MCQ on Power of Alternating Current: Introduction & Explanation

Jasmine Grover Content Strategy Manager

Content Strategy Manager

The power of alternating current is defined as the ratio of true power flowing through the circuit to the apparent power present in the circuit. The power is the product of voltage and time. The power of the alternating current cannot be calculated because in an AC circuit, the EMF and current change continuously with respect to time. Thus, to obtain the average power of an alternating current circuit, we need to calculate the instantaneous power of the circuit.

The formula for average power in an AC circuit is given by,

\(P_{av} = \int\limits_0^T VI\ dt/ \int\limits_0^T dt\)

Power of Alternating Current MCQs

Ques 1. The direction of current in an AC circuit is:

Always positive
Always negative
Either positive or negative
Changes constantly

Click here for the answer

Ans. (d). Changes constantly

Explanation: The direction of current in an AC circuit changes constantly. The voltage also reverses periodically in an AC circuit.

Ques 2. The average power supplied to an inductor over one complete alternating current cycle is:

0
IV²
VI²
IR²

Click here for the answer

Ans. (a). 0

Explanation: For a pure inductor circuit, ϕ = 90° (∵ current lags the voltage by 90° in the pure inductive circuit)

Cos ϕ = Cos 90° = 0

P = V_rmsI_rms 0

P = 0 W

Ques 3. In an AC circuit, V = 100 sint V and I = 100 sin (100t + \(\sqrt 2\)/3) A, then average power loss will be:

10⁴ W
2.5 kW
10 W
100 W

Click here for the answer

Ans. (b). 2.5 kW

Explanation: Here, potential difference (V) = 100 sint V and I = 100 sin (100t + \(\sqrt 2\)/3) A

So, peak voltage (E_o) = 100, peak current (I_o) = 100 and phase difference (θ) = \(\sqrt 2\)/3

The average power loss in an AC circuit is given by;

P = V_rmsI_rmsCosθ

P = V_o/\(\sqrt 2\) I_o/\(\sqrt 2\) Cosθ

P = 100/\(\sqrt 2\)2 100/\(\sqrt 2\)2 cos \(\sqrt 2\)/3

P = 2.5 x 10³ W

P = 2.5 kW

Ques 4. In a series LCR circuit, alternating emf (v) and current are given by v = vo sin t, i = io sin (t + ????/3). The average power dissipated in the circuit over a cycle of AC is:

v_oi_o/2
v_oi_o/4
3/2 v_oi_o
Zero

Click here for the answer

Ans. (b). v_oi_o/4

Explanation: v = v_o sin (t)

i = i_o sin (t + \(\sqrt 2\)/3)

Phase difference (θ) = \(\sqrt 2\)/3

Peak current (I) = i_o

Peak voltage (V) = v_o

rms current (I_rms) = I/\(\sqrt 2\) = i_o/\(\sqrt 2\)

rms voltage (V_rms) = V/\(\sqrt 2\) = vo/\(\sqrt 2\)

P_avg = V_rmsI_rmsCosθ

Average power (P_avg) = v_o/\(\sqrt 2\) i_o/\(\sqrt 2\) Cos\(\sqrt 2\)/3

P_avg = v_oi_o/2 x ½ = v_oi_o/4

Ques 5. A series LCR circuit with R = 40, L = 30 H and C = 25F is connected to a 200 V AC supply. At resonance what will be the average power transferred to the circuit in one complete cycle?

1000 W
2000 W
3000 W
4000 W

Click here for the answer

Ans. (a). 1000 W

Explanation: Given, R = 40 \(\Omega\), L = 30 H, C = 25\(\mu\)F and E_rms = 200 V

At resonance, X_L = X_C

Z = \(\sqrt R ^2\) + \(\sqrt {(X_L - X_C)^2}\) = \(\sqrt 40 ^2\) = 40

Average power dissipated per cycle is,

P_av = E²_rms / Z = (200)2/40 = 1000 W

Ques 6. A coil of inductive reactance 31 has a resistance of 8. It is placed in series with a condenser of capacitance reactance 25. The combination is connected to an AC source of 110 V. The power factor of the circuit is:

0.4
0.64
0.8
1.6

Click here for the answer

Ans. (c). 0.8

Explanation: R = 8\(\Omega\), XC = 25, XL = 31 and V = 110 V

Z = \(\sqrt R ^2\) + \(\sqrt {(X_L - X_C)^2}\)

Z = \(\sqrt 8 ^2\) + \(\sqrt {(31 - 25)^2}\) = \(\sqrt {(64+36)}\)

Z = 10\(\Omega\)

The power factor (P) of a series LCR-circuit is given by,

Cos ϕ = R/Z

Cos ϕ = 8/10 = 0.8

Ques 7. Average power in an LCR circuit depends upon:

Current
Current, emf, and phase difference
Emf
Phase difference

Click here for the answer

Ans. (b). Current, emf, and phase difference

Explanation: The average power dissipated in an LCR circuit is given by;

P_avg = V_rmsI_rmsCosθ

Thus, the average power depends upon current, emf, and phase difference.

Ques 8. Power loss in an AC circuit is measured by:

Power factor
Mean current
Impedance
None of these

Click here for the answer

Ans. (a). Power factor

Explanation: The power factor in an AC circuit is defined as the ratio of the actual power used by the circuit to the apparent power of the circuit. It is the amount by which power delivered in the circuit is less than the theoretical power of the circuit. Hence, it is a measure of loss of power in an AC circuit.

Ques 9. Which of the following quantities changes in an alternating current?

Direction
Magnitude
Magnitude and direction both
None of the above

Click here for the answer

Ans. (c). Magnitude and direction both

Explanation: In an alternating current, both the direction and the magnitude vary. The waveform of an AC power circuit is usually a sine wave. The sine wave oscillates periodically between positive and negative directions and leads to the most efficient transmission of energy.

Ques 10. The frequency of AC mains in India is:

30 Hz
50 Hz
60 Hz
120 Hz

Click here for the answer

Ans. (b). 50 Hz

Explanation: In most European countries, the frequency of AC mains is 50 Hz. In the USA, it is 60 Hz.

Ques 11. The AC produced in India changes its direction in every:

1/100 second
100 second
50 second
None of the above

Click here for the answer

Ans. (a). 1/100 second

Explanation: The frequency of AC voltage in India is 50 Hz. It means 50 waves are produced in 1 second. In a single wave, the direction changes 2 times. Thus, in 50 waves direction will be changed, 50 2 = 100 times.

This indicates that the direction is changed 100 times in 1 second. Thus, the direction is changed every 1/100 second.

For Latest Updates on Upcoming Board Exams, Click Here:https://t.me/class_10_12_board_updates

Check Also:

Class 12 Study Guides
Physics Study Guide	ChemistryStudy Guide	Biology Study Guide
NCERT Class 12 Physics Book	NCERT Class 12 Biology Book	NCERT Class 12 Maths Book
MCQs in Physics	MCQs in Chemistry	MCQs in Biology
CBSE Class 12 Notes	Topics for Comparison in Physics	Derivations in Physics
Physics Constants	SI Units in Physics	Formulas in Physics
Important Maths Formulas	Comparison Topics in Maths	Important Chemical Reactions
Periodic Table in Chemistry	Comparison Topics in Biology	NCERT Solutions for Class 12 Chemistry
NCERT Solutions for Class 12 Physics	NCERT Solutions for Class 12 Biology	NCERT Solutions for Class 12 Maths
Class 12 Chemistry Notes	Class 12 Biology Notes	Class 12 Maths Notes

MCQ on Power of Alternating Current: Introduction & Explanation

Power of Alternating Current MCQs

CBSE CLASS XII Related Questions

1.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.
An alpha particle and a deuterium ion are accelerated through the same potential difference. These are then directed towards a target nucleus to make a head-on collision. It is observed that their distance of closest approach is the same. Justify it theoretically.

6.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

MCQ on Power of Alternating Current: Introduction & Explanation

Power of Alternating Current MCQs

CBSE CLASS XII Related Questions

1.Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.An alpha particle and a deuterium ion are accelerated through the same potential difference. These are then directed towards a target nucleus to make a head-on collision. It is observed that their distance of closest approach is the same. Justify it theoretically.

4.A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle \(i\) such that it is refracted out grazing the face AD. Find the value of angle \(i\).

6.Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.
An alpha particle and a deuterium ion are accelerated through the same potential difference. These are then directed towards a target nucleus to make a head-on collision. It is observed that their distance of closest approach is the same. Justify it theoretically.

6.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).