CURRENT ELECTRICITY
Current : It is defined as rate of flow of charge
with respect to time. It is not a scalar quantity because
it has direction, it is not vector quantity because it
does not follow triangle law of vector addition hence
it is a tensor quantity.
The direction of current is always along the
direction of flow of positive charge or opposite to the
direcion of flow of electrons i.e. from highter potential
towards lower potential. Current is of two types :
(1) Instantaneous Current : The current flowing
across a conductor at a particular time is called as
instantaneous current and is given by
dq
dt
i
(2) Average Current : It is defined for a given
interval of time. If a charge  q flows across a given
cross-section in a time interval of  t then the average
current through the given cross-section is defined as :
iavg 
q
t
 q = Amount of charge
 t = Time interval
The area under the curve current versus
time is defined as the amount of charge
flowing through any cross-section in any
interval of time
Charge  Area under i Vs t curve
Unit : Amper or c/sec.
Dimension : [A]
CURRENT DUE TO
CIRCULATING CHARGE
Every charge moving along a circular path
behaves as a current carring loop.
Consider a point charge q moving along a circular
path with a uniform speed V. If radius of circular
path is R then the average current through any crosssection in the circuit is given by
i
q
2
i
qV
2R
q = Charge
R = Radius of circle
V = Linear velocity
  Angular velocity
Self Practice Problems
1.
The current i through a wire varies with time t
as shown. How many coulomb charge flows
through the wire in time interval t1 0 to t 2 3
second ?
(a) 13.5 C
(b) 27 C
(c) 37.5 C
(d) 54 C
Ans. (d)
CONDUCTORS IN THE
ABSENCE OF ELECTRIC FIELD
In the absence of any electric field, the free
electrons of the conductor behave like the molecules
of an ideal gas and move inside the conductor in a
random motion. The path followed by the electrons
is known as zig-zag path.
The speed of these electron is known as Root
mean square speed (R.M.S. speed). the electrons
moves along the zig-zag path with constant speed
which is given by
Vrms 
3kT
m
k = Boltzmann's constant, k = 1.38 × 10–23 J/k.
T = Temp. of the conductor in kelvin only
m = Mass of the electron
Root mean square velocity is also called as
thermal velocity of electrons.
Since, the free electrons are continuously in a
random motion and hence the net charge transfered
across a given cross-section is always zero. The
number of electrons crossing from left to right through
any cross-section is same as the number of electrons
crossing from right to left and Hence, no current flows
through a conductor in the absence of electric field.
Current Electricity # 1
Mean Free Path : Since, the free electrons inside
the conductor are continously in a randum motion and
collide with the other electrons inside the conductor.
It is defined as the distance travelled by the electron
between any two successive collisions ().
Relaxation Time : Since, the free electrons inside
the conductor are continously in a random motion and
collide with the other electrons inside the conductor.
The time taken by the electron in between two
successive collisions is known as relaxation time ().
Rela tio n between mean free path and
relaxation time :
For two success collisions.
Time = 
Distance = 
Speed = Vrms
Time 

distance
Speed

Vrms
  Relaxation time,   Mean free path,
Vrms  rms speed of electrons
DEPENDENCE OF RELAXATION
TIME ON TEMPERATURE
On increasing the temperature of the conductor
the r.m.s. speed of electrons increases and hence,
relaxation time decreases i.e. the collisions become
more frequent. (Number of collisions per sec
increases)
CONDUCTORS IN THE
PRESENCE OF ELECTRIC FIELD
Drift Velocity : When a potential difference is
applied across a conductor then the free electrons of
the conductor experience electrostatic force on them
due to which these electrons start accelerating inside
the conductor but due to continuous and repeated
collisions with the other electrons and atoms, these
electrons move with almost a constant velocity known
as drift velocity. It is also called as electrical velocity
of electrons.
Drift velocity of the electron is given by
eE
Vd 
m
 = Relaxation time
e = Charge of electron
E = E.F. inside the conductor
m = Mass of electron
The direction of drift velocity is always opposite
to the direction of E.F.
Since relaxation time is very small and
hence, instantaneous velocity is same as
average velocity.
Relation between Drift Velocity and Potential
Difference : When a potential difference is applied
across a conductor then the free electrons of the
conductor experience electrostatic force on them due
to which these electrons start accelerating inside the
conductor.
E
Vd 
V
l
eV
ml
V = P. D. across the conductor
l = Length of conductor
Effect of Temperature on drift Velocity
On increasing temperature, drift velocity
decreases because the relaxation time decreases.
Path of electron in the presence of E.F.
Due to the electric field, the path of electrons is
parabolic and in any small interval of time the
displacement of the electrons is against the direction
of electric field.
Theoretically the path is slightly curved but we
can neglect it because of small relaxation time.
In the absence of E.F. the free electrons of
the conductor move with a uniform speed
known as Vrms. But in the presence of E.F.
the free electrons of the conductor move
with both with drift velocity and r.m.s.
speed.
On increasing temperature Vrms of electrons
increases but drift velocity decreases.
6
  108 m ,   1014 sec , Vrms  10 m / sec
Vd  104 m / sec .
CURRENT DENSITY
When a potential difference is applied across a
conductor then the free electrons of the conductor
experience electrostatic force on them due to which
these electrons start accelerating inside the conductor,
as a result of which a current starts flowing across the
conductor.
Current Electricity # 2
It is defined as current flowing per unit area of
cross-section through a conductor.
J
i
A
i = Current
A = Area of cross section of the conductor i.e.
area to which current flows normally
For cylindrical conductor area of cross-section is
A  r 2
r = Radius of cross-section
Unit : Amp / m2
Dimension : [ J ]  [ AL2 ]
Relation between drift velocity and current
When a potential difference is applied across a
conductor then the free electrons of the conductor
experience electrostatic force on them due to which
these electrons start accelerating inside the conductor,
as a result of which a current starts flowing across the
conductor.
Vd 
i
Ane
i = Current flowing in the conductor
A = Area of corss section of conductor (the area
to which current flows normally)
n = Number of free electrons per unit volume of
conductor
e = Charge of electron (1.6 × 10–19)
Consider a current flowing in the conductor so
that in a small time dt, dq amount of charge passes
through a given corss-section and covers a smalldistance dx then
dq  n( A)( dx )× e
( q  ne)
dq
dx
 n( A)( e)
dt
dt
i  nAe (Vd )
Vd 
i
nAe
RELATION BETWEEN CURRENT
DENSITY AND DRIFT VELOCITY
When a potential difference is applied across a
conductor then the free electrons of the conductor
experience electrostatic force on them due to which
these electrons start accelerating inside the conductor,
as a result of which a current starts flowing across the
conductor.
Vd 
J
ne
J = Current density
n = Number of electrons per unit volume of
conductor
e = Charge of electron (1.6 × 10–19)
Mobility : It is defined as the ratio of drift velocity
and electric field. It is the measure of how fast the
electrons are moving on the application of a unit
electric field.

Vd
E

e
m
Vd = Drift velocity
E = Electric field intensity
 = Relaxation time
e = Charge of electron
m = Mass of electron
On increasing temperature relaxation time
decreases and hence mobility of electrons
decreases.
Relation between current density and E.F.
J  E
J = Current density
 = Conductivity of the conductor
E = Electric field

ne2 
m
m = Mass of electron
n = Number of electrons per unit volume
 = Relaxation time
e = Charge on electron
Conductivity of a conductor depends upon
following two factors :
(1) Upon the nature of material of conductor
[n = number of free electrons per unit volume]
(2) Temperature of the conductor. On increasing
temperature, conductivity of conductor decreases.
Resistivity (Specific resistance) : The reciprocal
of conductivity is defined as resistivity of the
conductor.
1
Current Electricity # 3
m
ne 2
 = Conductivity of conductor
m = Mass of electron
n = Number of electrons per unit volume
 = Relaxation time
e = Charge on electron
Resistivity of a conductor depends upon
following two factors :
(1) Upon the nature of material of conductor
[n = number of free electrons per unit volume]
(2) Temperature of the conductor. On increasing
temperature, resistivity of conductor increases.
RESISTANCE OF A CONDUCTOR
It is the property of nature of material of the
conductor, due to which it opposes the flow of charge
or current in it.
When a P.D. is applied across a conductor then
the free electrons of the conductor start moving with
a constant drift velocity, these electrons collied with
the other electrons and atoms of the conductor. Hence,
motion of electrons is opposed during collisions and
this opposition offered by the conductor to the flow
of electrons is called as resistance.
The reason behind resistance is collision of the
electrons. More the collisions of electrons, more will
be resistance of the conductor.
The resistance of the conductor is given by :
R
l
A
R
m l
ne2  A
Specific Resistacne / resistivity of conductor
l = Length of conductor (Length along the
direction of flow of current)
A = Area of cross-section of conductor (The area
to which current flow normally)
m = Mass of electron
n = Number of electrons per unit volume
 = Relaxation time
e = Charge on electron
Resistance of conductor depends uppon following
factors
(1) Upon the nature of material of conductor
[n = number of free electrons per unit volume]
(2) Length of conductor
(3) Area of cross section of conductor.
(4) Temperature of the conductor. On increasing
temperature, resistance of conductor increases.
Conductance : It is defined as reciprocal of
resistance.
C
1
R
C
ne2  A
m l
Conductance depends upon following factors :
(1) Upon the nature of material of conductor
[n = number of free electrons per unit volume]
(2) Length of conductor
(3) Area of cross section of conductor.
(4) Temperature of the conductor. On increasing
temperature, conductance of conductor decreases.
Ohm's Law
If the physical state (length, area, material,
temperature) of a conductor remains unchanged then
current flowing in the conductor is directly
proportional to the potential difference applied
across it.
iV
V  iR
V  iR
V = P.D. across conductor
i = Current flowing in the conductor
R = Resistance of the conductor
DERIVATION OF OHM'S LAW
Curve between current and P.D. across a
conductor :
The curve is a straight line passing through origin.
The slope of V versus i curve is defined as resistance
of the conductor.
Slope
dv
di
R
tan
  Angle made by straight line with positive x-
axis
Slope of current vs voltage curve 
1
R
Current Electricity # 4
Unit of resistance :
Volt
  (ohm)
Amp
Dimensions : [ R ]  [ ML2T 3 A2 ]
Derivation of Resistance using ohm's law :
Resistor : Any conductor having resistance is
known as resistor. It is analogous as a capacitor in
electrostatics.
Potential Difference across a resistor : P.D.
acorss a resistor is defined as the product of it's
resistance and current flowing in it.
COMBINATIONS OF RESISTORS
(1) Series Combination : Consider three conductors of resistance R1, R2 and R3 connected end to end
in series then their equivellent resistance is given by :
Reff  R1  R2  R3
In series combination equiivellent resistance
is maximum.
In series combination current in all the
resistances is same.
In series combination P.D. across each
resistor is directly proportional to it's
resistance.
V  iR
VA  VB  iR
VB  VA  iR
TEMPERATURE DEPENDENCE OF
RESISTANCE AND RESISTIVITY
Resistance and resistivity of a conductor increase
linearly with temperature according to
R  R0 (1  t )
0
(1
(2) Parallel Combination : For parallel
combination all the resistors must be joined across
the same two points.
consider three conductors of resistances R1, R2
and R3 are connected across the same points in parallel
then their equivalent resistance is given.
1
1
1
1



Reff
R1 R2 R3
t)
R0 = Resitance of conductor at zero degree celcius.
R = Resistance of conductor at any temp toC.
 = Temperature cofficient of resistance
t = Any general temp (in oC only)
Resistivity of conductor at 0o C
0
Resistivity of conductor at any temp toC.
Unit of  : /oC
In parallel combination equivalent resistance
is minimum.
In parallel combination P.D. across all the
resistances is same because all of them are
connected across the same two points.
In parallel combination current in each
resistor is inversily proportional to it's
resistance.
Self Practice Problems
The resistance of wire is 5  at 50oC & 6  at
100oC. The resistance of the wire at 0oC will be :
(a) 2 
(b) 1 
(c) 4 
(d) 3 
Ans. (c)
1.
KIRCHOFF'S LAWS
(1) Kirchoff's current law (K.C.L) : It is based
upon the principle of conservation of charge.
Total current entering the junction is equal to total
current leaving the junction.
(2) Kirchoff's Voltage Law (K.V.L): It is based
upon the principle of conservation of energy. The
sum of potential difference across all the elements
along a closed loop in a circuit network is always
zero.
Self Practice Problems
1.
Six equal resistances are
connected between points P,
Q and R as shown in the
figure. Then, the net
resistance will be maximum
between :
(a) P and Q
(b) Q and R
(c) P and R
(d) Any two points
Ans. (a)
Current Electricity # 5
METHOD TO DETERMINE
CURRENT AND P.D. ACROSS
THE RESISTORS IN A CIRCUIT
(1) Using equivalent resistance : Here we
determine the equivalent resistance across the points
where the battery is connected and we find out current
supplied by the battery to this single resistor and this
current supplied by the battery to the single resistor is
same as current supplied by the same battery to the
combination.

i
Reff
3.
4.
In the circuit element given here, if the potential
at points B, VB = 0, then the potentials of A and
D are given as :
(a) V A
1.5V , VD
2V
(b) V A
1.5V , VD
2V
(c) V A
1.5V , V D
0.5V
(d) V A 1.5V , VD – 0.5V
Calculate current in 6 V battery :
(a)
1
A
4
(b)
1
A
8
(c)
1
A
2
  E.M.F. of battery
Reff = Equivalent resistance across the points
where the battery is joined.
P.D. across a resistor is defined as the product of
current and resistance of the resistor.
(2) Using K.V.L. : Here we supply some current
from the battery to the circuit and distribute this current
in all the branches of the circuit. Using K.V.L. in
different loops we can determine current in all the
resistors.
P.D. across a resistor is defined as the product of
current and resistance of the resistor.
Self Practice Problems
1.
2.
In the circuit given here, the
points A, B and C are 70 V,
0, 10 V respectively. Then :
(a) The point D will be at a
potential of 60 V
(b) The point D will be at a potential of 20 V
(c) Current in the paths AD, DB and DC are in
the ratio of 9 : 2 : 3
(d) Current in the paths AD, DB and DC are in
the ratio of 3 : 2 : 1
The figure shows part of certain circuit, find :
Potential difference VC V B .
(a) 6 V
(b) 10 V
(c) 4 V
(d) 8 V
(d) None
5.
The value of R in the circuit
should be :
(a) 1 
(b) 1.5 
(c) 2 
(d) 2.5 
6.
In the circuit shown in figure potential difference
between point A and B is 16 V. Find the current
passing through 2 resistance :
(a) 3.5 A
(b) 0 A
(c) 2.5 A
(d) 4 A
Ans. 1. (d) 2. (a) 3. (d) 4. (c) 5. (d) 6. (a)
WHEATSTONE'S BRIDGE
It is a method to determine the equivelent
resistance. The general circuit diagram of wheatstone's
bridge is :
Wheatstone's bridge is balanced when
R1 R3

R2 R4
Current Electricity # 6
When the Wheatstone's bridge is ballanced then
potentials at points C and D are same and hence, P.D.
across resistor R5 become zero and hence, no current
flows in the resistor and hence it may be simply
removed from the circuit. The reduced circuit network
of a balllanced wheatstone's bridge is
EQUIVALENT RESISTANCE IN
SYMETRICAL CIRCUITS
To determine the equivalent resistance in
symetrical circuits we connect a battery across the
points where the equivalent resistance is to be
calculated and supply current from the battery and
distribute this current in all the branches of the circuit.
The battery supplies same current to the resistances
which are equal and which are symmetrically
connected with respect to the battery.
Self Practice Problems
The equivalent resistance across A and B is :
1
1
1


R AB R1  R2 R3  R4
The above method is applicable only when
the bridge is ballanced. i. e.
R1 R3

R2 R4
.
CONDITION OF BALANCED
WHEATSTONE'S BRIDGE
The circuit of wheatstone's bridge consists of a
battery connected across A and B and a galvanometer
is connected across C and D along with the resistor R5.
The bridge is said to be balanced, when there is no
deflection in galvanometer i.e. there is no charge or
current in the branch of galvanometer. Hence the
resistor R5 can simply be removed from the circuit.
EQUIVALENT RESISTANCE IN
UNBALLANCED WHEAT
STONE'S BRIDGE
If the bridge is unbalanced then to determine the
equivalent resistance we connect a battery across the
points where the equivalent resistance is to be
calculated and supply current from the battery and
distribute this current in all the branches of the circuit.
The battery supplies same current to the resistances
which are equal and which are symmetrically
connected with respect to the battery. Using K.V.L.
we find out current supplied by the battery to the
combination. Now we replace the entire circuit
network by a single resistor and find out current
supplied by same battery to this resistor. By equating
these two currents we can determine equivalent
resistance of the circuit.
1.
In the circuit shown P R, the reading of
galvanometer is same with switch S open or
closed, then :
(a) I R I G
(b) I P
IG
(c) I Q
IG
(d) I Q
Ans. (a)
IR
INTERNAL
RESISTANCE OF A BATTERY
The resistance offered by the electrolyte of a cell
to the flow of the current is called as internal resistance
of the battery. When a battery is connected to any
element then the direction of current is from positive
to negative terminal outside the battery and from
negative to positive terminal inside the battery. When
the current flows inside the battery then the ions of
the electrolyte collide with the other ions and atoms
of the electrolyte i.e. the electrolyte offeres a resistance
to the flow of current, this resistance is called as
internal resistance of the battery.
Charge carriers in metals or conductors are free
electrons and that in an electrolyte are positive and
negative ions.
r = internal resistance of battery.
Internal resistance of a battery depends upon the
following factors :
(1) Upon the concentration of the electrolyle
( r  concentration)
(2) Distance between the electrodes (r  distance )
(3) Upon the dipped area of the electrodes
FG r  1 IJ
H area K
Current Electricity # 7
F
IJ
1
(4) Upon temperature G r 
H temperature K
i.e. on
increasing temperature internal resistance of the
battery decreases.
TYPE OF BATTERIES
There are two types of batteries :
(1) Ideal battery
(2) Real battery (Non ideal)
(1) Ideal battery :
The battery having no internal resistance (zero)
is known as an ideal battery.
Terminal potential difference (T.P.D.) across an
ideal battery is same as its E.M.F.
(2) Real battery (Non Ideal Battery) :
The battery having some internal resistance is
known as a real battery. The internal resistance of the
battery is always connected in series with the battery.
Terminal potential difference (T.P.D.) across a
real battery may be less than, greater than or equal to
its E.M.F.
TERMINAL POTENTIAL DIFFERENCE
(T.P.D.) ACROSS A REAL BATTERY
Case I : If the battery is in an open circuit then
T.P.D. across the battery is some as its e.m.f. V  
Case II : If the battery is in discharging mode
then T.P.D. across the battery is less than its e.m.f.
V    ir
Case III : If the battery is in charging mode then
T.P.D. across the battery is greater than its e.m.f.
V    ir
Case IV : If the battery is short circuited then
T.P.D. across the battery is zero. V  Zero
POWER SUPPLIED
BY THE BATTERY
The rate at which the work is done by the battery
is called as power supplied by the battery. It is defined
as the product of current supplied by the battery and
e.m.f. of the battery.
P  i
 = E.M.F. of the battery
i = current supplied by the battery
Case I : It the battery is in discharging mode then
work done by the battery is positive and hence power
supplied by the battery is also positive. In this case
the energy is suppled by the battery i.e. chemical
energy of the battery is converting into electrical
energy.
P   i
Case II : If the battery is in charging mode then
work done by battery is negative and hence power
supplied by the battery is also negative. In this case
power is being consumed by the battery. In this case
the energy is not supplied by the battery but it is
consumed by the battery i.e. electrical energy is
converting into chemical energy stored in the battery.
P  i
HEATING EFFECTS OF CURRENT
(THERMAL EFFECT OF CURRENT)
ENERGY DISSIPATED IN A RESISTER
It is a phenomenon in which electrical energy is
converted into heat energy.
When an electric current is passed through a
conductor, it gets heated up, this effect of current is
called as heating effects of current.
When a current flows through an external
resistance, then due to collision of electron with the
other electrons and with other atoms of the conductor,
energy of the system is lost in the form of heat energy
known as jule heating.
When a charge flows through a conductor it's
energy decreases and this energy appears in the form
of heat.
If the current i flows through resistor R in a time
interval of t then the amount of heat lost in the
resistance is given by :
H  i 2 Rt
i = Current,
V = P.D. across resistor
R = Resistance
t = time interval
H
V2
t
R
H  Vit
Unit : Joule or K.W.H. (1 K.W.H. = 3.6 × 106 J)
Current Electricity # 8
The above results are applicable only when
current flowing in the conductor is constant.
i.e. if current is variable then heat produced
z
in the resistor is given by H  i 2 Rdt .
A capacitor stores energy in it's electric field
but a resistor only dissipates energy in the
form of heat.
POWER DESSIPATED
IN THE RESISTOR
Power lost in a resistor is defined as the rate at
which heat is desipated in the resistor.
H 2
V2
i R
 Vi
t
R
P
P
z i Rdt
2
0
t
MAXIMUM POWER DESSIPATED IN
AN EXTERNAL RESISTANCE
Consider an external resistance R connected to a
battery of E.M.F.  and internal resistance r. Then power
lost in the external resistance will be maximum,when
external resistance is equal to internal resistance of the
battery.
The value of maximum power lost is given by
Pmax 
 eff  1   2
reff  r1  r2
The above results are true in general for any
number of battries.
If polarity of one battery is reversed then
equivalent E.M.F. and equivalent internal
resistance is given by
  1  2
The above result is applicable only when
current in the circuit is constant.
If current is a function of time then power
dessipated in the resistor is given by .
t
COMBINATIONS OF REAL BATTERIES
Case I : Series Combination : Consider two
batteries of e.m.f. 1 and 2 and internal resistance r1
and r2 respectively connected in series. Then, their
equivalent e.m.f. and internal resistance are given by :
r  r1  r2
If polarity of each battery is same then the
polarity of their equivalent battery will also
be the same.
The polarity of the equivalent battery will
be governed by the battery of maximum
E.M.F.
Case II : Parallel Combination : Consider two
batteries of E.M.F. 1,  2 and internal resistances r1
and r2 connected across the same two points in
parrallel then their equivalent E.M.F. and equivalent
internal resistance are given by :
1  2

r
r2
 1
1 1

r1 r2
2
4r
1
 = E. M. F. of battery
r = Internal resistance of the battery
Self Practice Problems
1.
The charge supplied by
source varies with time t
as Q at bt 2 . The total
heat produced in resistor
2R is (Assume direction of current is not
changing)
(a) a R / 6b
(c) a 3 R / 3b
Ans. (b)
3
3
(b) a R / 27b
(d) None of these
reff

1 1

r1 r2
The above results are true in general for any
number of batteries.
If polarity of one battery is reversed then
equivalent E.M.F. and equivalent internal
resistance are given by
1  2

r1 r2

1 1

r1 r2
1
reff

1 1

r1 r2
Current Electricity # 9
If porarity of each battery is same then the
polarity of their equivalent battery will also
be the same.
The polarity of equivalent battery will be

source then the power desipated in the combination is
given by
1 1 1
 
P P1 P2

1
2
governed by  1 if r  r and by  2 if
1
2
In series combination the power desipated
is minimum.
In case of series combination the bulb having
larger resistance will glow more brightly.
 2 1

r2 r1
BULB AND IT'S COMBINATIONS
Bulb is an electrical device which consists of a
tungustun fillament which glows when some current
flows through it.
Vs  Specified voltage of the bulb i.e. the maximum voltage at which bulb can be operated safely.
PS  Specified power of the bulb i. e. maximum
power which can be desipated in the bulb.
PS 
R
(2) Parallel combination : Consider two bulbs
which when separately connected across the same
voltage source, desipate powers P1 and P2. If both the
bulbs are connected in parallel acorss the same voltage
source then the power desipated in the combination is
given by
P  P1  P2
VS2
R
In parallel combination the power desipated
is maximum.
In parallel combination the bulb having
smaller resistance will glow more brightly.
VS2
PS
R = Resistance of the bulb
Brightness of a bulb : It is defined as the power
desipated in a bulb. The more power desipated in the
bulb, the more will be it's brightness.
If a bulb is switched on for a long time then
due to heating temperature of it's filament
continously increases due to which it's
resistance also increases and hence, the
power desipated in the bulb decreases so that
brightness of the bulb continously decreases.
If the applied potential difference across a
bulb is greater than its specified voltage then
the amount of heat produced in the filament
is so large that temperature of the filament
exceeds its melting point and the bulb gets
fused. The condition for any bulb to get fused
(V  VS )
COMBINATIONS OF BULBS
(1) Series combination : Consider two bulbs
which when separately connected across the same
voltage source, desipate powers P1 and P2. If both the
bulbs are connected in series acorss the same voltage
Self Practice Problems
1.
2.
3.
Three identical bulbs are connected in series as
shown. When switch is closed :
(a) Both bulbs A and B become dimmer
(b) Both bulbs A and B become brighter
(c) Brightness of bulbs remains unchanged
(d) A becomes brighter and B becomes dimmer
If two bulbs of wattage 60 W and 100 W
respectively each rated at 110 V are connected
in series with the supply of 220 V, which bulb
will fuse ?
(a) 60 W bulb
(b) 100 W bulb
(c) Both bulbs
(d) Bulbs will not fuse
How many 60 W bulbs may be safely run on
220 V using a 5 A fuse ?
(a) 18
(b) 16
(c) 14
(d) 12
Current Electricity # 10
4.
The bulbs B1, B2 and B3
are connected to the
mains as shown in
figure. If B 3 is
disconnected from the
circuit by opening
switch S, then incandescence of bulb B1 will :
(a) Increase
(b) Decrease
(c) Become zero
(d) No change
Ans. 1. (b) 2. (a) 3. (a) 4. (b)
ELECTRICAL
MEASURING INSTRUMENTS
Galvanometer : It is an electrical device which
is sensitive to current. The current flowing in the
galvanometer is directely proportional to the angle of
deflection of the galvanometer i   .
It is used to measure current and potential
difference across any branch is any circuit.
Ammeter (Galvanometer as an ameter) or
Conversion of Galvanometer into an ammeter :
An ammeter is a device which measurs current in
any branch in any circuit. It is connected in series
across the branch where current is to be measured
because in series current is same.Error in the
measurement of current  Rg
Due to the introduction of the galvanometer there
is an error in the measurement of the current. Since
the error in measurment of current is directly
proportional to resistance of galvanometer. Hence, to
minimize the error in the measurerment of current or
to maximise the accuracy in the measurment of current,
a shunt of shunt resistance RS is connected in parallel
to the galvanometer because in parallel equivalent
resistance is minimum. Shunt resistance must be as
d
i
small as posible Rs  Rg so that accuracy in the
measurment of the current is maximum.
Hence, an ammeter consists of a galvanometer
having a shunt resistance connected in parallel with it.
FG
H
i  ig 1 
IJ
R K
Rg
S
i  Main current or range of the ammeter or
reading of ammeter
ig  Current in the branch of galvanometer or full
scale deflection current
Rg  Resistance of galvanometer
Rs  Shunt resistance.
The resistance of ammeter must be as small
as posible. The resistance of an ideal
ammeter must be zero.
If not specified then we assume the ammeter
to be ideal ( R  0) .
Voltmeter (Galvanometer as a voltmeter ) or
Conversion of Galvanometer into voltmeter :
Voltmeter is a device which measures potential
difference across any two points in a circuit. Voltmeter
is always connected in parallel across the points whose
potential difference has to be measured, because in
parallel potential difference is same.
Due to the introduction of galvanometer there is
an error in the measurment of potential difference.
The error in the measurment P.D. is inversely
proportional to the resistance of galvanometer.
Hence, to minimize error in the measurement of
potential difference or maximum the accuracy
resistance of the galvanometer must be very large.
Hence a shunt of shunt resistance RS ( Rs  Rg ) is
connected in series with the galvanometer, because
in series equivelent resistance is maximum.
Hence, a voltmeter consists of a galvanometer
having a shunt resistance connected in series with
it.
F
GH
V  Vg 1 
RS
Rg
I
JK
Vg  ig Rg
V = Potential difference across the voltmeter or
Reading of voltmeter or Range of voltmeter.
Vg  potential difference acoss galvanometer
ig  current in the branch of galvanometer or full
scale deflection current
Rs  Shunt resistance
Rg  Galvanometer resistance
The resistance of voltmeter must be as large
as posible. The resistance of an ideal
voltmeter must be infinite.
If not specified then we assume the voltmeter
to be ideal ( R   ) .
Current Electricity # 11
Comparison between ammeter and voltmeter
Ammeter
1. It is a device used to
measure current.
2. Ammeter is joined in
series with the branch
where the current is to
be measured
3. To convert galvanometer into ammeter,
shunt resistance is
connected in parallel
with it.
4. In ameter shunt
resistance must be
very much less than
that of galvanometer.
5. The resistance of an
ammeter is very small.
6. The resistance of an
ideal ammeter is zero.
Voltmeter
It is a device used to
measure potential
difference.
Voltmeter is joined in
parallel across the
branch whose potential
difference is to be
measured.
To convert galvanometer into voltmeter
shunt resistance is
connected in series
with it.
In voltmeter shunt
resistance must be very
much greater than that
of galvanometer.
The resistance of a
voltmeter is very large.
The resistance of an
ideal voltmeter is
infinite.
Self Practice Problems
1.
In the circuit shown, R1 is
increased. What happens to
the reading of the voltmeter
(ideal) ?
(a) Increases
(b) Decreases
(c) First increases then decreases
(d) Does not change
Ans. (d)
POTENTIOMETER
It is an electrical device which is used to measure
potential difference across any two points in any circuit
network.
It consists of a uniform wire known as potentiometer wire which is maintained at some constant
potential difference with the help of an external battery.
The basic principle of the potentiometer is based
upon the fact that the potential differnce across any two
points in a current carrying wire is diretly proportional
to the distance between the same two points.
To measure potential difference across any
element, one end of the element is directly connected
to point A of the potentiometer wire and the other
end is connected by galvanometer and is touched
through the jockey at different points on the
potentiometer wire. The point corresponding to zero
deflection in the galvanometer is known as null point
and distance of the null point from point A is known
as ballancing length corresponding to the given
element.
C = Null point
lAC = Ballancing length
At null point there is no interaction of current
between internal and external circuit of potentiometer.
Potential difference across any two points over
the potentiometer wire is directely proportional to the
distance between the same two points.
VPQ  VAB
l AC
l AB
VPQ = P.D. across the given element,
VAB = P.D. across the potentiometer wire (E.M.F.
of battery connected across A and B)
lAB = Length of potentiometer wire
lAC = Balancing length coresponding to the given
element
By putting the value of VAB ,lAB ,lAC we can
determine P.D. across the given element PQ.
Hence, Potential difference across any element is
directly proportional to it's corresponding balancing
length.
To measure potential difference
potentiometer arrangement is preffered
over a voltmeter because in potentiometer
we can accurately measure balancing length
while for an ideal voltmeter its resistance
must be infinite and practically it is not
posible to design a voltmeter having infinite
resistance.
Potentiometer works only when the
potential difference across the given
element is less then the P.D. across the
potentiometer wire i.e. If P.D. across the
element is greater then that across the
potentiometer wire then there will be no
null point.
Current Electricity # 12
If the jockey is pressed at any point other
than null point, then there will be interaction
of current between internal and external
circuit of potentiometer
APPLICATIONS
OF POTENTIOMETER
(1) To compare E.M.F. of two batteries :
Consider two ideal bateries of E.M.F. 1 and 2
which are connected separately acorss a potentiometer,
if their corresponding ballancing lengths are l1 and l2
then,
1 l1

 2 l2
In this case if the ideal battery is replaced
by real battery then the balancing length
corresponding to both the batteries will
remain unchanged because both the
batteries are in an open circuit so that there
is no current in the internal resistance of
the battery and T.P.D. of a battery in an
open circuit is same as E.M.F. and hence,
the ratio of their E.M.F. remains the same.
1 l1

 2 l2
(2) Determination of internal resistance of a
battery with the help of external resistance :
Consider a battery of E.M.F.  and internal
resistance r connected to the potentiometer
arrangement so that the balancing length is l1. If now
an external resistance R is connected across the
battery to complete the circuit and in this case the
balancing length is l2 then.
r
R ( l1  l2 )
l2
r = Internal resistance of battery
R = External resistance
l1 and l2 = Balancing lengths in two cases.
SENSITIVITY OF POTENTIOMETER
It indicates the smallest potential difference
which can be measured with it. It depends upon the
potential gradient (rate of change of potential of the
potentiometer wire with respect to distance) i.e.
smaller the potential gradient, the more will be the
sensitivity of potentiometer.
Potentiomter is said to be more sensitive
when it shows a significant balancing length
for a small potential difference.
Sensitivity can be increased by increasing
length of the potentiometer wire.
Self Practice Problems
1.
A potentiometer having a wire 10 m long
stretched on it is connected to a battery having a
steady voltage. A leclanche cell gives a null point
at 750 cm. If the length of potentiometer wire is
increased by 100 cm, find the new position of
null point :
(a) 750 cm
(b) 675 cm
(c) 825 cm
(d) 900 cm
Ans. (c)
METER BRIDGE
It is an electrical device which is used to measure
an unknown resistance with the help of a known
resistance.
It consists of a uniform wire which is maintained
at a constant potential difference with the help of an
external battery and the length of the wire is 1 m or
100 cm. Two resistances known and unknown are
connected across the two arms of meter bridge.
The basic principle of meter bridge is based upon
the phenomenon of a balanced wheatstone 'sbrigde.
When there is no deflection in the branch of
galvanometer then the current supplied by the battery
will be distributed in the meter bridge wire and the
external circuit so that there is a perfect balanced wheat
stone's bridge across A and B.
Acoording to the condition of balanced wheat
stone's bridge.
SR
b100  l g
l
R = Known resistance
S = Unknown resistance
l = Balancing length in cm.
In the meter bridge if the two resistances
known and unknown are interchanged then
the above the expression changes, never the
less the condition of balanced wheat stone's
bridge remains unchanged.
Current Electricity # 13
In meter bridge if battery and galvanometer
are interchanged then also position of null
point remains the same, because the
condition of balanced wheat stone's bridge
remains the same.
Self Practice Problems
1.
i
In the figure shown for gives values of R1 and
R2 the balance point
for Jockey is at 40 cm
from A. When R2 is
shunted
by
a
resistance of 10 ,
balanced shifts to 50
cm, R1 and R2 are (AB = 1m)
(a)
10
,5
3
(b) 5 ,
(c) 20 , 15
Ans. (a)
15
2
(d) 20 , 30
RC– CIRCUITS
CHARGING OF A CAPACITOR WITH
THE HELP OF AN EXTERNAL RESISTANCE
If a capacitor is directely connected to a battery
using conducting wires then the capacitor
spontaneously because there is no resistance offered
by the capacitor to the flow of the current.
Consider a capacitor of capacitance ‘C’ is
connected by a battery of E.M.F.  and an external
resister R, then due to the resistance offered by the
resistor charge stored on the capacitor increases
slowly according to the relation
q  c(1  e
 t / RC
Unit and Dimensions of time constant are same
as that of time.
Unit : second.
Dimensions : [ ] [T ]
Current in the circuit as a function of time
)
At any general time t = t if q be the charge on
the capacitor and i be the current in the circuits then
using KVL.
Initially At t = 0, q = 0
In the steady state (after a very long time)
t   qmax  c.
Hence, Theortically it takes an infinite amount
of time for the capacitor to get fully charged.
 t / RC
e
R
Initially at t  0 imax =
In the steady state (after a very long time)
when t   i  Zero
Initially charge on the capacitor is zero while
the rate of flow of charge is maximum and hence
current in the circuit is maximum while in the steady
state charge on the capacitor is maximum but rate
of flow of charge is zero and hence current in the
circuit is the zero.
Potential difference across resistor as a function of
time
V  e t / RC
Potential difference across the capacitor as a
function of time
VC 
q
C
e
V   1  et / RC
j
In the whole process (t = 0 to t =  )total work
done by the battery is given by
W  q
W  c 2
In the whole process, the energy stored in the
capacitor :
U
1 2
c
2
In the whole process of time total heat developed
in the resistor :

z
H  i 2 Rdt
Time constant of the circuit   RC
After one time constant 63.2 % of the maximum
charge would be stored on the capacitor.

R
0
H
1 2
C
2
Current Electricity # 14
Hence, 50% of work done by the battery in
charging the capacitor is stored in the form of
electrostatic energy of the capacitor while the
remaing 50% is lost as heat.
Initially at t = 0 the capacitor is uncharged
and does not offer any resistance to the
flow of current due to which current in
the circuit is maximum.
In the seady state when the
capacitor gets fully charged then it offers
an infinite resistance to the flow current and
doesn't allow any further current in it's
branch.
DISCHARGING OF A
CAPACITOR WITH THE HELP
OF AN EXTERNAL RESISTANCE
If a charged capacitor is directly connected to an
ideal conducting wire then the capacitor spontaneously
gets discharged.
Due to resistance offered by the resistor, the current
in the circuit decreases slowly and exponentially with
time.
Consider a capacitor of capacitance C having
initial charge q 0 connected with an external
resistance R then capacitor slowly starts discharging
and the charge remaining on the capacitor as a
function of time is given by :
P.D. across capacitor and resistor at any time t
are same because there connected in parallel.
Hence, Theortically it takes an infinite amount of
time for the capacitor to get fully discharged.
Initially At t = 0, Q = q0
In the steady state (after a very long time)
t   ,Q = 0
Hence, Theortically it takes an infinite amount of
time for the capacitor to get fully discharged.

Initially at t  0 imax =
R
In the steady state (after a very long time)
when t   i  Zero
Hence, in the whole process the total electrostatic
energy stored in the capacitor is lost in the from of
heat energy through the resistor. Amount of
electrostatic energy stored in the capacitor is equal to
amount of joule heating through the resistance.
H
Self Practice Problems
1.
Q  q0et / RC
If at any time t = t, q be the charge that has flows
from the capacitor then by using KVL.
q  q0 (1  et / RC )
q0 = Initial charge on the capacitor
q = Charge flown from the capacitor after time t
Q = Charge remaining on the capacitor after time t
Current is defined as rate of change of that charge
which flows in the circuit.
Current in the circuit as a function of time :
i
q0  t / RC
(e
)
RC
P.D. across capacitor at any time t V 
q0  t / RC
e
C
P.D. across resistor at any time t V  q0 e t / RC
C
q02
2C
2.
What is the charge stored
on each capacitor C1 and
C2 in the circuit shown in
figure.
(a) 6 C, 6 C
(b) 6 C, 3 C
(c) 3 C, 6 C
(d) 3 C, 3 C
When the switch is closed, then the initial current
through 1 resistor in figure is :
(a) 12 A
(b) 4 A
(c)
10
A
7
(d) 3 A
Ans. 1. (a) 2. (b)
CHEMICAL EFFECTS OF CURRENT
Electrolysis : It is a process in which electrical
energy is converted into chemical energy.
The process of decomposition of a compound by
the application of electric field is known as electrolysis.
The arrangement in which the process is carried
out is called as electrolytic cell or voltameter.
Current Electricity # 15
After the decomposition, both the charges positive
and negative move towards their respective electrodes
due to which a resultant current flows in the circuit.
Resultant current in the circuit is I  I   I 
I  current due to motion of + ve charge.
I   current due to motion of – ve charge.

Faraday's Laws of Electrolysis :
First Law : The amount of a substance deposited
or liberated at any electrode is directly proportional
to the amount of charge passed through the electrolyte.
m  z×q
z = Electrochemical equivallent of the subsance
(.c..) [It is the property of nature of mateiral of the
substance which is deposited or librated on any
electrode.
m = Mass of substance
only when q
are in series.
constant. i.e. when the voltameters
m1, m2 = Mass of substances deposited or librated
at the electrodes
z 1, z 2 = Electrochemical equivalents of the
substances.
THERMOELECTRICITY
Electrical energy can be converted into heat energy
but heat energy can also be converted into electrical
energy in thermoelectricity.
It is a process in which heat energy is converted
into electrical energy producing an electric current.
Seebeck effect : When two different metals are
joined and their junctions are kept different
temperatures then the current starts flowing in the
circuit. This phenomenon is known as seeback effect
and the arrangement of two metals is known a
thermocouple.
q = Charge passed through the electrolyte
z
Equivallent weight
Faraday
1 F = 96500 C
Unit : Z = gm/c or kg/c.
Dimensions : [MA–1T–1]
If a constant current i flows through the electrolyte
for time t then amount of charge passed through the
electrolyte is :
q = it

m  zit
The above formula is applicable only when
current is constant. If current in the circuit
z
is varriable then q  i dt .
Second Law : If the same amount of charge is
passed through two different electrolyles then the ratio
of masses deposited or librated at the respective
electetrodes will be in the ratio of their respective
electrochemical equivalents.
m1 Z1

m2 Z2
Every metal has large number of free electrons
but the free electron density (nuber of free electrons
per unit volume) is different in different metals. When
two different metals are joinded then due to difference
in their free electron density the electrons move from
the metal having higher electron density to metal
having low electron density due to which a current
flows in the circuit.
At the hot junction the kinetic energy acquired
by the electrons is greater then the K.E. acquired
by the electrons near the cold junctin and hence
the number of electrons diffusing at the hot
junction is greater than number of electrons
diffusing at the cold junction. Due to which the
potential difference near the hot junction is greater
than P.D. near cold junction and hence a resultant
EMF is developed in the loop and a current starts
flowing in the loop.
Current Electricity # 16
In a thermocouple if both the junctions
are maintained at the same temperature,
then no emf is produced in the loop and
h en ce no cu rren t flows in the loop
because in this case the kinetic energy of
electrons at hot and cold junctions will
be same as a result of which the number
o f electro ns d iffus ing near the hot
ju n ctio n i s eq u al to the number of
electrons diffusing near the cold junction.
Hence EMF developed near hot and cold
junctions will be same and the resultant
EMF in the loop becomes zero and there
will be no current the circuit.
THERMO ELECTRIC EMF
The EMF produced in a thermocouple is known
as thermo electric emf and is given by :
N  Positive
N
 is +ve and  is – ve]
Maximum thermoelectric emf is given by
 max  
2
2
max  Positive
d
 rate of change of thermoelectric emf with
d
respect to temperature and is also called as
thermoelectric power.
Inversion Temperature : The temperature of hot
junction at which the current in the circuit starts
reversing its direction is known as inversion
temperature or the temperature at which current and
emf becomes zero is called as inversion temperature.
2
H
1
2
H
2
 and  are constant for the given metal pair.
= temperature difference between hot and cold
junction.
2
i
2
0
i  Positive
 is + ve and  is – ve]
   H  C
Neutral temperature and Inversion
temperature are the property of metals used
in the thermocouple.
0
0
 H  temperature of hot junction
c  temperature of cold junction.
   (  H   C )   (  H  C ) 2
If temperature of cold junction is 0o i.e. cold
junction is assumed as a reference (C  0o )
2
H
2
H
Neutral Temperature : The temperature of hot
junction at which the thermo electric emf is maximum
is known as natural temperature.
d
   2H  0
d H
RETATION BETWEEN NEUTRAL
AND INVERSION TEMPERATURE
Case I : If temperature of cold junction (C  0o )
i  2 N
Case II : If temperature of cold junction is nonzero.
 N  C  i   N
N 
i  c
2
i  inversion temperature
 N  neutral temperature
c  temperature of cold junction.
Current Electricity # 17
CARBON CODE FOR RESISTORS
There are two types of resistors :
(1) Wire wound Resistor : They are made by winding the wires of an alloy, like nichrome, magnin, and
costantan. Their resistivity is less sensitive to temperature used in physics laboratory.
(2) Carbon Resistors : Here, carbon with a suitable binding agent is moulded into a cylinder. Lead wire
is attached to this cylinder and it is kept inside a ceramic jacket. Carbon resistors are compact and inexpensive.
Their values are given using a colour code. These resistors are used in electronic circuits, for Ex. Radio,
amplifires. The value of resistance is indicated by four, coloured bands, marked on the surface of cylinders :
Colour
Digit
Multiplier
Black
Brown
Red
0
1
2
1
10
102
Orange
Yellow
Green
Blue
Violet
Gray
3
4
5
6
7
8
103
104
105
106
107
108
White
Gold
Silver
9
109
10–1
10–2
Tolerance
5%
10 %
B B Roy great britain very good wife wearing gold silver Necklace.
Colour 1 – First significant figure
Colour 2 – Second significant figure
Colour 3 – Decimal multiplier
Colour 4 – Tolerance or possible variation in percentage. If tolerance band is missing from the code then
tolerance is assumed to be 20 %.
Self Practice Problems
1.
A carbon resistor has coloured strips as shown in figure. What is its resistance ?
(a) 410
Ans. (b)
2%
(b) 470
5%
(c) 420
3%
(d) 405
2%
Current Electricity # 18
MISCELLANEOUS SOLVED EXAMPLES
Example 1
How many electrons pass through a lamp in 1
min, if the current is 300 mA. Given, the charge
on an electron is 1.6 × 10–19C.
Sol. Given current, I = 300 mA = 300 × 10–3 A
Charge on an electron, e 1.6 10 19 C
Time, t = 1 min = 60 sec
Charge passing through a lamp in 1 min,
q = I ×t
= 300 × 10–3 × 60 s
Let n electrons pass through the lamp in 1 min.
Then,
q
ne
n
q
e
300 10 3 60
1.6 10 19
1.125 10 20 C
Example 2
The plot represents the flow of current through a
wire at three different times. The ratio of charges
flowing through the wire at different times is :
(a) 2 : 1 : 2
(c) 1 : 1 : 1
(e) 2 : 3 : 3
(b) 1 : 3 : 3
(d) 2 : 3 : 4
Sol. Charge for a given time = area under the currenttime graph for the given time
q1
2 1 2C
q2
1 2C
q3
1
2 2
2
q1 : q 2 : q3
Example 3
An aluminium wire of diameter 0.24 cm is
connected in series to a copper wire of diameter
0.16 cm. The wires carry an electric current of
10 A. Determine the current density in aluminium
wire.
Sol. Given, diameter = 0.24 cm
radius r
0.24 10 2 m
2
Current, I 10 A
I
A
Since, current density, J
I
r2
J
10 2 2
3.14 (0.24 10 2 ) 2
J
2.2 10 6 Am
2
Example 4
A resistor of 5  is connected in series with a
parallel combination of a number of resistors,
each of 5 . If the total resistance of the
combination is 6 , how many resistors are in
parallel ?
Sol. Let n resistors each of 5  be connected in
parallel, then their effective resistance is given
by,
1
RP
1
5
RP
5
n
1
5
1
5
......n times
n
5
As, the parallel combination of resistors is
connected in series with 5 resistor, then total
resistance of the combination is given by,
R
RP
5
5
5
n
5
5 6
n
2C
2 : 2 : 2 1:1:1
5
1
n
n 5
Current Electricity # 19
Example 5
The figure shows current in a part of electric
circuit. The current I is :
Sol. For the first balanced bridge situation,
R
S
33.7
100 33.7
33.7
66.3
.... (i)
When 12  resistance is connected in parallel
with S, the equivalent resistance is S eq
12 S
12 S
For the second balanced bridge situation,
(a) 1.7 A
(b) 1.3 A
(b) 3.7 A
(d) 1 A
Sol. According to Kirchhof's first rule.
S eq
51.9
48.1
51.9
48.1
...(ii)
Putting the value of
R
from Eq. (i) in Eq. (ii),
S
or
R(12 S )
12 S
51.9
100 51.9
we have
At junction P, I PQ
2 2
At junction Q, I QR
4 1 3A
At junction R, I
I QR 1.3
4A
I
E
R r
R
E
I
r
IR
51.9
48.1
From Eq. (i), we get
R
33.7
,S
66.3
33.7
13.5 6.86
66.3
Example 8
Calculate the value of the resitance R in the
circuit shown in the figure, so that the current in
the circuit is 0.2 A. What would be the potential
difference between points A and B ?
10
3 17
0.5
Sol. For BCD, equivalent resistance
and terminal voltage,
V
33.7
66.3
On solving, we get S 13.5
3 1.3 1.7 A
Example 6
A battery of emf 10 V and internal resistance
3 is connected to a resistor. If the current in
the circuit is 0.5 A, what is the resistance of
resistor ? What is the terminal voltage of the
battery, when the circuit is closed ?
Sol. Given, E 10V , r 3 , I 0.5 A
As
(12 S )
12
0.5 17 8.5V
Example 7
In a meter bridge, the null point is found at a
distance of 33.7 cm from A. If a resistance of
12  is connected in parallel with S, the null
point occurs at 51.9 cm. Determine the value of
R and S.
R1
5
5
10
Across BA, equivalent resistance R2
1
R1
1
10
1
1
30 15
3 1 2
30
R2
6
30
1
5
5
Potential differene,
VBA
V AB
I
R2
0.2 5 1V
1V
Current Electricity # 20
Example 9
A galvanometer of resitance 15  gives full scale
deflection for a current of 2 mA. Calculate the
shunt resistance needed to convert it to an
ammeter of range 0 to 5 A.
Sol. Given,
G 15
Ig
I
r r2
Potential difference across P and S,
VS V P
I r2
Charge on the capacitor,
2mA
3
2 10 A
I
E
Q
C (VS VP ) C I r2
r2
CE r2
r r2
5A
Shunt resistance, S
I gG
I
Ig
2 10 3 15
5 2 10 3
C
E
(r r2 )
For the network shown in the figure the value of
the current i is :
0.006
The resistance S = 0.006  is connected in
parallel with the galvanometer. The small
resistance is connected in parallel, because we
have to decrease the resistance of the galvanometer so that most of the current passes through
it and it gives the exact value of the current.
In the given circuit diagram, when the current
reaches steady state in the circuit, the charge on
the capacitor of capacitance C will be :
(a)
18V
5
(b)
5V
9
(c)
9V
35
(d)
5V
18
P
4
2
P
Q
R
, i.e., the Wheastone bridge is a
S
Sol. As Q
r
2 and
R
S
6
2
3
r
2
(a) CE (r r )
2
1
(b) CE (r r )
1
(c) CE
1
(d) CE (r r )
2
r
Sol. At steady state no current flows through capacitor
C. Thus the current through the network will be
as shown in figure. Then
balanced Wheastone bridge. Hence, the middle
resistance of 4  will be ineffective. The effective
resistance of the circuit is
R
(4 2) (6 3)
(4 2) (6 3)
I
V
R
18
5
5V
18
Current Electricity # 21
Sol. Mass deposited, m = volume × density
[2(6 6) 0.01] 10 gram =7.2 g
The thermo e.m.f. E in volts of a certain thermocouple is found to vary with temperature
difference  in oC between the two junctions
2
according to the relation E 30
15
30
dE
d
30
or
n
m
Zt
7.2
0.001 (60 60)
2A
15
2
15
At neutral temperature,
0 30
I
The circuit shown her is used to compare the
e.m.f. of two cells E1 and E2 (E1 > E2). The
null point is at C when the galvanometer is
connected to E1. When the galvanometer is
connected to E2, the null point will be :
2
E
ZI t
. The
neutral temperature for the thermo-couple will
be :
(a) 450 oC
(b) 400 oC
(c) 225 oC
(d) 30 oC
Sol. As,
m
n
,
dE
d
0
2 n
15
30 15
2
(a) To the left of C
(b) To the right of C
(c) At C itself
(d) Non where on AB
225 o C
A steel plate of size 6 cm × 6 cm is to be coated
by a metal on both side with a coating thickness
of 0.1 mm by electrolysis. If the density and
c of the metal are respectively 10 g cm–3 and
0.001 g C–1, then the strength of the current to
complete the process in one hour is :
(a) 1 A
(b) 0.5 A
(c) 6 A
(d) 2 A
Sol.
E1
E2
l1
l2
As E1 E 2 , therefore l1 l 2 . Therefore the null
point for the cell of e.m.f. E2 must be at shorter
length than that of cell E1. thus the null point
on potentiometer wire should shift towards left
of C.
Current Electricity # 22
MISCELLANEOUS SOLVED EXAMPLES
(JEE -ADVANCED )
Example 1
The ratio of masses and area and densities of
two wires are 1 : 1, 2 : 1 and 4 : 1, respectively.
If their resistivities are in ratio 1 : 4. The find the
ratio of their resistance.
Ans. 1 : 64.
M1
M2
Sol.
1
2
R
1 A1
,
1 A2
2 d1
,
1 d2
dR
dt
R0 (a 2bt )
dR
dt
R0 (a 2bt )
R0 (1 at bt 2 )
4
1
1
4
1
R
a 2bt
1 at bt 2
Example 3
l
A
Find the currents through the resistance in the ckt
shown in fig.
Mass= density × volume
m
d
l
m
dA
R
R
R1
R2
Al
m
dA
A
Sol. Using KLV in difference loops :
1
2
1
2
.... (1)
i R2
(i i1 ) R3
0
.... (2)
2
Equation (1) and (2)
iR1
Example 2
Resistance of a wire at temprature t 0 C is
R  R0 (1  at  bt 2 ) where R0 is temperature at
00C Then find out temperature coefficient of
resistance at temperature t.
R
0
2 1
1
64
Sol.
(i i1 ) R3
2
m1 d 2 A2
m2 d1 A1
1
1
1
4
4
1
iR1
m
dA 2
R0 (1 at bt
2
i1 R2
1
0
iR1
i1
1
2
.... (3)
R2
Putting i1 in equaiton (1)
iR1
2
i1 R2
1
1
i
0
2
i R1
Since temperature coefficient of resistance in
defined as fractional change in resistance per
degree rise in temperature
dR
R dt
R
R t
iR1
i R1
R3
1
iR3
R1 R3
R2
(
2
1
iR1
R2
(
2
1
) R3
(
2
0
iR1
R3
R2
R2
1
R3
1
0
) R3
R2
Current Electricity # 23
(
1
2
1
Example 5
)R3
R2
R1 R3
R3
R2
i
R1
Putting i in equation (3) we can determine the value
of i1
Hence, current flowing in
Two bulbs consume same power when operated at
200 V and 300 V respectively When these bulbs
are connected in series across a D.C. source of
500 V. Then :
(a) Ratio of P.D. across them is 3/2
* (b) Ratio of P.D. across them is 4/9
4
9
2
(d) Ratio of Power consumed across them is
3
Sol. Let resistances of bulbs are R1 and R2
* (c) Ratio of Power consumed across them is
R1
i
R2
i1
R3
i i1
Example 4
A resistance R carries a current I. The rate of heat
loss to the surroundings is ( T T0 ) where  is a
constant. T is the temperature of the resistance and
T0 is the temperature of the atmosphere. If the
cofficient of linear expansion is  the strain in the
resistance is :
(a) Protortional to the length of the resistance
wire.
* (b) equal to
Temp. diff
(T T0 )
t
i2R
T T0
i R
....(1)
Linear expansion :
2
2
 1 (1
t)
2
1 1
t
1

Strain
4
9
V1
V2
iR1
iR2
P1
P2
i 2 R1
i 2 R2
.... (1)
R1
R2
R1
R2
4
9
4
9
2
P
R
Maximum in series
P


2
2
3
Two identical battery each of emf 2V and internal
resistance 1 are available to produce heat in an
external resistance by passing a current through it.
What is the maximum power that can be developed
across the external resistance using these battery.
Sol. Maximum power is dessipated in series
combination
t
1
(300) 2
R2
R1
R2
t
1
Strain
P2
Example 6
T
2
T
P1
R1
R2
 2
I R

i2R
V2
R
(200) 2
R1
1 2
I R
(c) equal to
2
(d) equal to  ( IR )
Sol. Power dessipated :
P
P
Max
T
i 2R
Current Electricity # 24
2 2
Sol. Circuit at t = 0
4 volt
Using KVL
r 1 1 2
2
Pmax
(4) 2
4(2)
4r
i1 R1
16
8
2W
i1
Example 7
In the circuit shown in figure C1 = 2C2. Capacitor
C1 is charged to a potential of V, the current in the
circuit just after the switch S is closed is :
(a) Zero
(b)
2V
R
Sol.
C1
i 2 R2
i2
0
R2
Using KVL
iR1
V
2R
i
2C 2
Just after switch is closed second capacitor (C2)
is uncharged
Using KVL
q
C1
0
R1
Capacitor is fully charged hence current is R2 is
zero
(c) Using KVL is steady state
iR 0 iR
2iR
i
0
0
q
C1
q
C1 (2 R )
Energy
V
i
q1
C1
Example 8
(a) Initial current through each resistor.
(b)Steady state current through each resistance.
(c) Final energy stored in the capacitor.
(d)Time constant of the ckt when switch is closed
and when switch is reopened.
0
q
c
U
q2
2c
c2 2
2C
1
c
2
2
Reff
R2
Teff
R2C
R1 is short circuited
V
2R
For the ckt. shown in figure find :
q
c
(d) when switch is closed
Initial potential difference across C1 is V
Hence,
R1
(b) Circuit in steady state is
(c) Infinite
* (d)
0
Reff
R1
R2
Teff
( R1
R 2 )C
Example 9
In the ckt shown switch is closed
at t = 0 the correct statements
(a) Rate of increase of charge
is same in both capacitors.
* (b)Ratio of charge stored in
capacitor C and 2C at any time t would be
is ratio 1 : 2.
* (c) Time constant of both capacitance are equal.
* (d)Steady state charge in capacitance C and 2C
are in ratio 1 : 2
Current Electricity # 25
Sol. Deal both RC circuits seperately :
Sol. Both capacitors are identical (C = C0) consider
the circuit at a general time t when charge supplied
by battery is q and current in the ckt is i.
Using KVL
for upper ckt q1 c (1 e
For lower ckt q 2 2c (1 e
t /2 RC
)
t /2 RC
)
For 1st RC circuit
(2 R )C
2 RC
For 2nd RC circuit
(2C ) R
2 RC
q1
Ratio of charge q
2
q1
q2
c (1 e
2c (1 e
t /2 RC
)
t /2 RC
)
1
2
(Q0 q)
C
iR
Time constant of 1st ckt
eff
2 R (c )
Q0
2 RC
R (2C )
1
2
0
2q
iR
C
Time constant of 2nd ckt
eff
q
C
2 RC
Q0
iR
C
2RC
c
(Q0 2q )
Rc
dq
dt
c
(Q0 2q)
Rc
dq
(Q0 2q)
t
Steady state ckt
i
Using KVL
q1
c
0
q1
c
q2
2c
0
q2
q
0
1
ln[c
2
2c
q1
Ratio of steady state charges is q
2
c
c
2c
1
2
ln[c
(Q0
(Q0
The switch s is
closed at t = 0. The
capacitance C is
uncharged but C0 has
a charge Q0  2C
at t  0 If R  100, C  2F , C0  2F , E  4V
Calculate current as a function of time.
2q)]
2q)] ln(c
ln
c
2q
q
0
0
dt
Rc
t
Rc
t
Rc
Q0 )
Q0 2 q
c Q0
t
Rc
2q
Q0
t
Rc
ln 1
1
c
c
2q
Q0
e
t / RC
Current Electricity # 26
q
i
i
i
c
Q0
2
2 10
6
1
e
Rc
6
0.03e
2
(1 e
t / Rc
)
Find the current in each branch of the ckt.
dq
dt
c
2
dq
dt
2
t
6
e
100 2 10
6
Ans. (a)  2A (b)  1.5 A (c)  0.5 A
(d)  0.25 A
(e)  0.25 A
Sol.
10 4 t
(Q0
dt
dq
1
100 2 10
dq
(Q0 2q)
c
t
Q0
t / RC
4 2 10
2
i
c
2q)
Using KVL : i 1 21 5i 6(i i1 ) 0
6(i i1 ) 4i1 5 8(i1 i2 )
2 8(i1 i2 ) 16i2
0
.... (2)
.... (3)
On solving a bove equations :
We can determine i, i1 & i2
Current in diffrent branches is :
1 dt
2 t
5
1
ln t
2
1
ln[c
2
0
... (1)
(Q0
2q )]
&1
i
6
i i1
4
i1
8
i1 i2
16
i2
2A
1.5 A
0.5 A
0.25 A
0.25 A
Current Electricity # 27