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

12th Grade > Physics

ALTERNATING CURRENT MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21. One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has self inductance of value (f = 50 Hz)

0.052 H
2.42 H
16.2 mH
1.62 mH

Discuss Question

Answer: Option A. -> 0.052 H
:
A
Current through the bulb

i = \frac{P}{V} = \frac{60}{10} = 6 A

One 10 V, 60 W Bulb Is To Be Connected To 100 V Line. The Re...

V = \sqrt{V_{R}^{2} + V_{L}^{2}}

(100)^{2} = (10)^{2} + V_{L}^{2} \Rightarrow V_{L} = 99.5

Volt
Also

V_{L} = i X_{L} = i \times (2 π v L)

\Rightarrow 99.5 = 6 \times 2 \times 3.14 \times 50 \times L \Rightarrow L = 0.052

Question 22. An ac source of angular frequency

ω

is fed across a resistor r and a capacitor C in series. The current registered is I. If now the frequency of source is changed to

\frac{ω}{4}

(but maintaining the same voltage), the current in then circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency

ω

√35
√25
√15
12

Discuss Question

Answer: Option D. -> 12
:
D
At angular frequency

ω

, the current in RC circuit is given by

i_{r m s} = \frac{V_{r m s}}{\sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}} . . . . . . . (i)

Also

\frac{i_{r m s}}{2} = \frac{V_{r m s}}{\sqrt{R^{2} + {(\frac{1}{\frac{ω}{4} c})}^{2}}} = \frac{V_{r m s}}{\sqrt{R^{2} + \frac{16}{ω^{2} C^{2}}}} . . . . . . (i i)

From equation (i) and (ii) we get

3 R^{2} = \frac{12}{ω^{2} C^{2}} \Rightarrow \frac{\frac{1}{ω C}}{R} = \sqrt{\frac{3}{12}} \Rightarrow \frac{X_{c}}{R} = \sqrt{\frac{3}{12}}

Question 23. An LCR series circuit with a resistance of 100 ohm is connected to an ac source of 100 V (r.m.s.) and angular frequency 300 rad/s. When only the capacitor is removed, the current lags behind the voltage by . When only the inductor is removed the current leads the voltage by . The average power dissipated is

50 W
100 W
200 W
400 W

Discuss Question

Answer: Option B. -> 100 W
:
B

t a n ϕ = \frac{X_{L}}{R} = \frac{X_{C}}{R} \Rightarrow t a n 60^{\circ} = \frac{X_{L}}{R} = \frac{X_{C}}{R}

\Rightarrow X_{L} = X_{c} = \sqrt{3} R

i.e. Z

= \sqrt{R^{2} + (X_{L} - X_{c})^{2}} = R

So average power

P = \frac{V^{2}}{R} = \frac{100 \times 100}{100} = 100

Question 24. In a certain circuit current changes with time according to i = 2

\sqrt{t}

r.m.s. value of current between t = 2 to t = 4s will be

3A
3√3 A
2√3 A
(2−√2) A

Discuss Question

Answer: Option C. -> 2√3 A
:
C

^{2} = \frac{\int i^{2} d t}{\int d t} = \frac{\int_{2}^{4} (4 t) d t}{\int_{2}^{4} d t} = \frac{4 \int_{2}^{4} t d t}{2} = 2 {[\frac{t^{2}}{2}]}_{2}^{4} = {[t^{2}]}_{2}^{2} = 12

\Rightarrow i_{r m s} = \sqrt{^{2}} = \sqrt{12} = 2 \sqrt{3} A

Question 25. A bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased, while keeping the voltage of the source constant, then

Bulb will give less intense light
Bulb will give more intense light
Bulb will give light of same intensity as before
Bulb will stop radiating light

Discuss Question

Answer: Option B. -> Bulb will give more intense light
:
B
When a bulb and a capacitor are connected in series to an ac source, then on increasing the frequency the current in the circuit is increased, because the impedance of the circuit is decreased. So the bulb will give more intense light.

Question 26. In the circuit shown in the figure, the ac source gives a voltage V = 24cos(2000t). Neglecting source resistance, the voltmeter and ammeter reading will be