Question

The potential difference V and the current I flowing through an instrument in an AC circuit are given by $V=5\cos \omega tV$, $I=2\sin \omega tA$. The power dissipated in the circuit is

• A. 5 watt
• B. 10 watt
• C. 2.5 watt
• D. 0

Answer 1

The correct option is D

0

Step 1: Given

$I=2\sin \omega tA$

$V=5\cos \omega tV$

Step 2: Calculating power dissipated.

Power in AC circuit, $P=V_{rms}·I_{rms}·cos\varphi$

$\varphi$ = phase difference between the current and the voltage.

Power in the circuit can be calculated as,

$P=5\cos \omega t\times 2\sin \omega t\times \cos \varphi$

Here

$\omega =2\pi f$

$\varphi$ = $\frac{\mathrm{\pi }}{2}$

$\cos \varphi =\cos \frac{\pi }{2}=0$

$P=5\cos \omega t\times 2\sin \omega t\times 0$

$P=0$

Final Answer: Correct option is (d): 0.