P-80
Vidyasagar Classes
CURRENT ELECTRICITY
ii.
CURRENT
It is amount of charge passing through any cross-section
area in unit time.
∆q
I ave =
∆t
dq
∆q
=
∆t
dt
It is fundamental quantity.
S.I. unit → ampere
I inst =
lim
∆t→0
Definition of ampere : When 1 coulomb of charge flows
through a conductor in 1 second, the current flowing is said
to be l ampere.
CGG unit = emu of current and is called biot (Bi)
(1 / 10)emu of ch arg e
1
1C
=
=
Bi
S
S
10
Dimension : [A]
1A=
I
E
E
Current has both magnitude and direction, still it is scalar
because : (i) current doesn’t follow vector rule of addition.
i.e. if two currents i 1 and i 2 reaches a point we always have
i 1 + i 2 = i whatever the angle between i 1 and i 2 .
i1
θ
i2
(ii)
i = i1 + i2
Current is also defined as flux of current density i.e.
I=
∫
→
→
J ⋅ ds and dot product of two vectors is scalar.
Charge entering at one end per sec of a conductor is equal
to charge leaving other end per sec. (charge is conserved)
So,
i.
Current in different situations is calculated as follows :
a.
Due to translatory motion of charges :
i.
If n particles, each having a charge q, pass through a
given cross-sectional area S in time t :
nq
I=
t
ii.
If n particles, each having a charge q, pass per sec per
unit area through a cross-sectional area S.
I = nqS
iii. If there are n particles per unit volume, each having
a charge q and moving with speed v, the current
through cross-sectional area S:
I = nqvS
b.
Direction of current : In the direction of flow of positive
charge. (i.e. field)
OR
Opposite to the direction of flow of negative charge
I
Conductor remains uncharged (neutral) when current
flows through it. So, current carrying wire wont have
any conservative electric field associated with it.
If a point charge q is moving in a circle of radius r
with speed v.
q
qv
I=qf=
=
2πr
T
1
2πr
where, T = =
v
f
c.
When a voltage V is applied across a resistance R:
then current through resistance:
V
I=
R
w
If w = power then, I =
V
Classification of current:
i.
Direct current (dc): Magnitude and direction of
current does not vary with time.
ii.
Alternating current (ac) : Current is periodic with
constant amplitude and has half cycle positive and
half cycle negative.
I or V
I or V
+
O
–
t
(ac)
(ac)
I or V
Current doesn’t change with change in its crosssection along its length, for given conductor.
i1
I3
O
I or V
+
t
i2
i1 = i2 = i3
Physics (Current Electricity)
Periodic but not ac
(Amplitude is not constant)
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t
O
O
t
Periodic but not ac
(Direction is not changing)
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Vidyasagar Classes
I or V
I or V
t
O
j=
(ac)
(ac)
Current in different situations is due to motion of different
charge carriers.
i.
Current in conductors and vacuum tubes is due to
motion of electrons.
ii.
Current in electrolytic solution is due to motion of
available positive and negative ions.
iii.
Current in semiconductors is due to motion of
electrons and holes.
iv.
Current in discharge tube containing atomic gases is
due to motion of positive ions and negative electrons.
Note : In case of electrolytic solution and discharge tubes
though positive and negative charges are equal, the
currents due to them will be in same direction but
usually unequal due to different mobility of charge
carriers (heavier charge carriers lesser will be the
mobility).
(If j = constant, i α S)
(If S = constant, i α j)
Now :
∆i = j ∆S cos θ
→
di = jds cos θ
⇒i=
→
→
∫ j ⋅ ds (current is flux of current density)
Direction : In the direction of current.
Note :
1.
→
I = nqvS where n = no. of particle/volume each
having charge q and moving with speed v.
I
= nqv = vρ (ρ = charge density = nq
S
= charge/volume)
In case of conductors as :
∴j=
Definition : It is defined as a vector having magnitude
equal to current per unit area surrounding that point and
normal to the direction of charge flow and direction in
which current passes through that point.
2.
V
L
and R = ρ
L
A
E
L
I
So, EL = Iρ
⇒J=
=
= σE
ρ
A
A
V = IR and E =
It’s vector quantity. So for description of a vector quantity
we need magnitude and direction :
→
→
(in vector form J = σ E )
∆i
Magnitude : j =
∆S cos θ
In case of conductors current density is proportional to
→
where ∆i = current passing through area ∆S.
θ = angle between normal to the area and direction of
current.
∆S
∆I
→
⇒ ∆i = j ⋅ ∆s
Current density ( J ) :
i.
di
∆i
=
ds
∆s
If current i is uniformly distributed over an area S
i
and is perpendicular to it then, j =
S
(If i = constant, j α 1/S)
t
O
lim
∆s→ D
θ
electric field ( E ). This in turn implies that in case of
conductors.
→
i.
Direction of current density J is same as that of
→
→
electric field E .
J
→
If, θ = 0 : then Average current density will be
∆i
j=
∆s
and current density at any point will be :
Physics (Current Electricity)
ii.
If electric field is uniform ( E = constant), current
→
density J will be constant (as σ = constant).
iii.
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If electric field is zero inside conductor, current
density, hence current will be zero.
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Drift speed:
We have; J =
Metal consists of a ‘lattice’ of fixed positively charged ions
in which free electrons are moving randomly at speed
which at given temperature ‘T’ in accordance with kinetic
3kT
theory of gases is given by V rms =
m
Conductor = free electrons + lattice
So, V d =
I
= neV d
S
J
ne
Free electrons are also called as conduction electrons.
Derivation of an expression for drift velocity in terms of
atomic constants :
Lattice : It is collection of heavy positive ions.
Consider a free electron in the lattice to which an electric
Free electrons collide with lattice. Lattice (atoms – free
electrons) are always in a state of vibration because of
thermal energy. While moving, free electrons continuously
collide with vibrating lattice and their motion are opposed.
This opposition gives rise to resistance of conductors.
The speed and direction of electron changes randomly after
each collision. As a result, electrons move on zig-zag path.
field. E is applied.
→
Case I : No electric field inside the conductor :
Number of electrons passing any area of cross section will
be equal to no. of electrons passing from opposite direction
through that cross section at any given time interval.
So, current through the area is zero.
→
If m is mass of electron, its acceleration.
At each collision, it’s speed become zero. But due to
electric force it again
gets some velocity and drifts in
→
direction opposite to E .
Two processes collision and acceleration results in a
dynamic equilibrium in which a uniform average drift
velocity is achieved by electrons.
So, drift velocity is the average uniform velocity acquired
by free electrons inside a metal by application of an electric
field, which is responsible for current through it.
So, if V d is the average drift speed in a metal by the
application of an electric field which is responsible for
current through it.
I=
dq
= neV d S
dt
so, V d =
Physics (Current Electricity)
I
neS
→
→
→
→
F
− eE
=
a =
m
m
→
dV
− eE
=
dt
m
⇒
→
If V rms is thermal velocity of electron and t is the time
between two collisions:
Case II : Electric field inside the conductor (or potential
difference is applied across conductor)
Electric force eE acts on each electron in direction opposite
to the field. Electrons get biased in their random motion in
favour of the force. As a result, electrons drift in this
direction, and result into current.
→
Electron will experience a force F = (–e) E .
→
V
∫
→
→
t
− eE
dV =
m
→
∫ dt
0
V rms
⇒
→
→
V = V rms
→
eE
–
t
m
→
Averaging V over large number of electrons as :
→ 
 V rms  = 0 and t av = τ = relaxation time

 av
The average drift velocity
→
 eτ  →
Vd = –   E
m
So, V d α E and V d α τ
Average collision time (τ) is constant for given material
and at a given temperature.
 eτ 
V d =   E = kE
m
∴ K ⇒ depends as material of conductor and temperature.
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Relationship b/w current density (j) and drift speed (v d )
Vdt
P
→
A
E
E=
Volume of length V d t = AV d t
n = no. of free electrons per unit volume.
V
12
=
= 48 V/m

0.25
∴µ=
∴ no. of free electrons in AV d t volume = nAV d t
All these electrons will pass the cross-section P in time t.
∴ Charge passing this area in time t
vd
1.92 × 10 −4
=
= 4 × 10–6 m2/V-s
E
48
–6 2
Ans.: 4 × 10 m /V-s
Consider a conductor of length L and uniform crosssectional area A.
Q = (nAV d t)e
∴j=
A potential difference of 12 V is applied across a
conductor of length 0.25 m. If the drift velocity of electrons
is 1.92 × 10–4, then calculate the electron mobility.
Solution:
A = area of cross-section.
∴ current i =
Example 2:
Q
= (nAV d )e
t
Potential difference V is created across the length.
V
i
= nV d e
A
i
→
∴ j = nV d e
E
  eτ 
j = neV d = ne  E = σE
  m 
∴ j=σE
V

i
j=
A
j = σE
E=
…(1)
σ ⇒ depends on material of conductor and temperature.
σ = electrical conductivity
Resistivity of material : ρ =
1
σ
i
σV
i
V
=
(by putting value j = and E =
)
A

A

1 

⇒V=
i=ρ
i
σ A
A
⇒
Resistivity is reciprocal to conductivity.
→
Note : σ or ρ are independent of E .
⇒ V = Ri
SOLVED EXAMPLE
Example 1:
Where : R =
A current of 8 A is maintained on a conductor of
–4
2
cross-section 2.5 × 10 m . If the number density of free
28 3
electron be 12 × 10 /m , calculate the drift velocity of free
–19
electron. (e = 1.6 10 C)
…(2)
ρ
A
Where : R = Resistance of conductor (unit Ω)
Conductance G =
1
R
Equation (1) and (2) are two different forms of Ohm’s law.
Solution:
vd =

A
8
I
=
28
neA 12 × 10 × 1.6 × 10 −19 × 2.5 × 10 − 4
= 1.67 × 10–6 m/s
Ans.: 1.67 × 10–6 m/s
Physics (Current Electricity)
Unit of Resistance = Ohm (Ω)
Unit of resistivity (specific resistance) = Ohm meter (Ωm)
Unit of conductivity (σ) = (Ohm m)–1 = mhom–1.
We arrived at Ohm’s law by making several assumptions
about existence and behaviour of free electrons.
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These assumptions are not valid for semiconductors,
insulators, solutions etc. Ohm’s law can’t be applied in
such cases.
Note :
1.
Direction of drift velocity for electrons in a metal is
→
opposite to that of applied field E (and current
→
density J )
2.
Solution:
L
R=ρ
A
1
L′
L/2
=ρ
=
′
4
A
2A
R′ = ρ
λ
Vrms
Example 6:
where λ = mean free path
R=ρ
so τ will decreases and
so, V d will decrease
=

50 × 10 −2
= 5 × 10–7 ×
A
π × (0.2 × 10 −3 ) 2
25
250 × 10 −9
=
= 1.99 Ω
−8
12
.56
3.14 × 4 × 10
Ans.: 1.99 Ω
SOLVED EXAMPLES
Example 3:
Example 7:
A battery of emf 2.05 V is connected across the resistance
of 41 Ω. Find the current.
Solution:
2.05
V
=
= 0.05 A
R
41
An electric current of 5 A is divided into three branches
forming a parallel combination. The length of the wires in
three branches are in ratio 2 : 3 : 4 and their diameters are
in the ratio 3 : 4 : 5. Find the currents in each branch if the
wires are of the same material.
Solution:
Ans.: 0.05 A
Let length of wires be 2 , 3  and 4  and their radii be 3 r,
4 r an 5 r respectively, then
Example 4:
Electric iron draws a current of 5 A from 230 V supply line.
When connected to 110 V supply line, what current it
draws? Assume that its resistance does not change
significantly.
Solution:
V 1 = 230 V
A constantan wire of length 50 cm and 0.4 mm diameter is
used in making a resistor. If the resistivity of constantan is
–7
5 × 10 Ωm, calculate the value of the resistor.
Solution:
With rise of temperature :
V rms increases
λ decreases
8
 L R
= =2Ω
ρ  =
4
4
 A
Ans.: 2 Ω
Vd
στ
=
E
m
Drift velocity depends on nature of metal i.e.
relaxation time τ.
τ=
I=
A wire has resistance of 8 Ω. If its length is made half by
folding, find its resistance after the free ends are connected
to each other.
Mobility : Drift velocity per unit field.
µ=
3.
Example 5:
I1 = 5 A
R=
V1
230 V
=
= 46 Ω
I1
5A
I2 =
V2
110
=
= 2.39 A
R
46 Ω
Ans : 2.39 A
Physics (Current Electricity)
R=?
R1 = ρ
2
π(3r ) 2
R2 = ρ
3
π(4r ) 2
R3 = ρ
4
π(5r ) 2
2 3
4
:
:
9 16 25
The current must be in inverse ratio as p.d is constant in
parallel combination
9 16 25
I1 : I2 : I3 = :
:
= 54 : 64 : 75
2 3
4
54
64
I1 =
× 5 = 1.40 A
I2 =
× 5 = 1.66 A
193
193
R1 : R2 : R3 =
I3 =
75
× 5 = 1.94 A
193
Ans : 1.40 A, 1.66 A, 1.94 A
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Example 8:
TEMPERATURE DEPENDENCE OF RESISTANCE
A copper wire is stretched to make it 0.1% longer. What is
percentage change in its resistance?
Solution:
The resistance offered by any material depends upon its
temperature.
Let R 0 be the resistance of a metal conductor at 0° C. When
the conductor is heated, its resistance increases. Let R be
the resistance at t° C.

R=ρ
A
where, ρ = specific resistance of copper wire
The increase in resistance is directly proportional to
 = length of wire
A = area of cross section
i.
(R − R 0 ) ∝ R 0
V=A
ii.
where, V = volume of wire
∴ (R − R 0 ) ∝ R 0 t
R − R 0 = αR 0 t
R = R 0 (1 + αt)
∆R
∆
=2
R

∆R
0.1
∆
× 100 = 2 ×
× 100 = 2 ×
× 100 = 0.2%
R
100

Ans : Resistance increases by 0.2%
Example 9:
A 4.0 m copper wire of 40 Ω resistance has uniform
diameter of 0.30 mm. Find the resistivity and conductivity
of the copper wire. If the silver wire of same resistance and
same cross-sectional area is taken, whether the resistivity of
silver wire will be same as that of copper wire?
Solution:
 = 4.0 m
r=
0.3
mm = 0.15 mm = 0.15 × 10–3 m
2
ρ=
Rπ r
RA
40 × 3.142 × (0.15 × 10 )
=
=
4


R −R0
R0t
If R 0 = 1 ohm and t = 1 °C, then α = R − R 0 .
TEMPERATURE COEFFICIENT OF RESISTANCE:
The temperature coefficient of resistance of a material is
defined as the increase in resistance per unit original
resistance at 0° C per one degree rise in temperature.
1
1
Its unit is per ° C or per K (°C–1 or
or K–1 or
)
C°
K
∴α=
1
1
=
= 1.4 × 106 Ω–1 m–1
ρ
7.07 × 10 −7
The resistivity of silver wire will be different as the
resistivity is a property of material.
Physics (Current Electricity)
α=
R −R0
represents the rate of change of resistance
t
R −R0
dR
with temperature, therefore
can be written as
.
dt
t
= 7.07 × 10–7 Ω-m
–7
6 –1 –1
Ans : 7.07 × 10 Ω-m, 1.4 × 10 Ω m , No
It has positive value for metal conductors. So resistance of
metal conductors increases with increase in temperature.
As ratio
−3 2
2
where α is a constant depends on the material of the
conductor. It is called the temperature coefficient of
resistance of the material.
1  dR 


R 0  dt 
For metals, resistance increases
uniformly with temperature.
The graph of resistance against
temperature is almost linear.
We teach success
Resistance
R = 40 Ω
σ=
the rise in temperature
(R − R 0 ) ∝ t
2
∴R=ρ
V
Since ρ and V are constants, R ∝ 2
∴
the original resistance at 0 °C.
R0
Rt
t
Temperature (°C)
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Resistivity
TEMPERATURE DEPENDENCE OF RESISTIVITY:
The variation of resistivity of metallic conductors, an
alloys, semiconductors or an insulator with increase in
temperature is not same in all cases.
i.
For metal conductors, the resistivity increases with
increase in temperature. So conductors have positive
temperature coefficient. If ρ 0 and ρ t are resistivities
of material at 0 °C and t °C, then
ρ = ρ 0 (1 + αt)
Here α = temperature coefficient of resistivity.
For metals α lies between 10–2 to 10–4 /°C.
The graph of resistivity versus
temperature is almost linear at
high temperature but non-linear
at
low
temperature.
The
resistivity of pure metallic
conductor becomes zero, when
temperature approaches to 0 K.
Temperature (°C)
Temperature of conductor increases.
SOLVED EXAMPLES
Example 10:
A conductor has resistance of 15 Ω at 10 °C and 18 Ω at
400 °C. Find the temperature coefficient of resistance of
material.
Solution:
R 1 = R 0 (1 + αt 1 )
R 2 = R 0 (1 + αt 2 )
1 + αt 2
R2
=
1 + αt 1
R1
18 1 + 400α
=
15
1 + 10α
18 + 180 α = 15 + 6000 α
5820 α = 3
More frequency collision between lattice and electrons
i.e. average time τ between two successive collision
decreases.
Motion of electron larges opposed due to frequent collision
i.e. Drift speed decreases.
Resistivity increases or conductivity decreases
or resistance increases or conductance decreases.
ii.
In the case of alloys, such as nichrome and maganin,
resistivity is very large. But they have small
temperature coefficient (about 0.00001/°C).
It means that due to a small
change
in
temperature
resistivity of the alloys is not
affected much. Therefore,
they are used for construction
of standard resistance coils of
Temperature (°C)
high quality.
α=
iii.
In case of semiconductors
and insulators, materials
have negative temperature
coefficient. The resistivity
of
these
material
decreases exponentially
with the temperature.
Resistivity
Resistivity
Thermal agitation of lattice increases or lattice will vibrate
with larger amplitude.
Ans.: 5.15 × 10–4 /°C
Example 11:
A silver wire has resistance of 2.1 Ω at 27.5 °C.
–3
If temperature coefficient of silver is 3.94 × 10 / °C, find
the of silver wire resistance at 100 °C.
Solution:
R = R 0 (1 + αt)
2.1 = R 0 (1 + 3.94 × 10–3 × 27.5)
2.1 = R 0 (1.10835)
R0 =
2.1
= 1.8947
1.10835
R 100 = 1.8947 × (1 + 3.94 × 10–3 × 100)
= 1.8947 × 1.394
Temperature (0°C)
Silicon and germanium has temperature coefficients of
– 0.07/°C and – 0.05/°C respectively at room temperature.
At zero K, both the semiconductors and insulators have
infinite value of resistivity.
Physics (Current Electricity)
3
= 5.15 × 10–4 /°C
5820
= 2.64 Ω
Ans.: 2.64 Ω
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Solution:
If the second wire of resistance ‘R’ is connected in parallel
with the first wire, then using
1
1
1
+
=
RP
R1
R2
1
1
1
=
+
2
R
2.2
O Temperature
PTC thermistor
NTC thermistor
If potential difference is applied V
across the thermistor, it generate
current. The current is directly
proportional to the applied
potential difference for small
voltage. But with higher current,
more heat is generated in
thermistor.
∴ R = 22 Ω
2 =
O Temperature
V–I characteristic of Thermistor:
11− 10
1
1 10
1
=
–
=
=
2 22
R
22
22
ρ=
Resistance
The resistance of a wire of length 2 m and area of
2
cross-section 0.5 mm is 2.2 Ω. Another wire made up of
same material having same cross-sectional area is
connected to this wire in parallel will give an effective
resistance 2 Ω. Calculate the length of the second wire.
The resistance of PTC thermistor increases non-linearly
with temperature while resistance of NTC thermistor
decreases non linearly with temperature as shown in
following graph.
Resistance
Example 12:
R 1A
2.2 × 0.5 × 10 −6
=
= 5.5 × 10–7 Ω m
2
1
R 2A
22 × 0.5 × 10 −6
=
= 20 m
ρ
5.5 × 10 −7
I
Due to this heat, temperature of thermistor increases, which
decreases its resistance. It causes decrease in potential
difference.
Ans : 20 m
THERMISTOR
A thermistor is a alloys of semiconductor which is
thermally very sensitive. A small change in its temperature
causes a large change in its resistance. Therefore it is called
a thermal resistor or thermistor.
Symbol of thermistor:
Thermistors have resistance ranging from few ohms to
kilo-ohms (0.1 Ω to 10 kΩ) and have temperature range of
– 100 °C to 1100 °C.
Thermistors are made from oxides of copper, manganese,
nickel, cobalt, iron and lithium. The oxides are mixed in a
suitable proportion and ground into a fine powder which is
then compressed into desired shape and heated to high
temperature to form the ceramic body of the thermistor.
Thermistor may have positive temperature coefficient
(PTC) or negative temperature coefficient (NTC).
Magnitude of Temperature coefficient of resistivity is often
quite large for semiconductors. This fact is used to
construct thermometers to detect small changes in
temperatures.
Physics (Current Electricity)
Uses of Thermistor:
1.
A thermistor can be used to detect small changes of
temperature of 0 − 0.1 °C.
2.
Thermistors are used as temperature sensors for
accurate measurement of temperatures between
−100 °C and 300 °C.
Some thermistors have positive temperature
coefficient of resistance. There resistance increases
with increase in temperature.
3.
Thermistors are used in electronics industry.
4.
It is used for measurement of conductivity of gases.
It is also used to find the flow of gases and liquids.
5.
Thermistors are used in digital thermometer that has
very high accuracy.
6.
Thermistors are used in remote temperature sensing
automatic temperature control system, voltage
stabilization and protection of motor winding.
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SUPERCONDUCTIVITY
First ring : First significant figure.
In 1911, Kamerlingh Onnes discovered that the resistivity
of some metals and alloys becomes exactly zero below a
certain temperature. This temperature is called critical
temperature for this transition. The material in this state is
called superconductor.
Above the critical temperature, resistivity follows the trend
of normal metal.
Second ring : Second significant figure.
Third ring : Decimal multiplier
Fourth ring or no ring (no colour) : Tolerance or possible
variance in percentage.
2nd digit
st
1 digit
decimal
multiplier
Superconductor
Resistivity
Tc = critical temperature
Tc Temperature
Applications:
1.
Superconducting cables are
distribution without loss.
2.
To increase speed of computer.
3.
To produce strong magnet.
used
for
power
A device which offers resistance to the flow of current is
called a resistor. Resistors of different values are used in
electrical and electronic circuits. A standard colour code is
used to indicate the value of the resistance and its
percentage tolerance. Every resistor has a set of coloured
rings marked on it.
The first two rings from the end of the resistor denote the first
two digits in the value of the resistance. The third ring denotes
the decimal multiplier and fourth ring indicates the tolerance
of the resistor as a percentage of the indicated value.
The following table indicates the standard colour code of
resistors.
Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Gray
White
Gold
Silver
No
colour
NUMBER
MULTIPLIER
0
1
2
3
4
5
6
7
8
9
−
−
0
10 = 1
1
10
2
10
3
10
4
10
5
10
6
10
7
10
8
10
9
10
10–1
10–2
±5
± 10
−
−
± 20
Physics (Current Electricity)
We can learn the order of colours by the sentence B B
ROY of Great Britain has a Very Good Wife.
If the resistor has the following colour bands:
COLOUR CODE FOR RESISTORS
COLOUR
Tolerance
Note:
TOLERANCE(%)
1.
Brown
Green
Red
Silver
2
10%
1
5
10
2
Resistance = 15 × 10 ohm ± 10%
2.
Red
Red
Yellow
Gold
4
2
2
10
5%
4
Resistance = 22 × 10 ohm ± 5%
Resistance :
Property of a substance through which it opposes flow of
current through it.
For a given body : It is ratio of potential difference to the
resulting current.
R=
±2
V
I
1volt
, so if a
1ampere
potential difference of 1 volt across a conductor produces a
current of 1 ampere through it, then resistance of conductor
is 1 ohm.
[V] = [W ] = [ML2 T −2 ] = [ML2T–3A–2]
Dimension :
[I] qI
[AT][A]
Unit : Volt/Ampere, Ohm (Ω) ⇒ 1ohm =
Reciprocal of resistance is called conductance (G) :
G=
1
I
=
R
V
Dimension = [M–1L–2T3A2]
S.I. Unit = ohm–1 / mho (Ω–1) / Siemen (S).
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⇒
Ohm’s law :
⇒
Devices or substances which don’t obey ohm’s law
are called non-ohmic or non-linear conductors.
For these I versus V curve is not a straight line. But
still; ratio of V to I will be resistance R.
At constant temperature, resistance of a wire doesn’t
change with the change in potential difference across it.
OR
While rate of change of V to change in I will be
called as dynamic resistance (r) for voltage V.
Current is directly proportional to applied potential
difference across a wire at constant temperature.
OR
R=
As long as physical state (material, dimensions and
temperature) of a conductor remains same, electric
current flowing through a given conductor is directly
proportional to the potential difference applied across it.
R=
V
= constant
I
I
φ
θ
OR I α V
V
Crystal rectifier
i.e. Resistance is independent of magnitude and
polarity of applied potential difference.
⇒
1
V
1
dV
=
while, r =
=
tan φ
I
tan θ
dI
I
I
Substances which obey ohm’s law are called ohmic
or linear conductors.
Graph b/w I and V (called static characteristic) is a
straight line passing through origin, reciprocal of
whose slope gives resistance.
R=
V
1
=
I
tan θ
I
= constant
1
R=
tan θ
θ
V
V
V
⇒
Resistance R can never be negative, dynamic
resistance can be (e.g. tetrode).
So, V = IR is valid for ohmic and non-ohmic devices
both but V α I is only valid for ohmic devices.
Although resistance of ohmic conductors is
independent of magnitude and polarity of applied
voltage, for a given body it is not unique and depends
on how the potential difference is applied i.e. on
length and area of cross-section.
A2
⇒
⇒
1
=
Slope of IV characteristic
Semiconductor
Tetrode valve
→
At given temp., ρ is constant and doesn’t depend on E .
Eg. Silver, copper, mercury, nichrome, etc.
Note :
ξ
A1
R2
L1
I
ξ
R1
R 2 > R 1 as
V
I–V curve of good conductor
⇒

L2
> 1
A1
A2
Square plates of same thickness (t) and material but of
different size : Resistance is independent of side.
Current increases with potential difference, but when
current increases, temp. of conductor increases,
which in turn increases resistance of conductor and
hence slightly decrease in current. Ohm has not
considered this temperature dependence in his
relation.
Physics (Current Electricity)
2
i
(1)
R1 = R2
R=ρ
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t
a
i
(2)
i
t
b

a
ρ
=ρ
=
a
a×t
t
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Kirchhoff’s Rules :
Resistivity or Specific resistance : (ρ)
R=ρ
Junction : A junction is any point in the circuit where the
current can split.

A
A

Resistivity is numerically equal to resistance of a substance
which has unit area of cross-section and unit length.
⇒ρ=R
Loop : It’s any closed conducting path in a circuit.
Kirchhoff’s junction Rule : OR Current law
⇒
E
(Ratio of electric field to current density)
J
⇒
ρ=
⇒
Unit : Ohm m
i1
i1
3 –3 –2
Reciprocal of resistivity is called conductivity or
specific conductance (σ)
1
σ=
ρ
i1 + i2 = i3
i1 = i2 + i3
⇒
This rule is a statement of conservation of charge.
That is, whatever current enters a given point in a
circuit must leave that point, since charge cannot
build up at a point.
⇒
If we consider currents entering to be positive and
those leaving to be negative, the algebraic sum of
currents into a junction must be zero.
Dimension = [M–1L–3T3A2]
It depends on nature of material. It doesn’t depend on
shape and size of material (, A).
i3
i3
Unit = mho/m
⇒
i2
i2
Dimension : [ML T A ]
⇒
The sum of the currents entering any junction must
equal the sum of currents leaving that junction.
It also depends on temperature of body.
ρ = ρ 0 [1 + α ∆θ + β(∆θ)2 + …]
Kirchoff’s loop rule : OR Potential law
Where α, β are temperature coefficients of resistivity.
If ∆θ is small : ρ = ρ 0 [1 + α ∆θ]
⇒
With rise in temperature, resistivity of metals will
increase (because α is positive for metals) while it
will decrease for non metals (as α is negative for
non-metals).
Conductivity (σ) :
Reciprocal of resistivity is called conductivity of material.
1
S.I. unit = Siemen/meter
σ=
ρ
(Sum of changes in potential across each element
around any closed-circuit loop must be zero.)
⇒
This rule follows from conservation of energy. That
is, any charge that moves around any closed loop in a
circuit (it starts and ends at the same point) must gain
as much energy as it loses.
⇒
Applying loop rule, first we assume direction of the
current in each branch of circuit. Then starting at any
point in the circuit, we imagine, travelling around a
loop, adding change in potential across a battery and
resistances.
Depending upon resistivity, materials are classified as
i.
Conductors
ii.
Insulators
iii.
Semiconductors.
i.
Conductors : Those materials whose resistivity is
negligibly small are called conductors eg. silver,
copper, aluminium etc.
ii.
Insulators : Those materials whose resistivity is very
high are called insulators eg. glass, rubber etc.
iii.
Semiconductor : Those materials whose resistivity
lies between that of conductors and insulators are
called semiconductors. eg. silicon, germanium etc.
Physics (Current Electricity)
Sum of potential differences across each element
around any closed-circuit loop must be zero.
As an aid in applying second rule, following points should
be noted.
1.
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If a resistor is traversed in the direction of current,
the change in potential across the resistor is  IR.
i
a
b
∆V = V b − V a = − IR
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2.
If a resistor is traversed in the direction opposite the
current, the change in potential across the resistor is
+IR.
i
a
b
Example 14:
In the network shown in figure, E 1 = 6V, E 2 = 4V, R 1 =
2Ω, R 2 = 3Ω and R 3 = 5Ω. Find the currents passing
through the resistors R 1 , R 2 and R 3
∆V = V b − V a = + IR
3.
R1
A
R2
B
E2
E1
If a seat of emf is traversed in the direction of emf
(from − to + on the terminals), the change in potential
is + ξ.
→
E +
−
b
a
C
R3
I1
I2
I1 + I2
E
F
D
Solution:
∆V = Vb − Va = + ξ
4.
In closed path ABEFA of the circuit,
If a seat of emf is traversed in the direction opposite
the emf (from + to − on the terminals), the change in
potential is − ξ.
→
E
+ −
a
– E 1 + (I 1 + I 2 ) R 3 + I 1 R 1 = 0
I 1 R 1 + (I 1 + I 2 ) R 3 = E 1
2I 1 + (I 1 + I 2 ) × 5 = 6
7I 1 + 5I 2 = 6
b
…(1)
∆V = Vb − Va = − ξ
In closed path BCDEB of the circuit,
SOLVED EXAMPLES
I 2 R 2 + (I 1 + I 2 ) R 3 = E 2
E 2 – I 2 R 2 – (I 1 + I 2 ) R 3 = 0
Example 13:
3I 2 + (I 1 + I 2 ) × 5 = 4
Calculate the unknown current ‘I’ in the figure given
below.
10 A
2A
I
5I 1 + 8I 2 = 4
Multiplying the equation (1) by 8 and multiplying equation
(2) by 5 and subtracting, we get
56I 1 + 40I 2 – 25I 1 – 40I 2 = 48 – 20
31I 1 = 28
3A
I1 =
Solution:
10 A
2A
…(2)
Putting in equation (1), we get
i
B
I
7×
A
3A
28
A
31
28
+ 5I 2 = 6
31
5I 2 = 6 –
At junction A,
i=2+3=5A
I2 = –
196 186 − 196
10
=
=–
31
31
31
2
A
31
At junction B,
26
A
31
I = 5 + 10 = 15 A
I3 = I1 + I2 =
Ans. : 15 A
Ans. : 28/31 A, – 2/31 A, 26/31 A
Physics (Current Electricity)
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Example 15:
Example 16:
In the given circuit, find the potential difference across 8 Ω
resistor.
1.5 V, 2 Ω
The four resistances, P = 10 Ω, Q = 15 Ω, R = 25 Ω
and S = 20 Ω, are connected in arms AB, BC, CD and DA
respectively of Wheatstone’s network ABCD. The cell is
connected between points A and C. If the current through
the cell is 1 ampere, find the current through galvanometer
connected between B and D. The resistance of
galvanometer is 20 Ω.
2Ω
3Ω
8Ω
2Ω
3Ω
2 V, 1 Ω
Solution:
Solution:
2Ω
D
2Ω
2Ω
1.5 V
A
I1
F
3Ω
1Ω
G
I2
Let the current be shown as in figure.
nd
Applying Kirchhoff’s 2 Law to loop ABCDA,
2 I 1 + 3 I 1 + 8 (I 1 + I 2 ) + 2 I 1 = 1.5
…(1)
nd
Applying Kirchhoff’s 2 Law to loop GCDFG,
3 I 2 + 8 (I 1 + I 2 ) + 2 I 2 + I 2 = 2
8 I 1 + 14 I 2 = 2
…(2)
[eq. (1) × 4] – [eq. (2) × 7.5] gives
25 Ω
20 Ω
C
I
D I2+Ig
C
I1 + I2
15 I 1 + 8 I 2 = 1.5
I
I1 – Ig
15 Ω
Ig
G 20 Ω
I1
I2
3Ω
8Ω
2V
A
B
B
10 Ω
P
S  10 20 
≠
 i.e. ≠  , the network is not balanced.
Q
R  15 25 
It means current Ig flows through galvanometer from B to D.
As
Let ‘I’ be the current entering at junction A, I 1 be
the current through arm AB and I 2 be the current through
arm AD.
∴ I = I1 + I2
As I = 1 ampere, I 2 = 1 – I 1
nd
Applying Kirchhoff’s 2 Law to the loop ABGDA, we get
–10 I 1 – 20 I g + 20 I 2 = 0
–10 I 1 – 20 I g + 20 (1 – I 1 ) = 0
–I 1 – 2 I g + 2 – 2 I 1 = 0
3 I1 + 2 I g = 2
…(1)
60 I 1 + 32 I 2 = 6
Applying Kirchhoff’s 2nd Law to the loop BCDGB, we get
60 I 1 + 105 I 2 = 15
– –
–
–15 (I 1 – I g ) + 25 (I 2 + I g ) + 20 I g = 0
–3 (I 1 – I g ) + 5 (I 2 + I g ) + 4 I g = 0
–3 I 1 + 3 I g + 5 I 2 + 5 I g + 4 I g = 0
–3 I 1 + 12 I g + 5 (1 – I 1 ) = 0
–8 I 1 + 12 I g + 5 = 0
8 I 1 – 12 I g = 5
…(2)
– 73 I 2 = – 9
I2 =
9
A
73
Putting the value of I 2 in eq. (1), we get
I1 =
5
A
146
Potential difference across 8 Ω resistor is given by,
9 
 5
V = (I 1 + I 2 ) × 8 = 
+ × 8
 146 73 
=
23
× 8 = 1.26 V
146
Ans. : 1.26 V
Physics (Current Electricity)
[8 × eq. (1)] – [3 × eq. (2)]
24 I 1 + 16 I g = 16
24 I 1 – 36 I g = 15
–
+
–
52 I g = 1
1
Ig =
A
52
1
Ans. :
A
52
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Example 17:
Example 18:
A constant voltage of 50 V is maintained between points A
and B of circuit shown below. Find the current through CD.
In the circuit given below, calculate the current through
resistor 2 Ω.
C
R1 = 1 Ω
+
10 V, 3 Ω
R2 = 2 Ω
A
–
B
12 V, 2 Ω
R3 = 3 Ω D R4 = 4 Ω
50 V
2Ω
Solution:
Solution:
1Ω
E
2Ω
C
F
+
3Ω
I3
H
I
(I2 + I3)
3Ω
I2
–
I2
I
10 V
(I1 – I3)
I1
A
I1
A
D
4Ω
B
G
12 V
2Ω
I1 + I2
G
F
C
D
2Ω
Let current ‘I’ is flowed from A to B.
B
nd
Applying Kirchhoff’s 2nd law to loop ECDHE,
Use Kirchhoff’s 2 law to the loop ABDFA,
–1 × I 1 – 0 × I 3 + 3 I 2 = 0
3 I 1 + 2 (I 1 + I 2 ) = 10
I1 = 3 I2
…(1)
5 I 1 + 2 I 2 = 10
…(1)
Applying Kirchhoff’s 2nd law to loop CFGDC,
Use Kirchhoff’s 2 law to the loop GCDFG,
–2 (I 1 – I 3 ) + 4 (I 2 + I 3 ) + 0 × I 3 = 0
2 I 2 + 2 (I 1 + I 2 ) = 12
nd
2 (I 1 – I 3 ) – 4 (I 2 + I 3 ) = 0
2 I1 – 4 I2 = 6 I3
2 I 1 + 4 I 2 = 12
…(2)
I1 + 2 I2 = 6
From (1) and (2),
Eq. (1) – eq. (2) gives
I2 = 3 I3
∴ I1 = 9 I3
…(2)
4I 1 = 4
…(3)
Applying Kirchhoff’s voltage law between points A and B,
∴ I1 = 1 A
V AB = Σir – ΣE
Putting the value of I 1 in eq. (2), we get
– 50 = –1 × I 1 – 2 (I 1 – I 3 )
I 2 = 2.5 A
3 I 1 – 2 I 3 = 50
Current through 2 Ω resistor = I = I 1 + I 2
27 I 3 – 2 I 3 = 50
25 I 3 = 50
[from eq. (3)]
= 1 + 2.5 = 3.5 A
Ans. : 3.5 A
I 3 = 2 ampere
Ans. : 2 A
Physics (Current Electricity)
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Example 19:
2 I = –1 +
Find the current in 3 resistances connected in Y shape as
shown in figure. The terminals of R 1 , R 2 and R 3 are
maintained at potentials V 1 = 10 V, V 2 = 6 V and V 3 = 15 V
respectively. [Given: R 1 = 10 Ω, R 2 = 20 Ω, R 3 = 30 Ω]
C
A
=
− 110 + 114
110
=
2
4
=
55
110
+6 V
R2 = 20 Ω
I=
I
I1
B
R1 = 10 Ω
I2
R3 = 30 Ω
I1 =
+10 V
D
+15 V
1
A
55
2 + 19
1
19
21
+
=
=
A
110
55 110
110
Ans. :
Solution:
114
110
1
21
9
A,
A, –
A
55
110
110
Example 20:
Using Kirchhoff’s first law (Junction law) at junction B,
I = I1 + I2
I1 = I – I2
…(1)
By Kirchhoff’s voltage law,
Eight equal resistors, each of resistance ‘r’ are connected
along the edges of a pyramid OABCD having square base
ABCD as shown in figure below. Calculate the equivalent
resistance of network between the points A and B.
O
V AB = Σir – ΣE
–10 + 6 = (– I × 10) – 20 I 1
r
2 = 5 I + 10 I 1
2 = 5 I + 10 I – 10 I 2
2 = 15 I – 10 I 2
[from eq. (1)]
D
– 10 + 15 = – 10 I – 30 I 2
–5 = 10 I + 30 I 2
–1 = 2 I + 6 I 2
…(3)
C
r
r
…(2)
Also,
r
r
r
r
B
A
r
Solution:
Let R be equivalent resistance between A and B of the
network of resistors shown in figure.
O
[2 × eq. (2)] – [15 × eq. (3)]
30 I – 20 I 2 =
4
I3
30 I + 90 I 2 = –15
– –
+
19
A
110
A
Putting the value of I 2 in eq. (3), we get
2I–
6 × 19
= –1
110
2I–
114
= –1
110
Physics (Current Electricity)
C
D I2 – I3
I1
– 110 I 2 = 19
I2 = −
I3
I1
I2
I
I –I1 – I2
X
E
B
r
I2
I
Y
E
I
Suppose that current I, enters at point A, a part
I 1 flow along AO, part I 2 flow along AB and current
(I – I 1 – I 2 ) along path AB.
I is current drawn by the battery then R =
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Applying Kirchhoff’s 2nd law to closed loop ABOA,
– (I – I 1 – I 2 ) r + I 1 r + I 1 r = 0
3 I1 + I2 = I
…(1)
For closed loop ADOA,
– I2 r – I3 r + I1 r = 0
…(2)
I1 – I2 – I3 = 0
For closed loop CDOC,
(I 2 – I 3 ) r − I 3 r – I 3 r = 0
…(3)
I2 – 3 I3 = 0
I
∴ I3 = 2
…(4)
3
Substituting the value of I 3 in equation (2), we get
I
I1 – I2 – 2 = 0
3
4
I1 –
I2 = 0
3
4
I2
…(5)
∴ I1 =
3
Substituting the value of I 1 in equation (1), we get
4
(3 × I 2 ) + I 2 = I
3
5I 2 = I
I
∴ I2 =
…(6)
5
From eq. (5) and eq. (6), we get
4
I1 =
× I2
3
4
I
4
= × =
I
3
5 15
In closed loop ABYXA of the circuit,
(I – I 1 – I 2 ) r – E = 0
(I – I 1 – I 2 ) r = E
4
1
(I –
I – I) r = E
15
5
8
Ir=E
15
E
8
=
r
R eff =
I
15
Ans. :
8
r
15
Physics (Current Electricity)
Example 21:
Determine the current drawn from 12 V supply with
internal resistance of 0.5 Ω by infinite network of each 1 Ω
resistors as shown in figure below.
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
0.5 Ω
Also when current I 2 enters at point D, its part I 3 flows
along DO and remaining part (I 2 – I 3 ) will flow along DC.
Again by symmetry, the current along path OC will be I 3
and the current along path CB will be I 2 .
By symmetry the current ‘I’ that flows out from B will be
composed of part I 1 from OB, part I 2 from CB and part
(I – I 1 – I 2 ) from path AB.
1Ω
Solution:
Let resistance of entire circuit is equal to X, since it is
infinite network. Adding one more set of three resistances,
each of resistance 1Ω, across the terminal will not affect the
total resistance.
The above circuit becomes equivalent to following circuit.
1Ω
1Ω
X
1Ω
If R eq is equivalent resistance of this network work then
R eq = R + (resistance equivalent to parallel combination of
X and R) + R
=R+
XR
+R
R+X
XR
X+R
Since R eq = X (assumed)
=2R+
XR
X+R
Since R = 1 Ω
X = 2R +
X
2X + 2 + X
3X + 2
=
=
X +1
X +1
X +1
2
X +X=3X+2
X=2+
X2 – 2 X – 2 = 0
− (−2) ± (−2) 2 − 4 × 1× (−2)
=1± 3
2
Since value of resistance cannot be negative,
X=
X=1+
3 = 1 + 1.732 = 2.732 Ω
If ‘I’ is current drawn by network, then
I=
E
12
12
=
=
= 3.713 A
2.732 + 0.5
3.272
X+r
Ans. : 3.713 A
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E.M.F, INTERNAL RESISTANCE & POTENTIAL
DIFFERENCE
Potential difference of a cell OR terminal voltage:
⇒
Electrochemical cell:
⇒
An electrochemical cell is a device which, by
converting chemical energy into electrical energy,
maintains the flow of charge in a circuit.
⇒
It consists of two electrodes and an electrolyte.
⇒
The electrode at higher potential is called positive
terminal, or anode while at lower potential →
negative terminal or cathode.
⇒
Internal resistance of cell (r):
⇒
In case of a cell the opposition of constituent of cell
(electrolyte and electrodes) to the flow of current
through it is called internal resistance of the cell.
⇒
Internal resistance of a cell depends on
When a cell is in use, i.e., discharging, current
outside the cell is from anode to cathode while inside
it from cathode to anode.
Anode
Energy spent by source of emf in circulating unit
charge through external resistance is called potential
difference of the cell.
i.
Distance between electrodes (r α d)
ii.
Area of electrodes (r α 1/A)
iii.
Concentration of electrolyte (r α c)
iv.
Temperature of electrolyte (r α 1/temp.)
Note: It is independent of material of electrodes.
Cathode
Expression of Terminal Potential difference:
Consider a cell of emf E, internal resistance r and external
resistance R connected across the cell. Let V be terminal
potential difference and V r be potential difference across
internal resistance of the cell.
I
Electric cell
+
–
R
Vr
Cell
+
⇒
V
–
Equivalent circuit
In order to maintain charge flow through conductor
and cell, work has to be done in moving the charges.
Cell acts as a source of energy for motion of charges.
Electromotive force (EMF) : E :
⇒
⇒
The energy supplied (work done) by a cell to
circulate unit charge once round the complete circuit
is called EMF of the cell.
EMF is a misnomer, as it is not a force. It is work
done or energy spent per unit charge.
EMF =
Work done
W
⇒ E=
Q
Ch arg e moved
Total energy supplied by cell in circulating charge ‘q’
round the circuit is equal to sum of energies spent in overcoming external resistance and overcoming internal
resistance.
According to principle of conservation of energy:
Total energy =
spent by cell
Energy spent
Energy spent in
In overcoming +
over coming
External resistance
internal resistance
⇒ qE = qV + qVr
⇒ E = V + Vr
…(i)
By Ohm’s law, V = IR, V r = Ir
So, E = I (R + r)
or I =
S.I. unit of EMF = joule/Coulomb or Volt.
E
R+r
…(ii)
Potential difference (PD) V:
Now Eq (i) can be written as:
⇒
V = E – Vr
In reference to an electric circuit, potential difference
between two points is defined as the work done in
moving unit positive charge from one point to other.
Physics (Current Electricity)
⇒ V = E – Ir
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Relationship among E, V and r : (Where E = emf of cell,
r = internal resistance of cell, V = terminal voltage across
cell)
I
B
R
I
C
Solution:
W
240
=
= 12 V
V=
Q
20
I
A
I
E
r
Ans.: 12 V
I D
Example 24:
A voltmeter is connected across a battery of e.m.f. 12 V and
internal resistance of 10 Ω. If the voltmeter resistance is
230 Ω, what is the reading will shown by voltmeter?
Solution:
E
I=
R+r
12
1
12
=
=
=
A
230 + 10
240
20
V R = IR
1
=
× 230 = 11.5 V
20
⇒ Applying Kirchoff’s second law in loop ABCDA:
−IR − Ir + E = 0
⇒ E = IR + Ir
⇒ In terms of V (= IR) ‘terminal voltage’
⇒
E
IR
Ir
=
+
IR
V
IR
⇒
E
r

= 1 + 
V  R
⇒
E
r
=
−1
R
V
Ans.: 11.5 V
E 
⇒ r = R  −1
V 
Example 25:
Two wires, when connected in series have equivalent
resistance of 18 Ω and when connected in parallel an
equivalent resistance of 4 Ω. Find their resistances.
Solution:
Note : Term Ir is called internal drop.
SOLVED EXAMPLES
R S = R 1 + R 2 = 18 Ω
Example 22:
A primary cell has an emf of 1.5 V. When a 5 Ω resistor is
connected across it, the current is 0.2 A. Find the internal
resistance of the cell.
Solution:
I=
E
R+r
0.2 =
1.5
5+r
1 + 0.2 r = 1.5
0.2 r = 0.5
r=
Example 23:
Find the potential difference between terminals of the
battery, if 240 joule of work is required to transfer
20 coulomb of charge from one terminal to the other
terminal?
5
= 2.5 Ω
2
Ans.: 2.5 Ω
Physics (Current Electricity)
RP =
…(1)
R 1R 2
=4Ω
R1 + R 2
R 1R 2
= 4 ⇒ R 1 R 2 = 72
18
We know that
(R 1 – R 2 )2 = (R 1 + R 2 )2 – 4 R 1 R 2 = (18)2 – (4 × 72)
= 324 – 288 = 36
R1 – R2 = ± 6
Take R 1 – R 2 = 6
…(2) (as R 1 – R 2 ≠ –6)
Then from eq. (1) and eq. (2), we get
R 1 + R 2 = 18
R1 – R2 = 6
2 R 1 = 24
R 1 = 12 Ω, R 2 = 6 Ω
Ans : 12 Ω, 6 Ω
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So,
Resistors in series and in parallel :
(1) In series :
i
R1
V1 : V2 : V3 = R1 : R2 : R3
i
R2
Req
i
i
V
i
i ε
⇒
⇒
i
i


R1
 (V 1 + V 2 + V 3 )
∴ V 1 = 
 R1 + R 2 + R 3 
i
ε
i


R2
 (V 1 + V 2 + V 3 )
V 2 = 
 R1 + R 2 + R 3 
i
When two or more resistors are connected such that
they have only one common point per pair, they are
said to be in series. (No junction in between two
resistances).
Current is same through each resistor since any
charge that flows through R 1 must equal the charge
that flows through R 2 .


R3
 (V 1 + V 2 + V 3 )
V 3 = 
 R1 + R 2 + R 3 
⇒
In many circuits, fuses are used in series with other
circuit elements for safety purposes. The conductor
in fuse is designed to melt and open the circuit at
some maximum current. If fuse is not used,
excessive currents could damage circuit elements.
⇒
In modern home construction, circuit breakers are
used in place of fuses. When the current in a circuit
exceeds some value (typically 15 A), the circuit
breaker acts as a switch and opens the circuit.
Applying Kirchhoff’s law in figure (1)
− iR 1 − iR 2 + ε = 0
⇒ ε = iR 1 + iR 2
Parallel combination of resistors :
…(I)
When two or more resistors are connected together such
that they have two common points per pair, they are said to
be in parallel.
Applying Kirchhoff’s law in fig (II)
− iR eq + ε = 0
⇒ ε = iR eq
…(II)
Put the value of ε from (II) into (I)
iR eq = iR 1 + iR 2
⇒ R eq = R 1 + R 2
The resistance R eq is equivalent to the series combination
R 1 + R 2 in sense that circuit current is unchanged when R eq
replaces R 1 + R 2 .
The equivalent resistance of three or more resistors
connected in series is simply :
The electric potential difference is same across parallelconnected resistors. This is because the wire connecting
the resistors on the left makes them into a single,
continuous path and thus constrains them to be at same
electric potential. This is also true of the right side of
resistors.
R1
I1
I
I1
A
R2
I2
I2
⇒ Distribution of potential in series connections :
When more than one resistance are connected in series, the
current through them will be same and potential will be
distributed in the ratio of resistances :
V = iR
V α R (if i = constant)
Physics (Current Electricity)
B
I
I
I
I1
R eq = R 1 + R 2 + R 3 + ……
Therefore, the equivalent resistance of a series combination
of resistors is always greater than any individual resistance.
Req
ε
I
I
ε
I
I
When current I reaches point A (called a “junction”), it
splits into two parts, I 1 going through R 1 and I 2 going
through R 2 .
If R 1 is greater than R 2 , then I 1 will be less than I 2 .
(as potential difference across resistances R 1 and R 2 is
same)
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Now
I = I 2 + I 2 (from fig 1) [junction rule]
ε
ε
ε
⇒
=
+
[Ohm’s law]
R eq
R1
R2
SOLVED EXAMPLES
Example 26:
Find the steady state current through the 2 Ω resistors in the
circuit given below.
(from fig. 2) (from fig. 1)
R 1R 2
1
1
1
⇒
=
+
⇒ R eq =
R eq
R1 + R 2
R2
R1
2Ω
3Ω
An extension to this analysis to three or more resistors in
parallel gives the following general expression.
1
1
1
1
=
+
+
+ -------R eq
R1
R2 R2
Equivalent resistance of two or more resistors connected in
parallel is always less then the smallest resistance in the
group.
Household circuits are always wired such that all the
appliances are connected in parallel. In this manner, each
device operates independently of the others, so that if one is
switched off, the others remain on.
Distribution of current in parallel combination :
R3
I3
I1
I2
I2 =
I3 =
R 1 = 2 || 3 =
i=
6
3
=
4
2
Let,
9
3
i×3
3
=
×
=
= 0.9 A
3+ 2
2
5 10
R1
Ans.: 0.9 A
Grouping of cells OR combination of cells OR Battery
I
1
R1
 1
1
1 


+
+
 R1 R 2 R 3 
1
R2
 1
1
1 


+
+
R
R
R
1
2
3


1
R3
 1
1
1 


+
+
 R1 R 2 R 3 
6
= 1.2
5
R eff = 1.2 + 2.8 = 4 Ω
i2 =
V = IR
As V = constant in parallel combination
1
Iα
R
1
1
1
:
:
So, I 1 : I 2 : I 3 =
R1 R 2 R 3
I1 =
Solution:
R2
I
2.8 Ω
6V
⇒ Combination of cells is called battery.
(A)
Series grouping :
Suppose 3 cells of emf ε 1 , ε 2, ε 2 and internal
resistances r 1 , r 2 , r 3 respectively are connected in
series.
ε2 r2
ε3 r3
i ε1 r1
i
× (I 1 + I 2 + I 3 )
i
i
εeq
i
× (I 1 + I 2 + I 3 )
i
R
req
i
× (I 1 + I 2 + I 3 )
Physics (Current Electricity)
i
R
i
Net emf = ε 1 + ε 2 + ε 3
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PARALLEL GROUPING
Total resistance = r 1 + r 2 + r 3 + R
net emf
Current in circuit =
total resis tan ce
ε1 + ε1 + ε 2
i=
(r1 + r 2 + r3 ) + R
3 Possible cases :
Case 1 :
∴ net emf of battery ε eq = ε 1 + ε 2 + ε 3
Net internal resistance r eq = r 1 + r 2 + r 3
In general form it can be written :
i=
∑ε
R + ∑r
i
; E eq =
∑ε
i
, r eq =
∑r
i
Suppose n cells each of emf ε and internal resistance
r are connected in series, then
net emf = n ε
total resistance = nr + R
net emf
current in circuit i =
total resis tan ce
nε
⇒
i=
nr + R
∴ Net emf = nε, net internal resistance = nr.
Note :
1.
2.
nE
, i.e. current is n times that of
R
circuit current due to single cell (E/R).
E
If nr >> R ; I =
i.e. current in circuit is equal to
r
that of a single short circuited cell.
If : nr << R ; I =
Note :
1.
emf of combination of cells is higher than emf of
individual cell.
2.
Current of battery of series combination is same that
of one cell.
3.
Effective internal resistance increases.
For example :
ε3 r3 i
ε2 r2
i ε1 r1
i
R
net emf = ε 1 + ε 2 − ε 3
Total resistance = (r 1 + r 2 + r 3 ) + R
ε1 + ε 2 − ε 3
i=
R + (r1 + r2 + r3 )
ε eq = ε 1 + ε 2 − ε 3
r eq = r 1 + r 2 + r 3
Physics (Current Electricity)
E
r
E
r
i
i
⇒
r
E
i
net emf E eq = E
i
i
net internal resistances r eq =
r
n
net external resistance = R
r
Total resistance = R +
n
net emf
E
=
Current through R =
r
total resis tan ce
R+
n
Note :
1.
To get maximum current, cells must be connected in
series if effective internal resistance is lesser than
external and in parallel if effective internal resistance
is greater than external.
r
E
<< R; I =
; current in circuit is equal to
R
n
E
circuit current due to a single cell   .
R
r
E
If
>> R ; I = n ; current in circuit is equal to n
n
r
E
times that of single short circuited cell   .
r
2.
If
3.
Case : 2 : Application of Kirchhoff’s second law :
E1 r1
C
B
i1
i1
i2 E2 r2
i2
D
A
i
i3 E
3
r3
i3
i
E
i
i
R
Applying Kirchhoff’s law in loop ABCDEFA,
F
E 1 − iR − i 1 r 1 = 0
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E
R
+ 1
r1
r1
⇒ i1 = − i
Case 3 :
…(1)
Similarly,
i2 = −
E
iR
+ 2
r2
r2
i3 = −
E
iR
+ 3
r3
r3
1 1 1  E
E 
E
⇒ i = −iR  + +  +  1 + 2 + 3 
r2
r3 
 r1 r2 r3   r1
r2
r3
E1 E 2 E 3
−
+
r1
r2
r3
=
1 1 1
+ +
r1 r2 r3
R eq = R +
=
R
E1 E 2 E 3
−
+
r1
r2
r3
i=
1 1 1
1 + R  + + 
 r1 r2 r3 
E eq

 1 1 1 
E
E
E
⇒ i 1 + R  + +  = 1 + 2 + 3
r2
r3
r1
 r1 r2 r3 


 1 1 1 
1 + R  + + 
 r1 r2 r3 

E2
…(3)
1 1 1  E
E 
E
(i 1 + i 2 + i 3 ) = −iR  + +  +  1 + 2 + 3 
r2
r3 
 r1 r2 r3   r1
⇒i=
r1
E3
…(2)
Add (1), (2), (3)
E1 E 2 E 3
+
+
r1
r2
r3
E1
E eq
R eq
1
1 1
 r  +  r
 1  2
 1
 +  
  r3 
Mixed Grouping of cells :
Note :
Potential difference V across external resistance R is such
that E min < V < Emax where Emin and Emax respectively
means the minimum and maximum emf of cells to be
arranged in parallel.
There are n identical cells in a row and number of rows are
m. Emf of each cell is E and internal resistance is r.
E r
For general equation :
E
∑  r 
i=
1+ R
1
∑  r 
∑r
E
r eq =
total cells = mn = P
Total external resistance = R
∑r
1
R eq = R +
Net emf = n E
nr
Total internal resistance =
m
where,
E eq =
R
Current through R, i =
1
1
 
r
∑
1
1
 
r
∑
Physics (Current Electricity)
i=
E
nE
=
R r
nr
+
R+
n m
m
(
mnE
mR − nr
)
2
+ 2 mnrR
i is maximum when,
mR = nR
nr
⇒
R=
m
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OR
Some important points about all :
i will be max if
1.
R+
When cell is discharging :
Current inside cell is from cathode to anode.
nr
= minimum i.e.
m
Terminal voltage V = ε – Ir
d R r 
 +  =0
dm  n m 
d
dm
⇒
R
r
– 2 =0
p
m
⇒
r
R
= 2
mn
m
⇒
R
r
=
n
m
 ε 
So, V = ε – 
r
R+r
 R 
= ε

R+r
i.e. potential difference across cell is lesser than emf
of cell and greater the current drawn from cell lesser
will be the terminal voltage.
2.
When cell is charging :
Current inside cell is from anode to cathode.
Terminal voltage V = ε + Ir and I =
rn
m
So, current and hence power transferred to load is
maximum when load resistance is equal to internal
resistance. This is known as maximum power transfer
theorem.
i.e. potential difference across cell is greater than emf
of cell.
3.
When cell is in open circuit :
E
mE
nE
I max =
=
2r
2R
I=0
2
n E
 nE 
P max = I R = 
 R=
4R
 2R 
(load)
2
2
ε
R+r
 R + 2r 
 ε 
∴V=ε+ 

 r=ε 
R+r
 R+r 
i.e. total external resistance = total internal resistance.
2
r
k
R= ∞
External resistance between cathode and anode is infinite;
R ~ ∞.
OR
I=0
2
Pmax
( load ) =
ε
R+r
 Rm r 

+  = 0
m
 p
⇒
⇒R=
and I =
m2E2R
 mE 

 R=
4r 2
 2r 
V (terminal voltage) = E (emf)
4.
Note :
If in a problem of mixed grouping of cells for maximum
current (or power), n and m came out be fractional, then as
physically neither rows nor cells in a row can be fractional,
taking proper integral values of m and n nearest to
fractional values such that m × n = P, calculate I.
Proper choice will be integral combination of P and S
which provides the maximum current (or power).
When cell is short-circuited :
E r
External resistance R = 0
I=
E
(maximum current)
r
V=0
Physics (Current Electricity)
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Example 28:
SOLVED EXAMPLE
Example 27:
A battery of emf E is connected with three resistances R,
2R and 3R in series. The voltage across 2 R is measured by
a voltmeter whose resistance is 10R. What is percentage
error in measurement of voltage?
Solution:
The arrangement is shown in figure.
E
R A 2R
How many cells of 1.5 V/500 mA rating would be required
in series-parallel combination to provide 1500 mA at 3 V?
Solution:
To get the current of 1500 mA we required three cells in
parallel.
1500 mA
B 3R
1500 mA
1.5 V
V 10 R
The combination gives net e.m.f. of 1.5 V
The resistance of the circuit (Before connecting voltmeter) is
R′ = R + 2R + 3R = 6R
E
E
Current in the circuit, I =
=
R′
6R
The voltage across the 2R is given by,
E
E
× 2R =
V = I × 2R =
3
6R
If the voltmeter of resistance 10R is connected in parallel
with resistance 2R, then the resistance between A and B
decreases and is given as
2
 2R × 10R   20R + R
=


 2R + 10R   12R
  5R 
 =
  3 

17
5
R + 3R =
R
3
3
Hence, the current in the circuit is given by,
E
3E
=
I′ =
17
17 R
R
3
Voltage across A and B is given by,
5R
5
5
3E
×
=
E
V′ = I′ × R =
3
17
17 R
3
The error in the reading is
 E 5E   2 E 
∆V = V – V′ =  −
 =

 3 17   51 
∆V
Percentage error =
× 100%
V
To get the e.m.f. of 3 V, we connected such a two
combinations in series
E net = 2 × 1.5 = 3 V
∴ Total 6 cells are required to get 3 V and 1500 mA.
Ans.: 6 cells
Work done by electric current :
⇒
When potential difference is applied across a
conductor, electrons flow (current passes).
While flowing, electrons collide with atoms and other
electrons and loses its speed. In order to maintain
electron flow (or current), work has to be done.
This work done by source is converted into heat.
⇒
Let V be potential difference across resistance R and
I be current flowing through it. When electric charge
q moves against a potential difference of V, the
amount of work done is given by :
Then total resistance of the circuit = R +
W = QV
But we know, q = It
So, W = V I t
200
= 11.75 %
17
or W =
( V = IR)
V2t
R
This work done produces heat in the conductor,
∴ Heat = VIt = I2Rt =
Ans : 11.75 %
Physics (Current Electricity)
…(1)
and W = I2 Rt
 2E / 51 
=
 × 100
 E/3 
=
1.5 V
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Power in electric circuit :
⇒ The rate at which work is done by electric source is
called electric power, or the rate at which energy is
drawn from the electric source is called electric
power.
dw
Power =
dt
dw
dw
dw
P=
P=
P=
dt
dt
dt
=
dVIt
dt
=
dI 2 Rt
dt
= I2R
= VI
=
dV 2 t / R
dt
= V2/R
Unit of power : Watt, joule/second, volt-ampere.
d[E 2 R /( R + r ) 2 ]
=0
dR
d
⇒
[R (R + r)–2] = 0
dR
⇒ (R + r)–2 + (R) (–2) (R + r)–3 = 0
⇒
⇒ (R + r)–3 [r – R] = 0
⇒ R = r as (R + r)–3 ≠ 0 for finite R
So, power transferred to the load (resistance) by a cell is
E2
maximum when R = r and P max =
4r
This statement in generalised form is called “maximum
power transfer theorem”.
⇒ Electrical energy consumed or dissipated
= Electric power × time
Note :
i.
Total power an consumed in circuit is
E2
[not E2R / (R + r)2]
R+r
2
and will be maximum (= E / r) when R = min = 0
with I = E/r = max
Unit of electric energy = Joule or watt – second, or watt hr,
kWh (kilo watt hour)
ii.
1 kW = 103 watt
1 hp = 746 watt
⇒ 1 commercial unit of electric energy
= 1 kWh
= 1000 watt × 3600 sec.
= 3.6 × 106 joule
iii.
When we use an electric appliance of rating of power 1000
watt for 1 hour, the energy consumed is 1 kWh.
Note :
As resistance of a given electric appliance (eg. Bulb, heater,
etc.) is constant and is given by
 Vs  Vs 2
Vs
W
 =
R=
= 
(as I =
)
V
I
W
 W / Vs 
Where V s and W are voltage and wattage (power) specified
on the appliance. So if the applied voltage is different from
specified, the “actual power consumption” will be :
V
P= A
R
2
V
=  A
 Vs
2
2

V
 × W (as R = s )
W

Power transferred to the load by the cell :
E
E2R
2
( I =
)
P=I R =
2
R
+r
(R + r )
P (minimum) = 0, if R = 0 or ∞
E
E2
with I =
2r
2r
A cell is said to be ideal if the internal resistance is
zero and standard if its emf is precisely specified and
does not change with time.
If R = r, P =
SOLVED EXAMPLES
Example 29:
An electrical heater of 2 kW operates on 230 V supply.
Calculate current drawn by the heater and amount of energy
consumed in 2 hours.
Solution:
P
2000 W
=
= 8.69 A
V
230V
Energy consumed = Power × time = 2 kW × 2 hrs = 4 kWh
= 4 × 3.6 × 106 J = 14.4 × 106 J
∴I=
6
Ans.: 14.4 × 10 J
Example 30:
An electric heater takes 6 A current from 220 V supply line.
Calculate the power of heater and electric energy consumed
(in kWh) by it in 2 hour.
Solution:
P (maximum) when :
P = VI = 220 × 6 = 1320 W = 1.32 kW
W = Pt = 1.32 × 2 = 2.64 kWh
dP
=0
dR
Ans.: 1.32 kW, 2.64 kWh
Physics (Current Electricity)
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SOURCE OF EMF
Advantages of secondary cell:
Electric current is a flow of charge from one end to another
end of the conductor. Normally free electrons in conductor
are in random motion so net current is equal to zero. But if
potential difference is applied across two ends of
conductor, free electrons move from higher potential to
lower potential and constitute the electric current through it.
1.
It can use again and again by recharging it.
2.
Internal resistance of secondary cell is less compared
to primary cell.
1.
After construction, secondary cell has to charge
before use. It requires lot of time.
A device which can maintain potential difference across
a conductor is called source of emf.
2.
The cost of secondary cell is large as compared to
primary cell.
TYPES OF SOURCE OF EMF:
1.
Electrochemical cell: It converts chemical energy
into electrical energy.
LEAD ACCUMULATOR (SECONDARY CELL):
2.
Photo cell: It converts light energy into electrical
energy.
3.
Thermocouple: It
electrical energy.
4.
Electric generator: The mechanical energy used to
rotate conducting loop in magnetic field which is
converted into electrical energy.
5.
converts
heat
energy
into
Atomic cells: In some radioactive reaction, electrons
are emitted for a long periods. Using such a
radioactive substances, atomic cells are constructed
to generate electrical energy.
Disadvantages of secondary cell:
It is invented by French physicist, Gaston Planet in 1859.
It is most common type of storage battery used in
automobiles.
The capacity of an accumulator is 100 hours, means it can
supply a current of 1 ampere for 100 hours. Its charged
potential per cell is 2 V. It discharges to 1.8 V. The specific
gravity of dil. sulphuric acid is (1.25)
While discharging the cell, both e.m.f. of the cell and
specific gravity of electrolyte fall. Whereas, e.m.f. of cell
should not be allowed to fall below 1.8 V and specific
gravity of the electrolyte should not drop below 1.12.
CONSTRUCTION:
1.
The secondary or storage cells, can be recharged by
sending a current through them from the positive to
the negative electrode, by using a source of higher
emf. One such secondary cell, which is widely used,
is the lead accumulator.
2.
It consists of a vessel of an insulating material filled
with a 20% solution of sulphuric acid, which acts as
an electrolyte. Two specially prepared lead plates act
as the electrodes. The plates are in the form of
grids.(Fig.). The positive plate is stuffed with a paste
of lead peroxide, while the negative plate is filled
with a spongy mass of lead.
ELECTROCHEMICAL CELL
In electrochemical cells, chemical reaction takes place
at a steady rate. The total amount of electrical energy
provided by the cell is limited by amount of reactants
present in the cell.
The electrochemical cells are of two types: Primary cells
and Secondary cells.
PRIMARY CELLS:
After discharging, primary cell can not be use again by
passing electric current from external source.
The chemicals inside the cell have to replace completely
after discharging the cell.
e.g. Daniel cell, Leclanche cell, Simple voltaic cell.
CHEMICAL ACTION:
1.
SECONDARY CELLS:
After discharging, the secondary cell can be use again by
recharging it by passing electric current from an external
source. Chemical reactions take place inside a secondary
cells are reversible in nature. Since secondary cells store
energy when it is charged, it is also called as storage cell or
accumulator.
If the terminals of the cell are connected by a
conductor, the electrons start from the negative plate
to the positive plate through the conductor. Inside the
electrolyte, hydrogen ions move to the positive plate
and oxygen ions to the negative plate.
At the positive plate, we get
PbO 2 + H 2 → PbO + H 2 O
E.g. Lead accumulator, Edison alkali cell etc.
Physics (Current Electricity)
During Discharging:
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Further, PbO reacts with dilute sulphuric acid to form
lead sulphate.
PbO + H 2 SO 4 → PbSO 4 + H 2 O
At the negative plate, we get
2Pb + O 2 → 2PbO and further
PbO + H 2 SO 4 → PbSO 4 + H 2 O
Thus, when the cell gets discharged both the plates
contain PbSO 4 and some traces of PbO.
Anode
Cathode
WHEATSTONE’S BRIDGE
A Wheatstone bridge is an electrical circuit used to measure
an unknown electrical resistance by balancing two legs of a
bridge circuit, one leg of which includes the unknown
component. It was invented by Samuel Hunter Christie in
1833 and improved and popularized by Sir Charles
Wheatstone in 1843.
Construction:
Wheatstone’s bridge consists of four resistances R 1 , R 2 , R 3
and R 4 connected to form the four sides of quadrilateral
ABCD. A cell ‘E’ and key ‘K’ are connected in series
across A and C. A galvanometer ‘G’ is connected in
between B and D.
B
R1
A
Pb
R3
H2SO4
During Charging:
If, now, a current is driven through the accumulator
from the positive plate to the negative plate by a
source of higher emf then oxygen ions move to the
positive plate and hydrogen ions to the negative plate
Working:
When circuit is closed, current flow in different branches of
the bridge. The resistances are adjusted so that there is no
current in galvanometer.
i.e. i g = 0
In this case, bridge is said to be balanced.
This happens, when resistance satisfy the following
balance condition.
At the positive plate, we get,
2PbSO 4 + O 2 + 2H 2 O → 2PbO 2 + 2H 2 SO 4
R
R1
= 3
R2
R4
2PbO + O 2 → 2PbO 2
At the negative plate, we get
Pb + H 2 SO 4 → Pb + H 2 SO 4
PbO + H 2 → Pb + H 2 O
Thus after recharging, the positive plate again gets
covered with PbO 2 and the negative plate with
spongy lead. The cell can now be used again.
PROOF USING OHM’S LAW:
Let ‘i’ be the current from battery which arrives at A. It is
divided into i 1 and i 2 which flow along AB and AD
respectively.
For balance condition, B and D should be equipotential, i.e.
i g = 0. So current through BC is i 1 and through DC is i 2 .
Note:
B
New types of accumulators such as the nickel−cadmium
accumulator and the silver−cadmium accumulator have
been developed. They are spill proof, have a low weight to
capacity ratio, and a very long life. They can be charged
and discharged any number of times. On account of their
reliability, they are widely used in emergency devices such
as burglar and fire alarms. Also, because of their
compactness, they are used in missiles, satellites, and
underwater of equipment such as water-proof cameras.
Physics (Current Electricity)
R4
K
D
E
2.
C
G
glass
Pb coated with PbO2
R2
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R2
i1
A i2
R3
i
E
i1
G
i2
D
C
R4
i
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Let
Dividing equation (1) by equation (2),
VB = VD
∴
i 2 R 3 i1R 1
=
i 2 R 4 i1R 2
∴
R 3 R1
=
R4 R2
VA – VB = VA – VD
R1 i1 = R3 i2
… (1)
Similarly,
VB – VC = VD – VC
R2 i1 = R4 i2
R1 R 3
=
R2 R4
… (2)
Dividing eq. (1) by (2), we get
By knowing any three resistances, the fourth unknown
resistance can be easily calculated.
R1
R
= 3
R4
R2
PROOF USING KIRCHHOFF’S LAW:
The distribution of current in the network is as shown in the
figure below. The current entering at point A is i. Let i 1 be
the current through R 1 and i 2 be the current through R 3 and
i g is the current through the galvanometer from B to D.
Hence i 1 − i g is the current through R 2 and i 2 + i g is the
current though R 4 .
B
R1
A
i2
i
C
E
D
VA – VB = VA – VD
Apply ohm’s law to the resistance R 1 , R 3 and R 2 , R 4 we
get
ig
R3
We have V B = V D
(where V A , V B , V C and V D are potentials at points A, B, C
and D respectively)
R2
G
For same balancing condition for wheatstone’s bridge can
be obtained by using ohm’s law.
VB – VC = VD – VC
(i1 −ig)
ig
i1
If the position of cell and galvanometer are interchanged,
the same balancing condition is obtained. So branches AC
and BD are called conjugate arms. The meter bridge works
on principle of Wheatstone’s bridge.
R4
(i2 + ig)
V A – V B = I1 R 1
i
V B – V C = I1 R 2
K
V A – V D = I2 R 3
V D – V C = I2 R 4
Applying Kirchhoff’s second’s law in loop ABDA,
We get,
− i1 R1 − i gG + i 2 R3 = 0
I1 R 1 = I2 R 3
…(1)
But i g = 0
and I 1 R 2 = I 2 R 4
…(2)
∴ − i1 R1 + i 2 R3 = 0
Taking ratio of above two equations (1) and (2), we get
i 2 R3 = i1 R1
…(1)
Similarly applying Kirchhoff’s second law in loop BCDB,
– ( i 1 − i g )R 2 + (i 2 + i g )R 4 + i g G = 0
But i g = 0
∴ – i 1R2 + i 2 R4 = 0
i 2 R4 = i1 R2
Physics (Current Electricity)
…(2)
R
R1
= 3
R4
R2
The measurement of resistance by Wheatstone’s bridge
does not affected by internal resistance of the cell.
By interchanging the position of galvanometer & cell, the
condition for balance for a Wheatstone’s Bridge network
will not be changed.
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Ans : I = 0.6 A, I R1 = I R 2 = 0.4 A, I R 3 = I R 4 = 0.2 A, I g = 0
SOLVED EXAMPLES
Example 31:
In Wheatstone’s bridge network, resistances R 1 = 5 Ω, R 2
= 2 Ω, R 3 = 10 Ω, R 4 = 4 Ω and the galvanometer
resistance (G) = 6 Ω are used. Calculate the current in
different branches of the network, if emf of cell is 2.8 V
and internal resistance can be neglected.
Example 32:
Calculate the equivalent resistance between a and b of the
following network of conductors.
5Ω
3Ω
a
2Ω
Solution:
Balance condition for Wheatstone bridge,
R
R1
= 3
R2
R4
5 10
=
2
4
As bridge is balanced, the current through the galvanometer
is zero (I g = 0).
R eff = (R 1 + R 2 ) | | (R 3 + R 4 )
= (5 + 2) | | (10 + 4)
= 7 | | 14
=
I1
7 × 14
7 + 14
7 × 14
=
21
5Ω
2Ω
6Ω
I
I2 10 Ω
I2 + I3
d
Distribution of current in compliance with junction rule is
shown in above figure.
Applying Kirchhoff’s 2nd law to the loop acda,
2 I1 – 4 I2 – 9 I3 = 0
2.8
2.8 × 3
E
=
=
= 0.6 A
14 / 3
14
R eff
Current through R 1 and R 2 is I 1 and R 3 and R 4 is I 2 such
that
I 1 + I 2 = I = 0.6
I 1 (5 + 2) = I 2 (10 + 4)
2Ω
I 1 = 30 K, I 2 = 51 K, I 3 = –16 K
The p.d between ‘a’ and ‘b’ is given by,
V ab = R eq (I 1 + I 2 )
...(3)
where, R eq = equivalent resistance
V ab = V ac + V cb
= 5 I 1 + 2 (I 1 – I 3 )
I1 = 2 I2
= 7 I1 – 2 I3
3 I 2 = 0.6
I 2 = 0.2 A
∴ I 1 = 2 × 0.2 = 0.4 A
Physics (Current Electricity)
…(2)
Solving (1) & (2), we get
I3
I1
I2
=
=
18 − (−12)
6 − (−45)
(−20 − (−4))
I
I
I
∴ 1 = 2 = 3 = K (constant)
− 16
30
51
7 I 1 = 14 I 2
But I 1 + I 2 = 0.6
…(1)
–3 I 3 – 4 (I 2 + I 3 ) + 2 (I 1 – I 3 ) = 0
4Ω
14
Ω
3
Current through main circuit,
From figure,
4Ω
Applying Kirchhoff’s 2nd law to the loop cdbc,
=
I=
b
Solution:
This is a network of five conductors, as shown in figure
below.
c
2Ω
5Ω
I1 – I3
I3
I I1
a
b
3Ω 4Ω
I
I2
–5 I 1 – 3 I 3 + 2 I 2 = 0
5 I1 – 2 I2 + 3 I3 = 0
Ig
2Ω
…(4)
Equating eq. (3) and eq. (4), we get
R eq (I 1 + I 2 ) = 7 I 1 – 2 I 3
( 7 I1 − 2 I 3 )
[7 × 30K ] − [2 × (−16K )]
R eq =
=
30K + 51K
( I1 + I 2 )
210K + 32K
242
=
=
=3Ω
81K
81
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Ans : 3 Ω
Example 33:
Case III :
Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively
are connected in cyclic order to form a Wheatstone’s
network. Is the network balanced? If not, calculate the
resistance to be connected in parallel with the resistance of
10 Ω to balance the network.
Solution:
E
R3
R4
1.
B
P = 15 Ω
10
R
=
= 2.5
4
S
Q = 12 Ω
A
P
R
≠
Q
S
S=4Ω
D
i.e.

 R1
 V < 
 R1 + R 2


 V

R1 R4 > R2 R3
2.
While performing experiment with Wheatstone
bridge, the cell key K C should be pressed first and
then galvanometer key otherwise a momentary
deflection in galvanometer is produced even in
balanced bridge due to induced emfs in various
resistance coils as initially the current is changing
from 0 to I.
3.
Wheat stone’s bridge is not suitable for measurement
of very low resistances (such as copper rod) or very
high resistances of the order of mega ohms. Very low
resistances are determined with the help of ‘Kelvin’s
double bridge’ while very high resistance by
‘Leakage method’.
P
R′
=
Q
S
P
15
15
×S=
×4=
=5
Q
12
3
1
1
1
=
+
R′
R
RS
If wheat stone’s bridge is unbalanced (i.e. ig ≠ 0),
current in BD will be from D to B if
V D > V B i.e. V A – V D < V A – V B
 R3
i.e. 
 R3 + R4
C
R = 10 Ω
The bridge is not balanced.
R′ =
R1
Note :
P
15
Let
=
= 1.25
Q
12
Let
R2
G
1
1
1
1
1
2 −1
1
=
–
=
–
=
=
5 10
10
R′ R
10
RS
R shunt = 10 Ω
METER BRIDGE
Ans : 10 Ω
Different forms of wheat stone’s bridge:
Case I :
R1
R2
G
R3
R4
Case II :
R1
R2
R3
R4
PRINCIPLE: It works on the principle of Wheatstone’s
network.
CONSTRUCTION: A meter bridge is a practical form of
Wheatstone’s network. It consists of a wooden broad over
which a uniform wire AC of one metre is stretched along a
meter scale.
The ends A and C of the wire are connected to two thick
L shaped copper strips. Another thick straight copper strip
is placed between the end strips such that two gaps are
formed between it. The unknown resistance ‘X’ is
connected in one gap, while a known resistance ‘R’
(usually a resistance box) is connected in the other gap.
The mid-point D of the straight copper strip is connected to
a jockey (J) through a sensitive galvanometer (G).
G
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The ends of the wire AC are connected in series with a
cell (E), a plug key (K) and a variable resistance (R h )
(usually a rheostat).
CIRCUIT DIAGRAM:
X
R
D
G
i1
i2
A
J
B
C
100 – 

E
i1
K
Rh
PROCEDURE:
1.
The plug key ‘K’ is closed and current ‘i’ is passed
through the circuit.
2.
A contact is made with the jockey at the ends of the
wire AC and the rheostat is adjusted such that the
deflections in the galvanometer G are within the scale.
3.
A point B is obtained on the wire AC with the help of
the jockey and the galvanometer (G) such that when
contact is made with the jockey at B, there is no
deflection in G. This point B is called a null point.
THEORY:
Let R AB be the resistance of the wire AB and R BC be the
resistance of the wire BC. Now when null point B is
obtained on the wire AC, the Wheatstone’s bridge is
balanced. X, R and resistance of wire AD and wire CD
forms four arms resistance of Wheatstone’s Bridge.
Hence according to the condition of balance, we get
Resistance of wire AD of length  1
X
=
R
Resistance of wire DC of length  2
R
X
= AB
…(1)
R
R BC
Let i be the current through the circuit and σ be the
resistance per unit length of wire. Let  1 and  2 be the
distance of point B (balancing lengths) measured in
centimeter from ends A and C of wire AC respectively.
For uniform wire, the resistance between any two points on
it is directly proportional to the length of wire between two
points. So,
R AB = σ 1
R BC = σ 2
R AB
σ 1
1
=
=
R BC
σ 2
2
Substituting this in equation (1), we get

X 1
=
…(3)
=
100 − 
R 2
 1 

X = R 
 100 −  1 
Knowing R and by measuring , the unknown resistance X
can be determined.
ERRORS AND CORRECTIONS:
1.
The wire of the bridge may not be uniform.
The error due to this can be minimised by
interchanging the position of unknown and known
resistances.
2.
The ends of wire may not coincide with ends of the
scale. The contact between the wire and the strips
may not be perfect. This introduces a contact
resistance. Also resistance of the strip and connecting
wires is ignored.
These errors can be minimised by arranging the null
points in the middle-third of the wire. This is done by
adjusting known resistance R.
MERITS OF WHEATSTONE BRIDGE METHODS
OVER OTHER METHODS:
1.
It is null method. The measurement of resistance
made by this method is not affected by internal
resistance of the battery.
2.
The measurement is not affected by resistance of
voltmeter and ammeter, as no measurement of
current or potential difference is involved.
3.
The value of unknown resistance can be measured to
a very high degree of accuracy by increasing the ratio
of the resistance in arms P and Q.
SOLVED EXAMPLES
Example 34:
Two equal resistances are introduced in two gaps of a meter
bridge. Find the shift in the null point if an equal resistance
is connected in series with resistance in left gap.
Solution:
First case: R L = R R = R
Therefore balance point is occurred at middle i.e.  = 50 cm
Second case: R L = R + R = 2R, R R = R
According to balance condition,
RL

=
100 − 
RR
2R

=
R
100 − 
200 – 2 = 
3 = 200
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When a resistor 12 Ω is connected in parallel with S, then
the value of resistance in branch CD becomes S′
200
= 66.70 cm
3
Shift in balance length = 66.70 – 50 = 16.70 cm
=
S′ =
Ans : 16.70 cm
Example 35:
Now at balancing point, ′ = 51.9 cm
With two resistance wires in the two gaps of a meter bridge,
the balance point was found to be (1/3) m from the zero
end. When a 6 Ω coil is connected in series with the smaller
of the two resistances, the balance point is shifted to
(2/3) m from the same end. Find the resistance of the
two wires.
Solution:
New balancing condition becomes,
R
′
51.9
=
=
S′ 100 −  ′ 100 − 51.9
∴
R
51.9
=
S′
48.1
33.7
S
51.9
66.3
=
12 S
48.1
(12 + S)
R 2 = 2R 1 ⇒ R 1 < R 2
33.7
(S + 12)
51.9
×
=
66.3
12
48.1
In second case,
R1 + 6
2/3
=
1/ 3
R2
S + 12 =
R1 + 6 = 2 R2
R1 = 2 Ω
R2 = 4 Ω
Putting the value of S in eq. (1), we get
Ans : 2 Ω, 4 Ω
R=
Example 36:
33.7
× 13.5 = 6.86 Ω
66.3
Ans : 6.86 Ω, 13.5 Ω
In a meter bridge shown in figure, the null point is found to
be at a distance of 33.7 cm from A. If now a resistance of
12 Ω is connected in parallel with S, the null point occurs at
51.9 cm from the same end. Determine value of R and S.
D
S
R

G
B
C
Solution:
Balance condition in Meter Bridge,

R
33.7
33.7
=
=
=
100 − 
S
100 − 33.7
66.3
33.7
×S
66.3
Physics (Current Electricity)
…(1)
Example 37:
Two resistances of 30 Ω and 45 Ω are connected in left and
right gap of Meter Bridge. Find the shift in null point, when
resistance of 30 Ω is shunted by another resistance of 30 Ω.
Solution:
100 –
K}
( )
R=
51.9 12 × 66.3
×
= 25.5 Ω
48.1
33.7
S = 25.5 – 12 = 13.5 Ω
R1 + 6 = 4 R1
A
…(2)
From eq. (1) and (2),
2
1
1
m,  2 = 1 –
= m
3
3
3
1/ 3
1
=
=
2/3
2
In first case,  1 =
R1

= 1
R2
2
12 S
12 + S
X = 30 Ω
R = 45 Ω
X

=
R
100 − 
where,  = balancing length from left
30

=
45
100 − 
3 = 200 – 2 
5 = 200
200
=
= 40 cm
5
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When 30 Ω is shunted by 30 Ω, the resistance in the left
gap becomes
3 = 100
∴=
2
1
1
1
=
+
=
30
Xp
30
30
100
= 33.3 cm
3
Ans : 33.3 cm, 20 Ω, 30 Ω
∴ X p = 15 Ω
Let ′ be the balancing length from left end.
X
′
=
R
100 − ′
15
′
=
45
100 − ′
1
′
=
3 100 − ′
KELVIN’S METHOD
Kelvin’s method is for the determination of the resistance
of a galvanometer. The galvanometer, whose resistance
‘G’ is to be measured, is connected in one of the gaps.
A resistance box ‘R’ is introduced in the other gap. A
jockey is directly connected by a wire to the point C
between G and R.
3′ = 100 – ′
THEORY:
4′ = 100
1.
A resistance R is introduced from the resistance box
but the jockey is not touching the wire. The current
‘i’ from the battery is divided into i 1 and i 2 at the
point A. The current i 1 flows through G and R and
current i 2 flows directly along the wire from A to B.
The galvanometer shows a deflection θ.
When two unknown resistances X and Y inserted in left and
right branch of Meter Bridge respectively, the null point is
obtained at 40 cm from left end. When a resistance of 10 Ω
is connected in series with X, the null point shifts by 10 cm.
Find the position of null point, when the 10 Ω resistance is
connected in series with resistance Y instead of X.
Hence determine the value of resistances X and Y.
2.
When jockey is touched at a point D on the wire,
there will be a flow of current between C and D
(provided C and D are not equipotential). The
distribution of current in other branches also changes.
Hence the current in galvanometer changes and
deflection in galvanometer θ also changes.
Solution:
3.
If C and D are equipotential, there will be no flow of
current across C and D and the distribution of current
remains unaltered and deflection θ will remain same.
4.
Since across C and D, there is no flow of current,
′ = 25 cm from left
Shift in null point =  – ′ = 40 – 25 = 15 cm towards left
Ans : 15 cm towards left
Example 38:
When the null point is at 40 cm,
40
X
40
2
=
=
=
60
Y
100 − 40
3
2
Y
3
When 10 Ω resistance is connected in series with X,
Let  G and  R : distances of point D from end A and
C of wire AC respectively.
X + 10
50
=
=1
Y
100 − 50
X + 10 = Y
G, R : resistance of wire  g and  r form four arms of
balanced Wheatstone’s bridge network.
2
Y + 10 = Y
3
∴ Y = 30 Ω
Resistance of wire ADof length  g
G
=
R
Resistance of wire CD of length  r
X=
σ : resistance per unit length of wire AC.
From balancing condition,
X = 20 Ω
=
When 10 Ω is connected in series with Y,

X
20
1
=
=
=
100 − 
Y + 10
30 + 10
2
Physics (Current Electricity)
σ g
σ r
G = R⋅
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=
G
R
G
G
=R⋅
R
100 −  G
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Using this, the galvanometer resistance G is
determined
In this experiment jockey is touched on the wire and the
neutral point is obtained. (galvanometer deflection remains
at θ). Hence it is also called constant deflection method.
CIRCUIT DIAGRAM:
It states that “For a uniform wire carrying steady
current, the potential difference between any two points
on it is directly proportional to the length of the wire
between these two points.”
i.e. V ∝ 
i.e.
V
= constant

CONSTRUCTION:
G
R
C
i1
i1
i2
A
J
D
G
i
E
B
R
K
A potentiometer consists of a long and uniform wire AB
stretched between the two terminals on a wooden board in
turns between two thick copper strips as shown. The wire is
about 10 m. The wire is usually made of constantan or
manganin. A meter scale is fixed on wooden frame so as to
measure the length of wire.
Wooden frame
Rh
Note:
Uniform wire
A
Resistance (R) from the resistance box should be such that
the neutral point should come within middle one third of
the meter bridge.
ERRORS AND CORRECTION:
B
In practice, the wire AB is not uniform and hence there is
always some error introduced in measurements of
 G and  R .
To eliminate this error the lengths  G and  R are observed
again on interchanging the gaps of G and R. The mean,
which is obtained from these results, is used to determine
the resistance of galvanometer G.
POTENTIOMETER
It is an electrical device by means of which potential
differences can be measured accurately. It may be the
Metre scale
THEORY:
Consider the potentiometer wire of a uniform cross-section
area of length ‘L’ and resistance ‘R’. The source of e.m.f. E
and negligible internal resistance is connected to the wire
through key. Let σ be the resistance per unit length.
R = σL
Let V AB be the P.d. across wire, there is uniform fall of
potential along the wire from A to B. By ohm’s law the
current passing through the wire is given by,
1.
potential difference between the ends of a current
carrying conductor.
2.
potential difference across any branch of an electric
circuit.
i=
3.
terminal potential difference of a cell.
Let P be any point on the wire, such that AP = 
4.
e.m.f of a cell etc.
Resistance of wire AP of length  is
PRINCIPLE:
VAB
V
= AB
R
σL
R AP = σ
So P.d. V AP between point A and P is given by,
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V AP = IR AP = Iσ =
VAB
⋅ σ
σL
USES OF POTENTIOMETER:
1.
V AB and L are constant
To compare the e.m.f ’s of two cells
V AP ∝  = K = constant
K called potential gradient of wire.
2.
OR
Potential gradient is the potential drop per unit length.
S.I unit of potential gradient of a wire is V/m.
This shows that the principle of a potentiometer can also be
stated as “For a uniform wire carrying steady current,
the potential gradient along the wire is constant.”
While working with potentiometer, the terminals of wire
AB (resistance R W , length L) are connected in series with
battery (of emf E and internal resistance r) and load
resistance (R L ).
Note: The wire AB used has large specific resistance and
low temperature coefficient of resistance. The cell used is
such that it sends a steady current through the wire AB.
+V–
+
A
A
A potentiometer wire of length 100 cm has a resistance of
10 Ω is connected in series with a resistance and a cell of
emf 2 V of negligible internal resistance. A source of emf
of 10 millivolt is balanced against a length of 40 cm of the
potentiometer wire. What is the value of external
resistance?
Solution:
2
E
=
R + 10
R +RW
20
 2 
V AB = IR W = 
 × 10 =
R + 10
 R + 10 
p. g. =
VAB
20
 20  1
=
V/m
× =
L
 R + 10  1 R + 10
V AP = (p. g.) 
 20 
10 × 10–3 = 
 × 0.4
 R + 10 
8
R + 10
R + 10 = 800
J
P
K
RL
by sum and difference method.
10–2 =
–
E
ii.
Example 39:
I=
WORKING:
by individual cell method.
To determine the internal resistance of a given cell.
SOLVED EXAMPLES
Potential gradient (p.g.):
The rate of fall of potential with length of a wire in the
direction of flow of current through the wire is known as
potential gradient of the wire.
i.
R = 790 Ω
B
Ans. : 790 Ω
Then, current in the wire is given by, I =
RW
E
+ RL + r
Potential difference across wire


E
 R W
V AB = I R W = 
 RW + RL + r 
Potential gradient =

 RW
VAB
E

= 
L
 RW + RL + r  L
If P is any arbitrary point on wire,
Example 40:
A battery of emf 2 V and internal resistance 1 Ω is used to
send a current through a potentiometer wire of length
200 cm and resistance 4Ω. What length of the
potentiometer wire will be required to balance a Daniell
cell of ‘emf’ 1.08 V?
Solution:
I=
E
2
=
= 0.4 A
4 +1
R+r
If σ is Resistance per unit length of wire,
∴ V AP = (p.g.) 
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σ=
Arrangement:
4
= 2 Ω/m
2
The cell of emf ‘E’ is connected between two ends, ‘A’ and
‘B’, of the potentiometer wire through Rheostat (R h ) and
single key (K). E 1 & E 2 are two cells whose e.m.f is to be
compared (E > E 1 , E 2 ). The positive terminals of two cells
are connected to terminal A of potentiometer. The negative
terminals of the cells are connected to terminals ‘a’ and ‘b’
of two-way key. The common terminal ‘c’ is connected to
the jockey (J) through galvanometer (G).
Circuit diagram:
e.m.f. =  × σ × i
1.08 =  × 2 × 0.4
∴=
1.08 10.8
=
= 1.35 m
0.8
8
Ans. : 1.35 cm
Example 41:
A 5 wire potentiometer each of having 1 m length is
connected to a storage cell of steady emf 2 V and 1 Ω
resistance. A primary cell is balanced against 3.5 m of it.
What resistance will be required in series with storage cell
to push the null point to the centre of the last wire, i.e.
4.5 m? (The wire has 3 Ω resistance per metre)
E1
a
c
+
A
E2
–
G
J
b
A
E
D
Solution:
C
K
Potentiometer has 5 wire each is length 1 m. Therefore, the
length of potentiometer wire = 5 m
B
Resistance per metre = σ = 3 Ω/m
Rh
∴ R W = 15 Ω
Let, I =
E
RW
2
2
1
=
=
A
=
8
15 + 1 16
+r
First case: E 1 =  1 × σ × I = 3.5 × 3 ×
1
10.5
=
= 1.31 V
8
8
Second case: To obtain the null point at 4.5 m, resistance
‘X’ is connected in series.
∴ I1 =
RW
2
2
E
=
=
15 + 1 + X 16 + X
+ r + RS
 2 
1.31 = 4.5 × 3 × 

 16 + X 
20.96 + 1.31 X = 27
∴
1.31 X = 6.04
…(2)
E1  1
=
E2  2
Thus E 1 and E 2 can be compared by knowing  1 and  2 .
6.04
= 4.61 Ω
1.31
Ans. : 4.61 Ω
APPLICATIONS OF POTENTIOMETER
COMPARISON OF E.M.F.’S OF TWO CELLS BY
INDIVIDUAL CELL METHOD:
Physics (Current Electricity)
E 1 = p.d. across AC = i R AC = i σ  1
…(1)
Now open ac and close bc, so E 2 is in circuit. Locate the
point D on the wire AB so that the deflection in the
galvanometer is zero. Measure the balancing length
AD =  2
E 2 = p.d. across AD = i R AD = i σ  2
where i = current through the circuit
σ = resistance per unit length
Eq. (1) by Eq. (2) gives
E 1 = 2 × σ × I1
∴X=
Working:
Close ac by keeping bc open. So E 1 is in circuit. Locate the
point C on the wire AB so that the deflection in the
galvanometer is zero. Measure the balancing length
AC =  1
SOLVED EXAMPLE
Example 42:
In a potentiometer arrangement, the cell of emf 1.5 V gives
a balance point at 42 cm length of the wire. If the cell is
replaced by another cell, the balance point is shifts to
72 cm. What is emf of the second cell?
Solution:
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Locate point D on the wire AB so that the galvanometer
deflection is zero. Measure the balancing length, AD = L 2
E1

= 1
E2
2
∴ E 1 − E 2 = IσL 2
42
1.5
=
72
E2
…(2)
Eq. (1) ÷ Eq. (2) gives
∴
E 1 + E 2 L1
=
E1 − E 2 L 2
Ans. : 2.57 V
∴
E1
L + L2
= 1
E2
L1 − L 2
COMPARISON OF E.M.F.’S OF TWO CELLS BY
SUM AND DIFFERENCE METHOD:
Thus E 1 and E 2 can be compared by knowing L 1 and L 2 .
SOLVED EXAMPLE
Example 43:
72
E2 =
× 1.5 = 2.57 V
42
Arrangement:
Connect the circuit as shown in a figure. The cell of emf
‘E’ is connected between two ends, ‘A’ and ‘B’, of the
potentiometer wire through Rheostat (R h ) and single key
(K). E 1 & E 2 are two cells whose e.m.f is to be compared
(E > E 1 , E 2 ). The positive terminal of E 1 is connected to
terminal A of potentiometer. The negative terminal of E 1 is
connected to either positive or negative terminal of
E 2 (< E 1 ) via four-way key. The common terminal ‘c’ is
connected to the jockey (J) through galvanometer (G).
Two cells of emf E 1 and E 2 (< E 1 ) are connected as shown
in figure below.
A
B
C
E1
E2
When the potentiometer is connected between points A and
B, the balancing length is 300 cm. On connecting the same
potentiometer between points A and C, the balancing length
is 100 cm. Calculate the ratio of the emfs of the two cells.
Circuit diagram:
Solution:
When potentiometer is connected between A and B,
E1 ∝ 1
When potentiometer is connected between A and C,
E2
d
E1
b
a c
+
A
G
J
E1 – E2 ∝ 2
E1 − E 2

100
= 2 =
300
E1
1
E
1
1– 2 =
3
E1
E2
1
2
=1– =
3
3
E1
E1
3
=
2
E2
–
∴
A
E
D C
K
B
Rh
Working:
Close da and bc and keep ac and db
E1
E2
open. Then cells, E 1 and E2 are in
– + –
+
assistance i.e. (E1 + E 2 ) are in circuit.
Locate the point C on the wire AB so that the galvanometer
deflection is zero. Measure the balancing length, AC = L 1
∴ E 1 + E 2 = IσL 1
…(1)
where σ = resistance per unit length
Close ac and db and keep da and bc
open. Then cells E 1 and E 2 are in
opposition i.e. (E 1 − E 2 ) are in circuit.
E1
+ –
–
E2
+
Ans. : 3 : 2
Example 44:
Two cells of emf E 1 and E 2 are connected together in two
ways as shown here
E2
E1
E2
E1
The balance points in a given potentiometer experiment for
these two combinations of cells are found to be
351.0 cm and 70.2 cm respectively. Calculate the ratio of
the emf of two cells.
Solution:
Let K be the potential gradient of potentiometer.
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First combination: E 1 + E 2 = K  1
Second combination: E 1 – E 2 = K  2
But V = IR
and E 1 = I (R + r)
E1
R+r
=
…(4)
V
R
Comparing eq. (3) and (4), we get

R+r
= 1
R
2

r
1+
= 1
R
2
Eq. (1) ÷ eq. (2) gives

351
E1 + E 2
= 1 =
=5
70.2
E1 − E 2
2
5 +1
6
3
E1
=
= =
4
2
E2
5 −1
Ans. : 3 : 2
DETERMINATION OF INTERNAL RESISTANCE
OF CELL:
Arrangement:
Connect the circuit as shown in a figure. The cell of emf ‘E’
is connected between two ends, ‘A’ and ‘B’, of the
potentiometer wire through Rheostat (R h ) and single key (K 1 ).
Let we have to calculate internal resistance ‘r’ of the cell
whose emf is ‘E 1 ’. A resistance box (R) and plug key (K 2 )
are connected in parallel with the cell of e.m.f ‘E 1 ’ as
shown in figure below. The negative terminal of E 1
connected to the jockey through the galvanometer (G).
E must be greater than E 1 so that V AB > E 1
Circuit diagram:
R.B.
E1
+
A
A
Example 45:
A cell of emf 2 V is connected across a wire of
potentiometer. It is used to determine internal resistance of
1.5 V cell. The balance point of the cell in open circuit is
76.3 cm. When a resistor of 9.5 Ω is used in the external
circuit of the cell, the balance point shifts to 64.8 cm length
of the potentiometer wire. Determine the internal resistance
of the cell.
Solution:


r =  1 −1 R =
2

 76.3 
− 1 × 9.5

 64.8 
Ans. : 1.68 Ω
G
P
K
B
Rh
Working:
Keep the key ‘K 2 ’ open. If  1 is the corresponding
balancing length, then
...(1)
E 1 = Iσ 1
where σ = resistance per unit length
Now some resistance R is taken from the resistance box.
Close the key K 2 . The cell E 1 delivers a current I through the
resistance R. Hence the terminal potential difference of the
cell is now V. If corresponding balancing length is  2 , then
…(2)
V = Iσ 2
Eq. (1) ÷ Eq. (2) gives
E1

…(3)
= 1
V
2
Physics (Current Electricity)
SOLVED EXAMPLE
= (1.177 – 1) × 9.5= 0.177 × 9.5= 1.68 Ω
K2
–
E


∴ r = R  1 − l 
 2

PRECAUTIONS TO USE POTENTIOMETER:
1.
The emf of source battery (driver cell) must be
greater than emfs of cell used for comparison
i.e. E > E 1 , E > E 2 , E > E 1 + E 2
2.
The positive terminal of E 1 or E 2 or the
combinations E 1 + E 2 and E 1 − E 2 must be
connected to that end where positive terminal of
driving cell is connected.
3.
The wire of potentiometer must be uniform
4.
The resistance of potentiometer wire should be high.
ADVANTAGES OF POTENTIOMETER OVER
VOLTMETER:
1.
A potentiometer measures p.d very accurately
because it does not draw any current from the source
of which the p.d. is to be measured. A voltmeter
always measures p.d. less than the true value because
it draws some current from the source of which the
p.d. is to be measured.
2.
A potentiometer measures terminal p.d. of a cell as
well as a e.m.f. of a cell very accurately. A voltmeter
measures terminal p.d. only.
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3.
4.
Very small e.m.f’s such as thermoelectric e.m.f’s and
very small potential differences can be measured
accurately by means of a potentiometer. A voltmeter
cannot be used for this purpose.
A potentiometer can be used to calibrate an ammeter
and a voltmeter. A voltmeter cannot be used for this
purpose.
DISADVANTAGES OF POTENTIOMETER OVER
VOLTMETER:
1.
While the voltmeter is a direct reading instrument,
potentiometer is not.
2.
Voltmeter is portable instrument, potentiometer is
bulky.
Physics (Current Electricity)
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