The potential difference V and the current i | vedclass

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(A) The average power dissipated in an $AC$ circuit is given by the formula $P = V_{rms} I_{rms} \cos \phi$,where $V_{rms}$ and $I_{rms}$ are the root

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(A) The average power dissipated in an $AC$ circuit is given by the formula $P = V_{rms} I_{rms} \cos \phi$,where $V_{rms}$ and $I_{rms}$ are the root-mean-square values of voltage and current,and $\phi$ is the phase difference between them.Alternatively,the instantaneous power is $p(t) = V(t) \cdot i(t)$. The average power is the integral of $p(t)$ over one complete cycle divided by the period $T$.Given $V = 5 \cos \omega t$ and $i = 2 \sin \omega t$.We know that $\cos \omega t = \sin(\omega t + \frac{\pi}{2})$.Thus,$V = 5 \sin(\omega t + \frac{\pi}{2})$.The phase of voltage is $\phi_V = \frac{\pi}{2}$ and the phase of current is $\phi_i = 0$.The phase difference $\phi = \phi_V - \phi_i = \frac{\pi}{2}$.The average power dissipated is $P = V_{rms} I_{rms} \cos \phi$.Since $\cos(\frac{\pi}{2}) = 0$,the average power dissipated $P = 0 	ext{ W}$.

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

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