Alternating Current(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

ALTERNATING CURRENT MCQs

Total Questions : 30 | Page 1 of 3 pages

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

$V = \sqrt{V_{R}^{2} + V_{L}^{2}}$
$(100)^{2} = (10)^{2} + V_{L}^{2} \Rightarrow V_{L} = 99.5$ Volt
Also $V_{L} = i X_{L} = i \times (2 π v L)$
$\Rightarrow 99.5 = 6 \times 2 \times 3.14 \times 50 \times L \Rightarrow L = 0.052$ H

Question 2.

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 3.

In the circuit shown below, what will be the readings of the voltmeter and ammeter

800 V, 2A
220 V, 1.1A
220 V, 2.2 A
100 V, 2A

Discuss Question

Answer: Option B. -> 220 V, 1.1A
:
B

$V = V_{R}^{2} + (V_{L} - V_{c})^{2} \Rightarrow V_{R} = V = 220 V$
Also $X_{L} = X_{c},$ Since $(V_{L} - V_{c})$ and current is same

Also $i = \frac{220}{200} = 1.1 A$

Question 4.

In the circuit given below, what will be the reading of the voltmeter

300 V
900 V
200 V
400 V

Discuss Question

Answer: Option C. -> 200 V
:
C

$V^{2} = V_{R}^{2} + (V_{L} - V c)^{2}$
Since $V_{L} = V_{c}$ hence $V = V_{R} = 200$ V

Question 5.

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

$\frac{ω}{4}$
$\frac{ω}{2}$
$ω$
2 $ω$

Discuss Question

Answer: Option D. -> 2

ω

:
D

The instantaneous values of emf and current in inductive circuit are given by E= $E_{0}$ sin $ω$ t and $i = i_{0}$ sin $(ω t - \frac{π}{2})$ respectively.
So, $P_{i n s t} = E_{i} = E_{0} s i n ω t \times i_{0} s i n (ω t - \frac{π}{2})$
$= E_{0} i_{0} s i n ω t (s i n ω t c o s \frac{π}{2} - c o s ω t s i n \frac{π}{2})$
$= E_{0} i_{0} s i n ω t c o s ω t$
$= \frac{1}{2} E_{0} i_{0} s i n 2 ω t$ $(s i n 2 ω t = 2 s i n ω t c o s ω t)$
Hence, angular frequency of instantaneous power is 2 $ω$ .

Question 6.

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 $\frac{100}{\sqrt{2}}$ 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 7.

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

0V, 0.47A
1.68V, 0.47A
8.46V, 1.4 A
5.6V, 1.4 A

Discuss Question

Answer: Option C. -> 8.46V, 1.4 A
:
C

$Z = \sqrt{(R)^{2} + (X_{L} - X_{c})^{2}}$
$R = 12 Ω, X_{L} = w L = 2000 \times 5 \times 10^{- 3} = 10 Ω$
$X_{c} = \frac{1}{w C} = \frac{1}{2000 \times 50 \times 10^{- 6}} = 10 Ω$ i.e.Z = 10 $Ω$
Maximum current $i_{0} = \frac{V_{0}}{Z} = \frac{24}{12} = 2 A$
Hence $i_{r m s} = \frac{2}{\sqrt{2}} = 1.4 A$
and $V_{r m s} = 6 \times 1.41 = 8.46$ V

Question 8.

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 9.

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
$3 \sqrt{3}$ A
$2 \sqrt{3}$ A
$(2 - \sqrt{2})$ A

Discuss Question

Answer: Option C. ->

2 \sqrt{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 10.

Match the following
$\begin{matrix} C u r r e n t s & r . m s . v a l u e s (1) x_{0} s i n ω t & (i) x_{0} (2) x_{0} s i n ω t c o s ω t & (i i) \frac{x_{0}}{\sqrt{2}} (3) x_{0} s i n ω t + x_{0} c o s ω t & (i i i) \frac{x_{0}}{(2 \sqrt{2})} \end{matrix}$

1. (i), 2. (ii), 3(i)
1. (ii), 2. (i), 3. (ii)
1. (ii), 2. (iii), 3. (i)
1. (ii), 2. (ii), 3. (i)

Discuss Question

Answer: Option C. -> 1. (ii), 2. (iii), 3. (i)
:
C

1. rms values $= \frac{x_{0}}{\sqrt{2}}$
2. $x_{0} s i n ω t c o s ω t = \frac{x_{0}}{2} s i n 2 ω t \Rightarrow$ rms value = $\frac{x_{0}}{2 \sqrt{2}}$
3. $x_{0} s i n ω t + x_{0} c o s ω t \Rightarrow$ rms value = $\sqrt{{(\frac{x_{0}}{\sqrt{2}})}^{2} + {(\frac{x_{0}}{\sqrt{2}})}^{2}}$
$= \sqrt{x_{0}^{2}} = x_{0}$