(B) The instantaneous power in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$, where $\phi$ is the phase difference between voltage and c

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(B) The instantaneous power in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$, where $\phi$ is the phase difference between voltage and current.Given voltage $V = V_{0} \sin \omega t$ and current $I = I_{0} \sin (\omega t - \frac{\pi}{2})$.The phase difference $\phi = \frac{\pi}{2}$.The power consumed is $P = V_{rms} I_{rms} \cos \phi = \frac{V_{0}}{\sqrt{2}} \cdot \frac{I_{0}}{\sqrt{2}} \cdot \cos(\frac{\pi}{2})$.Since $\cos(\frac{\pi}{2}) = 0$, the power consumed $P = 0$ watt.

Class 12 Physics Alternating Current Power in AC and Power Factor

Easy

An alternating voltage of $V = V_{0} \sin \omega t$ is applied across a circuit. As a result, a current $I = I_{0} \sin (\omega t - \frac{\pi}{2})$ flows in it. The power consumed per cycle is . . . . . . .

GUJCET 2006 All GUJCET

A
$1.919 V_{0} I_{0}$ watt
B
$0$ watt
C
$0.5 V_{0} I_{0}$ watt
D
$0.707 V_{0} I_{0}$ watt

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Alternating Current

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Power in AC and Power Factor

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GUJCET 2006 Paper All GUJCET years →

On which of the following does the power consumed in an $L-C-R$ series $AC$ circuit depend?

Difficult

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In an $AC$ circuit $E = 50 \sin(500t)$ $V$, $I = 600 \sin(500t + \frac{\pi}{3})$ mA. What is the power dissipated in the circuit (in $\text{ W}$)? [$\cos 60^{\circ} = 0.5$]

DifficultMHT CET 2026

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In an $AC$ circuit $E = 200 \sin(50t) \text{ V}$ and $I = 100 \sin(50t + \frac{\pi}{3}) \text{ mA}$. The power dissipated in the circuit is (Given: $\sin 30^{\circ} = \cos 60^{\circ} = 0.5$,$\sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$) (in $watt$)

EasyMHT CET 2024

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In an $AC$ circuit,$V$ and $I$ are given below. Find the power dissipated in the circuit:
$V = 50 \sin(50t) \ V$
$I = 50 \sin(50t + \frac{\pi}{3}) \ mA$ (in $W$)

EasyAP EAMCET 2015

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The $r.m.s.$ current in an $ac$ circuit is $2 \ A$. If the wattless current is $\sqrt{3} \ A$,the power factor is:

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An alternating voltage of $V = V_{0} \sin \omega t$ is applied across a circuit. As a result, a current $I = I_{0} \sin (\omega t - \frac{\pi}{2})$ flows in it. The power consumed per cycle is . . . . . . .

Similar Questions

On which of the following does the power consumed in an $L-C-R$ series $AC$ circuit depend?

In an $AC$ circuit $E = 50 \sin(500t)$ $V$, $I = 600 \sin(500t + \frac{\pi}{3})$ mA. What is the power dissipated in the circuit (in $\text{ W}$)? [$\cos 60^{\circ} = 0.5$]

In an $AC$ circuit $E = 200 \sin(50t) \text{ V}$ and $I = 100 \sin(50t + \frac{\pi}{3}) \text{ mA}$. The power dissipated in the circuit is (Given: $\sin 30^{\circ} = \cos 60^{\circ} = 0.5$,$\sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$) (in $watt$)

In an $AC$ circuit,$V$ and $I$ are given below. Find the power dissipated in the circuit: $V = 50 \sin(50t) \ V$ $I = 50 \sin(50t + \frac{\pi}{3}) \ mA$ (in $W$)

The $r.m.s.$ current in an $ac$ circuit is $2 \ A$. If the wattless current is $\sqrt{3} \ A$,the power factor is:

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In an $AC$ circuit,$V$ and $I$ are given below. Find the power dissipated in the circuit:
$V = 50 \sin(50t) \ V$
$I = 50 \sin(50t + \frac{\pi}{3}) \ mA$ (in $W$)