Length of wire bet. D&B
Jockey:- The jockey is a metallic rod whose one end has a knife edge which can
slide over the wire AB to make electrical connection(contact)
Fig.9.5

Using the conditions for the balance , we get
Where, RAD &RDB are resistance of the parts AD & DB of the
wire resistance of the wire AB.
If l - Length of the wire,
A- Area of cross section of wire,
ρ – Specific resistance of wire,
Then

Therefore,
Knowing, R, lx & lR, the value of the unknown resistance X can
be determined.
Source of errors:
1. The cross section of the wire may not be uniform.
2. The ends of the wire are soldered to the metallic strip where contact
resistance is developed , which is not taken into account.
3. The measurement of lx & lR may not be accurate.
4. These contact resistance affect the null deflection point & introduce an
error in ‘X’

To minimize the errors:
1. The value of R is so adjusted that the null point is obtained to middle one
third of the wire (between 34 cm & 66 cm) so that percentage error in the
measurement of lx & lR are minimum and nearly the same.
2. The experiment is repeated by interchanging the positions of unknown
resistance X & known resistance box R.
3. The jockey should be tapped on the wire & not slided. We use jockey to
detect whether there is a current through the central branch .this is possible
only by tapping the jockey.

2. Kelvin’s Method:
To determine the resistance of galvanometer (G) by using meter bridge.
Circuit Diagram:
G- Galvanometer,
R- Known resistance,
AC - Metal wire(1m),
Rh- Rheostat,
lg- Length of wire bet.
A&D,
lr- Length of wire bet.
D&C,
E-Battery,
K-key.
Rh

Let RAD & RDC be the resistance of the two parts of the wire AD & DC
respectively.
Since bridge is balanced
Using this formula, the unknown resistance of the galvanometer can be calculated.

3.Post Office Box (PO Box) :
A post office box (PO Box) is a practical form of Wheatstone bridge as
shown in the figure.
P, Q & R – Known Resistances,
X – Unknown Resistance,
G – Galvanometer,
E – Battery,
K1 & K2 – Tap Keys.

P & Q contains resistances 10 Ω, 100 Ω & 1000 Ω each,
R contains resistances from 1 Ω to 5000 Ω,
X forms the fourth resistance.
P & Q are fixed to desired ratio ,
R is adjusted so that the galvanometer shows no deflection
So bridge is balanced,
The unknown resistance
X = R Q/P
If L- Length of the wire ,
r – Radius of the wire then
The specific resistance of the material of the wire is given by

UNIT- (1) POTENTIOMETER
(2) GALVANOMETER

9.4 Potentiometer:-
 device used for accurate measurement of potential difference
 used to measure the e.m.f. of a cell
 used for comparison of e.m.f. of two cells
 used to measure the internal resistance of a cell
9.4.1 Construction and Principle
 Potentiometer consists of a long and uniform wire AB stretched on rectangular wooden
board into number of segments 100cm each
 L is length, R is resistance , A is cross sectional area of wire
 A cell of emf ɛ having internal resistance r is connected across AB as shown in fig.
Fig.9.6

Potential difference per unit length of the wire is
For a point C on wire, the potential
difference between A and C is
Potential gradient can be defined as
potential difference per unit length of wire
Principle of potentiometer:
Potential difference between two points on
the wire is directly proportional to length of
wire between them. ( provided, wire-
uniform, current- same, temp.-constant )
VAC α l

9.4.2 Uses of Potentiometer
A) To compare emf of cells
1) By individual method
Thus, emf of two cells can be compared

2) By sum and difference method
By using this formula, emf of two cells can
be compared

The key k1 is closed and k2 is open then,
B) To Find Internal Resistance (r) of a Cell :
Both the keys k1 and k2 are closed then,
Consider the loop PQSTP
and
This equation gives the internal
resistance of cell

C) Application of Potentiometer
1) Voltage Divider:
2)Audio Control
3) Potentiometer as a sensor
9.4.3 Advantages of Potentiometer Over Voltmeter:
1) Potentiometer used to measure potential difference as well as emf of cell.
A voltmeter measures terminal potential difference and can not used to measure emf of cell.
2) Potentiometer measures P.D. or emf very accurately. A very small P. D. can be measured.
3) Internal resistance of a cell can be measured .
Disadvantages or Demerits of Potentiometer:
1) Potentiometer is not portable.
2) Direct measurement of potential difference or emf is not possible.

9.5 Galvanometer
• Device used to detect very small electric current.
A moving coil galvanometer can be
used as Ammeter and Voltmeter
9.5.1 Galvanometer as an Ammeter:
Ammeter is current measuring instrument.
The necessary modifications are:
1)Its effective current capacity must be increased to desired high value
2) Its effective resistance must be decreased
3) It must be protected from damages due to excess electric current
This is achieved by connecting low resistance in parallel with galvanometer. This resistance
is shunt. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 34

A shunted permanent magnet moving coil galvanometer is called as an ammeter
Uses of shunt:
1)Reduces effective resistance of an ammeter
2) Increases the range of instrument
3)Provides an alternative path for excess current , which protects the galvanometer from
the damage
Expression for shunt resistance:
Since S and G are parallel
Ig –current through galvanometer G
Is – current through shunt S
Total current
This equation gives the value of shunt resistance

i) If I = n Ig then
ii) If Is is current through shunt resistance, ( I – Is ) will flow through galvanometer.
Hence,
OR
This is the required shunt to increase the
range n times
This equation gives the fraction of the total current
through the shunt resistance.
By solving above equations, we get the equations for current through galvanometer ( Ig)
and current through shunt resistance (Is) given as below,

9.5.2 Galvanometer as an Voltmeter
A voltmeter is an instrument used to measure potential difference between two points in
an electrical circuit.
The necessary modifications are:
1)Its voltage measuring capacity must be increased to desired higher value
2) Its effective resistance must be increased
3) It must be protected from damages due to excess applied potential difference
This is achieved by connecting high resistance (X) in series with galvanometer.
Expression for resistance X :
.
Ig – current through galvanometer
X- resistance connected in series with galvanometer
V- voltage to be measured
This equation gives the value of resistance X
If
is the factor by which the voltage
range is increased then,

Comparison of an ammeter and a voltmeter
AMMETER
1. Measures current
2. Connected in series
3. Is an MCG (moving coil
galvanometer) with low resistance.
(Ideally zero )
4. Smaller the shunt, greater will be the
current measured
5. Resistance of ammeter is
VOLTMETER
1. Measures potential difference
2. Connected inn parallel
3. Is an MCG ( moving coil
galvanometer) with high resistance.
(Ideally infinite)
4. Larger its resistance, greater will be
the potential difference measured
5. Resistance of voltmeter is

PR0BLEMS (CURRENT ELECTRICITY)
(10) A battery of emf 4 volt and internal resistance 1 Ω is
connected in parallel with another battery of emf 1 volt and
internal resistance 1 Ω ( with their like poles connected together )
.The combination is used to send current through an external
resistance 2 Ω .Calculate the current through the external
resistance.
Solution : -------

Applying Kirchhoff’ second law to the loop ABCDCFA, we get
- 2 ( I 1+ I 2 ) – I 1 + 4 = 0
3 I 1 + 2 I 2 = 4 ------------------ (1)
For loop BCDEB we get,
- 2 ( I 1+ I 2 ) - I 2 + 1 =0
2 I 1 + 3 I 2 = 1 ------------------( 2)

I 1 – I 2 = 3
I 1 = 3 + I 2 -------------------( 3 )
Putting value of I 1 in equation ( 2 ), we get,
2 ( 3 + I 2) + 3 I 2 = 1
6 + 5 I 2 = 1
I 2 = - 1 A -----------------------( 4 )

Subtract equation (2) from (1) ,we ge
From ( 3 )
I 1 = 3 – 1 = 2 A ----------------( 5 )
The through extertanal resistance 2 ohm from ( 4) and (5) is ,
I 1 + I 2 = 2 – 1
= 1 A ---------- answer

(11) Two cells of emf 1.5 volt and 2 volt having respective
internal resistances of 1 Ω and 2 Ω are connected in
parallel so as to send current in the same direction
through an external resistance of 5 Ω . Find current
through the external resistance.
Solution : --

Applying Kirchhoff’s second law to the loop ABCDCFA, we get
- 5 ( I 1+ I 2 ) – I 1 + 1.5 = 0
6 I 1 + 5 I 2 = 1.5 ------------------ (1)
For loop BCDEB we get,
- 5 ( I 1+ I 2 ) - 2 I 2 + 2 =0
5 I 1 + 7 I 2 = 2 -------------------( 2)
∴

Subtract equation (2) from (1) ,we get
I1- 2 I2 = - 0.5
I1 = -0.5 + 2 I2------------( 3)
Putting this value in equation (2) ,we get,
5 ( - 0.5 + 2 I2 ) + 7 I2 = 2
- 2.5 + 17 I2 = 2
17 I2 =2 + 2.5 =4.5
I2=
4.5
17 -------------------( 4)

From equation (3) ,
I1 = - 0.5 + 2 x
𝟒.𝟓
𝟏𝟕
=
1
34
A ------------( 5)
Therefore current through external resistance is,
I1 + I2 =
1
34
+
𝟒.𝟓
𝟏𝟕
=
10
34
=
5
17
A ------- answer

(12) A voltmeter has a resistance 30 Ω . What will be its reading,
when it is connected across a cell of emf 2 V having internal
resistance 10 Ω ?
Solution:-
Given : R = 30 Ω , E = 2 volt , r = 10 Ω, V = ?
I =
𝐸
𝑅+𝑟
=
2
30+10
=
1
20
A
V = I R =
1
20
x 30
=
3
2
=1.5 volt

(13) A set of three coils having resistances 10 ohm ,12 ohm
and 15 ohm are connected in parallel.This combination is
connected in series with series combination of three coils
of same resistances . Calculate the total resistance and
current through the circuit , if battery of emf 4.1 Volt is
used for drawing current.
Solution :- Figure-

1
Rp
=
1
10
+
1
12
+
1
15
=
6+5+4
60
=
15
60
∴ Rp = 4 ohm
Total resistance R = 4 + 10 + 12 + 15
= 41 Ohm
Current I =
𝑉
𝑅
=
4.1
41
∴ I = 0.1 A

(14) A potentiometer wire has a length of 1.5 m and
resistance of 10 ohm .It is connected in series with
the cell of emf 4 Volt and internal resistance 5
ohm . Calculate drop per centimetre of the wire.
Solution:- Given – L =1.5 m =150 cm , R = 10 ohm,
E = 4 v ,r = 5 ohm.
𝑉
𝐿
=?
𝑉
𝐿
=
𝐸.𝑅
𝐿 (𝑅+𝑟)

=
4 𝑥 10
150(10+5)
=
40
150𝑥15
=
4
225
= 0.01777
𝑉
𝑐𝑚
= 0.0178
𝑉
𝑚

(15) When two cells of emfs E1 and E2 are connected in series so
as to assist each other ,their balancing length on the
potentiometer is found to be 2.7 m. When cells are
connected in series so as to oppose each other, the
balancing length is found to be 0.3 m . Compare the emfs of
two cells.
Solution:-
𝐸1
𝐸2
=
𝐿1
+𝐿2
𝐿1
−𝐿2
=
2.7+0.3
2.7−0.3
=
3
2.4
=1.25 Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 52

(16) The emf of cell is balanced by a length of 120 cm of
potentiometer wire .When the cell is shunted by resistance
of 10 ohm, the balancing length is reduced by 20 cm . Find
the internal resistance of cell.
Solution :- Given- L1=120cm=1.2m,R=10 ohm
L2=120-20=100cm= 1m, r =?
r =
𝑅(𝐿1
−𝐿2
)
𝐿2
=
10(1.2−1)
1
=10 (0.2)
= 2 ohm Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 53

(17) A potential drop per unit length along a wire is 5 x 10 -3 𝑉
𝑚
.
If the emf of cell balances against length 216 cm of the
potentiometer wire ,Find the emf of cell.
Solution :- Given-
𝑉
𝐿
= 5 x 10 -3 𝑉
𝑚
, L=216cm = 2.16 m, E=?
𝑉
𝐿
= 5 x 10 -3
V=5 x 10 -3 x L
= 5 x 10 -3 x 2.16
= 0.01080 volt
As emf of cell balances against length 216 cm, then V=E
E= 0.01080 volt

(18) The resistance of potentiometer wire 8 ohm and its length
8 m. A resistance box and a 2 volt battery are connected in
series with it .What should be the resistance in box ,if it
desired to have a potential of 1 μ
𝑉
𝑚𝑚
?
Solution:- Given- R=8 ohm, L =8 m, E = 2 v.
𝑉
𝐿
= 1 μ
𝑉
𝑚𝑚
= 10-3 𝑣
𝑚
=0.001
Let R1 is the resistance in resistance box
𝑉
𝐿
=
𝐸.𝑅
𝐿(𝑅+𝑅1
)
0.001 =
2𝑥8
8(𝑅+𝑅1
)

R+R1 =
𝟐
𝟎.𝟎𝟎𝟏
= 2000
R1 = 2000 – R
= 2000 – 8
= 1992 ohm

(19) Find the equivalent resistance between the terminals
A and B in the network shown in figure below given that
the resistance of each resistor is 10 ohm.
Solution:- Applying Kirchhoff’s voltage law to loop
ABHGA we get,
4I2 – I1 =I ----------------(1)
Applying Kirchhoff’s voltage law to loop

GHFG We get,
-I2 +4I1 = 2I - -------------------(2)
Solving equation (1) and (2) we get
I1 =3/5 I And I2 = 2/5 I } ----------------(3)
Applying Kirchhoff’s voltage law to loop BCDEFHB we get.
-20 (I – I1 ) – 10 ( I – I2 ) + E =0
Putting the values of I1 and I2 from equation (3), we get

E =14 I -------------------(4)
If R is the equivalent resistance between A
and B then
E = I.R ---------------(5)
From equation (4) and (5) we have,
I.R = 14 I
R = 14 ohm
Therefore the equivalent resistance between
A and B = 14 ohm.

( 20) A voltmeter has a resistance of 100 ohm . What will be its
reading when it is connected across a cell of emf 2 volt
and internal resistance 20 ohm ?
Solution:- Given- R= 100 ohm,E=2 volt, r =20 ohm, V = ?
Current, I =
𝐸
𝑅+𝑟
=
2
100+20
=
2
120
=
1
60
A.
V = I.R
=
1
60
x 100
= 1.66 volt
Therefore reading shown by voltmeter =1.66 volt

QUESTION BANK
One mark questions:
1. What is mean by shunt resistance?
2. How Galvanometer is converted into voltmeter?
3. What is potential gradient?
Two mark questions:
1. State kirchhoff's laws for electrical circuit.
2. Explain the principle of a Potentiometer.
3. State any two sources of error in meter-bridge experiment.
4. State the uses of a potentiometer.
5. What are the disadvantages of a Potentiometer?
6. Distinguish between a potentiometer and voltmeter.
7. Distinguish between ammeter and voltmeter.
8. State the uses of the shunt resistance.

Three mark questions:
1. With the help of neat circuit diagram, obtain the balancing condition for
Wheatstone’s network.
2. Explain with neat circuit diagram, how the unknown resistance is determined by
using a meter-bridge.
3. Describe Kelvin's method to determine the resistance of a galvanometer by using a
meter-bridge.
4. Describe how a potentiometer is used to compare the EMFs of two cells by
connecting the cells individually.
5. Describe how potentiometer is used to compare the EMFs of two cells by sum and
difference method.
6. Describe with the help of a neat circuit diagram, how the internal resistance of a cell
is determined by using a Potentiometer. Derive the necessary formula.
7. Explain how a moving coil galvanometer is converted into an ammeter. Derive the
necessary formula.
8. Explain how a moving coil galvanometer converted into voltmeter . Derive the
necessary formula

There is no deleted portion due to
COVID-19 for Feb/Mar 2021 HSC
Exam. from this chapter.

E-CONTENT DEVLOPED BY……..
(1) Shri.D.B. Jadhav ( Kirchhoff’s Laws, Problems)
(2) Sou.C.C.Patil(Potentiometer,Galvanometer,Q.BanK)
(3) Miss D.P. More (Wheatstone Bridge)
Dattajirao Kadam Arts, Commerce And Science College,
Ichalkaranji.

9. Current Electricity

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