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(A) The instantaneous power $P$ in an $AC$ circuit is given by $P = V \cdot I$. Given $V = 5 \cos(\omega t)$ and $I = 2 \sin(\omega t)$. We can rewrit

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(A) The instantaneous power $P$ in an $AC$ circuit is given by $P = V \cdot I$. Given $V = 5 \cos(\omega t)$ and $I = 2 \sin(\omega t)$. We can rewrite $V$ as $V = 5 \sin(\omega t + \pi/2)$. The phase difference $\phi$ between voltage and current is $\pi/2$. The average power dissipated in an $AC$ circuit is given by $P_{avg} = V_{rms} I_{rms} \cos(\phi)$. Since $\phi = \pi/2$,the power factor $\cos(\phi) = \cos(\pi/2) = 0$. Therefore,the power dissipated $P_{avg} = V_{rms} I_{rms} 	imes 0 = 0 \, W$.

The potential difference $V$ and the current $I$ flowing through an instrument in an $AC$ circuit of frequency $f$ are given by $V = 5 \cos(\omega t)$ volts and $I = 2 \sin(\omega t)$ amperes (where $\omega = 2 \pi f$). The power dissipated in the instrument is......$W$

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