CURRENT ELECTRICITY
ELECTRIC CURRENT
Electric current across an area held perpendicular to the direction of flow of charge is defined to be
the amount of charge flowing across the area per unit time. If charge Q
Δ passes through the area in time
interval t
Δ , at uniform rate the current I is defined by
t
Δ
Q
Δ
I  ...(i)
If rate of flow of charge is not steady then instantaneous current is given by
t 0
Q
I lim
t
 



dQ
dt
 ...(ii)
The unit of current is ampere. Ampere (A) is the S.I. base unit, defined in terms of its magnetic
effect. Smaller currents are more conveniently expressed in milliampere (1mA = 10-3
A) or microampere
 .
A
10
A
μ
1 6

 .
Illustration: Two boys Aand B are sitting at two points in a field. Both boys are sitting near assemblence
of charged balls each carrying charge +3e.
A throws 100 balls per second towards B while B throws 50 balls per second towards A.
Find the current at the mid point of A and B.
Solution: Let mid point be C as shown
A C B
100e
50e
Charge moving to the right per unit time = 100 × 3e = 300e
Charge moving to the left per unit time = 50 × 3e = 150e
Movement of charge per unit time is 300e –150e = 150e towards right
I = 150e = 150  1.6  10–19
A.
Illustration: Flow of charge through a surface is given as 2
Q 4t 2t
  (for 0 to 10 sec.)
(a) Find the current through the surface at t = 5sec.
(b) Find the average current for (0 – 10 sec)
Solution: (a) Instantaneous current
2
dQ d
I (4t 2t) 8t 2
dt dt
    
at t = 5 sec;
I = 8 5 2 42
   Amp
(b) Average current
2
Q Q 4 (10) 2 10 420
I 42
t t 10 10
   
    

Amp.
Brain Teaser: In an electrolyte, the positive ions move from left to right and the negative ions from right to
left. Is there a net current? If yes, in what direction?

ELECTROMOTIVE FORCE(EMF)
To maintain a steady electric current, the conductor cannot be isolated; it must be part of a closed
circuit that includes an external agency or device (figure). This device is required to transport the positive
charge from B back to A, i.e., from lower to higher potential and thus maintain the potential difference
between A and B. The external device will need to do work for transporting positive charge from lower to
higher potential. Such a device is the source known as electromotive force abbreviated as emf. It is the
analogue of the pump in the water flow circuit.
A B
R
I
+
-
The external source, as said above, does work on taking a positive charge from lower to the higher
potential.
A natural way of characterizing the external source of energy is in terms of the work that it needs to
do per unit positive charge in transporting it from lower to higher potential. This is known as electromotive
force or emf of the device, denoted by ε .
open
V
ε 
The emf of a source is thus the potential difference between its two terminals in open circuit i.e.
when no external resistances are connected.
RESISTANCESAND RESISTIVITY
OHM’SLAW
Consider a closed circuit having a source of emf and a conductor in the external circuit. Let the
voltage drop across the ends of the conductor be V and let I be the steady current flowing through the
conductor. The quantity V/I is a measure of the resistance offered by the conductor for the steady flow of
charge through it. The more the resistance, the less is the current I for a given voltage difference, V.
For many conductors, it is found experimentally that theratio V/I is a constant at constant temperature,
the constant is called the resistance of the conductor and is denoted by R.
IR
V
or
R
I
V

 ...(iii)
The constancy of R implies that V and I are linearly related – a graph between measured values of
V and I is a straight line. The unit of resistance is ohm  
1
VA
1
Ω
1 
 . Ohm’s law is only an empirical law that
holds approximately for many substances over certain ranges of V and I.
I
V

RESISTIVITY
The resistance of a resistor (an element in a circuit with some resistance R) depends on its geometrical
factors (length, cross-sectional area) as also on the nature of the substance of which the resistor is made. It
is convenient to separate out the ‘size’ factors from the resistance R so that we can define a quantity that is
characteristic of the material and is independent of the size or shape. Consider a rectangular slab of length
l and area of cross section A. For a fixed current I, if the length of the slab is doubled, the potential drop
across the slab also doubles. (It is the electric field that drives the current in the conductor and potential
difference is electric field times the distance). This means that resistance of the slabs doubles with the
doubling of its length. That is, R l. Next, imagine the slab as being made of two parallel slabs, each of area
2
A
. If for a given voltage V, the current I flows across the full slab, it is clear that through each half-slab, the
current flowing is
2
I
. Thus, the resistance of each half-slab is twice that of the full slab. That is,
A
1
R  .
Combining the two dependences, we get
A
R
l
 ...(iv)
or
A
ρ
R
l
 ...(v)
where ρ is a constant of proportionality called resistivity. It depends only on the nature of the
material of the resistor and its physical conditions such as temperature and pressure. The unit of resistivity is
ohm m ( m
Ω ). The inverse of ρ is called conductivity, and is denoted by σ . The unit of σ is   1
m
Ω

or mho
m-1
or siemen m-1
.
l l
(a)
(b)
A/2
A/2
Aperfect conductor would have zero resistivity and a perfect insulator would have infinite resistivity.
Though these are ideal limits, the electrical resistivity of substances has a very wide range. Metals have low
resistivity of 10-8
m
Ω to 10-6
m
Ω , while insulators like glass or rubber have resistivity, some 1018
times (or
even more) greater. Generally, good electrical conductors like metals are also good conductors of heat, while
insulators like ceramic or plastic materials are also poor thermal conductors.
ORIGIN OF RESISTIVITY
The motion of charge carriers (electrons) in a conductor is very different from that of charges in
empty space. In the latter case, under an external electric field, the charge carriers would accelerate. In a

conductor, on the other hand, when the current is steady, the charge carriers move with a certain average
drift velocity.
At any temperature, the electrons in a metal have a certain distribution of velocities. When there is
no external field, all directions are equally likely, and there is no overall drift. In the presence of an external
field, each electron experiences an acceleration of
m
eE
opposite to the field direction. But this acceleration is
momentary, since electrons are continually making random collisions with vibrating atoms or ions or other
electrons of the metal. After a collision, each electron makes a fresh start, accelerates only to be deflected
randomly again.
A
B
B’
Brain Teaser: One of your friends argues that he has read in previous chapters that there can be no
electric field inside a conductor. And hence there can be no current through it. What is the
fallacy in this argument?
We next define a physical quantity called current density vector, denoted by j. The direction of j is
the direction of flow of positive charge (or opposite to the direction of drift of electrons in a metal). The
magnitude of j is the amount of charge flowing per unit cross sectional area per second. Thus, if S is an
area element, j. S is the amount of charge flowing across the area element per second. If A is taken to be
the cross-sectional area of a wire (with the direction of A along the conventional current), j. A is nothing but
the current through the wire. In this case, j is parallel to A, so
ds
.
j
dI


ds
.
j
I



I = jA
If the drift speed of electrons is '
d
υ the amount of charge flowing across a unit cross-sectional area
in unit time is contained in a cylinder of base of unit area and height d
υ i.e., in a volume 1 × d
υ = d
υ (Fig.). If
n is the number density of electrons in the metal, i.e., the number of electrons per unit volume, the total
magnitude of charge contained in the cylinder of volume d
υ is n e d
υ . Therefore,
d
υ
e
n
j 
and d
I n e
  A
Here, e is the magnitude of electronic charge.

A
E
x=d t
Then we can rewrite
E
σ
ρ
E
j 

The more familiar form V = IR is in terms of directly measurable quantities like V and I, while the form j

= σ E

relates the basic vector quantities of the problem, namely, current density vector and the electric field
vector.
Illustration: Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional
area 1.0 × 10-7
m2
carrying a current of 1.5 A. Assume that each copper atom contributes
roughly one conduction electron. The density of copper is 9.0 × 103
kg/m3
, and its atomic
mass is 63.5 amu
Solution: The direction of drift velocity of conduction electrons is opposite to the electric field direc-
tion, i.e., electrons drift in the direction of increasing potential. The drift speed d
υ is given by
 
neA
/
I
υd  .
Now, e = 1.6 × 10-19
C, A = 1.0 × 10-7
m2
, I = 1.5 A. The density of conduction electrons, n
is equal to the number of atoms per cubic metre (assuming one conduction electron per Cu
atom as is reasonable from its valence electron count of one). A cubic metre of copper has
a mass of 9.0 × 103
kg. Since 6.0 × 1023
copper atoms have a mass of 63.5 g.
6
23
10
0
.
9
5
.
63
10
0
.
6
n 



= 8.5 × 1028
m-3
which gives
7
19
28
d
0
.
1
0
.
1
10
6
.
1
10
5
.
8
5
.
1
υ 







= 1.1 × 10-3
m s-1
.
Illustration: A potential difference of 100 V is applied to the ends of a copper wire one metre long.
Calculate the average drift velocity of the electrons. Compare it with thermal velocity at
270
C. (use the results of Previous Illustration). given 7 1 1
5.81 10 m
 
   
Solution: Since m
1
,
V
100
V
Δ 
 l .
 electric field
1
Vm
100
1
100
V
Δ 



l
Also, conductivity 1
1
7
m
Ω
10
81
.
5
σ 



N = 8.5 × 1028
m-3


7
d 19 28
5.81 10 100
E
en 1.6 10 8.5 10

  
  
  
=0.43 m s-1
m
T
k
3
υ B
rms 
 
kg
10
1
.
9
K
300
JK
10
38
.
1
3
31
1
23








= 1.17 × 105
m s-1
Brain Teaser: When electrons drift in a metal from lower to higher potential, does it mean that all the ‘free
electrons of the metal are moving in the same direction?
Brain Teaser: A steady current flows in a metallic conductor of non-uniform crosssection. Which of these
quantities is constant along the conductor: current, current density, electric field, drift speed?
TEMPERATUREDEPENDENCEOFRESISTIVITY
The resistivity of all metallic conductors increases with temperature. Over a limited temperature
range that is not too large, the resistivity of metallic conductor can often be represented approximately by a
linear relation:
 
 
0
0
T T
T
α
1
ρ
ρ 

 (i)
where 0
ρ is the resistivity at a reference temperature 0
T and T
ρ its value at temperature T. The
The
factor α is called the temperature coefficient of resistivity and has dimensions of per degree Celsius.
RESISTORS IN SERIESANDIN PARALLEL
Figure illustrates four different ways in which three resistors having resistances, R1
, R2
and R3
might
be connected between points x and y. Figure(a) shows the resistances in series that means one after another.
The current is the same in each element.
a b
x y
I
I
R1 R2 R3
(a)
R1
R2
R3
y
x
(b)
I I
R2
R1
R3
(c)
I I
I I
(d)
R3
R2
R1
x y
The resistors in Figure(b) are said to be in parallel between the points x and y. Each resistor provides
an alternative path between the points. The potential difference is the same across each element.

In Figure(c) resistors, R2
and R3
are in parallel with each other and this combination is in series with
resistance R1
. In Figure(d), R2
and R3
are in series and this combination is in parallel with R1
.
It is always possible to find a single resistor that could replace a combination of resistors in any given
circuit and leave unchanged the potential differences between the terminals of the combination and the
current in the rest of the circuit. The resistance of this single resistor is called equivalent resistance of the
combination. If any one of the networks were replaced by its equivalent resistance R, we could write
I
V
R
or
IR
V
xy
xy 

where xy
V is the potential difference between the terminals x and y of the network and I is the
current at point x or y. Hence, the method of computing an equivalent resistance is to assume a potential
difference xy
V across the actual network, compute the corresponding current I, and take the ratio xy
V /I.
If the resistors are in series as in Figure(a), current in each one must be the same and equal to the line
current I.
Hence,
3
by
2
ab
1
xa IR
andV
IR
V
,
IR
V 


by
ab
xa
xy V
V
V
V 


= I (R1
+ R2
+ R3
)
3
2
1
xy
R
R
R
I
V



But
I
Vxy
is, by definition, the equivalent resistance R. Therefore,
3
2
1 R
R
R
R 


The equivalent resistance of any number of resistors in series equals the sum of their individual resistances.
If the resistors are in parallel as in Figure (b), the potential difference between the terminals of each
must be same and equal to xy
V . If the current in each are denoted by 3
2
1 I
and
I
,
I respectively, then
3
xy
3
2
xy
2
1
xy
1
R
V
I
,
R
V
I
,
R
V
I 


since charge is not accumulated at x, it follows that














3
2
1
xy
3
2
1
R
1
R
1
R
1
V
I
I
I
I
or
3
2
1
xy R
1
R
1
R
1
V
I



But R
1
V
I
xy
 , so that

3
2
1 R
1
R
1
R
1
R
1



for any number of resistors in parallel, the reciprocal of the equivalent resistance equals the sum of
the reciprocal of their individual resistances.
For special case of two resistors in parallel,
2
1
1
2
2
1 R
R
R
R
R
1
R
1
R
1 



and
2
1
2
1
R
R
R
R
R


also, Since 2
2
1
1
xy R
I
R
I
V 

1
2
2
1
R
R
I
I

and the current carried by two resistors in parallel are inversely proportional to their resistances.
The equivalent resistance of the networks in Figure(c) and Figure(d) could be found by the same
general method, but it is simpler to consider them as combination of series and parallel arrangements. Thus,
in figure(c) the combination of R2
and R3
in parallel is first replaced by its equivalent resistance which then
forms a simple series combinations with R1
. In Figure(d), the combination of R2
and R3
in series forms a
simple parallel combination with R1
. Not all networks, however, can be reduced to simple series-parallel
combinations and special methods must be used for handling such networks.
F C
D
A B
6
3
3
3
3
3
6
6
E
Illustration: Find the effective resistance between the point A and B.
Solution: Resistors AF and FE are in series with each other. There-
fore, network AEF reduces to a parallel combination of
two resistors of 6 each. Req.
6 6
3
6 6

  

.
Similarly, the resistance between A and D is given
6 6
3
6 6

 

.
Now, resistor AC is in parallel with AD. Therefore, the resistance between A and C
is
6 3
2
9

  .
Now AC and BC are in series with eq
R 5
  and this R is in parallel with AB
eq.
5 3 15
R
8 8

  
ELECTRICCIRCUITSANDKIRCHHOFF’SRULES
Most electrical circuits consist not merely a single source and a single external resistor, but comprise
of a number of sources, resistors or other elements such as capacitors, motors, etc. interconnected in a
complicated manner. The general term applied to such a circuit is called network.

INTERNALRESISTANCE
Suppose for flow of unit charge in the circuit.
potential difference across the load is V
By energy conservation
'
eq vq w
 
with q = 1 unit
then '
V w
  
V

where w is the part of the total work done by source of emf per unit charge  
ε used up in the
source itself . In a cell, this happens because in moving the positive ions from lower to higher potential (or
negative ions from the higher to lower potential), the other ions and atoms of the electrolyte offer resistance.
Now, by Ohm’s law V = IR. Assuming that Ohm’s law is also valid for the flow of current in the source, we
can assign an internal resistance r to the source and write '
W = Ir,
,
 
r
R
I
Ir
IR
Ir
V
ε 





this can also be written as
Ir
ε
V 

showing that the external voltage is less than the emf of the source by the quantity Ir. It is as if an
internal resistance r combines in series with the external resistance R to determine the current in the circuit
for a given source of emf. Clearly, If I = 0, i.e., the circuit is open, V = Vopen
.
Brain Teaser: The current passing through a battery of emf  and internal resistance r is decreased by
some external means. Does the potential difference between the terminals of the battery
necessarily decrease or increase? Explain.
KIRCHHOFF’SRULES
Using the fact that there is no net current at the junction i.e., the incoming current equals the outgoing
current and if we complete the circuit via a path the total potential change is zero. These facts, called
Kirchhoff’s rules, are very useful for many electrical circuit problems; they are discussed in detail below :
(a) Junction rule: This rule is based on the fact that charge cannot accumulate at any point in a conductor
in a steady situation. It states that ‘at any junction of several circuit elements, the sum of currents
entering the junction must equal the sum of currents leaving it’. Unless this happens, net positive
or negative charge will accumulate at the junction at a rate equal to the net electrical current at the
junction. Consider for example, the circuit in figure given below. At the junction ‘a’, I3
flows in and I1
and I2
flow out. We must then have
2
1
3 I
I
I 
 ...(i)
Applying this to another junction of this circuit leads to nothing new equation. For further progress in
analyzing this circuit, we need another rule, given below:
(b) Loop rule : ‘The algebraic sum of changes in potential around any closed resistor loop must
be zero’. Otherwise, one can continuously gain energy by circulating charge around a closed loop in
a particular direction. So, this rule is based on energy conservation. Now consider the loop ‘ahdcba’
in Figure.
we have, from Kirchhoff’s second rules,
-30I1
, -41I3
+ 45 = 0 ...(ii)
For the second loop, the circuit ‘ahdefga’ is taken. We have

-30I1
, +21I2
– 80 = 0 [iii]
Calculate Using Eq. (i), (ii) & (iii) we I1
= 0.86A, I2
= 2.59A and I3
= 1.73A, respectively.
g I2
d
30
b c
h
40
1
20
e
I2
I1
I3
f
80 V
45 V
a
1
SIGN CONVENTION INAPPLYINGKIRCHHOFF’SRULES
First, all quantities, known as unknown, should be labelled carefully, including an assumed sense of
directions for each unknown current or emf. The solution is then carried out using the assumed directions, and
if the actual direction of a particular quantity is opposite to the assumed direction, the value of the quantity will
emerge from the analysis with a negative sign. Hence, Kirchhoff’s rules, correctly used, give the direction
and magnitude of unknown currents and emf’s.
Usually in labelling currents it is advantageous to use the point rule immediately to express the
currents in terms of as few quantities as possible. For example, Figure(a) shows a circuit correctly labelled,
and Figure(b) shows the same circuit, relabelled by applying the point rule to point a to eliminate I3
.
The following guidelines will help with the problems of signs:
1. Choose any closed loop in the network, and designate a direction (clockwise or counter clockwise) to
transverse the loop in applying the loop rule.
2. Go around the loop in the designated direction, adding emf’s and potential differences. An emf is
counted as positive when it is traversed from (-) to (+) and negative when transformed from (+) to
(-). An IR term is counted negative if the resistor is traversed in the same direction of the assumed
current, and positive if in the opposite direction.
I1
I3
I2
R2
R1
R3
Figure (a)
I1
I1+I2
I2
R2
R1
R3
1,r1 2,r2
Figure (b)
a

3. Equate the sum of step (2) to zero.
4. If necessary, choose another loop to obtain different relations between the unknowns, and continue
until there are as many equations and unknowns or until every circuit element has been included in at
least one of the chosen loops.
Illustration: Find the potential differnce A B
V V
 and the rate of production of heat in 1
R in the network
shown in fig.
C
A B
D
E
F H
G
2
2
2
1
i3
i1
i2 R2
R4
R1
R3
2F
1
2
with 1
 =12 V and 2
 = 3 V
Solution: At the steady state, the capacitor branch CDGH acts as an open branch, i.e., of infinite
resistance with zero current in it.
If 1 2
i ,i and 3
i be the currents in the ED, AB and GF branches respectively, then from the
Kirchhoff’s first law, we get at point D
1 2 3
i i i 0
   or 3 1 2
i i i .
 
Assuming the sources of negligible resistance and applying Kirchhoff’s second law to the
meshes ABCDEA and EDGFE, we get,
2 2 1 1 2 1
i R i R
    
and  
1 1 1 2 3 1
i R i i R
      
2 3 1 1 3 1
i R i R R
   
solving these equations for 2
i , and 1
i , we get
 
1
12 2 2 3 2
i 3.5
4 4 4
  
 
 
amp
and
 
2
3 2 2 12 2
i 1
4 4 4
  
  
 
amp
 A B
V V 3 1 2 5 volts.
      
Rate of production of heat in  
2
2
1 1 1
R i R 3.5 2 24.5 joules.
   

Brain Teaser: When a current is established in a wire, the free electrons drift in the direction opposite to
the current. Does the number of free electrons in the wire continuously decrease?
ENERGY, POWERANDHEATINGEFFECT
When a current I flows for time t from a source of emf E, then the amount of charge that flows in
time t is Q = I t.
Electrical energy delivered W = Q. V = V I t
Thus, Power given to the circuit, = W/t =VI or V2
/R or I2
R
In the circuit
E. I = I2
R + I2
r, where
E I is the rate at which chemical energy is converted to electrical energy, I2
R is power supplied to the
external resistance R and I2
r is the power dissipated in the internal resistance of the battery.
An electrical current flowing through conductor produces heat in it. This is known as Joule’s effect. The heat
developed in Joules is given by H = I2
.R.t
I
 r
R
Illustration : An electric bulb rated 220V and 60 W is connected in series with another electric bulb rated
220V and 40 W. The combination is connected across 220 volt source of emf which bulb will
glow more ?
Solution :
P
V
R
2

 Resistance of first bulb is
1
2
1
P
V
R 
And resistance of the second bulb is
2
2
2
P
V
R 
In series same current will pass through each bulb
 Power developed across first is
1
2
2
1
P
V
I
'
P 
and that across second is
2
2
2
2
P
V
I
'
P 
1
2
2
1
P
P
'
P
'
P



as 1
P
P
P
P
1
2
1
2 


2
1
2
1
'
P
'
P
1
'
P
'
P




The bulb rated 220 V & 40 W will glow more.
Brain Teaser: Two unequal resistances 1
R and 2
R are connected across two identical batteries of emf 
and internal resistance r (figure). Can the thermal energies developed in 1
R and R2
be equal
in a given time. If yes, what will be the condition?
r

R1
r

R2
MAXIMUM POWERTRANSFERTHEOREM
In a circuit, for what value of the external resistance the maximum power drawn from a battery? For
the shown network power developed in resistance R equals
 
2
2
2
E .R E
P ( I and P I R)
R r
R r
  



I
 r
R
Now, for dP/dR = 0 (for P to be maximum 0
dR
dP
 )
    
 
0
r
R
r
R
R
2
r
R
.
E 4
2
2






  R
2
r
R 


or R = r
The power output is maximum, when the external resistance equals the internal resistance. R = r
Illustration : A copper wire having a cross-sectional area of 0.5 mm2
and a length of 0.1 m is initially at
250
C and is thermally insulated from the surroundings. If a current of 10 A is set up in this
wire
(a) Find the time in which the wire starts melting. The change of resistance of the wire with
temperature may be neglected.

(b) What will this time be, if the length of the wire is doubled ?
Density of Cu = 9 × 103
Kg m-3
specific heat of Cu = 9 × 10-2
Cal Kg-1 0
C-1
,
M.P. (Cu) 10750
C and specific resistance = 1.6 × 10-8
m
Ω .
Solution : (a) Mass of Cu = Volume × density
= 0.5 × 10-6
× 0.1 × 9 × 103
= 45 × 10-5
Kg.
Rise in temperature = θ = 1075-25 = 1050 0
C.
Specific heat = 9 × 10-2
Kg-1 0
C × 4.2 J
R
.
I
θ
mS
t
θ
mS
Rt
I 2
2




but Ω
.
10
2
.
3
10
5
.
0
1
.
0
10
6
.
1
A
L
ρ
R 3
6
8










s
558
10
2
.
3
10
10
09
.
0
1050
10
2
.
4
45
t 3
5









 

(b) When the length of wire is doubled, R is doubled, but correspondingly mass is also
doubled. Therefore, wire will start melting in the same time.
BATTERY COMBINATION :
(i) Series Combination :-
eq
= n
req
= nr I =
L
R
nr
n



(a) RL
>>> nr I =
R
n
.
This combination is useful as current across load resistance can be varied by varying nunmber of
cells in the circuit.
(b) RL
<<< nr  I =
r

. This combination is not prefered
(c) In general, equivalent emf of the battery is worked out using
Thevenin theorem
(i) VAB
= 1
+ 2
RAB
= r1
+ r2

eq


(ii)
A B
RL
 VAB
= |1
– 2
|
RAB
= r1
+ r2
(ii) PARALLEL COMBINATION :-

eq
=  Req
=
n
r
,  I =
R
n
r


I =
nR
r
n


Illustration: Find equivalent eq
 of the battery as shown in figure and A B
V V
 ?
(i) (ii)
Solution: (i) Current in the circuit 1 2
1 2
i
r r
  


...(i)
eq. 1 2
    
and VA
– 2
– r2
2
1
2
1
r
r 



= V
VB
 VA
– VB
=
2
1
1
2
2
1
r
r
r
r




(ii) Current 1 2
1 2
i
r r
  


...(i)
and VA
– 1
+ r1 











2
1
2
1
r
r = V
VB
eq


VA
– VB
=
2
1
1
2
2
1
r
r
r
r




.
(iii) MIXED COMBINATION

req
=
m
nr
, I =
m
nr
R
n


 I =
mR
nr
mn


EFFECTIVEGROUPING OFCELLS
For effective grouping current should be maximum
 mR + nr should minimum
Now, mR + nr =       mnrR
2
nr
mR
nr
mR
2
2
2




The second term is nonzero the term in the parenthesis
 
2
0
mR nr
 
 mR = nr
 r
m
n
R 
Illustration: Length of AB = 6 m
Find the position of Null pt ? Given AB
R 15r

J
Soln. AJ AJ
15r
V iR .x
16r 6
  
   
 
...(i)
also for J to be null point
 AJ
V
2

 ...(ii)

equating (i) and (ii)
x 320cm

Illustration : There are 27 cells with an internal resistance 0.4 Ω and an external resistance 2.4 Ω . What
is the most effective way of grouping them ?
Solution : Let there be n cells series of m parallel branches
mn = 27 ...(i)
mR = nr ...(ii)
1.2m = n (0.4)
 3m = n ...(iii)
from equation (i) and (iii), we get
m = 3 & n = 9
SUPERPOSITIONPRINCIPLE
Concepts :
Whenever a circuit has more than one cell or battery, the superposition principle may be
used to find current and voltages. This principle is based on the fact that every cell or battery acts
independent of the presence of others.
According to this principle, the total current I in the circuit equals the algebraic sum of current
I1
, I2
, ............... In
produced by each source (cell or battery) taken one at a time.
Mathematically, I= I1
+ I2
+ .......... + In
The superposition splits the original two - source problem into two one - source problems. Instead of
solving a difficult two source problem, we can solve to simple problems.
Determine the current I in the 2 resistor
I = I1
+ I2
=
2
6
+
2
4
= 3 + 2 = 5 A.
A.
AMMETER
It is an instrument used to measure currents. It is put in series with the branch in which current is to
be measured. An ideal Ammeter has zero resistance. A galvanometer with resistance G and current rating ig
can be converted into an ammeter of rating I by connected a suitable resistance S in parallel to it. (The
resistance connected in parallel to the ammeter is called a shunt.)
Thus S(I – ig
) = ig
G


g
g
i G
S
I i


G
A
I I
S
I
ig
I - ig
Illustration : A galvanometer having a coil resistance of 100Ω gives a full scale deflection when a current
of 1 mA is passed through it. What is the value of the resistance which can convert this
galvanometer into ammeter giving full scale deflection for a current of 10A ?
Solution :
  
  99
.
9
1
.
0
A
10
10
Ω
100
A
10
i
i
G
.
i
S 3
3
g
g




 

 Ω
10
Ω
99
.
99
1
S 2



VOLTMETER
It is an instrument to find the potential difference across two points in a circuit.
It is essential that the resistance Rv
of a voltmeter be very large compared to the resistance of any circuit
element with which the voltmeter is connected. Otherwise, the voltmeter itself becomes an important circuit
element and alters the potential difference that is measured.
Rv
>> R
For an ideal voltmeter Rv
=  .
V
G
R1
R1
Rv
G V
ig
Rv

G
i
V
R
V
)
R
G
(
i
g
g 




Illustration : A galvanometer having a coil resistance 100 Ω gives a full scale deflection when a current
of 1 mA is passed through it. What is the value of the resistance which can convert this
galvanometer into a meter giving full scale deflection for a potential difference of 10V ?
Solution : V = lg
[G+Rv
]
10=(10-3
) (100 + Rv
)
 Ω
K
9
.
9
Ω
900
,
9
100
10
10
R 3
v 








 
Brain Teaser: An ammeter is always connected in series and voltmeter connected to parallel to any circuit
element, why ?
WHEATSTONEBRIDGE
Figure shows the fundamental diagram of wheatstone bridge. The bridge has four resistive arms,
together with a source of emf (a battery) and a galvanometer. The current through the galvanometer depends
on the potential difference between the point c and d. The bridge is said to be balanced when the potential
difference across the galvanometer is 0V so that there is no current through the galvanometer. This condition
occurs when the potential difference from point c to point a, equals the potential difference from point d to
point a; or by referring to the other battery terminal, when the voltage from other point c to point b equals the
voltage from point d to point b. Hence, the bridge is balanced when
1 1 2 2
I R I R
 …(i)
if the galvanometer current is zero, the following conditions also exist:
3
1
3
1
R
R
ε
I
I


 …(ii)
and
4
2
4
2
R
R
ε
I
I


 …(iii)
G
a
b
c d
R1 R2
R3 R4
I1 I2
I3
I4
Unknown
Standard
arm

Combining Eqs. (i), (ii) and (iii) and simplifying, we obtain
4
2
2
3
1
1
R
R
R
R
R
R


 …(iv)
from which we get
3
2
4
1 R
R
R
R  …(v)

Equation (v) is the well known expression for balance of the wheatstone bridge. If three of the resistances
have known values, the fourth may be determined from Equation (v). Hence, if R4
is the unknown resistor,
its resistance can be expressed in terms of remaining resistors
1
2
3
4
R
R
R
R  …(vi)
Resistance R3
is called the standard arm of the bridge and resistors R2
and R1
are called the ratio arms.
Illustration: The four arms of a wheatstone bridge (figure) have the following resistances :
AB = 100 Ω , BC = 10 Ω , CD = 5 Ω and DA = 60 Ω .
The galvanometer of 15Ω resistance is connected across BD. Calculate the current through
the galvanometer when at potential difference of 10 V is maintained across AC.
Solution: Considering the mesh BADB, we have 100 I1
+ 15 Ig
– 60 I2
= 0
or 20 I1
+ 3 Ig
– 12 I2
= 0 …(i)
Considering the mesh BCDB, we have
– 10 (I1
– Ig
) + 15 Ig
+ 5 (I2
+ Ig
) = 0
 10 I1
– 30 Ig
– 5 I2
= 0
2 I1
– 6 Ig
– I2
= 0 …(ii)
Considering the mesh ADCEA,
– 60 I2
– 5 (I2
+ Ig
) + 10 = 0
 65 I2
+ 5 Ig
= 10
13 I2
+ Ig
= 2 …(iii)
10
60 5
100 
I2
I +
2 Ig
I –
1 Ig
I1
I2
Ig
15
D
A
C
B
10 V
E
I
I
Multiplying Equation (ii) by 10
20 I1
– 60 Ig
– 10 I2
= 0 …(iv)
From (iv) and (i) we have
63 Ig
– 2 I2
= 0
2 g g
63
I I 31.5I
2
  …(v)
Substituting the value of I2
into Equation (iii), we get
13 (31.5 Ig
) + Ig
= 2
410.5 Ig
= 2
Ig
= 4.87 mA.

METRE BRIDGE – Special case of Wheatstone Bridge
This is the simplest form of wheatstone bridge and is specially useful for comparing resistances
more accurately. The construction of the metre bridge is shown in the Figure. It consists of one metre
resistance wire clamped between two metallic strips bent at right angles and it has two points for connection.
There are two gaps; in one of them a known resistance and in second an unknown resistance whose value
is to be determined is connected. The galvanometer is connected with the help of jockey across BD and the
cell is connected across AC. After making connections, the jockey is moved along the wire and the null point
is from two resistances of the wheatstone bridge, wire used is of uniform material and cross-section. The
resistance can be found with the help of the following relation:
 
1
1
R
S 100


 


1
1
R
S 100




1
1
100
S
R
l
l

 . . . (1)
where σ is the resistance per unit length of the wire and l1
is the length of the wire from one end
where null point is obtained. The bridge is most sensitive when null point is somewhere near the middle point
of the wire. This is due to end resistances.
G
100–l1
l1
R S
B
C
A
Metre scale
 K1
END CORRECTION :
Sometimes at the end points of the wire, some length is found under the metallic strips and as a result,
in addition to length l1
or (100 –l1
) some additional length should be added for accurate measurements. The
resistance due to this additional length is called end resistance. If the end resistance is small it can be
determined by first introducing known resistances P and Q in the gap and obtaining the null point reading l1
,
then interchanging P and Q and obtaining the null point reading l2
. Let  and  be the lengths on the
respective end under the metallic strips, then we have
1
1
100
l
P
Q l




  ...(i)
2
2
Q
P 100
l
l
 

   …(ii)
Solving Equations. (i) and (ii) for β
and
α , we have
Q
P
P
α 2
1



l
Ql
…(iii)

100
Q
P
Q
P
β 2
1




l
l
…(iv)
hence the value of β
,
α can be calculated and suitably accounted for when accurate measurements
are required.
Illustration: Figure (a) shows a metre bridge (which is nothing but a practical Wheatstone bridge) con-
sisting of two resistors X and Y together in parallel with a metre long constantan wire of
uniform cross section. With the help of a movable contact D, one can change the ratio of the
resistances of the two segments of the wire, until a sensitive galvanometer G connected
across B and D shows no deflection. The null point is found to be at a distance of 33.7 cm
from the end A. the resistance Y is shunted by a resistance Y of 12.0 Ω [Figure (b)] and the
null point is found to shift by a distance of 18.2 cm. Determine the resistance of X and Y.
G
X Y
A C
D
B
(a)
G
X Y
A C
D
B
(b)
12 
Solution: Since the wire is of uniform cross-section, the resistances of the two segments of the wire
AD and DC are in the ratio of the lengths of AD and DC. Using the null-point conditions of
a wheatstone bridge, we have
(X/Y) = (33.7/66.3) (1)
When Y is shunted by a resistance of 12.0 Ω , net resistance changes
Y’ = 12 Y (Y + 12)
Since Y’ is less than Y, the ratio X/Y’ is greater than X/Y’. Thus the null point must shift
towards the end C, i.e.,
(X/Y’) = (51.9/48.1)
or X (Y + 12) / 12 Y = (51.9 / 48.1)
i.e.
7
.
33
3
.
66
1
.
48
9
.
51
12
12
Y









which give Y = 13.5 Ω and X = 6.86 Ω , using (1)
POTENTIOMETER
1
3
2
J
B
A


1
2
G
R

K
This instrument is identical to the meter bridge but more sensitive
than meter bridge because resistance wire is of more than meter in length.
It obeys the rule R   (for constant cross section area of wire and tem-
perature)
Let wire AB has total resistance 0
R
Y’

Current I in the circuit =
 
0
R R


p.d. across AB
 
 
0 0
0
V R
R R



p.d. per unit length 0
0
R 1
R R AB
 
  
    
  
 
It is used in several process like comparing emfs of two batteries and to calculate resistnace of unknown
resistor. 1 3
2
J
B
A


1
2
G
R

K
C
Let 1
 is connected by connecting 1 3
 and a null point is observed at
‘C’ on AB
then
 
0
1
0
R AC
R R AB
  
   
  
...(i)
and similarly C is null point for 2
 in circuit 1
3
2
J
B
A


1
2
G
R

K
C’
then
0
2
0
R AC
R R AB
  

   

 
...(ii)
Using (i) and (ii)
1
2
AC
AC




A
B
C
V
R
R0
Illustration: A resistance of R Ω is powered from a poten-
tiometer of resistance Ω
R0 (figure). A volt-
age V is supplied to the potentiometer. Derive
an expression for the voltage fed into the
branch BC when the slide is in the middle of
the potentiometer.
Solution: While the slide is in the middle of the potentiometer only half of its resistance  
2
/
R0 will be
between the points A and B. Hence, the total resistance between A and B, say, R1
, will be
given by the following expression :
 
2
/
R
1
R
1
R
1
0
1


R
2
R
R
R
R
0
0
1


The total resistance between A and C will be sum of resistance between A and B and B and
C, i.e.,

2
/
R
R 0
1 
 The current flowing through the potentiometer will be
0
1
0
1 R
R
2
V
2
2
/
R
R
V
I




The voltage V1
taken from the potentiometer will be the product of current I and resistance
R1
,
0
1
1
R
R
R
2
V
2
IR
V 










 0 0
2
2
Substituting for R1
, we have
2
R
R
R
2
R
R
R
2
V
2
V 0
0
0
0
1 












Brain Teaser: Would you prefer a voltmeter or a potentiometer to measure the emf of a battery?
RC-CIRCUIT
Charging: Let us assume that the capacitor in the shown network is uncharged for t < 0. The switch is
connected to position 1 at t = 0.
Now, ‘C’ is getting charged.
If the charge on capacitor at time ‘t’ is q.
Writing the loop rule,
0
E
IR
C
q



1
2
I
I
I
R
S
E C

c
q
E
dt
dq
R 

 q
EC
dt
dq
RC 

 dt
RC
1
q
EC
dq


Integrating  


q
0
t
0
dt
RC
1
q
EC
dq
- ln | EC - q t
.
RC
1
|q
0 

RC
t
EC
q
EC
ln



 t / RC
q EC[1 e ]

 
 At t = 0, q = 0
and at t =  , q = E C (the maximum charge.)
thus,  
RC
/
t
max e
1
q
q 


t
q

qmax = EC
q = 0.63 EC

RC
/
t
RC
/
t
max
e
R
E
e
RC
q
dt
dq
i 




R
E
i
where
e
i
i max
RC
/
t
max 
 
t
i
imax = E/R
TIMECONSTANT  
τ
It is the time during which the charging would have been completed, had the growth rate been as it
began initially. Numerically it is equal to RC.
DISCHARGING
Consider the same arrangement as we had in previous case with one difference that the capacitor
has charge q0
for t<0 and switch is connected to position 2 at t = 0. If the charge on capacitor is q at any later
moment t then the loop equation given as
Flip the switch to 2
0
IR
C
q


C
S
R

1
1
2

dq q
R
dt C


 dt
RC
1
q
dq 

Integrating, at t = 0, q = q0
t t, q q
 
  


q
q
t
0
0
dt
RC
1
q
dq

RC
/
t
0
0
e
.
q
q
or
RC
t
q
q
ln 



RC
/
t
0
e
RC
q
i 

RC
/
t
e
RC
EC
i 


t / RC
0
i i e
 
‘-ve’ sign indicates that the discharging current flows in a direction opposite to the charging current.

q
t
O
O
t
i
-imax =/R
* Just as the flow of charge starts to the plates of capacitor with no initial charge, potential difference
across the capacitor is zero. But at that moment current is not zero.
* When the capacitor is fully charged that state is known as steady state. Flow of charge i.e. Current
through capacitor is zero in steady state but potential difference across capacitor is non zero.
Illustration: Corresponding to the shown circuit find the current through battery
(a) at t = 0 just after closing the switch.
(b) After a long time find the current through the battery and charge at capacitor.
R
C
S

R
Solution:(a) Just after closing the switch potential difference across capacitor is zero.
Let current be as shown.
Applying kirchhoff’s loop law(KLL) to circuit ABEFA
R
C
S

R
C
B
A
F
D
I1
I2
E
I2
I1
I + I
1 2
1
RI 0
   …(i)
Applying KLL to circuit ACDFA
2
RI 0
   …(ii)
Current through battery 1 2
2
I I
R R R
  
   
R
S

R
C
B
A
F
D
I1
I2
E
I + I
1 2
+q -q
(b) After long time, capacitor is in steady state fully charged, cur-
rent through it is zero only current is through CD. 1
I 0

Applying KLL to ACDFA
2 0
RI
  
 2
I
R



Current through battery 1 2
I I 0
R R
 
   
Let charge at capacitor be q
Applying KLL to ABEFA
0
C
q
I
R 1 




 q C
  .
Brain Teaser: Does the time required for the charge on a capacitor in an RC circuit to build up to a given
fraction of its final value depend on the value of the applied emf.? Does the time required
for the charge to change by a given amount depend on the applied emf?
Brain Teaser: A capacitor is connected across the terminals of a battery. Does the charge that eventually
appears on the capacitor plates depend on the value of the internal resistance of the battery?