If a current I given by I = I_0 sin(omega t - | vedclass

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(D) The given potential is $E = E_0 \sin(\omega t)$.The given current is $I = I_0 \sin(\omega t - \frac{\pi}{2})$.The phase difference $\phi$ between

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$P = 0$

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(D) The given potential is $E = E_0 \sin(\omega t)$.The given current is $I = I_0 \sin(\omega t - \frac{\pi}{2})$.The phase difference $\phi$ between the voltage and the current is $\phi = \omega t - (\omega t - \frac{\pi}{2}) = \frac{\pi}{2}$ or $90^{\circ}$.The average power consumption in an $AC$ circuit is given by the formula $P = E_{rms} I_{rms} \cos \phi$.Since $\phi = 90^{\circ}$,the power factor $\cos \phi = \cos(90^{\circ}) = 0$.Therefore,the power consumption $P = E_{rms} I_{rms} 	imes 0 = 0$.

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

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