Q.In an AC circuit, e and i are given by $e = 150\sin(150t)$ V and $i = 150\sin(150t + \frac{\pi}{3})$ A. The power dissipated in the circuit is (A) 106 W (B) 150 W (C) 5625 W (D) Zero

The average power in an AC circuit is not simply $V_{rms}I_{rms}$ but includes the power factor $\cos\varphi$ accounting for the phase lag between current and voltage.

In a series LCR circuit driven by $v = V_0\sin\omega t$, the current is $i = I_0\sin(\omega t - \varphi)$, where $\varphi$ is the phase angle between voltage and current, $\tan\varphi = (X_L-X_C)/R$.

Instantaneous power:

$p = vi = V_0 I_0 \sin\omega t\,\sin(\omega t - \varphi)$

Using $\sin(\omega t - \varphi) = \sin\omega t\cos\varphi - \cos\omega t\sin\varphi$:

$p = V_0 I_0\left[\sin^2\omega t\cos\varphi - \sin\omega t\cos\omega t\sin\varphi\right]$

Averaging over a full cycle: $\langle\sin^2\omega t\rangle = \tfrac12$ and $\langle\sin\omega t\cos\omega t\rangle = 0$, so

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