1a.current Electricity Synopsis (1 32)
1a.current Electricity Synopsis (1 32)
1a.current Electricity Synopsis (1 32)
CURRENT ELECTRICITY
SYNOPSIS v is linear velocity of the charge q
r is radius of the circular path
Strength of Electric Current 5. If in a discharge tube n1 protons are moving from
The strength of electric current is defined as rate of
flow of charge through any cross section of a left to right in t seconds and n2 electrons are moving
conductor. simultaneously from right to left in t seconds, then
1. The instantaneous current is defined by the the net current in any crossection of the discharge
equation, tube is
Q dQ (n1 n2 )e
I Lt I (from left to right)
t 0 t dt t
q here e is the magnitude of charge of electron (or)
Average current i proton.
t
Ampere : If one coulomb of charge passes DRIFT VELOCITY : Drift velocity is the
through a cross-section of the conductor per average velocity acquired by free electrons
second then the current is one ampere. inside a metal by the application of an electric
1coulom b field which results in current.
1 ampere = J I
1sec ond Drift velocity vd
current is a scalar quantity. ne Ane
Applications on electric current where, J = I/A is current density
1. If the current is varying with time t, then the charge n is number of free electrons per unit volume
e is charge of electron
flowing in a time interval from t1 to t 2 is
The drift velocity is related to relaxation time
t2
is
q Idt
t1 eE
vd
2. If n particles, each having a charge q, pass m
through a given cross sectional area in time t, Note : 1.The drift velocity of electrons is of the order
nq of 104 ms 1 .
then average current is i =
t 2. Greater the electric field, greater will be the
3. If a point charge q is revolving in a circle of drift velocity vd E
radius r with speed v then its time period is
3. The direction of drift velocity for electrons in a
metal is opposite to that of electric field applied
V
r
q
E
T (2 r / v) W.E-1: In a hydrogen atom, electron moves in an
orbit of radius 5 × 10-11 m with a speed of 2.2 ×
106 m/s. Calculate the equivalent current.
4. The average current associated with this
revolving charge is v
Sol: Current i f .e .e
q vq 2 r
I fq q
T 2 2 r 2.2 106
Where f is the frequency of revolution in Hz. = × 1.6 × 10–19
2 5 1011
is the angular frequency in rad/sec = 1.12 × 10–3 amp = 1.12 mA.
NARAYANA GROUP 1
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
1
m = mass of electron V
V
T
= relaxation time. 2
2
2 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
a) Diode b) 1
conductance, G .
R
V V
The S.I unit of conductance is mho or siemen or
I
ohm-1.
Resistivity: As we know, that the resistance of
the conductor is directly proportional to its
c) length and inversely proportional to its area of
cross section, we can write
V
l l
Resistance-Definiton : The resistance of a R R
A A
conductor is defined as the ratio of the potential where is specific resistance or resisitivity
difference ‘V’ across the condutor to the current of the material of the conductor.
‘i’ flowing through the conductor.
NARAYANA GROUP 3
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
Note: h
1. Resistitivity is the specific property of a material Resistance across EF, REF
but Resistance is the bulk property of a l b
conductor. If l b h , then
2. Resistivity is independent of dimensions of the l h
conductor such as length, area of the cross Rmax Rmin
bh l b
section. 5. If a wire of resistance R is stretched to ‘n’ times
3. Resistivity depends on the nature of the material its original length, its resistance becomes n2 R.
of the conductor, temperature and impurities. 6. If a wire of resistance R is stretched until its
4. Resistivity of any alloy is more than resistivity
of its constituent elements. 1
radius becomes th of its original radius then
n
i) R alloys R conductors ii) metals alloys
its resistance becomes n4R.
Special Cases : 7. When a wire is stretched to increase its length
1. The alternate forms of resistance is by x% (where x is very small) its resistance
increases by 2x %.
l2 l 2 d V m 8. When a wire is stretched to increase its length
R 2
V m A d A2 by x% (where x is large) its resistance increases
Where d is density of material of conductor 2
V is volume of the conductor 2x x
by 100 .
m is mass of the conductor
2. If a conductor is streched or elongated or drawn 9. When a wire is stretched to reduce its radius
or twisted, then the volume of the conductor is byx% (where x is very small), its resistance
constant. Hence increases by 4x%.
l 2 W.E-4: A rectangular block has dimensions 5 cm
a. R R l 2 × 5 cm × 10 cm. Calculate the resistance
V
measured between (a) two square ends and
V 1 1 (b) the opposite rectangular ends. Specific
b. R 2 R 2 4
A A r resistance of the material is 3.5 105 m .
l2
c. Interms of mass of the wire R Sol: a) Resistance between two square ends R1
m A
m m 3.5 10 5 10 10 2
and R R1 4
1.4 103
A2 r 4 5 5 10
3. For small changes in the length or radius during
the stretching
R l R A r
2 ; 2 4
R l R A r 5 cm
4. In case of a cuboid of dimensions l b h is
F 5 cm 10 cm
h
C
b A
F
B b) Resistance between the opposite rectangular
E
D ends R2
E
A
l
3.5 10 5 5 10 2
l R2 1.4 10 4
Resistance across AB, RAB 5 10 10 4
b h
Conductivity: Conductivity is the measure of
b the ability of a material to conduct electric
Resistance across CD, RCD
lh current through it. It is reciprocal of resistivity.
4 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
at temperature t 2 ,then
du
on
rc
2 1 1 t 2 t1
fo
NARAYANA GROUP 5
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
T1
6 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
W.E-12: There are two concentric spheres of W.E-15: How many number of turns of nichrome
radius a and b respectively. If the space wire of specific resistance 106 m and
between them is filled with medium of
diameter 2mm that should be wound on a
resistivity , then the resistance of the
cylinder of diameter 5cm to obtain a resistance
intergap between the two spheres will be
of 40 ?
(Assume b > a)
Sol: Consider a concentric spherical shell of radius Sol: If R is the radius of the cylinder
x and thickness dx, its resistance is r is the radius of the wire
N is the number of turns
dx
dR, dR =
4 x 2 ρ r 2 p R N
then R ' R '
Total resistance A pr2
106 22.5102N
b b
dx 1 1
R = dR R 4 x 2 4 a b 40 = N 800
a a 1106
W.E-13: A hollow copper cylinder is of inner Thermistor: A thermistor is a heat sensitive
radius 4cm and outer radius 5cm. Now and non-ohmic device.
hollow portion is completely filled with This is made of semiconductor compounds as
suitable copper wires. Find percentage oxides of Ni, Fe, Co etc.
change in its electric resistance. This will have high +ve (or) -ve temperature
Sol: A hollow cylidner of inner radius ‘r’ and outer coefficient of resistance.
radius ‘R’ has specific resistance ' ' . If its Thermistor with -ve ‘ ’ are used as resistance
length is ' l ' then its resistance themometers which can measure low temperature
of order
–3
of 10K and small changes of in the order
of 10 K.
l Having -ve , these are widely used in
measuring the rate of energy flow in micro wave
R2 r2 beam.
l l k
r R R1 = 52 42 9 9 Thermistor can also be used to serve as
thermostat.
Final Resistance Resistor Colour codes
l l k Colour Number Multiplier Tolerance(%)
R2 = 52 25 25 Black 0 × 10°
Brown 1 × 101
Percentage of change = R2 R1 100
R1 Red 2 × 102
Orange 3 × 103
k k
Yellow 4 × 104
25 9 100 64%
k Green 5 × 105
9 Blue 6 × 106
W.E-14: If resistivity of the material of a Violet 7 × 107
conductor of uniform area of cross-section Gray 8 × 108
varies along its length as 0 1 x . Find White 9 × 109 –
then the resistance of the conductor if its –1
Gold – ×10 5%
lengths is ‘L’ and area of cross-section is ‘A’ Silver – ×10–2 10%
0 1 2 No clour – 20%
A L 2 L digit 2 tolerence
digit 1 multiplier
L
dx dx
Sol: dR = 0 1 x ; R dR ]
A A 0 wire
lead
NARAYANA GROUP 7
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
1 21 2
R1
2 R s R s2 4R s R p If A1 = A2 then = .
1 2
1
2
R 2 R s R s2 4R s R p and conductivity 1
2
2
.
6. If a uniform wire of resistance R is, stretched 10. If ‘n’ wires each of resistance ‘R’ are connected
to ‘m’ times its initial length and bent into a to form a closed polygon, equivalent resistance
regular polygon of ‘n’ sides
a) Resistance of the wire after stretching is n 1
across two adjacent corners is Reff R
2 ' 2 n
R1 m R ( R l )
W.E-17: For a circuit shown in Fig find the value
m2 R of resistance R 2 and current I 2 flowing
b) Resistance of each side R2 through R2
n
c) Resistance across diagonally opposite points I=10 P
I1 I2
n +
2 R2 - 50V R1 10 R2
m2 R
R0 R0
2 4 Q
Sol: If equivalent resistance of parallel
d) Resistance across one side combination of R1 and R2 is R, then
R1R2 10R2
(n 1) (n 1)m 2 R R
R3 R2 R1 R2 10 R2
n n2
7. 12 wires each of resistance ‘r’ are connected to V
According to Ohm’s law, R
form a cube. Effective resistance across I
5r 50 10R2
a) Diagonally opposite corners = . R 5 5 R 10 .
2
10 10 R2
6
The current is equally divided into R1 and R2.
3r
b) face diagonal . Hence I2 = 5A.
4
W.E-18: Two wires of equal diameters of
7r
c) two adjacent corners . resistivities 1 and 2 and length x1 and x2
12
respectively are joined in series. Find the
8. If two wires of resistivities 1 and 2 , lengths equivalent resistivity of the combination.
l1 and l2 are connected in series, the equivalent
Sol: Resistance, R1 11 ;R2 2 2
resistivity A1 A2
1 l1 2 l 2 1 x1 , 2 x2
. As the wires are of equal diameters A1 A2 A .
l1 l2
1 2 x1 x x
R1 , R2 2 ; R
If l1 l 2 then . A A2 A
2
where x = x1 + x2; R = R1 + R2
21 2
If l1 l 2 then conductivity = . x 1x1 2 x2
1 2 ; x 1x1 2 x2
A A A
9. If two wires of resistivities 1 and 2, Areas (x1 x2 ) 1x1 2 x2 [ x x1 x2 ]
of cross section A1 and A2 are connected in
parallel, the equivalent resistivity x1 x 2
1 2 A1 A2
x 2 x2
1 1 1
also s1 s2
= x1 x2
1 A2 2 A1
. s x1 x 2
NARAYANA GROUP 9
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
W.E-19: Find equivalent resistance of the R R
network in Fig. between points (i) A and B R1 . r ; Resistance of section PTQ
2 r 2
and (ii) A and C.
10 Rr 2
R2 ; R 2
A B C
2 r R2
30 7.5 2
As R1 and R2 are in parallel
R1 R2 R
Sol: (i)The 10 and 30 resistors are connected in
So, Req 2 2
parallel between points A and B. The equivalent R1 R2 4
resistance between A and B is W.E-22: Determine the current drawn from a 12V
10 30 supply with internal resistance 0.5 Ω . by the
R1 ohm = 7.5
10 30 infinite network shown in Fig. Each resistor
(ii) The resistance R1 is connected in series with has 1 Ω . resistance.
resistor of 7.5 , hence the equivalent resistance 1 1 1
between points A and C is, R2=(R1+7.5) ohm =
(7.5+7.5) ohm = 15 .
12V, 0.5 1 1
W.E-20: Find potential difference between points
A and B of the network shown in Fig. and
distribution of given main current through 1 1 1
different resistors. Sol: First calculate net resistance of network
1
4 6
I=2.7A
8 X 1 x
A 9 B
I
1
Sol: Between points A and B resistors of 4 , 6 x
and 8 resistances are in series and these are x=2+ ; 2 ;
in parallel to 9 resistor.. x+1 x 2x 2 0
Equivalent resistance of series combinaiton is on solving, x 1 3 2.73
R1 = (4 + 6 + 8 ) ohm = 18 Total resistance = 2.73 + 0.5 = 3.23 Ω
If equivalent resistance between A and B is
R = 9 × 18 / (9 + 18) ohm = 6 12
I= =3.73A
Potential difference between A and B is 3.23
V = IR = 2.7 × 6V = 16.2V JOULE’S LAW: According to Joule’s law,
Current through 9 resistor = 16.2/9=1.8A the current passing through a conductor produces
Current through 4 ,6 and 8 resistors = heat.
2.7 – 1.8 = 0.9A. W = vit
W.E-21: P and Q are two points on a uniform Now, work done, W = (iR) i t
ring of resistance R. The equivalent resistance 2
between P and Q is W = i2 R t = v t = v i t
P R
This work is converted into energy in the
conductor.
O
Q Thermal energy produced, Q = i2 Rt in Joules
10 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
Electrical Energy: W.E-23: A fuse wire with radius of 0.2mm blows
The electric energy consumed in a circuit is off with a current of 5 Amp. The fuse wire of
defined as the total workdone in maintaing the same material, but of radius 0.3mm will blow
current in an electric circuit for a given time. off with a current of
V2t 3 5 3
Electrical Energy = Vit = Pt = i2Rt
R 1) 5 Amp 2) Amp
2 2
S.I. unit of electric energy is joule
1 K.W.H. = 36 × 105 J 27
3) 5 Amp 4) 5 Amp
Electrical Power: 8
The rate at which work is done in maintaining 3 3
the current in electric circuit. Electrical power i1 r1 2 0.2 2
i 2 r2
Sol: i2 r 3 ;
W V2 0.3
P Vi i2R watt (or) joule / sec
t R
Heat energy produced due to the electric current 27
i2 5 Amp
8
W Pt E it i 2 Rt E 2t If radiation losses are neglected, due to
H=
J J J J RJ heating effect of current the temperature of
H mst fuse wire will increase continuously, and it
Where s 4200J /Kg0C melt in time ‘t’ such that
where J is mechanical equivalent of heat. I 2 Rt
Fuse wire: A fuse wire generally prepared H ms ; ms( mp r )
J
from tin - lead alloy (63% tin + 37% lead). It
should have high resistivity, low melting point. 2 r 4 s mp r J
Let R be the resistance of fuse wire. t 2
;t r 4
I
rL i.e., in absence of radiation lossess, the time
We know that R 2
pr in which fuse will melt is also independent on
(L and r denote length and radius) length and varies with radius as r4.
The heat produced in the fuse wire is Note :
a) If resistances are connected in series , i.e.., I is
i2rL same
H i2R
pr 2 P R with V a R [ as V IR ]
If H0 is heat loss per unit surface area of the i.e.., in series potential difference and power
fuse wire, then heat radiated per second is = consumed will be more in larger resistance.
H 0 2rL At thermal equilibrium, However, if resistances are connected in
parallel, i.e.., V is same
i2 r2 L i2 r 1 1
H0 2prL (or) H0 Pa with I a [ as V IR ]
pr 2 2p2r 3 R R
According to Newton’s law of cooling. i.e.., in parallel current and power consumed
H 0 K will be more in smaller resistance. This in turn
implies that more power is consumed in larger
Where is the increase in temperature of fuse
resistance if reistances are in series and in
wire and K is a constant. smaller reistance if reistances are in parallel.
i 2r b) A reistance R under a potential difference V
q dissipates power.
2p2r3K
Here is independent of length L of the fuse P V 2 / R
wire provided i remains constant. So If the resistance is changed from R to (R/n)
For a given material of fuse wire i2 r 3 . keeping V same, the power consumed will be
NARAYANA GROUP 11
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
2
V2 V2 V
P1 n nP P A W
( R / n) R VS
i.e.., if for a given voltage, resistance is changed 2
from R to (R/n) , power consumed changes from so, P 100 1000 250W
P to nP. 200
c) If n equal resistances are connected in series c) The total electrical energy consumed by an
with a voltage source, the power dissipated will electric appliance in a specified time is given
be by,
V2 W1h1
Ps [ as Rs nR ] E kWh
nR 1000
And if the same resistances are connected in
parallel with the same voltage source so, E 1000 (10 30) 300kWh
1000
V2 nV 2 Bulbs connected in Series:
Pp [as Rp ( R / n)]
( R / n) R If Bulbs (or electrical appliances) are connected
Pp in series, the current through each resistance is
So, n 2 i.e.., P n 2 P . same. Then power of the electrical appliance
Ps P S
P R & V R P i2Rt
i.e.., power consumed by n equal resistors in
i.e. In series combination; t he potential
parallel is n 2 times that of power consumed in
difference and power consumed will be more
series if V remains same. in larger resistance.
d) As resistance of a given electric appliance
When the appliances of power P1, P2 , P3 .... are
( e.g.., bulb , heater, geyser or press ) is constant
and is given by, in series, the effective power consumed (P) is
2 1 1 1 1
V VS V W ......... i.e. effective power is
R S s [ as I ] P P1 P2 P3
I (W / VS ) W V less than the power of individual appliance.
Where Vs and W are the voltage and wattage If ‘n’ appliances, each of equal resistance ‘R’
specified on the appliance. So if the applied are connected in series with a voltage source
voltage is different from specified, the ‘ actual V2
power consumption’ will be ‘V’, the power dissipated ‘ Ps ’ will be Ps .
nR
2 2
2 Bulbs connected in parallel:
VA VA
P W [ as R VS ]. If Bulbs (or electrical appliances) are connected
R VS W in parallel, the potential difference across each
W.E-24: A 1 kW heater is meant to operate at 1 1
resistance is same. Then P and I .
200 V. (a) What is its resistance ? (b) How R R
much power will it consume if the line voltage i.e. The current and power consumed will be
drops to 100 V ? (c) How many units of more in smaller resistance.
electrical energy will it consume in a month When the appliances of power P1, P2 , P3 .... are
(of 30 days) if it operates 10 hr daily at the in parallel, the effective power consumed(P) is
specified voltage ?
P P1 P2 P3 .........
Sol:a) The resistance of an electric appliance is given i.e. the effective power of various electrical
VS
2 2
200 40 appliance is more than the power of individual
by , R so, R appliance.
W 1000 If ‘n’ appliances, each of resistance ‘R’ are
b) The ‘ actual power ‘ consumed by an electric connected in parallel with a voltage source ’V’,
appliance is given by , the power dissipated ‘Pp’ will be
12 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
NARAYANA GROUP 13
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
E EV
V V = E – i r ; Hence E V
i ........... (1) Power delivered will be maximum when R = r.
R
E E2
and also i ........... (2) So Pmax
R r 4r
This statement in generalized form is called
iR + ir = E, V +ir = E, V = E – ir ‘maximum power transfer theorem’
Lost volts: It is the difference between emf 2
and P.D. of a cell It is used in driving the current Pmax = E /4r
between terminals of the cell. P
Lost volts E - V = i r
Note: Formulae related with cells
E V R=r
i ..................(A)
r
R
NARAYANA GROUP 15
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
Here the % of energy lost and energy useful are W.E-33: When a battery is connected to the
each equal to 50% resistance of 10 the current in the circuit is
Back EMF: When current flows through the 0.12A. The same battery gives 0.07A current
electrolyte solution, electrolysis takes place with with 20 . Calculate e.m.f. and internal
a layer of hydrogen and this hinders the flow of resistance of the battery.
current. In the neighbourhood of both electrodes, Sol: We know that E Ir IR
the concentrations of ions get altered. This I2 R2 I1R1
opposing EMF is called back EMF and the I1r I1R1 I2 r I2 R2 ; r
I1 I2
phenomenon is called Electrolytic polarisation.
To reduce back emf manganese dioxide (or) 0.07 20 0.12 10 1.4 1.2 0.2
r 4
potassium dichromate is added to electrolyte of 0.12 0.07 0.05 0.05
cell. Internal resistance r 4
W.E-31: When a current drawn from a battery is e. m. f E Ir IR
0.5A, its terminal potential difference is 20V. 0.12 4 0.12 10 0.48 1.2 ; E = 1.68 volt.
And when current drawn from it is 2.0A, the GROUPING OF CELLS
terminal voltage reduces to 16 V. Find out. 1. Electric Cells in Series: When ‘n’ identical
e.m.f and internal resistance of the battery. cells each of EMF ‘E’ and internal resistance
Sol: We know ‘r ’ are connected in series to an external
V = E --- Ir ; I = 0.5 A, V = 20 Volt, we have resistance ‘R’, then
20 = E – 0.5r ....... (i) total emf of the combination = n E
I = 2 A, V = 16 Volt, we have effective internal resistance = n r
16 = E-0.2r ....... (ii)
From eqs (i) and (ii) nE
Current through external resistance i =
2E – r = 40 and E – 2r = 16 R nr
Solving we get E = 21.3V, r = 2.675 .
W.E-32: An ideal battery passes a current of 5A If R << n r then i = = current from one cell
through a resistor. When it is connected to r
another resistance of 10 in parallel, the nE
If R>> n r then i =
current is 6A. Find the resistance of the first R
resistor. If two cells of different emf’s are in series
R1 R2 = 10 E1 E2
Eeq = E1 + E2 ; req = r1 + r2 ; i r r R
1 2
R1
5A E1 E2
Sol: 6A
r1 r2
V
V
Current through R1 in the first case i1 = 5A R
Current in the second case i2 = 6A T.P.D across the first cell V1 = E1 - ir1
Effective resistance in the second case T.P.D across the second cell V2 = E2 - ir2
R1R2 RR If one of the cell is in reverse connection
1 2
R ; V I1R1 and V I2 R R ( E1 > E2 ) then Eeq = E1 - E2
R1 R2 1 2
E1 E2
I1R1 I2
R1R2
I1 I2
R2 req = r1 + r2 ; i r r R
R1 R 2 R1 R2 1 2
E1 E2
10
5 6 5(R1 10) 60
R1 10
r1 r2
5R1 + 50 = 60, 5R1 = 10
10
R1 2 R1 2 .
5 R
16 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
First cell is discharging then V1 = E1 - ir1 W.E-36: In the given circuit as shown below,
Second cell is charging then V2 = E2 + ir2 calculate the magnitude and direction of the
cell having less emf in charging state. current
WRONGLY CONNECTED CELLS A 2 e 2 B
By mistake if ‘m’ cells out of ‘n’ cells are 10V 5V
wrongly connected to the external resistance ‘R’
(a) total emf of the combination = (n – 2m) E D 1 C
(b) total internal resistance = n r
(c) total resistance = R + n r Sol: Effective resistance of the circuit is
n 2mE Reff 2 2 1 5
(d) current through the circuit (i) =
R nr V V
total current in the circuit is i 1 2
W.E-34:Two cells A and B with same e.m.f of 2 V Reff
each and with internal resistances
rA 3.5 and rB 0.5 are connected in series 10 5
i= 1A
with an external resistance R 3 . Find the 5
terminal voltages across the two cells. Since the cell of larger emf decides the direction
Sol: Current through the circuit of flow of current, the direction of current in the
circuit is from A to B through e
22 4
i
R r
3 3.5 0.5
7 W.E-37: A voltmeter resistance 500 is used to
measure the emf of a cell of internal resis-
i) R 3, rA 3.5 , E 2V
tance 4 . The percentage error in the read-
Terminal voltages A, VA E ir ing of the voltmeter will be
4 Sol: V = E – ir
2 3.5 0 volt
7 E ir
ii) rB 0.5 , R 3 , E2V Percentage error 100 100
E E
Terminal voltage at B, VB Eir
E
r
R r
4
2 0.5 1.714 volts. r
100 100
R r
7
W.E-35: Two cells A and B each of 2 V are E
connected in series to an external resistance 4
100 0.8%
R=1 ohm . The internal resistance of A is rA 500 4
=1.9 ohm and B is rB =0.9 ohm. Find the
potential difference between the terminals of ELECTRIC CELLS IN PARALLEL
A. When ‘n’ identical cells each of EMF ‘E’ and
Sol: internal resistance ‘r’ are connected in parallel
voltage to an external resistance ‘R’, then
Total current through the circuit i total emf of the combination = E
Total resist an ce
r
r = 1.9 0.9 effective internal resistance =
n
A B r
total resistance in the circuit = R +
n
current through the external resistance
R =1
E nE
4 4 i= r
A nR r
R
1 1.9 0.9 3.8 . n
potential difference at A, VA ir , r E
If R >> , then i = = current from one cell.
4 n R
2 1.9 = 2 – 2 = 0.
3.8
NARAYANA GROUP 17
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
r nE nr
If R << , then i = Condition for maximum power R and
n r m
If two cells of emf E1 and E2 having internal
E2
resistances r1 and r2 are connected in parallel Pmax mn
to an external resistance ‘R’, then 4r
r1 E1 Condition for maximum current
i1 + - i
R r
E2 + =minimum
- i n m
i2 +r
i 2 R d mR r N
dm N m
0; n m
E 1 r 2 E 2r 1
the effective emf, E = r 1 r2 R r R r
2 0; i.e., (N = n x m)
N m n m
r1 r2 So in case of mixed grouping of cells, current
the effective internal resistance, reff r r
1 2 R r
in the circuit will be maximum when
E n m
Current through the circuit, i r R
eff nE mE
I max
i = i1 + i2 2 R 2r
E1 i R E2 iR Total number of cells = m n
i1 = r and i 2 r KIRCHHOFF’S LAWS
1 2
Potential difference across R, i.e terminal When the circuit is complecated to find current
kirchhoff’s laws are formulated.
ER
potential of the cells is V iR R r (i) Kirchhoff`s First Law (Junction Law or
eff Current law) : It states that the sum of the
When the cell E2 is reversed in polarity then currents flowing into a junction is equal to the
sum of the currents flowing out of the junciton .
we should use - E2 in all the above equations. Or
Mixed Grouping: If n identical cell’s are “The algebraic sum of currents at a junction is
connected in a row and such m rows are zero”.
connected in parallel as then
E,r E,r E,r
R4
1
1 2 n I1 I4
2 R1 R3
A I3
i
m I2 R2
V
NARAYANA GROUP 19
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
VA VB V1 ir1 [ V1 cell is discharging] Sol: Applying Kirchhoff’s first law at the junction B
we have i1 + i2 = i3 ....... (1)
V V2
or VA VB V1 1 r1 Applying Kirchhoff’s second law to loop
r1 r2 ABEFA
– 12 + i2 x 1.5 – i1 x 1 + 8 = 0
V1r2 V2 r1 i1 – 1.5 i2 = – 4. ....... (2)
or VA VB
r1 r2 From loop BCDEB
Equivalent emf of the battery = V – (i2 x 1.5) – (i3 x 9) + 12 = 0
1.5 i2 + 9i3 = 12 ....... (3)
V1r2 V2 r1
V r r
on solving i1 = – 1A
1 2 and i3 = 1 A
(ii) Internal resistance of equivalent battery. r1 and WHEATSTONE BRIDGE
r2 are in parallel. B
I3
1 1 1 rr P Q
or r 1 2 I1 I
r r1 r2 r1 r2 A G C
I2
W.E-39: In the given circuit values are as follows
R S
1 2V, 2 4V, R1 1 and R2 R3 1 . I4
Calculate the Currents through R1, R2 and R3. I D
A B C ( )
E1 E2 E K
R1 R2 R3 Condition for balancing of bridge :
Applying Kirchhoff`s first law at junction B and
D we get I1 – I3 – IG = 0 ; and I2 + IG – I4 = 0
G E D
Applying Kirchhoff’s second law for closed
A -B C
loop ABDA, I1P IGG I2 R 0
i1 E E2
1 Applying Kirchhoff’s second law for closed
R1 R2 R3 i
Sol:
2 loop BCDB , I3Q I4S IGG 0
(i1+i2) The values of P, Q, R, S are adjusted such that
G +F D IG becomes zero. At this stage the bridge is set
Let i1 , i2 are currents across R1 and R3. to be in balance condition.
(i1 + i2) is current across R2. i.e., In balanced condition of bridge IG = 0
Their direction are taken as shown In balanced condition the above equations
From Kirchhoff’s second law for AGFBA loop respectively become
i1R1 (i1 i2 ) R2 E1 0 ; i1 i1 i2 2 I1 = I3 ........(1)
and I2 = I4 ........(2)
2i1 i2 2 (1)
From Kirchhoff’s second law for BCDEB loop I1P I 2 R ...........(3)
i2 R3 (i1 i2 ) R2 E2 0 ; i2 i1 i2 4 I 3Q I 4 S ............(4)
Dividing equation (3) by equation (4)
i2 2i2 4 (2)
Solving equation (1) and (2) we get i1 = 0A, i2 = 2A I1P I 2 R
Thus currents across R1 is 0, while across R3 I 3Q I 4S
and R2 are 2A each. Using eqns (1) and (2) we get
20 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
R X
i J
PQ R S
QR PS A l1 l2 B
H.R G
if QR PS , VB VD current flows from B to D
When the Meter bridge is balanced then
QR PS , VB VD current flows from D to B R l1 l1
QR PS , VB VD Balanced bridge X l2 100 l1
NARAYANA GROUP 21
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
J
E1 E2 xl1 E1 E2 xl2
A B
G E1 E2 l1 E1 l1 l2
E (or) E l l
E1 E2 l2
( )
2 1 2
R1 K1
NARAYANA GROUP 23
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
A + B
+ - E1 1
2 G
1 3
Then iR1 xl1 and A
+ -
K2 Rh
R2 l2 l1 + - ()
i R1 R2 xl2
R1 l1 For the calibration of an ammeter, 1 resistance
To determine thermo emf: coil is specifically used in the secondary circuit
e r
of the potentiometer, because the potential
K Rh
+ - difference across 1 is equal to the current
()
R A following through it i.e V i
A HRB B If the balancing length for the emf
G
+ -
E0
E0 1 2 3
G E0 is l0 then E0 xl0 x l (Process of
0
Cold ice Hot sand standardisation)
The value of thermo-emf in a thermocouple for Let i ' current flows through 1 resistance giving
ordinary temperature difference is very low
potential difference as V ' i ' 1 xl1 where l1
10 6
volt . For this the potential gradient x is the balancing length. so error can be found as
must be also very low 104 V / m . Hence a E
i i i ' i xl1 i 0 l1
high resistance (R) is connected in series with l0
the potentiometer wire in order to reduce current Here i is ammeter reading
in the primary circuit Calibration of voltmeter:
The potential difference across R must be equal Checking the correctness of voltmeter readings
to the emf of standard cell with the help of potentiometer is called
calibration of voltmeter.
E
i.e iR E0 i 0 If l0 is balancing length for E0 the emf of
R
The small thermo emf produced in the standard cell by connecting 1 and 2 of bi-
thermocouple e xl directional key, then x E0 / l0
K1 Rh
iR| iR| I + -
()
x i e
L L
where L = Length of potentiometer wire, A + B
+ - E0 1
=resistance per unit length, l balancing V
2
3
G
length of e and RB
K2
+ -
R| = Resistance of potentiometer wire ()
24 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
The balancing length l1 for unknown potential
difference V ' is given by (closing 2 and 3) C. U. Q
E ELECTRIC CURRENT AND DRIFT
V | xl1 0 l1
l0 VELOCITY
1. If n, e, , m, are representing electron density,,
If the voltmeter reading is V then the error will
charge, relaxation time and mass of an
be V V | which may be +ve, -ve or zero electron respectively then the resistance of
W.E-44:The length of a potentiometer wire is 1m wire of length l and cross sectional area A is
given by
and its resistance is 4 . A current of 5 mA is
flowing in it. An unknown source of e.m.f is ml 2mA ne 2 A
balanced on 40 cm length of this wire, then 1) 2) 3) ne 2 A 4)
ne 2 τA ne 2 2m
find the e.m.f of the source. 2. Among the following dependences of drift
R 5 4 velocity vd on electric field E, Ohm’s Law
Sol: x I I L 1 20 mV
obeyed is
E = 1 x = 0. 40 x 20 = 8 mV
1) vd E 2) vd E 2
W.E-45: A cell of e.m.f 2 volt and internal
resistance 1.5 is connected to the ends of 3) vd E 4) vd constant
1m long wire. The resistance of wire is 3. A steady current is passing through a linear
0.5 / m . Find the value of potential gradient conductor of nonuniform cross-section. The
on the wire. net quantity of charge crossing any cross
section per second is
I R E R 2 0.5 1) independent of area of cross-section
Sol: X =0. 5 V/m
L R r L 0.5 1.5 2) directly proportional to the length of the
W.E-46: In a potentiometer experiment the conductor
balancing length with a cell is 560 cm. When 3) directly proportional to the area of cross
an external resistance of 10 is connected in section.
4) inversely proportional to the area of the
parallel to the cell, the balancing length conductor
changes by 60 cm. Find the internal resistance 4. Given a current carrying wire of non-uniform
of the cell. cross section. Which of the following quantity
Sol: Balancing length 1 560 cm or quantities are constant throughout the
Change in balancing length (1 2 ) 60 cm length of the wire?
560 2 60 1) current, electric field and drift speed
2) drift speed only
2 500 cm
3) current and drift speed
4) current only
rR 1 2 r10
60 6
1.2 .
2 500 5 5. When electric field ( E ) is applied on the ends
W.E-47: In a potentiometer experiment when a of a conductor, the free electrons starts
battery of e.m.f. 2V is included in the moving in direction
secondary circuit, the balance point is 500cm. 1) similar to E 2) Opposite to E
Find the balancing length of the same end
3) Perpendicular to E 4) Cannot be predicted
when a cadimum cell of e.m.f. 1.018V is 6. The drift speed of an electron in a metal is of
connected to the secondary circuit. the order of
Sol: E 1) 10–13 m/s 2) 10–3 mm/s
E1 1 –4
3) 10 m/s 4) 10–30 m/s
E2 2 7. In metals and vacuum tubes charge carriers
E2 1.018
are
2 1 500 254.5cm . 1) electrons 2) protons
E1 2
3) both 4) positrons
NARAYANA GROUP 25
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
8. The electric intensity E, current density j and 17. i-v graph for a metal at temperatures t1, t2, t3
conductivity are related as : are as shown. The highest temperature is
1) j E 2) j E / 3) jE 4) j 2 E t3
t2
9. Electric field (E) and current density (J) have i
t1
relation
1 2 1
1) E J 1 2) E J 3) E 2 4) E V
J J 1) t1 2) t2 3) t3 4) t1 t2 t3
10. Assertion : A current flows in a conductor only
18. A certain piece of copper is to be shaped into
when there is an electric field within the
conductor. a conductor of minimum resistance. Its length
Reason : The drift velocity of electron in and cross sectional area should be
presence of electric field decreases. 1) L and A 2) 2L and A/2
1) Both (A) and (R) are true and (R) is the correct 3) L/2 and 2A 4) 3L and A/3
explanation of A. 19. When light falls on semiconductors, their
2) Both (A) and (R) are true but (R) is not the resistance
correct explanation of A. 1) decreases 2) increases
3) (A) is true but (R) is false 3) does not change 4) can’t be predicted
4) (A) is false but (R) is true 20. With the increase of temperature, the ratio of
OHM’S LAW & FACTORS conductivity to resistivity of a metal conductor
EFFECTING RESISTANCE 1) Decreases 2) Remains same
11. In an electric circuit containg a battery, the 3) Increases 4) May increase or decrease
charge (assumed positive) inside the battery 21. The conductivity of a super conductor, in the
1) always goes form the positive terminal to the super conducting state is
negative terminal
2) may move from the positive terminal to the 1) Zero 2) Infinity
negative terminal 3) Depends on temp
3) always goes from the negative terminal to the 4) Depends on free election
positive terminal 22. When a piece of aluminium wire of finite length
4) does not move . is drawn through a series of dies to reduce its
12. From the following the quantity which is diameter to half its original value, its resistance
analogous to temperature in electricity is will become
1) potential 2) resistance 1) Two times 2) Four times
3) current 4) charge
13. The flow of the electric current through a 3) Eight times 4) Sixteen times
metallic conductor is 23. Metals have
1) only due to electrons 1) Zero resistivity 2) High resistivity
2) only due to +ve charges 3) Low resistivity 4) Infinite resistivity
3) due to both nuclei and electrons. 24. Consider a rectangular slab of length L, and
4) can not be predicted. area of cross-section A. A current I is passed
14. For making standard resistance, wire of through it, if the length is doubled the potential
following material is used drop across the end faces
1) Nichrome 2) Copper
1) Becomes half of the initial value
3) Silver 4) manganin
15. Material used for heating coils is 2) Becomes one-forth of the initial value
1) Nichrome 2) Copper 3) Becomes double the initial value
3) Silver 4) Manganin 4) Remains Same
16. A piece of silver and another of silicon are heated 25. A metallic block has no potential difference
from room temperature. The resistance of applied across it, then the mean velocity of
1) each of them increases free electrons is (T = absolute temperature
2) each of them decreases of the block)
3) Silver increases and Silicon decreases 1) Proportional to T 2) Proportional to T
4) Silver decreases and Silicon increases 3) Zero
4) Finite but independent of temperature.
26 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
26. The resistance of a metal increases with 33. Fuse wire is a wire of
increasing temperature because 1) low melting point and low value of a
1) The collisions of the conducting electrons 2) high melting pointand high value of a
with the electrons increases.
2) The collisions of the conducting electrons 3) high melting point and low value of a
with the lattice consisting of the ions of the metal 4) low melting point and high value of a
increases 34. Assertion : Material used in the construction
3) The number of the conduction electrons of a standard resistance is constantan or
decreases. manganin.
4) The number of conduction electrons increase. Reason : Temperature coefficient of constantan
27. In the absence of applied potential, the electric is very small.
current flowing through a metallic wire is zero 1) Both (A) and (R) are true and (R) is the correct
because explanation of A.
1) The average velocity of electron is zero 2) Both (A) and (R) are true but (R) is not the
2) The electrons are drifted in random direction with correct explanation of A.
a speed of the order of 10-2cm/s. 3) (A) is true but (R) is false
3) The electrons move in random direction with a
speed of the order close to that of velocity of light. 4) (A) is false but (R) is true
4) Electrons and ions move in opposite direction. 35. Assertion (A) : Bending of a conducting wire
28. A long constan wire is connected across the effects electrical resistance.
terminals of an ideal battery. if the wire is Reason (R) : Resistance of a wire depends on
cut in to two equal pieces and one of them is resistivity of that material.
now connected to the same battery, what will 1) Both (A) and (R) are true and (R) is the correct
be the mobility of free electrons now in the explanation of A.
wire compared to that in the first case? 2) Both (A) and (R) are true but (R) is not the
1) same as that of previous value correct explanation of A.
2) double that of previous value 3) (A) is true but (R) is false
3) half that of previous value 4) (A) is false but (R) is true
4) four times that of previous value 36. Assertion (A) : When the radius of a copper
29. Ohm’s law is not applicable for wire is doubled, its specific resistance gets
1) insulators 2) semi conductors increased.
3) vaccum tube 4) all the above Reason (R):Specific resistance is independent
30. V - I graphs for two materials is shown in the of cross-section of material used
figure. The graphs are drawn at two different 1) Both (A) and (R) are true and (R) is the correct
temperatures. explanation of A.
Y 2) Both (A) and (R) are true but (R) is not the
T2
T1 correct explanation of A.
I
3) (A) is true but (R) is false
4) (A) is false but (R) is true
THERMISTOR
V X 37. The thermistors are usually made of
1) T1 T2 cot 2q 2) T1 T2 sin 2q 1) metals with low temperature coefficient of
resistivity
3) T1 T2 tan 2q 4) T1 T2 cos 2q
2) metals with high temperature coefficient of
31. Wires of Nichrome and Copper of equal resistivity.
dimensions are connected in series in 3) metal oxides with high temperature coefficient
electrical circuit. Then of resistivity
1) More current will flow in copper wire
2) More current will flow in Nichrome wire 4) semiconducting materials having
3) Copper wire will get heated more low temperature coefficient of resistivity
4) Nichrome wire will get heated more 38. For a chosen non-zero value of voltage, there
32. At absolute zero silver wire behaves as can be more than one value of current in
1) Super conductor 2) Semi conductor 1) copper wire 2) thermistor
3) Perfect insulator 4) Semi insulator 3) zener diode 4) manganin wire
NARAYANA GROUP 27
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
ELECTRIC POWER 45. Two metallic wires of same material and same
39. A heater coil is cut into two equal parts and length have different diameters. When the
only one part is used in the heater. Then the wires are connected in parallel across an ideal
heat generated becomes battery the rate of heat produced in thinner
1) become one fourth 2) halved wire is Q1 and that in thicker wire is Q2. The
3) doubled4) become four times correct statement is
40. Two lamps have resistance r and R, R being
greater than r. If they are connected in parallel 1) Q1 = Q2 2) Q1 < Q2 3) Q1 > Q2
in an electric circuit, then
1) the lamp with resistance R will shine more 4) It will depend on the emf of the battery
brightly 46. There are two metalic wires of same material,
2) the lamp with resistance r will shine more brightly same length but of different radii. When these
3) the two lamps will shine equal brightly are connected to an ideal battery in series, heat
4) the lamp with resistance R will not shine at all
41. Two bulbs are fitted in a room in the domestic produced is H1 but when connected in parallel,
electric installation. If one of them glows heat produced is H2 for the same time. Then
brighter than the other, then the correct statement is
1) the brighter bulb has smaller resistance
2) the brighter bulb has larger resistance 1) H1 H2 2) H1 H2
3) both the bulbs have the same resistance 3) H1 H2 4) No relation
4) nothing can be said about the resistance unless
other factors are known 47. Two electric bulbs rated P1 watt and V volt ,
42. Three identical bulbs P, Q and R are connected are connected in series, across V-volt supply.
to a battery as shown in the figure. When the The total power consumed is
circuit is closed P1 P2 P1 P2
1) 2) P1 P2 3)
P P
4) P1 P2
P
K
2 1 2
28 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
CELL-INTERNAL RESISTANCE 62. The resistance of an open circuit is
EMF 1) Infinity 2) Zero
52. Back emf of a cell is due to 3) Negative 4) cann’t be predicted
1) Electrolytic polarization 63. According to joule's law if potential difference
2) Peltier effect across a conductor having a material of specific
3) Magnetic effect of current resistance , remains constant, then heat
4) Internal resistance produced in the conductor is directly
53. The direction of current in a cell is proportional to
1) ve pole to ve pole during discharging 1 1
1) 2) 2 3) 4)
2) ve pole to ve pole during discharging
64. Internal resistance of a cell depends on
3) Always ve pole to ve pole 1) concentration of electrolyte
4) always flows from (+)ve ploe to (-) ve pole 2) distance between the electrodes
54. When an electric cell drives current through 3) area of electrode
load resistance, its Back emf, 4) all the above
1) Supports the original emf 65. When cells are arranged in series
2) Opposes the original emf 1) the current capacity decreases
3) Supports if internal resistance is low 2) The current capacity increases
4) Opposes if load resistance is large 3) the emf increases 4) the emf decreases
55. The terminal voltage of a cell is greater than 66. To supply maximum current, cells should be
its emf. when it is arrange in
1) being charged 2) an open circuit 1) series 2) parallel 3) Mixed grouping
3) being discharged 4) it never happens 4) depends on the internal and external resistance
56. What is constant in a battery ( also called a 67. The terminal Pd of a cell is equal to its emf if
source of emf) ? 1) external resistance is infinity
1) current supplied by it 2) internal resistance is zero
2) terminal potential difference 3) both 1 and 2
3) internal resistance
4) emf 4) internal resistance is 5
57. From the following the standard cell is 68. The electric power transfered by a cell to an
1) Daniel cell 2) Cadmium cell external resistance is maximum when the
3) Leclanche cell 4) Lead accumulator external resistance is equal to ...(r internal
58. A cell is to convert resistance)
1) chemical energy into electrical energy r
2) electrical energy into chemical energy 1) 2) 2r 3) r 4) r2
2
3) heat energy into potential energy 69. Which depolarizers are used to neutralizes
4) potential energy into heat energy hydrogen layer in cells
59. ‘n’ identical cells, each of internal resistance 1) Potassium dichromite 2) Manganese dioxide
(r) are first connected in parallel and then 3) 1 or 2 4) hydrogen peroxide
connected in series across a resistance ( R). 70. Assertion : Series combination of cells is used
If the current through R is the same in both when their internal resitance is much smaller
cases, then than the external resistance.
1) R = r/2 2) r = R/2 3) R = r 4) r = 0
60. The value of internal resistance of ideal cell is n
Reason : I where the symbols have
1) Zero 2) infinite 3) 1 4) 2 R nr
61. In a circuit two or more cells of the same emf their standard meaning,in series connection
are connected in parallel in order 1) Both (A) and (R) are true and (R) is the correct
1) Increases the pd across a resistance in the circuit explanation of A.
2) Decreases pd across a resistance in the circuit 2) Both (A) and (R) are true but (R) is not the
3) Facilitate drawing more current from the correct explanation of A.
battery system 3) (A) is true but (R) is false
4) Change the emf across the system of batteries 4) (A) is false but (R) is true
NARAYANA GROUP 29
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
71. Assertion (A) : To draw more current at low 78. In a wheatstone's bridge three resistances
P.d; parallel connection of cells is preferred. P,Q,R connected in three arms and the fourth
Reason (R) : In parallel connection, current arm is formed by two resitances S1,S2 con-
nE nected in parallel.The condition for bridge to
i , if r >> R. be balanced will be
r
1) Both (A) and (R) are true and (R) is the correct P R P 2R
explanation of A. 1) Q S S 2) Q S S
1 2 1 2
2) Both (A) and (R) are true but (R) is not the
correct explanation of A. P R ( S1 S 2 ) P R ( S1 S 2 )
3) (A) is true but (R) is false 3) Q S S 4) Q 2S S
1 2 1 2
4) (A) is false but (R) is true 79. Assertion : At any junction of a network,
KIRCHHOFF’S LAWS algebraic sum of various currents is zero
WHEATSTONE BRIDGE Reason : At steady state there is
72. Kirchoff’s law of meshes is in accordance with no accumulation of charge at the junction.
law of conservation of 1) Both (A) and (R) are true and (R) is the
1) charge 2) current correct explanation of A.
3) energy 4) angular momentum 2) Both (A) and (R) are true but (R) is not the
correct explanation of A.
73. Kirchoff’s law of junctions is also called the
3) (A) is true but (R) is false
law of conservation of
4) (A) is false but (R) is true
1) energy 2) charge
3) momentum 4) angular momentum METERBRIDGE
74. Wheatstones’s bridge cannot be used for 80. Metal wire is connected in the left gap, semi
measurement of very ——— resistances. conductor is connected in the right gap of
1) high 2) low 3) low(or) high 4) zero meter bridge and balancing point is found.
75. In a balanced Wheatstone’s network, the Both are heated so that change of resistances
resistances in the arms Q and S are in them are same. Then the balancing point
interchanged. As a result of this : 1) will not shift
1)galvanometer and the cell must be interchanged 2) shifts towards left
to balance 3) shifts towards right
4) depends on rise of temperature
2) galvanometer shows zero deflection
81. A metre bridge is balanced with known
3) network is not balanced
resistance in the right gap and a metal wire
4) network is still balanced
in the left gap. If the metal wire is heated
76. If galvanometer and battery are interchanged the balance point.
in balanced wheatstone bridge, then 1) shifts towards left
1) the battery discharges 2) shifts towards right
2) the bridge still balances 3) does not change
3) the balance point is changed 4) may shift towards left or right depending on
4) the galvanometer is damaged due to flow of high the nature of the metal.
current 82. In metre bridge experiment of resistances, the
77. Wheatstone bridge can be used known and unknown resistances are inter-
1) To compare two unknown resistances. changed . The error so removed is
2) to measure small strains produced in hardmetals 1) end correction
3) as the working principle of meterbridge 2) index error
4) All the above 3) due to temperature effect
4) random error
30 NARAYANA GROUP
JEE-ADV PHYSICS- VOL- III CURRENT ELECTRICITY
83. In a metre-bridge experiment, when the 89. If the emf of the cell in the primary circuit is
resistances in the gaps are interchanged, the doubled, with out changing the cell in the
balance-point did not shift at all. The ratio of secondary circuit, the balancing length is
resistances must be 1) Doubled 2) Halved
1) Very large 2) Very small 3) Uncharged 4) Zero
3) Equal to unity 4) zero
84. Assertion (A) : Meterbridge wire is made up 90. The potential gradients on the potentiometer
of manganin wire are V1 and V2 with an ideal cell and a
Reason (R) : The temperature coeffiecient real cell of same emf in the primary circuit
of resistance is very small for manganin then
1) Both (A) and (R) are true and (R) is the
1) V1 V2 2) V1 V2 3) V1 V2 4) V1 V2
correct explanation of A.
2) Both (A) and (R) are true but (R) is not the 91. If the current in the primary circuit is
correct explanation of A. decreased, then balancing length is obtained
3) (A) is true but (R) is false at
4) (A) is false but (R) is true 1) Lower length 2) Higher length
POTENTIOMETER 3) Same length 4) 1/3rd length
85. A potentiometer is superior to voltmeter for 92. Temperature coefficient of resistance ' ' and
measuring a potential because resistivity ‘ ’ of a potentiometer wire must
1) voltmeter has high resistance be
2) resistance of potentiometer wire is quite low
1) high and low 2) low and high
3) potentiometer does not draw any current from
the unknown source of emf. to be measured. 3) low and low 4) high and high
4) sensitivity of potentiometer is higher than that 93. A series high resistance is preferable than
of a voltmeter. shunt resistance in the galvanometer circuit
86. In comparing emf’s of 2 cells with the help of of potentiometer. Because
potentiometer, at the balance point, the 1) shunt resistances are costly
current flowing through the wire is taken from 2) shunt resistance damages the galvanometer
1) Any one of these cells.
3) series resistance reduces the current through
2) both of these cells
galvanometer in an unbalanced circuit
3) Battery in the primary circuit
4) From an unknown source 4) high resistances are easily available
87. A potentiometer wire is connected across the 94. The sensitivity of potentiometer wire can be
ideal battery now, the radius of potentiometer increased by
wire is doubled without changing its length. 1) decreasing the length of potentiometer wire
The value of potential gradient 2) increasing potential gradient on its wire
1) increases 4 times 2) increases two times
3) increasing emf of battery in the primary circuit
3) Does not change 4) becomes half
88. In a potentiometer of ten wires, the balance 4) decreasing the potential gradient on its wire
point is obtained on the sixth wire. To shift 95. A cell of emf ‘E’ and internal resistance ‘r’
the balance point to eighth wire, we should connected in the secondary gets balanced
1) increase resistance in the primary circuit. against length ‘ ’ of potentiometer wire. If a
2) decrease resistance in the primary circuit. resistance ‘R’ is connected in parallel with
3) decrease resistance in series with the cell whose the cell, then the new balancing length for
emf. has to be measured. the cell will be
4) increase resistance in series with the cell whose R
1)
Rr R R
emf. has to be measured. l 2) 4)
R l 3) r
l
R r Rr
NARAYANA GROUP 31
CURRENT ELECTRICITY JEE-ADV PHYSICS- VOL- III
96. Potentiometer is an ideal instrument, because 103. The quantity that cannot be measured by a
1) no current is drawn from the source of unknown potentiometer is ...........
emf 1) Resistance 2) emf
2) current is drawn from the source of unknown 3) current in the wire 4) Inductance
emf 104. Assertion : Potentiometer is much better than
3) it gives deflection even at null point a voltmeter for measuring emf of cell
4) it has variable potential gradient Reason : A potentiometer draws no current
97. On increasing the resistance of the primary while measuring emf of a cell
circuit of potentiometer, its potential gradient 1) Both (A) and (R) are true and (R) is the
will correct explanation of A.
1) become more 2) become less 2) Both (A) and (R) are true but (R) is not
3) not change 4) become infinite the correct explanation of A.
98. If the value of potential gradient on potentiometer 3) (A) is true but (R) is false
wire is decreased, then the new null point will 4) (A) is false but (R) is true
be obtained at 105. A : The emf of the cell in secondary circuit
1) lower length 2) higher length must be less than emf of cell in primary circuit
3) same length 4) nothing can be said in potentiometer.
99. A cell of negligible internal resistance is R : Balancing length cannot be more than
connected to a potentiometer wire and length of potentiometer wire.
potential gradient is found. Keeping the 1) Both (A) and (R) are true and (R) is the
len gt h as const an t , i f t h e radi u s of correct explanation of A.
potentiometer wire is increased four times, 2) Both (A) and (R) are true but (R) is not
the potential gradient will become (no series the correct explanation of A.
resistance in primary) 3) (A) is true but (R) is false
1) 4 times 2) 2 times 3) half 4) constant 4) (A) is false but (R) is true
100. For the working of potentiometer, the emf of cell
in the primary circuit (E) compared to the emf of
C. U. Q KEY
the cell in the secondary circuit (E1) is 1) 1 2) 1 3) 1 4) 4 5) 2 6) 3
1) E > E1 2) E < E1 7) 1 8) 1 9) 2 10) 3 11) 2 12) 1
3) Both the above 4) E = E1 13) 1 14) 4 15) 1 16) 3 17) 1 18) 3
101. The balancing lengths of potentiometer wire 19) 1 20) 1 21) 2 22) 4 23) 1 24) 3
are l1 and l2 when two cells of emf E1 and E2 25) 3 26) 2 27) 1 28) 1 29) 4 30) 1
are connected in the secondary circuit in 31) 4 32) 1 33) 4 34) 1 35) 2 36) 4
series first to help each other and next to 37) 3 38) 2 39) 3 40) 2 41) 1 42) 2
E1 43) 2 44) 4 45) 2 46) 2 47) 3 48) 4
oppose each other E is equal to (E1>E2). 49) 4 50) 1 51) 1 52) 1 53) 1 54) 2
2
55) 1 56) 4 57) 2 58) 1 59) 3 60) 1
l1 l1 l2 l1 l2 l2 61) 3 62) 1 63) 4 64) 4 65) 3 66) 4
1) l 2) l l 3) l l 4) l
2 1 2 1 2 1 67) 3 68) 3 69) 3 70) 1 71) 1 72) 3
102. At the moment when the potentiometer is 73) 2 74) 2 75) 3 76) 2 77) 4 78) 3
balanced, 79) 1 80) 3 81) 2 82) 1 83) 3 84) 1
1) Current flows only in the primary circuit 85) 3 86)3 87) 3 88) 1 89) 2 90) 2
2) Current flows only in the secondary circuit 91) 2 92) 2 93) 3 94) 4 95) 4 96) 1
3) Current flows both in primary and secondary 97) 2 98) 2 99) 4 100)1 101)3 102) 1
circuits 103) 4 104) 1 105) 1
4) current does not flow in any circuit
32 NARAYANA GROUP