UNIT – II
CURRENT ELECTRICITY
A)
Contents
Electric current ,flow of electric charges in a metallic conductor, drift velocity ,mobility &
their relation with electric current; Ohm’s law, electrical resistance, V-I characteristics (linear
& non linear), electrical energy & power, electrical resistivity & conductivity .Carbon
resistors, colour code for carbon resistors; series & parallel combination of resistors;
temperature dependence of resistance.
Internal resistance of a cell , potential difference & emf of a cell , combination of cells in
series & in parallel.
Kirchoff ’s laws & simple applications. Wheatstone bridge , meter bridge.
Potentiometer - principle & its applications to measure potential difference & for comparing
emf of two cells; measurement of internal resistance of a cell.
B)
Concept and principles /laws involved in units
Electric current, resistance, resistivity or specific resistance, conductance, conductivity or
specific conductance, drift velocity and relaxation time , colour code for resistors, temperature
coefficient of resistance, resistances in series, resistances in parallel, division of current in
resistors joined in parallel, emf., internal resistance, relations between emf(E),terminal
potential difference(V) and internal resistance(r).cells in series, cells in parallel, cells in mixed
grouping, wheat stone bridge, slide wire bridge or meter bridge. Potentiometer
Principle - principle of potentiometer.
Laws-
ohm’s law,
Kirchoff ’s laws
i.
1st law or junction law or law of conservation of charge or law of current.
ii.
2nd law or loop law or law of conservation of energy or law of voltage.
C ) Important terms , their definitions , units/dimensions of physical quantities associated
with the terms.
1) Electric current-It is the rate of flow of charge through a given conductor.
I = q/t
or
SI unit of current is ampere (A)
1ampere = 1colomb/1second
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
I = dq/dt
Or
1A = 1CS-1
Dimension of charge
= [M0L0TA]
Dimension of electric current = [M0L0T0A]
2) Ohm’s law- The current flowing through a conductor is directly proportional to the
potential difference across its ends, provided the temperature and other physical conditions
remain unchanged.
VαI
Or V = RI
Dimension of p.d. = [ML2T-3A-1]
Or V/I = R
3) Resistance -It is the property by virtue of which a conductor opposes the flow of charges
through it.It depends on the length (l) and area of cross-section (A) of the conductor.
R=  l/A
 = resistivity of the material
SI unit of resistance is ohm.
1 ohm = 1volt / ampere
Or, 1Ω = 1VA-1
Dimension of resistance = [ML2T-3A-2]
4) Resistivity or specific resistance – It is the resistance offered by a unit cube of the
material of a conductor.
 = RA / l
dimension of resistivity = [ML3 T -3A-2]
SI unit of  is Ω m
5. Conductance – It is the reciprocal of resistance.
Conductance = 1/R
SI unit of conductance = ohm -1 or mho or siemen.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Dimension of conductance =M-1L-2T3A2
6. Conductivity or specific conductance
It is the reciprocal of resistivity.
Conductivity = 1/ Resistivity
Or ,
 = 1/
SI unit of conductivity = ohm-1m-1
Or mho m-1
Dimension of conductivity =M-1L-3T3A2
7. Drift velocity & relaxation time - The average velocity acquired by the free electrons of
a conductor in the opposite direction of the externally applied electric field is called drift
velocity (vd). The average time elapsed between the two successive collision of an electron
is called relaxation time ( τ ) .
d = eEτ /m
I = nAvde
R = ml/nAe2τ  = m/ne2 τ
Here n = no. of free electrons per unit volume or free electron density
& m = mass of an electron
8. Colour code for resistors
colour
number
black
0
100
brown
1
101
red
2
102
orange
3
103
yellow
4
104
green
5
105
blue
6
106
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
multiplier
violet
7
107
grey
8
108
white
9
109
How to remember colour codes
B
B
ROY of Great Britain had a Very Good
Wife
↓
↓
↓↓↓
↓
↓
↓
↓
↓
0
1
234
5
6
7
8
9
Tolerance:
Gold
±5%
Silver
± 10%
No colour
± 20%
A set of coloured concentric bands is printed on the resistor which reveals the following facts
i)
The first band indicates the 1st significant figure.
ii)
The second band indicates the 2nd significant figure.
i)
How to read the resistance when three colours are there with out Gold / Silver.
ii)
How to read a resistor with three colours with Gold/Silver at one of the end.
iii)
The third band indicates the power of ten with which the above two significant figures
must be multiplied to get the resistance value in ohms.
iv)
The last band indicates the tolerance in percent of the indicated value.
9) Temperature coefficient of resistance (). It is defined as change in resistance per unit
original resistance per degree rise in temperature. It is given by
=R2-R1/R1 (t2-t1)
If t1=00C & t2=toC
then
=Rt-Ro/Rot
or,
Rt =Ro(1+ t)
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
The unit of  is oC-1 or K-1
10) Resistance in series
When a number of resistances are connected in series, their equivalent resistance (RS ) is
equal to the sum of the individual resistances.
RS=R1+R2+R3+.......
11) Resistances in parallel
When a number of resistances are connected in parallel, the reciprocal of their total
resistance (RP) is equal to the sum of the reciprocals of the individual resistances.
1/RP=1/R1+1/R2+1/R3+…..
For two resistances in parallel,
RP=R1R2/(R1+R2)
12) Division of current in resistors joined in parallel
The current is divided in resistors, connected in parallel, in the inverse ratio of their
resistances.
I1=R2 I / (R1+R2)
I2=R1 I / (R1+R2)
13) EMF (E) (Misnomer)
It is the work done by the source in taking a unit charge once round when cell is not used
in circuit or when no current is drawn from it. It is equal to the potential difference
measured in open circuit.
EMF= Work done / Charge
E= W / q
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
SI unit of EMF is volt .
14) Internal resistance (r) – It is the resistance offered by the electrolyte of a cell when an
electric current flows through it.
15) Relations between emf(E), terminal p.d. (V) &internal resistance (r)
When ?
E=V + Ir
V=IR=ER/(R+ r)
r =(E-V) /I = R (E-V)/V
I=E/(R+ r)
Imax = E/r
Terminal P.d. of acell when it is being charged is V=E+Ir
So V>E
16) Cells in series
If n cells of emf E & internal resistance r each are connected in series, then current
drawn through external resistance R is
I = n E/ (R+ nr)
17) Cells in parallel
If n cells are connected in parallel, then current drawn through external resistance R is
I = n E / (n R + r)
18) Cells in mixed grouping
If n cells are connected in series in each row and m such rows are connected in parallel,
then current drawn through an external resistance R is
I= mnE /( m R + nr)
For maximum current, the external resistance must be equal to the total internal
resistance;
i.e. R = n r/ m
or, m R=n r
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
19) Kirchoff ‘s Rules
Ist Rule (Junction Rule)
At any junction of circuit, the sum of currents entering the junction is equal to the sum
of currents leaving that junction.
or
The algebriac sum of the currents meeting at a junction is zero.
Sign conventions
a)
b)
The currents flowing towards the junction are positive.
The currents flowing away from the junction are negative.
 IJ  0
U
I1 + I2 - I3 - I4 = 0
I1 + I2 = I3 + I4
2nd Rule ( Loop Rule)
Around any closed loop of a network, the algebraic sum of the emfs must be equal to the
algebraic sum of the products of current and resistances of various parts of the loop.
Mathematically,
 IR +  E = 0
n
n
j 1
j 1
  I j Rj   E j  0
Sign Convention for applying loop rule.
I.
Any direction ( clock wise or anti clock wise ) may be taken as the direction of traversal.
II.
IR product is positive if current through the resistor is in the direction of traversal,
otherwise negative.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
III.
The emf of a cell is positive if the direction of traversal is from negative to positive
electrode through the electrolyte, otherwise negative
As shown in figure, traversing in the clock wise direction, we see from Kirchoff’s loop
law, that
1)
For loop 1, E2 = I2 R2 + (I1 + I2 )R3
For loop 2, E1 – E2 = I1R1 – I2R2
I
2)
V = – IR
R
V=IR
3)

4)

20. Wheat stone’s bridge
It is an arrangement of four resistance P,Q,R,S joined to form a quadrilateral ABCD
with a battery between A & C and a sensitive galvanometer between B & D. The
resistances are so adjusted that no current flows through the galvanometer.The bridge is
then said to be balanced.
In the balanced condition
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
P/Q=R/S
Knowing any three resistance, the fourth resistance can be computed.
21. Slide wire bridge or meter bridge.
Principle-It is based on the principle balanced Wheat stone bridge principle.
It is an application of wheat stone bridge in which R is fixed and balanced point is
obtained by varying P & Q i.e. by adjusting the position of jockey on a hundred cm long
resistance wire stretched between two terminals. If the balanced point is obtained at
length l,then P / Q = R / S = l / ( 100 – l ) ; p  l ; Q 100 – l  P  l . r Q 100 – l  r
where r is resistance per unit length.
22. PotentiometerIt is a device used to compare the emf of two cells. Its working is based on the principle
that when a constant current flows through a wire of uniform cross sectional area, the
p.d. across any length of the wire is directly proportional to that length
Vl
V
V
K;
= Potential gradient  Fall of Potential
l
l
along whole length of wire of potentiometer.
V = kl 
Where k is the potential drop per unit length which is called potential gradient.
Potentiometer has two main uses
(i) To compare the emfs of two cells
If l1 and l2 are the balancing lengths of the potentiometer wire for the cells of emfs E1
& E2 respectively,
Then E2 / E1 = l2 / l1
(ii) To find the internal resistance r of a cell
If l1 is the balancing length of the potentiometer wire without shunt & l2 is the
balancing with shunt across the cell, then internal resistance of the cell will be
r = ( E-V) R / V
or r = ( l1-l2 ) R / l2
23. Electric energy
It is the total work done in maintaining an electric current in an electric circuit for a
given time
Electric energy = Electric power x Time
W = Pt = VIt = I2Rt = V2t / R
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
( in joule )
Units of electric energy = 1 KWh = 1000 Wh = 3.6x106 J
24. Electric Power
It is the rate at which an electric appliance converts electric energy into other form of
energy.
Electric power = P = W / t = VI= I2R= V2R
S.I. unit of power is watt
1 watt = 1 joule / second = 1 volt x 1 ampere
1 kilowatt = 1000 watt
25. Power rating
The electric power consumed by a circuit element is known as its power rating.
P = V2 / R = I2R = VI.
(D) Important Formulas Relations / Physical Constant of the Unit ( SI only )
A) 1.
Electric Current
Or
I
= charge/time
= q/t= dq/dt = t a
2.
q=ne
So I=ne/t where n is the no. of electrons.
3.
If an electron revolving in a circle of radius r with speed ,the Period of revolution
of the electron is
T = 2r/v
4.
Frequency of revolution
5.
Current at any
 = 1/T = v/2r.
point of the orbit is
I = e / T = ev / 2r
UNITS USED
Electric charge in coulomb (c)
Time in sec (s)
Current in ampere (A)
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Constant used – Charge of an electron, e=1.6 x 10-19c
6.
Ohm’s Law
or
R
V
= V/I
= IR
7.
Resistance of a uniform conductor
8.
Resistivity  =RA/l
9.
Conductance =1/R
10.
R = l /A
Condictivity==1/=l /RA
UNITS USED
Potential difference V in Volt (V)
Resistance R in ohm ()
Resistivity  in m
Conductance in ohm-1 or mho or siemen (s)
Conductvity in  -1 m-1
11.
Current in terms of drift velocity (vd ) is
I = nAvd e
12.
Resistance
R = ml/nAe2
13.
Resistivity
 = m/ne2 
UNITS USED
Drift Velocity vd in m/s
Free electron density (n) per m3
Cross Section area A in m2
14.
Temperature coefficient of resistance
 = (R2 – R1) /R1( t2- t1)
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
If
t1 = 00c & t2 = t0 c
 =( Rt – R0) /Rot
15.
Resistance of wire at t0c is Rt = Ro (1+ t)
UNITS USED
Temperature in 0c or K
16.
The equivalent resistance Rs of a number of resistances connected in series is given by
Rs = R1 + R2 + R3 + ………………
17.
The equivalent resistance Rp of a number of resistances connected in parallel is given by
1/Rp = 1/ R1 +1/ R2 +1/ R3 + ………………
18.
For two resistances in parallel current through the two resistors will be
I1 = R2I /(R1+R2) & I2 = R1I /(R1+R2)
19.
Emf of a cell
E = W/q
20.
For a cell of internal resistance r, the Emf is
E = V+ Ir = I(R+r) ; r 
21.
E
–R
I
Terminal potential difference of a cell is
V = IR = ER/(R+r)
22.
Internal resistance of a cell ,
r = R ( E –V)/V
23.
For n cells in series,
I = nE / (R+nr)
24.
For a n cells in parallel,
I = nE /(nR+r)
25.
For mixed grouping ,
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
I = mnE/(mR+nr)
Where,
n - no. of cells in series in one row.
m - no. of rows of cells in parallel
26.
For maximum current , the external resistance must be equal to total internal resistance :
i.e. nr/m = R
or , nr = mR
27.
I = 0
or total incoming current equal to total outgoing current
28.
E = IR
29.
For comparing emf of two cells
E2/E1 = l2/l1
30.
For measuring internal resistance of a cell
r = (l1 - l2 ) x R/ l2
31.
Potential gradient of the Potentiometer wire
k = V /l
32.
SI unit is Vm -1
Unknown emf balanced against length,
E = kl
33.
For a balanced wheat stone bridge,
P/Q = R/S
If X is unknown resistance then,
P/Q = R/X
Or X = RQ/P
34.
In a slide wire bridge, if balance point is obtained at l cm from the zero end, then
P/Q = R/X = l /(100-l)
Or X = ( 100- l )R/ l
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
35.
Heat produced by electric current
H = I2Rt joule = I2Rt / 4.18 Calorie
36.
Or H = Vit joule = VIt /4.18 Calorie
Electric power
P = W/t = VI = I2R = V2/R
37.
Electric energy W = Pt = VIt = I2Rt
UNITS USED
Electric energy in Joule or in KwH
1 unit =1 KwH
1 KwH =3.6x10 6 J
G.Very Short answer type Question ( 1 mark each )
1.
Write a relation to show the factors on which the resistance of a conductor depends ?
R =l / A
Where R → Resistance of the conductor
l → length of the conductor
A → cross section of the conductor
→ resistivity of it’s material
2.
State two factors on which resistivity of a metal depends ?
Two factors are
3.
(i)
Nature of metal
(ii)
Temperature of the conductor
Define emf of a cell.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
It is defined as the work done per unit charge in moving the charge completely through the
circuit including cell.
4.
Why is electric current not a vector quantity ?
Because electric current does not obey the laws of vector addition. (Conventional Current)
5.
Is a wire carrying current charged ?
No, because the wire has as much negative charge as positive.
6.
How does resistance of a conductor depends upon temperature ?
It increases with increase in temperature.
7.
Name the metal with highest conductivity.
Silver.
8.
Establish the S.I. unit of resistivity.
R = l / A or  = R A/l =  m2 / m
S.I. unit of  is  m.
9.
Two wires of equal length one copper and other manganin have same resistance. Which
one is thicker ?
We have, R =l / A
and we know that m >c
 Am > Ac
10.
so manganin wire is thicker.
Two wires of equal thickness one copper and other manganin have same resistance.
Which one is longer ?
We have, R =  l / A
and we know that m >c
lcopper > l manganin
11.
A copper wire of resistivity  is stretched to reduce its diameter to half. What will be its
new resistivity ?
Resistivity does not depend upon thickness. So no effect.
12.
Why are copper wires used as connecting wires ?
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Electrical conductivity is high and resistivity low for copper.
13.
Out of copper and rubber which has larger resistivity ?
Rubber.
14.
Two wires A and B of same metal, same cross section have their lengths in the ratio 2 :
1, what will be the ratio of current through them when connected in parallel across the
same voltage source ?
R =l / A, RA / RB = 2 / 1,
I = V/ R
As V is same so IA/ IB = RB / RA = 1 / 2
15.
Of metals and alloys which has a larger value of temperature coefficient of resistance ?
Metals have larger temperature coefficient of resistance as compared to alloys.
16.
Why are constantan and manganin used for making standard resistors ?
These materials have
i. High resistivity
iii.
17.
Low temperature coefficient of resistance.
Why don’t we consider the drift velocity of positive ions ?
The positive ions are heavy and tightly bound in the metal, they are hardly able to move
making the drift velocity of the ions negligibly small.
18.
How does electric field vary with drift velocity ?
Electric field intensity is directly proportional to drift velocity. E α vd
19.
For a good electrical conductor how does the relaxation time varies with temperature ?
With the increase in temperature ,relaxation time decreases.
20.
If the pd across the conductor is doubled, how will it affect the drift velocity of
electrons?
As vd α V , drift velocity is doubled.
21.
It is easier to start a car engine on a warm day than on chilly day. Why?
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Internal resistance of a car battery decreases on a warm day due to increases in temperature
allowing easier starting of a car.
22.
Mention some non-ohmic device.
Semiconductor diode, transistor, thyristor, etc.
23.
A Wire of resistance 4R is bent in the form of a circle. What is the effective resistance
between the ends of a diameter ?
‘R’
24.
Name the principle, on which meter bridge works.
A meter bridge works on Wheat stone’s bridge principle.
25.
Why is copper wire not used in potentiometers and meter bridge ?
It has large temperature coefficient of resistance and small resistivity.
26.
Why should the jockey not be rubbed against potentiometer or meter bridge wire ?
Rubbing hard with jockey may change the areas of cross section of the wire, thereby affecting
the fall of potential along the wire.
27.
Why do we prefer potentiometer for measuring emf of a cell rather than a voltmeter ?
A potentiometer measures emf without drawing any current from the cell, where as some
current is drawn by the voltmeter is slightly less than the actual emf.
28.
On what basic principle is Kirchhoff’s first law based.
First law is based on conservation of charges.
29.
Why should the potentiometer wire be of uniform cross section ?
Then only the potential difference across any portion of the wire will be proportional to the
length of that portion.
30.
How can we increase the sensitivity of a potentiometer ?
To improve the sensitivity of a potentiometer the potential gradient has to be reduced. This is
done by
(a)
Increasing the length of potentiometer wire.
(b)
Reducing the current in potentiometer wire.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
31.
Why are the copper strips on a meter bridge kept thick ?
Thick copper strips offer negligibly small resistance.
32.
When is Wheat stone’s bridge said to be balanced ?
When no current flows through the galvanometer.i.e the ratio of resistance of adjacent arms
are equal.
H. Short Answer Type Question ( 2 Marks Each )
1.
2.
Why does (i) Resistance increases in series combination (ii) Resistance decreases in
parallel combination
(i)
In series combination the effective length increases and R α l.
(ii)
In parallel combination the area of cross section increases and R α 1 / A.
A given copper wire is stretched to reduce its diameter to half of its previous value.
What will be its new resistance ?
Keeping the volume constant ( d2 / 4 ) l
= l1 ( d / 2 )2 / 4
Earlier
l1 = 4l
A = d2 / 4
R=l/A
A1 = A / 4
R1 =  l1 / A1 =  4 l /( A / 4)
R1 = 16  l / A
3.
,
R1 = 16 R
Resistance becomes 16 times.
State the conditions under which Ohm’s law is not obeyed by a conductor.
Ohm’s law is not obeyed if
(i) The temperature is not constant.
(ii) The conductor is not ohmic.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
4.
The voltage (V) current (I) graph for the parallel and series combination of two metallic
resistors is shown. Which graph shows the parallel combination ? Justify your answer.
Slope of V – I graph gives resistance
Or slope ofA > slope of B
5.
Slope of V – I graph gives resistance
or slope of B > slope ofA
‘ A’ represents series combination.
B Represent series combination
‘ B ‘ represents parallel combination
A represent parallel combinatin
Define terminal pd of a cell , under what condition its value can be greater than the emf
of cell.
It is the work done by the chemical energy of the cell to move a charge of one coulomb in the
external circuit.
V = E – Ir
Normally, V < E. However if the current is given to the cell that is when a cell is charged
V = E + Ir and V > E.
6.
Name of one material with small value of temperature coefficient of resistance. State one
use of this material.
Constantan ( Alloy of Cu and Ni )
Or Manganin ( Alloy of Cu, Ni, and Mn )
It is used for making standard resistors.
7.
The V –I graph of a metallic wire at two temperatures T1 and T2 is shown, which of the
two temperatures is higher and why ?
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Slope V – I graph is resistance of the conductor
Slope 1 > slope 2
R1 > R2
But resistance α Temperature

8.
T1 > T2
A pd ‘ V ‘ is applied across a conductor of length ‘ l ‘ and diameter ‘ d ‘ . What is the
effect on the drift velocity of electrons when, (i) ‘ V ‘ is doubled (ii) ‘ I ‘ is doubled (iii) ‘ d
‘ is doubled.
Drift velocity
vd = e V/ml
And
vd = I / neA
A = d2 / 4
(i)
When ‘ V ‘ is doubled, drift velocity is also doubled
(ii)
When ‘ I ‘ is doubled, drift velocity is also doubled.
(iii) When ‘ d ‘ is doubled A becomes 4 times , and drift velocity becomes one fourth.
I.
1.
Short Answer Type Question ( 3 Marks each )
Define EMF of a cell. Show that voltage drop across a resistor connected in parallel with
a cell is different from the emf of the cell.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
The current in the circuit
I=E/(R+r)
The p.d. across R is V = IR
V = ER / ( R + r )
As
R<R+r
V<E
2.
Deduce Ohm’s law using the concept of drift velocity.
We know Vd = eVτ/ml
And
I = neAVd
(1)
(2)
Where Vd → drift velocity
I → current
Putting Vd for ( 1 ) in ( 2 )
A → cross section of conductor
I = neA (eV/ml)
τ → relaxation time
l---length of conductor
I = nAe2τV /ml
n→ number of electrons per unit volume
For a given conductor at a constant temperature n, e , τ, m, l, and A are constant
 IαV
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Or V = IR This is Ohm’s law.
3.
A pd ‘ V ‘ is applied across a conductor of length ‘ L ‘ and diameter ‘ D ‘. How are the
electric field (E) and the resistance ( R ) of a conductor affected when (i) ‘ V ‘ is halved
(ii) ‘ D ‘ is doubled. Justify your answer in each case.
E = V / L and R = L / A = 4 L /  D2
i.e. R α L / D2
(i)
When ‘ V ‘ is halved
‘ E ‘ is halved and ‘ R ‘ remains unaffected
(ii)
When ‘ L ‘ is halved.
‘ E ‘ is doubled ‘ R ‘ is halved
(iii) When ‘ D ‘ is doubled
‘ E ‘ remains unaffected ‘ R ‘ becomes 1 / 4th .
4.
Define resistivity of a substance, give its unit. How does it vary with temperature for
(i) a good conductor (ii) an ionic conductor and (iii) a semi conductor
With the increases in temperature, the resistivity of
(i)
a good conductor increases
(ii)
an ionic conductor decreases
(iii) a semiconductor decreases
5.
A student has two wires of iron and copper of equal length and diameter. He first joins the two
wires in series and passes electric current through the combination which increases gradually.
After that he joins the two wires in parallel and repeat the process of passing current. Which
wire will glow first in each case and why?
Ans) Using R=ρ(l/a).for copper ,
Rc=cl/a
Since ρi is more than ρc therefore
and for iron, Ri= ρil/a
Ri>Rc
In series ,same current flows through iron and copper wire. since P=I2 R. therefore more heat
is produced in iron wire which makes it glow first.
In parallel ,same voltage is applied to both wires. Since P=V2/R. therefore more heat is
produced in copper wire which makes it glow first.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
A set of n identical resistors, each of resistance R ohm, when connected in series have an
effective resistance X ohm and when the resistors are connected in parallel, their effective
resistance is Y ohm. Find the relation between R, X and Y.
6.
State kirchhoff’s laws for an electrical network.
Ans) (1) junction Rule:the sum of incoming current at a point (junction) is equal to the
outgoing current at that point(junction).
(2) loop rule: the sum of all potential difference across each element of a closed loop is
zero.
7.
Draw the diagram of Wheatstone bridge. Why does no current flow through the galvanometer
when the bridge is balanced?
Ans. In the circuit shown(fig.3) R is the unknown resistance. In the balance condition of the
bridge, potential at A becomes equal to potential at B that is there is no potential
difference across the galvanometer hence no current flows through it (Ig =0 )
Here I1R=I2P and I1S=I2Q
Therefore P/Q=R/S or R=PS/Q
The above condition is possible when Ig becomes zero (fig.)
8.
Explain the principle on which the working of a potentiometer is based. Why is the use of a
potentiometer preferred over that of a voltmeter for measurement of emf of a cell?
Ans.
A potentiometer works on the principle that potential difference across any part of a
uniform wire is directly proportional to the length of that part when steady current
flows through the wire.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Applying ohm’s law V=IR. But R=ρ(l/A) where ρ is the resistivity, l is length & A is
the area of the wire.
Therefore
Where
V=I(ρ l/A)=(I ρ/A)=k l
k=l ρ/A, a constant
=> V l provided I, ρ, & A are constant
This method is null method ,in which no current is drawn by the galvanometer from
the source so emf measured is independent of internal resistance of source as such
measurement made with the help of a potentiometer is accurate as compared to a
simple voltmeter.
9.
Define ‘relaxation time’ of electrons in a conductor. Explain how it varies with increase in
temperature of a conductor. State the relation between resistivity and relaxation time.
Ans. . relaxation time or mean free time is the small interval of time between two
successive collisions between electron and ions in the lattice of a conductor.
relaxation time =mean free path /r.m.s. velocity of electrons=λ/Cr.m.s.
Cr.m.s. increases with increases in the temperature so relaxation decreases with increase in
temperature .
Relation between resistivity ρ and τ is given by the following expression:
Q.Why resistor of semiconductor decreses with the inc of temp. ρ =m/ne2 τ
Where n is the electron density ,m is the mass of electron and e is the charge on an electron.
10.
Write down simple relationship to show the variation of resistance with temperature. Define
temperature coefficient of resistance.
Ans. R=R0 (1+αT)
Where R is the resistance at temperature T and R0 is the resistance at 0˚C. here α is the
temperature
coefficient of resistance.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
It is defined as the increase in resistance per unit resistance of the conductor at 0˚C per
degree increase of
temperature i.e. α=(R-R0)/R0T.
11.
Draw a circuit diagram of a metre bridge to determine the resistance of a wire. Give the
formula used.
Ans. let X be the unknown resistance to be measured fig. obtain in null position by adjusting
R and
the position of the jockey. In the balanced condition, X =(100-l1)R/l1.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
J. Numerical Problems(formula based) carrying 2 marks
1.
A wire of resistance 4R is bent in the form of a circle. What is the effective resistance between
the ends of diameter?
Ans. Equivalent resistance is given by Req =R1R2/R1+R2=2R X 2R/2R+2R=R
A
2.
B
A 10Ω thick wire is stretched so that its length become 3 times. Assuming that there is no
change in its density on stretching, calculate the resistance of the new wire.
Ans.
Area of cross section becomes 1/3 of the original value when becomes 3 times.
Therefore original resistance, R= l/A
New resistance, R’= 3 l/A/3=9 l/A=9R=9 X 10=90 Ω
3.
Two wires A&B are of same metal, have the same area of cross section & have there lengths
in the ratio 2:1. What will be the ratio of current flowing through them respectively when the
same potential difference is applied across length of each of them?
Ans.
Resistance R  length therefore RA/RB = 2/1
Current is inversely proportional to resistance therefore IA/IB=1/2
4.
A student obtains resistance 3,4,12&16 ohms using only two metallic resistance wires either
separately or joined together. What is value of resistance of each of these wires?
Ans.
The lowest value is 3Ω &the highest is 16.
Let 4Ω & 12Ω be the resistances of individual wires then in series combination,
equivalent resistance is given by Rs=4 +12= 16Ω
Similarly in parallel combination , Rp=12 X 4/12+4=48/16=3Ω.
5.
Calculate conductivity of the material of a conductor of length 3 meter, area of cross section
0.02mm2 having a resistance of 2 ohm.
Ans.
Electrical conductivity is given by =1/=l/RA
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Here l = 3 meter , A =0.02mm2=0.02 X 10-6m2 & 2 ohm therefore
 = 3/2(0.02 X 10-6)= 3 X 108/4= 7.5 X 107 S/m
6.
A dry cell of emf 1.6 volt & internal resistance 0.10 ohm is connected to a resistor of a
resistance R ohm. The current drawn from the cell is 2A .
i) what is the voltage drop across R?
ii) what is the energy dissipation in the resistor?
Ans.
Using, E= V+Ir,
where E= 1.6V, r=0.1 ohm & I=2A
i) Voltage drop across R i.e V= E – Ir =1.6 – 2 X 0.1 = 1.4 v
ii) Energy dissipated in a unit time = VI =1.4 X 2 =2.8 j/s
7.
A current of 5A flows through an electric press of resistance 11 ohm . calculate the energy
consumed by press in 5 min.
Ans.
8.
Electric energy is consumed is given by E= I2Rt =52 X 11 X 300 = 8.25 X 104j
Two resistor of 2 ohm & 4 ohm are connected in parallel to a constant DC voltage, in which
resistor is more heat produced.
Ans.
Heat produced in 1s , H = V2/R therefore H1=V2/R1 & H2=V2/R2
or
H1/ H2 = V2/R1 X R2/ V2 = R2/R1 =4/2= 2
Heat produced is more in 2 ohm resistor.
9.
Two wires A & B of the same material & having same lengths, have their cross sectional area
in the ratio 1:4. What would be the ratio of heat produced in these wires when same voltage is
applied across each?
Ans.
Using R =  l/A & H = V2 t/R
R1 =  l/A , R2 =  l/4A & H1 = V2 tA/ l
H2 = V2 t X 4A/ l therefore H1/ H2 = 1/4
10.
Two electric bulbs A & B are marked 220V , 40 W & 220V , 60 W respectively . which of
these bulbs has a higher resistance?
Ans . using P = V2/R i.e R = V2/P
For 40W lamp, R= 220 X 220 /40 = 1210 ohm
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
For 60W lamp, R’= 220 X 220 /60 = 807 ohm
So 40W lamp has higher resistance
Numerical Problems(formula based) carrying 3 marks
1.
Three identical cells each of emf 2V and internal resistance 0.2 ohm are connected in
series to an external resistance of 7.4 ohm. Find the current in the circuit.
Ans . I = E / ( R + r ) = 6 / ( 7.4 + 0.6 )
= 6/8
I =( 3 / 4) A
(Because, effective emf, E = 2 + 2 + 2 = 6 V
0.2 + 0.2 + 0.2 = 0.6 )
2.
&
effective internal resistance
r=
Show that one ampere of current is equivalent to a flow of 6.25 X 10 18 elementary
charges per second.
Ans . I = q / t = ne / t
or
n = It / e = ( 1 X 1) / 1.6 X 10-19
n = 6.25 X 1018 electrons.
3.
A pd of 30 V is applied across a colour coded carbon resistor with rings of blue, black
and yellow colours. What is the current through the resistor ?
Ans . From the colour code R = 60 X 104 
V = 30 V
I = V / R = 30 /( 60 X 104 )
I = 5 X 10-5 A
4.
Calculate the electrical conductivity of the material of a conductor of length 3m, cross
section 0.02 mm2 having a resistance of 2 .
Ans .
 = 1 /  and  = RA / l

 = l / RA = 3 / 2 x 2 x 10-8 = 7.5 x 107
( 1 mm2 = 10-6 m2 )
 = 7.5 x 107 m -1
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
5.
What is the equivalent resistance in the following network between ‘ P ‘ and ‘ Q ‘ ?
Ans .
R1 and R2 are in series
R12 = 2 + 2 = 4 
R3 and R12 are in parallel
1 / R = ( 1 / 2)+ 1 / 4 ) = 3 / 4
R=4/3 Ω
6.
A storage cell of emf 10V , internal resistance 1 is being charged by a 50V dc source
using a series resistor of 19, calculate (i) current in the circuit and (ii) terminal voltage
across the cell during charging.
Ans .
(i)
I=(V–E)/(R+r)
V = 50V, E = 10V , r = 1, R = 19
I = ( 50 – 10 ) / ( 1 + 19 ) = 2A
(ii)
Potential during charging
V = E + Ir
= 10 + 2 x 1
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
= 12 volt
7.
In a potentiometer for a cell of emf 1.5V, the balance point is obtained at 42.0 cm length.
If the cell is replaced by another cell for which balance point is obtained at 63.0 cm
length. Find the emf of second cell.
Ans .
E1 = 1.5V
l1 = 42.0 cm
E1 / E2 = l1 / l2
E2 = ?

l2 = 63.0 cm
E2 = E1l2 / l1 = ( 1.5 x 63 / 42 )
E2 = 2.25V
8.
If no current flows through 3 resistance find resistance ‘ X ’ and current drawn from
5V cell.
Ans .
The circuit is equivalent to a balanced Wheat stone’s bridge
2 / 8 = X / 16
For the current drawn from cell neglecting 3 resistance
R = 10 x 20 / (10 + 20) = 200 / 30 = 20 / 3 
I = V / R = 5 / ( 20 / 3) =3 / 4 A
9.
The length of wire AB is 400 cm, where should the free end of the galvanometer be put
on AB so that galvanometer shows zero deflection.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Ans . Let the balancing length AJ = l, then JB = ( 400 – l ).
Now for the balanced bridge
8 / 12 = l / ( 400 – l )
or, l = 160 cm.
10.
Find the equivalent between ‘ P ‘ and ’ Q ‘ if each resistance is 10.
Ans . The arrangement is equivalent to as shown.
It is a balanced Wheat stone’s bridge.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
R12 = 10 + 10  = 20
R45 = 10 + 10 = 20
R = ( 20 x 20 ) / ( 20 + 20 ) = 10.
K. Long Answer type Question ( 5 marks each )
1.
Define resistivity of a conductor, How does it vary with temperature. Derive an
expression for the resistivity of a wire in terms of material parameter number density of
free electrons and collision time.
Ans . Expression for resistivity
Let there be a conductor of length ‘l’ across which a pd of ‘V’ volts is applied. If a ‘A’ be the
cross sectional area of the conductor and ‘n’ be the number of electrons per unit volume, then
drift velocity of electrons
Vd = e / m( V / l ) τ
(1)
τ→ relaxation time
also, I = neAVd
Putting Eq. (1) in (2)
I = neA e / m( V / l ) τ
I = ( ne2 AV / ml ) τ
Or, V / I = m / ne2 τ ( l / A )
Or, R = m / ne2 τ ( l / A )
Or, R =  ( l / A )
 → Specific resistance or resistivity.
Where
 = m / ne2 τ
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
(2)
2.
State Ohm’s law and deduce it from the knowledge of drift velocity of free electrons in a
conductor carrying current.
Ans .
I = neAVd
(1)
A → Area of cross section.
Vd = e / m ( V / l ) τ
(2)
V → pd across conductor
l → length of conductor
τ → relaxation time
Putting Eq. (2) in (1)
I = neA e / m ( V / l ) τ
I =( ne2 τ / m ) V ( A / l)
For a given conductor m, n, e, τ, A and l are constants

Or
I α V
V / I = Constant ( R )
R → resistance of the conductor.
3.
Explain with the help if a circuit diagram, the use of potentiometer for determination of
internal resistance of a primary cell. Derive the necessary mathematical expression.
Ans. The circuit is as shown in the figure . After checking the connections close key K1 only and
locate point J on the wire such that galvanometer shows no deflection . now
E=Kl1
……………………………………………(i)
Now press K2 such that current is drown by resistance R from the cell and its emf falls to
voltage V then
V=Kl2
……………………………………………………………………………..(ii)
Dividing eqn (i) & (ii)
E/V=l1/l2
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
But internal resistance of a cell is given by
r=(E/V-1)R=(l1/l2-1)R
thus r is determined.
4.
Explain the principle on which the working of a potentiometer is based. Why is the use of a
potentiometer preferred over that of a voltmeter for measurement of emf of a cell?
Ans. A potentiometer works on the principle that potential difference across any part of a
uniform wire is directly proportional to the length of that part when steady current flows
through the wire.
Applying ohm’s law V=IR. But R=ρ(l/A) where ρ is the resistivity, l is length & A is
the area of the wire.
Therefore
Where
V=I(ρ l/A)=(I ρ/A)=k l
k=l ρ/A, a constant
=> V l provided I, ρ, & A are constant
This method is null method ,in which no current is drawn by the galvanometer from the
source so emf measured is independent of internal resistance of source as such measurement
made with the help of a potentiometer is accurate as compared to a simple voltmeter.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
L. Question for practice ( both 2 marks & 3 marks)
1.
A student has two wires of iron and copper of equal length and diameter. He first joins the two
wires in series and passes electric current through the combination which increases gradually.
After that he joins the two wires in parallel and repeat the process of passing current. Which
wire will glow first in each case and why?
2.
A set of n identical resistors, each of resistance R ohm, when connected in series have an
effective resistance X ohm and when the resistors are connected in parallel, their effective
resistance is Y ohm. Find the relation between R, X and Y.
3.
n identical cells of emf. E and internal resistance r are connected in series to a resistor R.
i)Deduce and expression for the internal resistance r of one cell in terms of the current I
flowing through it. ii)How does the internal resistance of the cell vary with temperature?
4.
When a battery of emf E and internal resistance r is connected to a resistance R, a current I
flows through it. Derive a relation between E,I,r and R.
5.
Draw the diagram of Wheatstone bridge. Why does no current flow through the galvanometer
when the bridge is balanced?
6.
Draw a circuit diagram of a metre bridge to compare two resistance. Explain the principle of
the experiment, and give the formula used.
7.
How will you use a metre bridge to measure an unknown resistance? Draw the necessary
circuit diagram. Explain the principle of the experiment. Give the formula used.
8.
Draw a circuit diagram of a metre bridge to compare two resistances. Write the formula used.
Why is this method suitable only for two resistances of the same order of magnitude?
9.
Explain the principle on which the working of a potentiometer is based. Why is the use of a
potentiometer preferred over that of a voltmeter for measurement of emf of a cell?
10.
Explain whit the help if a circuit diagram, the use of potentiometer for determination of
internal resistance of a primary cell. Derive the necessary mathematical expression.
11.
With the help of a circuit diagram, explain how would you compare the emf of two primary
cells using a potentiometer. State the formula used.
Length of a conductor is increased to three times. What is its effect on (i)drift speed of
electrons (ii)resistance and (iii)resistivity of potential across the conductor is kept . constant.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
12.
Define ‘relaxation time’ of electrons in a conductor. Explain how it varies with increase in
temperature of a conductor. State the relation between resistivity and relaxation time.
13.
Write down simple relationship to show the variation of resistance with temperature. Define
temperature coefficient of resistance.
14.
Deduce dimensional formula for potential difference.
15.
Deduce Ohm’s law using the concept of drift velocity.
16.
Given n resistors each of resistance R. how will you comment them to get the maximum and
minimum effective resistance? What is the ratio of the maximum to the minimum resistance?
17.
What happens in a meter bridge if the galvanometer and cell are interchanged at the balance of
the bridge? Would the galvanometer show any current.
18.
Would the potentiometer circuit work for determining extremely small e.m.f.s(a few mV)?
19.
Derive an expression for current in the circuit having mixed grouping of cells. What is the
condition for maximum current in the circuit?
M. Numerical problems for practice, minimum 20 to 25
1.
The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5
steam point is 5.23 Ω. When the thermometer is inserted in a hot bath, the resistance of the
platinum
wire is 5.795 Ω. Calculate the temperature of the bath.
Ans. 345.65 ˚C
2.
A network of resistors is connected to a 16 V battery with internal resistance of 1 Ω, as shown
in Fig.
(a) Compute the equivalent resistance of the network. (b) Obtain the current in each resistor.
(c)obtain
the voltage drops VAB, VBC and VCD
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Ans. a) 7Ω
b)I=2A, I1=I2=1A, I3=2/3A, I4=4/3A
c) VAB=4V, VBC=2V, VCD=8V
3.
A battery of 10 V and negligible internal resistance is connected across the diagonally
opposite corners of a
cubical network consisting of 12
equivalent resistance of
the network and the current along each edge of the cube.
Ans. Req=5/6R, For R=1 Req=5/6A , I=4A
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
4.
Determine the current in each branch of the network shown in Fig..
Ans. I1=2.5A, I2=5/8A, I3=15/8A
AB : 5/8A,CA : 5/2A, DEB : 15/8A,
AD : 15/8A, CD : 0A, BC : 5/2A
5.
The four arms of a Wheatstone bridge (Fig. ) have the following resistances:
Ans. Ig=4.87 mA
6.
7.
In a metre bridge (Fig. ), the null point is found 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. Determine
the values of R and S.
Ans. R=6.86Ω, S=13.5Ω
A resistance of R Ω draws current from a potentiometer. The potentiometer has a total
resistance R0 Ω
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
(Fig. ). A voltage V is supplied to the potentiometer. Derive an expression for the voltage
across R when the
sliding contact is in the middle of the potentiometer.
8.
Ans. V1= 2VR/R0+4R
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4
Ω, what is the maximum current that can be drawn from the battery?
Ans.30A
9.
in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the
battery when the circuit is closed ?
Ans.17Ω, 8.5V
10.
At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the
temperature of the element if the resistance is found to be 117 Ω, given that the temperature
coefficient of the material of the resistor is 1.70 × 10–4 °C–1.
Ans.
1027°C
11.
Determine the temperature coefficient of resistivity of silver.
Ans.
12.
0.0039°C–1
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2
A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature
of the heating element if the room temperature is 27.0 °C? Temperature coefficient of
resistance of nichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1.
Ans.
867°C
13. Determine the current in each branch of the network shown in Fig .
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Ans. Current in branch AB = (4/17)A
In BC = (6/17)A, in CD=(-4/17)A
In AD = (6/17)A, in BD=(-2/17)A
Total current = (10/17)A
14.
(a) n a metre bridge [Fig.], the balance point is found to be at 39.5 cm from the end A, when
the resistor Y is of 12.5 Ω.
(a) etermine the resistance of X. Why are the connections between resistors in a Wheatstone
or meter bridge ade of thick copper strips?
(b) etermine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the
bridge? Would the galvanometer show any current?
Ans. a) X=8.2Ω; to minimize resistance of the connection which are not accounted for in the
bridge formula.
b) 60.5 cm from A
c) the galvanometer will show no current
15.
dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during
charging? What is the purpose of having a series resistor in the charging circuit?
Ans. 11.5V , the series resistor limits the current drawn from the external source. In its
absence , the current will be dangerously High.
16.
17.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length
of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what
is the emf of the second cell?
Ans. 2.25V
Two wires of equal length, one of aluminium and the other of copper have the same
resistance. Which of the two wires is lighter? Hence explain why aluminium wires are
preferred for overhead power cables.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
( Al = 2.63 × 10–
8.9.)
Ans.
18.
Cu
= 1.72 × 10–
the mass ratio of copper to alluminium wire is (1.72/2.63) × (8.9/2.7) = 2.2 , since
alluminium is lighter , it preferred for long suspension of cables.
(a) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent
resistance of
(i)
b) Determine the equivalent resistance of networks shown in Fig.
Ans. a) i) in series , ii) all in parallel, n2
b) i) join 1 ohm , 2 ohm in parallel & the combination in series with 3 ohm
ii) parallel combination of 2 ohm & 3 ohm in series with 1 ohm
iii) all in series
iv) all in parallel
c) i) (16/3) ohm, ii) 5R
19.
Figure shows a potentiometer with a cell of 2.0 V and internal resistance
a potential drop across the resistor wire AB. A standard cell which maintains a constant emf
of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length
of the wire. To ensure very low currents drawn from the standard cell, a very high resistance
cell is then replaced by a cell of unknown emf 
turns out to be at 82.3 cm length of the wire.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
(a) What is the value ε?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had
an emf of 1.0V instead of 2.0V?
(f ) Would the circuit work well for determining an extremely small emf, say of the order of a
few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the
circuit?
Ans. a) ε = 1.25V
b) to reduce current through galvanometer when the moveable contact is far from the
balance point
c) No
d) No
e) No, if ε is greater than the emf of the driver cell of the potentiometer, there will be
no balance point on the wire AB
f) No, the circuit is modified by putting a suitable resistor R in series with the wire
AB.
20.
Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5
V cell. The balance point of the cell in open cir
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.
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
Ans. 1.7Ω

B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR
B.SESHA SAI, PGT, (PHY),K.V.NO1 ,BBSR