In an AC circuit,the current flowing is I=5 s | vedclass

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(D) Given:$I = 5 \sin \left(100 t - \frac{\pi}{2}\right) A$$e = 200 \sin (100 t) V$Comparing these with the standard equations $I = I_0 \sin(\omega t

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(D) Given:$I = 5 \sin \left(100 t - \frac{\pi}{2}ight) A$$e = 200 \sin (100 t) V$Comparing these with the standard equations $I = I_0 \sin(\omega t + \phi_1)$ and $e = E_0 \sin(\omega t + \phi_2)$,we get the phase of current $\phi_1 = -\frac{\pi}{2}$ and the phase of voltage $\phi_2 = 0$.The phase difference $\phi = \phi_2 - \phi_1 = 0 - (-\frac{\pi}{2}) = \frac{\pi}{2}$.The average power consumption in an $AC$ circuit is given by $P = V_{	ext{rms}} I_{	ext{rms}} \cos(\phi)$.Since $\phi = \frac{\pi}{2}$,we have $\cos(\phi) = \cos(\frac{\pi}{2}) = 0$.Therefore,$P = V_{	ext{rms}} I_{	ext{rms}} 	imes 0 = 0 W$.

In an $AC$ circuit,the current flowing is $I=5 \sin \left(100 t-\frac{\pi}{2}\right) A$ and the potential difference is $e=200 \sin (100 t) V$. The power consumption is equal to (in $W$)

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