Current Electricity(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

CURRENT ELECTRICITY MCQs

Electricity And Circuits

Total Questions : 60 | Page 3 of 6 pages

Question 21. Three resistances of equal value are connected in four different combinations as shown in figure. Arrange them in increasing order of power dissipation

Discuss Question

Answer: Option B. -> II
:
B
Let R be the value of each resistance. The resistances of combinations I, II, III and IV are 3R,

\frac{2 R}{3}

\frac{R}{3}

and

\frac{3 R}{2}

respectively. Now power dissipation is inversely proportional to resistance

Question 22. In the circuit shown in figure the current through

the current in 3 ohm resistor is 0.5 A
the current in 3 ohm resistor is 0.25 A
the current in 4 ohm resistor is 0.5 A
the current in 4 ohm resistor is 0.25 A

Discuss Question

Answer: Option D. -> the current in 4 ohm resistor is 0.25 A
:
D
The equivalent resistance between points A and B to the right of AB is 4 ohm. Therefore total resistance =3+4+2=9ohm. Current I

= \frac{9}{9}

=1A. This current is equally divided in the 8 ohm resistor between A and B and the remainder 8 ohm resistor. Hence current in AC =0.5A. This current is equally divided between the 8 ohm resistor in CD and the circuit to the right of CD. Therefore current in the 4 ohm resistor = 0.25 A

Question 23. The effective resistance between points P and Q of the electrical circuit shown in figure is

2Rr(R+r)
8R(R+r)(3R+r)
2r + 4R
5R2+2r

Discuss Question

Answer: Option A. -> 2Rr(R+r)
:
A

The branches ABPQ and PQCD are a balanced Wheatstone's bridge. Therefore resistances between E and E and between F and G do not contribute and the circuit simplies to the one show in figure. The effective resistance

R_{e}

between P and Q is given by

\frac{1}{R_{e}} = \frac{1}{4 R} + \frac{1}{2 r} + \frac{1}{4 R}

which gives

R_{e} = \frac{2 R r}{(R + r)}

Question 24. A steady current flows in a metallic conductor of non uniform cross-section. The quantity that remains constant along the length of the conductor is

current, electric field and drift speed
drift speed only
current and drift speed
current only

Discuss Question

Answer: Option D. -> current only
:
D
The drift speed depends on A the cross sectional area of the conductor but the current is independent of A

Question 25. A 100 W bulb

B_{1}

and two 60 W bulbs

B_{2}

and

B_{3}

are connected to a 250 V source as shown in figure.

W_{1}

W_{2}

and

W_{3}

are the output powers of the bulbs

B_{1}

B_{2}

and

B_{3}

respectively then

A 100 W Bulb B1 And Two 60 W Bulbs B2 And B3 Are Connec...

W1> W2 = W3
W1> W2> W3
W1
W1

Discuss Question

Answer: Option D. -> W1
:
D
The resistances of bulbs

B_{1}, B_{2}

and

B_{3}

respectively are

R_{1} \frac{v^{2}}{w_{1}} = \frac{(250)^{2}}{100} = 625

ohm

R_{2} = \frac{(250)^{2}}{60} = 1042

ohm =

R_{3}

Voltage across

B_{3}

V_{3} = 250 V

Voltage across

B_{1}

V_{1} = \frac{V R_{1}}{(R_{1} + R_{2})} = \frac{250 \times 625}{(625 + 1042)} = 93.7 V

Voltage across

B_{2}

V_{2} = 250 - 93.7 = 156.3 V

Power output

W_{1} = \frac{V^{2}}{R_{1}} = \frac{(93.7)^{2}}{625} = 14 W

W_{2} = \frac{(156.3)^{2}}{1042} = 23 w

W_{3} = \frac{(250)^{2}}{1042} = 60 w

Hence

W_{1} < W_{2} < W_{3}

Question 26. Eight cells marked 1 to 8 each of emf 5 V and internal resistance 0.2 ohm are connected as shown in figure. What is the reading of the ideal voltmeter V?

40 V
20 V
5 V
0

Discuss Question

Answer: Option D. -> 0
:
D

The emfs of cells connected in reverse polarity cancel each other. Hence cells marked 2, 3 and 4 together cancel the effect of cells marked 5, 6 and 7 and the circuit reduces to that shown in figure. Now cells 1 and 8 are in reverse polarity. Hence the voltmeter reading = 5 – 5 = 0 V

Question 27. What is the equivalent resistance of the network shown in figure? The numbers indicate resistance in ohms

8 ohm
163 ohm
4 ohm
165 ohm

Discuss Question

Answer: Option B. -> 163 ohm
:
B
The network consists of four loops connected in series. Each loop consists of two 1 ohm resistors in series connected in parallel with two ohm resistors in series. Thus the effective resistance R of each loop is given by

\frac{1}{R} = \frac{1}{2} + \frac{1}{4}

giving

R = \frac{4}{3}

ohm. Therefore the equivalent resistance of the network = resistance of fourloops connected in series each having an effective resistance of

\frac{4}{3}

ohm =

4 \times \frac{4}{3} = \frac{16}{3}

ohm

Question 28. The voltage across a bulb is decreased by

2 % .

Assuming that the resistance of the filament remains unchanged the power of the bulb will

decrease by 2%
increase by 2%
decrease by 4%
increase by 4%

Discuss Question

Answer: Option C. -> decrease by 4%
:
C
Power

P = \frac{v^{2}}{R}

Therefore

△ P = \frac{2 V △ V}{R}

Hence

\frac{△ P}{P} = \frac{2 △ V}{V}

Now

\frac{△ V}{V} = - 2 %

Therefore

\frac{△ P}{P} = - 2 \times 2 = - 4 %

i.e. the power decreases by

4 %

Question 29. A wire of resistance 5 ohm is drawn out so that its new length is three times its original length. What is the resistance of the new wire ?

53 Ω
5 Ω
15 Ω
45 Ω

Discuss Question

Answer: Option D. -> 45 Ω
:
D
We know that

V o l u m e = \frac{M a s s}{d e n s i t y}

Let m be the mass of the given wire, d its density and

ρ

its resistivity.
These quantities remain unchanged when the wire is drawn out.
Let

l_{1}

be the length of the smaller wire and

r_{1}

its corresponding radius of cross section. If

l_{2}

and

r_{2}

represent the length and radius of cross section of the elongated wire,

⟹ m = π r_{1}^{2} l_{2} d

for smaller wire and for elongated wire,

m = π r_{2}^{2} l_{2} d

.
Since mass and density are constant

⟹ r_{1}^{2} l_{1} = r_{2}^{2} l_{2}

R_{1}

is the resistance of the smaller wire and

R_{2}

is the resistance of theelongated wire, we have,

R_{1} = \frac{ρ l_{1}}{π r_{1}^{2}}

R_{2} = \frac{ρ l_{2}}{π r_{2}^{2}}

Now,

\frac{R_{2}}{R_{1}} = \frac{l_{2}}{l_{1}} \times \frac{r_{1}^{2}}{r_{2}^{2}}

It is given that,

l_{2} = 3 l_{1} .

Therefore

r_{2}^{2} = \frac{r_{1}^{2}}{3}

Substituting the values,

\frac{R_{2}}{R_{1}} = 9

Since,

R_{1} = 5 Ω

R_{2} = 45 Ω

Question 30. Figure shows a network of resistances connected to a 2 V battery. If the internal resistance of the battery is negligible current I in the circuit is

0.25 A
0.5 A
0.75 A
1 A

Discuss Question

Answer: Option B. -> 0.5 A
:
B
Because

\frac{P}{Q} = \frac{R}{S}

the circuit is balanced Wheatstone's bride.Therefore potential at B =potential at D. Hence no current flows through the 5 ohm resistor. It is therefore not effectiveThe equivalent resistance between A and C is the resistance of a parallel combination of P + R = 2 + 4 = 6ohm and Q + S = 4 + 8 = 12ohm which is given by