Solution

CURRENT ELECTRICITY

Class 12 - Physics
Section A
1.
(d) 8E, 10r
Explanation: 8E, 10r
2.
(d) Charge
Explanation: Charge
3.
R1 R2
(c) R3
=
R4

Explanation: For null point condition, no current should flow between junction of R1 - R3 and R2 - R4. This implies bridge
balance condition.

s
Bridge balance condition is given by ,

sse
R1 R2
=
R3 R4

4.
(b) R
Explanation: In the balanced condition, the resistance R of the galvanometer is ineffective. We now have (R + R) and (R +
R) resistances in parallel.
cla
∴ R = α
=R
2R × 2R

2R + 2R

5. (a) E1 + E2
n
Explanation: The equivalent emf of two voltages in series with the cathode of one connected to the anode of the other is equal
to the sum of the emfs.
itio

Therefore
ns

E = E1 + E2

6.
(c) 0.5 A
tra

Explanation: The series combination of 1 Ω and 2 Ω is in parallel with 3 Ω resistor.

3 × 3 9 3
∴ R= 3 + 3
=
6
=
2
Ω

V 1 − 5
I= R
=
3
= 1.0 A

Current through the 3 Ω resistor = 3

6
× 1 = 0.5 A
7.
(c) maximum when the battery is connected across 1cm × 1

2
cm
Explanation: As we know R = ρ ⋅ l

A
, where length is constant l=10cm.So resistance is directly proportional to 1/A. The
maximum resistance will be when the value of A
1
is maximum, i.e., ‘A’ must be minimum, it is minimum when the area of
cross-section is 1cm × cm 1

8.

(b)

Explanation:

1/6
 I= nε

nr
=
ε

∴ I is independent of n and it is constant.

9. (a) I R 2

Explanation: The power dissipated

P = V × I

Since V = IR

2
P = I R

10.

(b)

s
sse
Explanation: cla
Current in the circuit, I = ε

R+r

Potential difference across R,

n
V = IR = ( ε

R+r
)R

V= ε
itio

r
1+
R

When R = 0, V = 0
R = ∞, V = ε
ns
tra

11.
(d) Both A and R are false.
Explanation: Actually the bulbs fuse when they are switched on. As the bulb is switched on, it lights up. Its temperature
increases. Due to which the resistance of the filament of the bulbs increases. This happens very quickly. After many cycles, the
filament of the bulb become thin and the bulb is at the verge of burning out. When such a bulb is switched on, its initial
resistance being low, there will be a sudden, rush of current as a result of which the filament burns out.
12. Electromotive force is defined as the electric potential produced by either an electrochemical cell or by changing the magnetic
field.
The unit for electromotive force is Volt.
13. In series,
εeq = ε + ε = 2ε
2ε
Is =
2r+R

In parallel,
εeq = ε
r
req =
2
ε
Ip = r
+R
2

2/6
 ∵ IP = IS
2ε ε
= r
2r+R +R
2

2 1
= r
2r+R +R
2

2
r

2
+ 2R = 2r + R
r=R
14. Relaxation Time (t): The average time elapsing between two successive collisions of an electron inside a conductor under the
application of an external electric field is called the relaxation time. It is measured in second.
Relaxation time = mean free path of electron/drift velocity.
15. Here V = 10 V, R = 50 Ω , t = 1 h = 3600 s
Heat energy produced in 1 h is
2

H= V

R
t
=
10 × 10 × 3600

50
= 7200 J
The source of this energy is the chemical energy stored in the battery.
16. Eight cell arrangement cannot provide the high current needed to start the car engine because it has a high internal resistance of
about 10 Ω while the internal resistance of a car battery is just 0.1 Ω .
17. Resistivity of materials is the resistance to the flow of an electric current with some materials resisting the current flow more than
others.

s
For same length and same radius, resistance of wire

sse
R∝ ρ [where ρ is resistivity]
As, ρ > ρ
nichrome copper

Hence, resistance of nichrome section is more.

In series, same current flows through both sections and heat produced = I2Rt. So, more heat is produced in nichrome section of
wire.
cla
n
itio

18. Yes, It is under the influence of the electric field only that the free electrons flow inside the conductor and constitute the electric
current.
19. I = 300 mA = 300 × 10-3 A,
t = 1 minute = 60 s, e = 1.6 × 10-19C
ns

q
As I = = t
ne

∴ Number of electrons,
tra

= 1.125 × 1020
−3

n= It

e
=
300 × 10 × 60

−19
1.6 × 10

20. Total resistance between the points P and Q,

R= +4=2+4=6Ω
4 × 4

4 + 4

Current in the circuit, I = 1 A

Power dissipated in the circuit,
P = I2R = 12 × 6 = 6 W
21. i. As Resistance is the voltage divided by current, so region having negative slope will have negative resistance. So, DE is the
region of negative resistance because the slope of curve in this part is negative.
ii. BC is the region where Ohm's law is obeyed because in this part, the current varies linearly with the voltage.
Section B
22. The equivalent circuit for the given network is shown in Figure.

R 2R
Clearly, 2R
=
4R

3/6
 Thus the part of the circuit on the right side is a balanced Wheatstone bridge. Hence resistance 6R in arm BD is ineffective. The
equivalent circuit then reduces to the circuit shown in Figure.
23. The equivalent resistance of the 5 Ω and 20 Ω resistances connected in parallel = = 4Ω . This resistance is connected in
5 × 20

5 + 20

series with the rheostat whose minimum and maximum resistances are 0 Ω and 30 Ω .
When the rheostat is adjusted at the minimum resistance of 0 Ω , current will be maximum.
Imax = = 1.5 A6 V

4Ω

When the rheostat is adjusted at the maximum resistance of 30 Ω , current will be minimum.
6 V
Imin = (4 + 30)Ω
= 0.18 A
24. Let emf of the source = V volts
Resistance of each resistor = P ohms
In series, total resistance = 3 P
2 2
V V
∴ Power dissipated = 3R
= X or R = 3R

In parallel, total resistance P ' is given by

3 R
1
=
′
+ +
1

R
= or R =
1

R
1

R R
′

3
R
2 2 2
V 3V 3V
∴ Power dissipated = ′
=
R
=
2
= 9X
R V

3X

25. The internal resistance is defined as, "the resistance offered by the electrolyte and electrodes of the cell, when electric current

es
flows through it."
When the key is closed, the current flows through the circuit as shown in the figure. Now, the reading of the voltmeter is equal to

s
terminal potential difference (V) of the cell.
We can write
E - V = Ir
⇒ r =
E−V

But from Ohm's law, I =

I
...(i)
V
las
nc
R

Substituting the value of I in equation (i), we get

E−V

r= R
V

(E−V )
io

⇒r= V

R
=( E

V −1
R
)

26. 300 K = 300 - 273 = 27°C

sit

∴ R27 = R0 (1 + α × 27) = 1 Ω

and Rt = R0 (1+ α × t) = 2 Ω
1 + αt 2
∴ =
n

1 + 27α 1

or 1 + α t = 2 + 54α
tra

1 + 54α 1 + 54 × 0.00125
or t = =
α 0.00125
= 854° C
= 854 + 273 = 1127 K
Section C
27. When I = 0,
total emf = terminal voltage
∴ 4 ε = 5.6 V

or ε = 1.4 V
When I = 1.0 A, V = = 0.7 V 2.8

Internal resistance
ε − V 1.4 − 0.7
r= I
=
1.0
= 0.7Ω

The output power is maximum when external resistance = internal resistance = 4 r

Imax = =
Total emf

Total resistance
4ε

4r + 4r

= ε

2r
=
1.4

2 × 0.7
= 1A

28. a. Electrical resistance Resistivity

It is the property of material due to which it opposes the The resistivity is defined as the resistance of a material of 1 metre
flow of electricity through the conductor. length and 1 square metre area of cross section.

4/6
 Unit : Ohm Unit : Ohm-metre

Symbol : R Symbol : ρ

Depends on length, cross-section area of conductor and

Depends on temperature and material of the conductor.
temperature.
b. For resistor R1,
Resistivity = ρ 1

Length = L
Area = A1
L
∴ R1 = ρ1 ×
A1

For resistor R2,

Resistivity = ρ 2

Length = L
Area = A2
L
∴ R2 = ρ2 ×
A2

For the equivalent resistor,

s
Resistivity = ρ

sse
Length = L
Area = A = A1 + A2
L
∴ R= ρ ×
A

Since, the resistors are connected in parallel,

Equivalent resistance = R =
R1 R2

R1 + R2
cla
ρ L ρ L
1 2
×
A A

or, ρ × L

A
=
1

ρ L
2

ρ L
1 2
+
A A
1 2
n
ρ ρ ( A1 + A2 )
∴ ρ = Effective resistivity = 1 2

ρ1 A2 + ρ2 A1
itio

29. Difference between emf (ε) and terminal voltage (V)

emf Terminal voltage

It is the potential difference between two terminals of the cells when It is the potential difference between two terminals when
no current is flowing through it. current passes through it.
ns

It is the cause. It is the effect.

Following plot is showing variation of terminal voltage versus the current.
tra

Note:- Negative slope gives internal resistance, where

V = ε − Ir
ε−V
Or r = I

Section D
30. (b) 0.888A
mE
Explanation: I = mR+r
,m = number of cells = 4
E = 2V, R = 2Ω , r = 1Ω
8 8
I= 8+1
=
9
= 0.888 A
31. (d) nr - 4r
Explanation: nr - 4r

5/6
32. (d) (ii) is correct but (i) is wrong.
Explanation: Let two cells of emfs ε and ε and of internal resistance r1 and r2 respectively are connected in parallel.
1 2

The equivalent emf is given by

ε1 p + ε2 γ
εeq =
2

n+r2
... (i)
The equivalent internal resistance is given by
γ r2
1

req
=
1
+
r1
or r =
1

r2
... (ii)
cq
1

r1 + r2

Let us consider, two cells connected in parallel of same emf ε and same internal resistance r.
εr+εr
From equation (i), we get ε = = ε
eq
r+r

From equation (ii), we get

2
r r
req = =
r+r 2

33. (a) Depends upon the relative values of internal and external resistances

es
Explanation: Depends upon the relative values of internal and external resistances
34. (c) ε

s
Explanation: For parallel combination of n cells, ε eq = ε

las
io nc
n sit
tra

6/6