In an A.C. circuit,the e.m.f. (varepsilon) an | vedclass

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(B) The instantaneous power $p$ in an $A.C.$ circuit is given by the product of instantaneous $e.m.f.$ and instantaneous current:$p = E \cdot i = (E_o

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$P = \frac{E_o I_o}{2} \cos \phi$

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(B) The instantaneous power $p$ in an $A.C.$ circuit is given by the product of instantaneous $e.m.f.$ and instantaneous current:$p = E \cdot i = (E_o \sin \omega t) \cdot (I_o \sin(\omega t - \phi))$Using the trigonometric identity $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$,we get:$p = \frac{E_o I_o}{2} [\cos \phi - \cos(2\omega t - \phi)]$The average power $P_{av}$ over one complete cycle $T$ is the average of the instantaneous power:$P_{av} = \frac{1}{T} \int_0^T p \, dt = \frac{1}{T} \int_0^T \frac{E_o I_o}{2} [\cos \phi - \cos(2\omega t - \phi)] \, dt$Since the average of $\cos(2\omega t - \phi)$ over a complete cycle is $0$,we have:$P_{av} = \frac{E_o I_o}{2} \cos \phi$Thus,the average power is $\frac{E_o I_o}{2} \cos \phi$.

In an $A.C.$ circuit,the $e.m.f.$ $(\varepsilon)$ and the current $(i)$ at any instant are given respectively by $E = E_o \sin \omega t$ and $i = I_o \sin(\omega t - \phi)$. The average power in the circuit over one cycle of $A.C.$ is:

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