In an AC circuit,the emf (e) and the current | vedclass

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(C) The instantaneous power $p$ in an $AC$ circuit is given by the product of instantaneous $emf$ and current:$p = e \cdot i = (E_{0} \sin \omega t) \

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$\frac{E_{0} I_{0}}{2} \cos \phi$

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(C) The instantaneous power $p$ in an $AC$ circuit is given by the product of instantaneous $emf$ and current:$p = e \cdot i = (E_{0} \sin \omega t) \cdot (I_{0} \sin (\omega t - \phi))$Using the trigonometric identity $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$,we get:$p = \frac{E_{0} I_{0}}{2} [\cos \phi - \cos(2\omega t - \phi)]$The average power $P_{av}$ over one complete cycle $T$ is the average of $p$ over time $T$:$P_{av} = \frac{1}{T} \int_{0}^{T} p \, dt = \frac{E_{0} I_{0}}{2T} \int_{0}^{T} [\cos \phi - \cos(2\omega t - \phi)] \, dt$Since the average of $\cos(2\omega t - \phi)$ over a full cycle is $0$,the expression simplifies to:$P_{av} = \frac{E_{0} I_{0}}{2} \cos \phi$Here,$\cos \phi$ is known as the power factor of the $AC$ circuit.

In an $AC$ circuit,the $emf$ $(e)$ and the current $(i)$ at any instant are given respectively by:
$e = E_{0} \sin \omega t$
$i = I_{0} \sin (\omega t - \phi)$
The average power in the circuit over one cycle of $AC$ is:

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In an $AC$ circuit,the $emf$ $(e)$ and the current $(i)$ at any instant are given respectively by:$e = E_{0} \sin \omega t$$i = I_{0} \sin (\omega t - \phi)$The average power in the circuit over one cycle of $AC$ is: