1

CURRENT ELECTRICITY
ELECTRIC CURRENT :
The rate of flow of charge is called electric current .
Electric current ,
I=q
t

It is a scalar quantity . It's S.I. unit is Ampere or C/s

DRIFT VELOCITY :
When some potential difference is applied across the ends of a conductor , free electrons inside conductor start moving
towards positive end of conductor with a particular average velocity , which is known as drift velocity ( vd ) .
Suppose V is the potential difference applied across the ends of a conductor of length l then electric field set up on the
conductor ,
E=V
l
Due to this electric field force on each free electron inside conductor ,
F=eE ( E = F/q )
Now acceleration of each free electron ,
a=F
m
a=eE ……………(1)
m
Due to this acceleration each free electron acquires additional velocity component apart from it's initial thermal velocity .
If u1 is the initial thermal velocity of an electron then velocity acquired by electron on being accelerated by external potential
difference ,
v 1 = u 1 + a t1
Here t1 is the time difference between two successive collision of free electron with ions / atoms of conductor .
Similarly velocities acquired by other electrons ,
v 2 = u 2 + a t2
v 3 = u 3 + a t3 …………. and so on
v n = u n + a tn

Now average velocity of n free electrons i.e. drift velocity

vd = v1 + v2 + v3 + ………………….. + vn
n
vd = ( u1 + a t1 ) + ( u2 + a t2 ) + …………………….. + ( un + a tn )
n
vd = ( u1 + u2 + u3 + ………… + un ) + a ( t1 + t2 + t3 + …………. + tn )
n n
But average of initial thermal velocities of n free electrons is always zero ,
u1 + u2 + u3 + ……………… + un = 0
n
and t1 + t2 + t3 + ……………….. + tn = τ ( average time of relaxation )
n
Now vd = 0 + a τ
vd = a τ
vd = e E τ ( from eq. 1 )
m

MOBILITY :
Mobility of charge carrier is the magnitude of drift velocity of charge per unit electric field applied .
μ = vd
E
μ=qEτ/m
E
μ=qτ
m
S. I. unit of mobility is C m / N s or m2 / V s
 2
RELATION BETWEEN CURRENT AND DRIFT VELOCITY :
Consider a conductor of length l and uniform area of cross-section A . If n is the electron density ( i.e. numbers of electrons per
unit volume ) then total numbers of free electrons inside the conductor = n A l
If e is the charge of an electron then total charge inside conductor ,
q=nAle
Suppose vd is the drift velocity of free electrons on applying
potential difference across the ends of conductor then time
taken by free electrons to cross the conductor ,
t=l
vd
Now electric current ,
I=q
t
I=nAle
l / vd
I = n e A vd

OHM'S LAW :
The electric current flowing through a conductor is directly proportional to the potential difference across the ends of
conductor provided that physical conditions like temperature of the conductor remains constant .
If I is current flowing through a conductor at potential difference V then ,
V I
V=IR
R=V
I
R is called resistance or electrical resistance of conductor . It's S. I. unit is Ω .

DEDUCTION OF OHM'S LAW BY DRIFT VELOCITY :

Current flowing through a conductor ,
I = n e A vd
But drift velocity vd = e E τ = e V τ ( E=V)
m ml l
now I =n e A ( e V τ )
ml
V= ml
I n e2 τ A
Here , m l = constant ( R ) , therefore
n e2 τ A
V=R
I
This is Ohm's law .

SPECIFIC RESISTANCE OR RESISTIVITY OF A CONDUCTOR :

Resistance of a conductor depends upon ,
(1)R l ( length of conductor )
(2)R 1 ( area of cross-section )
A
combining both ,
R l
A
R=ρl ………..( 1 )
A
ρ is called specific resistance or resistivity of the conductor .
If l = 1 ; A = 1 then ρ = R
Hence specific resistance of a material is equal to the resistance of a conductor of unit length and unit area of cross-section of
that material . It's S. I. unit is Ω m .
By Ohm's law R=ml ……….( 2 )
n e2 τ A
Comparing eq. ( 1 ) and ( 2 )
ρ=m
n e2 τ
 3
CURRENT DENSITY :
At a point inside a conductor the amount of current flowing normal to per unit area of cross-section of conductor is called
current density .
It is a vector quantity , it's direction is in the direction of flow of positive charge .
If I is the current flowing through a conductor of area of cross-section A then current density ,
j=I
A
It's S. I. unit is A / m2

CONDUCTANCE :
The reciprocal of resistance is known as conductance .
G=1
R
It's S. I. unit is Ω-1 or Mho or Siemen

ELECTRICAL CONDUCTIVITY :
The reciprocal of specific resistance or resistivity is known as electrical conductivity .
σ=1
ρ
It's S.I. unit is Ω-1m-1 or Mho / m or S / m

RELATION BETWEEN j , σ and E :

As , I = n e A vd
but vd = e E τ
m
Now , I=neA(eEτ)
m
I = n e2 τ E
A m
Put I = j and m = ρ
A n e2 τ
j=E
ρ
j=σE 1=σ
ρ

EFFECT OF TEMPERATURE ON RESISTANCE :

The resistance of a metal conductor is
R=ml
n e2 τ A
for a given conductor m , l , n , e and A are constants , therefore
R 1
τ
When the temperature of metal conductor is raised the atoms / ions of metal vibrate with greater amplitude and greater
frequency . Now the frequency of collision of free electrons with atoms / ions also increase which reduce the average relaxation
time τ . Hence the value of resistance R increase with rise of temperature .
The resistance of a metal conductor at temperature t
Rt = R0 ( 1 + α t )
Here α is called temperature coefficient of resistance .
For metals α is positive therefore resistance of metals increases with rise in temperature .
For alloys α is positive but very small so the resistance of alloy increases with rise in temperature but this increase is small as
compared to pure metals .
For insulators and semiconductors α is negative therefore the resistance decreases with rise in temperature .

VARIATION OF RESISTIVITY WITH TEMPERATURE :

Resistivity of a material
ρ=m
n e2 τ
In metals free electrons density n does not change with temperature but with increase in temperature the relaxation time τ
decreases hence resistivity increases with increase in temperature .
 4
In semiconductors and insulators the electron density n increases with rise in temperature and relaxation time τ decreases .
The increase in n compensates more than decrease in τ therefore resistivity of semiconductor decreases with rise in
temperature .
ρ ρ ρ

alloy

pure semiconductor &

metal insulator

O T O T O T

NON OHMIC CONDUCTORS :

Such conductors which do not obey Ohm's law are known as non Ohmic conductors .For ex. vacuum tube , semiconductor
devices etc .

I I

Ohmic conductor non Ohmic

conductor

O V O V

COLOUR CODING FOR CARBON RESISTORS :

COLOUR NUMBER MULTIPLIER COLOUR TOLERANCE

Black 0 100 Gold 5%
Brown 1 101 Silver 10 %
Red 2 102 No colour 20 %
Orange 3 103
Yellow 4 104
Green 5 105
Blue 6 106
Violet 7 107
Grey 8 108
White 9 109
Gold 10-1
Silver 10-2

In colour coding following sentence gives great help ( where bold letters stand for colours )

B B ROY Great Britain Very Good Wife wearing Gold Silver Necklace .

In this system
(1) Colour of first strip A gives the first significant figure
of resistance in Ohm .
(2) Colour of second strip B gives second significant
figure of resistance in Ohm .
(3) Colour of third strip C indicates the multiplier i.e.
number of zeros following after two significant figures .
(4) Colour of fourth strip R indicates the tolerance limit
i.e. percentage accuracy of resistance .
 5
RESISTANCES IN SERIES :
Suppose three resistances R1 , R2 and R3 are connected in series . If same current I is flowing through the three resistances in
series then V1 , V2 and V3 are the potential differences across them , then total potential difference
V = V1 + V2 + V3
V = I R1 + I R 2 + I R3
V = I ( R1 + R2 + R 3 ) ……..(1)
If R is the equivalent resistance of series combination
By Ohm's Law
V=IR …….(2)
Comparing eq (1) and (2)
R = R1 + R2 + R3

RESISTANCE IN PARALLEL :
Suppose three resistances R1 , R2 and R3 are connected in parallel . On applying same potential difference V across the three
resistances suppose I1 , I2 and I3 are the currents flowing through them , then total current
I = I1 + I2 + I3
I=V + V + V
R1 R2 R3
I=V 1 + 1 + 1 ….(1)
R1 R2 R3
If R is the equivalent resistance of parallel combination
By Ohm's law
I=V ……(2)
R
Comparing eq (1) and (2)
1 = 1 + 1 + 1
R R1 R2 R3
If we have only two resistances then
R = R1 R2
R1 + R 2

ELECTROMOTIVE FORCE ( E M F ) OF CELL :

The E.M.F. of a cell is equal to the work done by the cell to drive unit positive charge once around the complete circuit .
The E.M.F.
E=W
q0
It is a scalar quantity . It's S.I. unit is Volt .

INTERNAL RESISTANCE OF CELL :

The resistance offered by the electrolyte and electrodes of a cell in the path of current is called internal resistance( r ) of cell .

TERMINAL POTENTIAL DIFFERENCE OF CELL :

The potential difference between two electrodes of a cell in closed circuit is called terminal potential difference of cell .
When the key K is closed the current flowing through the circuit
I=E
r+R
potential difference across internal resistance
Vin = I r

Now terminal potential difference while

discharging the cell
V=E-Ir
In open circuit as I = 0
V=E
By Ohm's law V=IR
V=ER
r+R
Vr+VR=ER
Vr=R(E-V)
r=R(E-V)
V
 6
r=R E-1
V
While charging the cell terminal potential difference
V=E+Ir

TWO CELLS IN SERIES : E 1 , r1 E2 , r2

Terminal potential difference of first cell + - + -
V1 = E1 - I r1 I
Terminal potential difference of second cell
V2 = E2 - I r2
Total terminal potential difference
V = V1 + V2
V = E1 - I r1 + V2 - I r2
V = E 1 + E 2 - I ( r1 + r 2 ) ……(1)
If E is the equivalent E.M.F. and r is the equivalent internal resistance of series combination then ,
V=E-Ir …….(2)
Comparing eq. (1) and (2)
E = E1 + E2
r = r1 + r2 E1 , r1 E 2 , r2
Special case : + - - +
If the terminals of second cell is reversed then
E = E1 - E2
r = r1 + r2

TWO CELLS IN PARALLEL

For 1st cell V = E1 - I1 r1 E1 , r1
I1 = E 1 - V + -
r1 I1
For 2nd cell I2 = E 2 - V
r2 I2
Now total current + -
I = I1 + I2 E 2 , r2
I = E1 - V + E2 - V
r1 r2
I = E1 r2 - V r2 + E2 r1 - V r1
r1 r2
I = E1 r2 + E2 r1 - V ( r1 + r2 )
r1 r2
I r1 r2 = E1 r2 + E2 r1 - V ( r1 + r2 )
V ( r1 + r2 ) = E1 r2 + E2 r1 - I r1 r2
V = E1 r2 + E2 r1 - I r1 r2
r1 + r2 r1 + r2 ……(1)
If E is the equivalent E.M.F. and r is the equivalent internal resistance of parallel combination then
V=E-Ir …….(2)
Comparing eq(1) and eq(2)
E = E1 r2 + E2 r1 …….(3)
r1 + r2
r = r1 r2 …….(4)
r1 + r2
Dividing eq(3) by eq(4)
E = E1 r2 + E2 r1
r r1 r2
E = E1 + E2
r r1 r2
 7
KIRCHHOFF'S LAW :
(1) The algebraic sum of various currents meeting at a junction in a closed circuit is always zero .
ΣI=0
sign convention :
The currents flowing towards the junction are taken as positive and I3
the currents flowing away from the junction are taken as negative I2
O
Ex. I1 + I2 - I3 - I4 - I5 = 0
I1 + I2 = I3 + I4 + I5 I1
I4
I5
(2) The algebraic sum of all the potential differences in a closed lop of a closed circuit is always zero .
ΣV=0
sign convention :
First define a loop , then
(1) The currents flowing in the direction of loop are taken as positive while currents flowing in the opposite direction of loop are
taken as negative .
(2) While traversing the loop if positive terminal of cell is encountered first , it's emf is taken as positive and if negative terminal
of cell is encountered first , it's emf is taken as negative .

Ex. At point B , I1 + I2 = I3 ….(1)

In loop ABEFA
I1 R1 + ( I1 + I2 ) R3 - E1 = 0
I1 R1 + ( I1 + I2 ) R3 = E1 …..(2)
In loop CBEDC
I2 R2 + ( I1 + I2 ) R3 - E2 = 0
I2 R2 + ( I1 + I2 ) R3 = E2 ….(3)

WHEASTONE BRIDGE PRINCIPLE :

If we arrange four resistances P, Q, R and S in the form of a bridge ( quadrilateral ) in such a manner that a galvanometer is
connected across one diagonal and a cell is connected across other diagonal of bridge , in balancing condition P = R no
current flows through the galvanometer , and galvanometer shows no deflection . Q S

Proof :
Applying Kirchhoff's 2nd law in closed loop ABDA
I1 P + Ig G - ( I - I1 ) R = 0 ....(1)
Applying Kirchhoff's 2nd law in closed loop BCDB
( I1 - Ig ) Q - ( I - I1 + Ig ) S - Ig G = 0 …..(2)
In balancing condition Ig = 0
From eq(1) I1 P - ( I - I1 ) R = 0
I1 P = ( I - I1 ) R …..(3)
From eq(2) I1 Q - ( I - I1 ) S = 0
I1 Q = ( I - I1 ) S …..(4)
Divide eq(1) by eq(2)
I1 P = ( I - I1 ) R
I1 Q ( I - I1 ) S
P=R
Q S
This is balancing condition .

SLIDE WIRE BRIDGE OR METER BRIDGE :

It is a practical form of Wheatstone bridge which is used to determine unknown resistance of a wire .
A meter bridge has two gaps . Across left gap a resistance box R and across right gap an unknown resistance S are connected .
Close key K and take out a suitable resistance R from resistance box . Now adjust the position of jockey on meter bridge wire AB
and find a point J where galvanometer shows no deflection . In that case the bridge is balanced , now by Wheatstone bridge
principle
P=R ……(1)
Q S
Here P = resistance of length l of wire AB ; Q = resistance of length ( 100 - l ) of wire AB
 8

If r is the resistance per cm length of wire AB , then R S

P=lr
Q = ( 100 - l ) r
Put these values in eq(1)
lr =R G
( 100 - l ) r S
S = R ( 100 - l )
l A J B
By knowing only l we can determine S .
By this method we can also determine ()
resistivity of an unknown wire . + - K
E
PRINCIPLE OF POTENTIOMETER :
The potential difference across any portion of the wire is directly proportional to the length of that portion provided that wire
has uniform area of cross-section and a constant current flows through it .
By Ohm's law V=IR
But R = ρ l
A V=Iρl
A
Here I ρ = constant ( K )
A
V=Kl
V l
here K is called potential gradient i.e. rate of fall of potential per unit length of wire .

COMPRISON OF EMF's OF TWO CELLS USING POTENTIOMETER :

Close key K and adjust a suitable constant current in the potentiometer
with the help of rheostat . Insert the plug in the gap between the
terminals 1 and 3 of two way key so that cell E1 comes in the circuit .
Adjust the position of jockey on potentiometer wire AB and find a point
J1 where galvanometer shows no deflection . Note the length AJ1 = l1 .
In this condition emf of cell E1
E1 = K l1 ……(1)
K = potential gradient of potentiometer wire
Now remove the plug from the gap between 1 and 3 and insert it in the
gap between 2 and 3 of two way key so that cell E2 comes in the circuit .
Again find a point J2 on potentiometer wire where galvanometer shows
no deflection . Note the length AJ2 = l2 .
In this condition emf of cell E2
E2 = K l 2 …….(2)
Divide eq(1) by eq(2)
E1 = K l 1
E 2 K l2
E1 = l1
E2 l2

DETERMINATION OF INTERNAL RESISTANCE OF A CELL BY POTENTIOMETER :

Initially close key K and K1 remains open . Adjust the position of jockey on potentiometer wire and find a point J1 where
galvanometer shows no deflection . Note the length AJ1 = l1 . In this condition emf of cell
E = K l1 ……….(1)
K = potential gradient of potentiometer wire
Now close key K1 , by taking out suitable resistance R from resistance
box again find a point J2 on potentiometer wire where
galvanometer shows no deflection .Note the length AJ2 = l2 . In this
condition terminal potential difference of cell
V = K l2 ……….(2)
Divide eq(1) by eq(2)
E = K l1
V K l2
 9
E = l1
V l2
Now internal resistance of cell
r=R E-1
V
r = R l1 -1
l2

SENSITIVENESS OF POTENTIOMETER :
The sensitiveness of a potentiometer can be increased by decreasing it's potential gradient . To decrease potential gradient ,
(1) Increase the length of potentiometer keeping its area of cross-section uniform .
(2) Decrease the current in potentiometer wire with the help of rheostat .

JOULE'S LAW OF HEATING :

The amount of heat produced in a conductor of resistance R in time t when a current I flows through it
H I2 R t

HEAT PRODUCED BY ELECTRIC CURRENT :

Suppose a current I is flowing through a conductor in time t , total charge produced in the conductor
q=It
Now work done in carrying charge q at potential difference V
W=Vq
W=VIt
This work done is the electric energy consumed in the circuit .
By Ohm's law V = I R
W = I2 R t
Again put I = V/R
W = V2 t
R
Now heat produced
H = W = V I t = I2 R t = V2 t
4 18 4 18 4 18 4 18

ELECTRIC POWER :
The rate of dissipation of electric energy in a circuit is called electric power .
P=W
t
It's S.I. unit Joule/s or Watt . Another unit of power is H.P. ( horse power ) ; 1 H.P. = 746 W
P=VIt
t
P=VI
By ohm's law V = I R
P = I2 R
Again put I = V/R
P = V2
R
COMMERCIAL UNIT OF ELECTRIC ENERGY :
The commercial unit of electric energy is called B.T.U. ( Board of trade unit ) .
1 B T U = 1 KWH = 1000 × 3600 Joule = 3 6 × 106 Joule

MAXIMUM POWER TRANSFER THEOREM :

The power output across a certain load due to a cell or battery is maximum if the load resistance is equal to the effective
internal resistance of cell or battery .