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(D) The given current is $i = 5 \sin \left( 100 t - \frac{\pi}{2} \right)$ and the potential is $V = 200 \sin (100 t)$.Comparing these with the standa

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(D) The given current is $i = 5 \sin \left( 100 t - \frac{\pi}{2} ight)$ and the potential is $V = 200 \sin (100 t)$.Comparing these with the standard forms $i = I_0 \sin(\omega t + \phi_1)$ and $V = V_0 \sin(\omega t + \phi_2)$,we get the phase of current $\phi_1 = -\frac{\pi}{2}$ and the phase of voltage $\phi_2 = 0$.The phase difference $\phi$ between voltage and current is $\phi = \phi_2 - \phi_1 = 0 - \left( -\frac{\pi}{2} ight) = \frac{\pi}{2}$.The average power consumption in an $ac$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.Since $\phi = \frac{\pi}{2}$,the power factor $\cos \phi = \cos \left( \frac{\pi}{2} ight) = 0$.Therefore,the power consumption $P = V_{rms} I_{rms} 	imes 0 = 0 	ext{ watts}$.

In an $ac$ circuit,the current is given by $i = 5 \sin \left( 100 t - \frac{\pi}{2} \right)$ and the $ac$ potential is $V = 200 \sin (100 t) \text{ volts}$. Then the power consumption is $....... \text{ watts}$.

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