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

12th Grade > Physics

ALTERNATING CURRENT MCQs

Total Questions : 30 | Page 2 of 3 pages

\frac{25}{π} μ F

capacitor and 3000-ohm resistance are joined in series to an ac source of 200 volt and

50 s e c^{- 1}

frequency.The power factor of the circuit and the power dissipated in it will respectively

0.6, 0.06 W
0.06, 0.6 W
0.6, 4.8 W
4.8, 0.6 W

Discuss Question

Answer: Option C. -> 0.6, 4.8 W
:
C
z=

\sqrt{R^{2} + {(\frac{1}{2 π v C})}^{2}} = \sqrt{(3000)^{2} + \frac{1}{{(2 π \times 50 \times \frac{2.5}{π} \times 10^{- 6})}^{2}}}

\Rightarrow Z = \sqrt{(3000)^{2} + (4000)^{2}} = 5 \times 10^{3} Ω

So power factor

c o s ϕ = \frac{R}{Z} = \frac{3000}{5 \times 10^{3}} = 0.6

and
Power P=

V_{r m s} i_{r m s} c o s ϕ = \frac{V_{r m s}^{2} c o s ϕ}{Z} \Rightarrow P = \frac{(200)^{2} \times 0.6}{5 \times 10^{3}} = 4.8 W

Question 12. The vector diagram of current and voltage for a circuit is as shown. The components of the circuit will be

LCR
LR
LCR or LR
LC

Discuss Question

Answer: Option C. -> LCR or LR
:
C
From phasor diagram it is clear that current is lagging with respect to

E_{r m s}

. This may be happen in LCR or LR circuit.

Question 13. What is the r.m.s. value of An alternating current which when passed through a resistor produces heat which is thrice of that produced by a direct current of 2 amperes in the same resistor

6 amp
2 amp
3.46 amp
0.66 am

Discuss Question

Answer: Option C. -> 3.46 amp
:
C
Heat produced by ac = 3 X Heat produced by dc

∴ i_{r m s}^{2} R t = 3 \times i^{2} R t \Rightarrow i_{r m s}^{2} = 3 \times 2^{2}

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

= 3.46A

Question 14. What will be the self inductance of a coil, to be connected in a series with a resistance of

π \sqrt{3} Ω

such that the phase difference between the emf and the current at 50 Hz frequency is

30^{\circ}

0.5 Henry
0.03 Henry
0.05 Henry
0.01 Henry

Discuss Question

Answer: Option D. -> 0.01 Henry
:
D

t a n ϕ = \frac{X_{L}}{R} = \frac{2 π v L}{R} \Rightarrow t a n 30^{\circ} = \frac{2 π \times 50 \times L}{π \sqrt{3}} L = 0.01 H

Question 15. If A and B are identical bulbs which bulbs glows brighter

(o m e g a = 100 r a d / s)

If A And B Are Identical Bulbs Which Bulbs Glows Brighter(om...

A
B
Both equally bright
Cannot say

Discuss Question

Answer: Option A. -> A
:
A

(X_{C}) >> (X_{L})

As the impedance is more in capacitor, bulb B will be glows less brighter compared to Bulb A

Question 16. The self inductance of a choke coil is 10 mH. When it is connected with a 10V dc source, then the loss of power is 20 watt. When it is connected with 10 volt ac source loss of power is 10 watt. The frequency of ac source will be

50 Hz
60 Hz
80 Hz
100 Hz

Discuss Question

Answer: Option C. -> 80 Hz
:
C
With dc :

P = \frac{V^{2}}{R} \Rightarrow R = \frac{(10)^{2}}{20} = 5 Ω

;
With ac:

P = \frac{V_{r m s}^{2} R}{Z^{2}} \Rightarrow Z^{2} = \frac{(10)^{2} \times 5}{10} = 50 Ω^{2}

Also

Z^{2} = R^{2} + 4 π^{2} v^{2} L^{2}

\Rightarrow 50 = (5)^{2} + 4 (3.14)^{2} v^{2} (10 \times 10^{- 3})^{2} \Rightarrow v = 80

Hz.

Question 17. In an LCR circuit ohm. When capacitance C is removed, the current lags behind the voltage by

\frac{π}{3}

. When inductance L is removed, the current leads the voltage by

π

. The impedance of the circuit is

50 ohm
100 ohm
200 ohm
400 ohm

Discuss Question

Answer: Option B. -> 100 ohm
:
B
When C is removed circuit becomes RL circuit hence

t a n \frac{π}{3} = \frac{X_{L}}{R} . . . . . . (i)

When L is removed circuit becomes RC circuit hence

t a n \frac{π}{3} = \frac{X_{c}}{R} . . . . (i i)

From equation (i) and (ii) we obtain

X_{L} = X_{c} .

This is the condition of resonance and in resonance Z=R=100

Ω

Question 18. A telephone wire of length 200 km has a capacitance of 0.014

μ

F per km. If it carries an ac of frequency 2.5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum

0.35 mH
35 mH
1.4 mH
Zero

Discuss Question

Answer: Option C. -> 1.4 mH
:
C
Capacitance of wire

C = 0.014 \times 10^{- 6} \times 200 = 2.8 \times 10^{- 6} F = 2.8 μ

F
For impedance of the circuit to be minimum

X_{L} = X_{c} \Rightarrow 2 π v L = \frac{1}{2 π v C}

\Rightarrow L = \frac{1}{4 π^{2} v^{2} C} = \frac{1}{4 (3.14)^{2} \times (2.5 \times 10^{3})^{2} \times 2.8 \times 10^{- 6}}

= 1.4 \times 10^{- 3}

H = 1.4mH

Question 19. The voltage of an ac source varies with time according to the equation V = 200sin

(100 π t) . c o s (100 π t)

where t is in seconds and V is in volts. Then

The peak voltage of the source is 100 volts
The peak voltage of the source is 50 volts
The peak voltage of the source is 100√2 volts
The frequency of the source is 50 Hz

Discuss Question

Answer: Option A. -> The peak voltage of the source is 100 volts
:
A

V = 100 \times 2 s i n 100 π t c o s 100 π t = 100 s i n 200 π t

\Rightarrow V_{0} = 100

Volts and Frequency =100Hz

Question 20. An alternating e.m.f. of angular frequency is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency

Discuss Question

Answer: Option D. -> 2ω
:
D
The instantaneous values of emf and current in inductive circuit are given by E=