Question

If a current $I$ is given by $I=I_0\sin (\omega t-\frac{\pi }{2})$ flows in an a.c. circuit across which an a.c. potential of $E=E_0\sin \omega t$ has been applied, then the power consumption $P$ in the circuit will be-

• A. $P=\frac{E_0{I}_0}{\sqrt{2}}$
• B. $P=\sqrt{2}{E}_0{I}_0$
• C. $P=\frac{E_0{I}_0}{2}$
• D. $P=0$

Answer 1

The correct option is D $P=0$

Given, $I=I_0\sin (\omega t-\frac{\pi }{2})E=E_0\sin \omega t\text{Phase difference between V and I},\varphi =\frac{\pi }{2}$

Power delivered in an a.c. circuit is,

$P=V_{rms}{I}_{rms}\cos \varphi$

$P=0[\because \varphi =\frac{\pi }{2}={90}^{\circ }]$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.