In an AC circuit the emf (e) and the current (i) at any instant are given respectively by e = E₀ sin ωt, i = I

Question

In an AC circuit the emf (e) and the current (i) at any instant are given respectively by e = E₀ sin ωt, i = I₀ sin(ωt - ϕ). The average power in the circuit over one cycle of AC is

Accepted Answer

The instantaneous power is dissipated as $p = ei = E_0 \sin(\omega t) 	imes I_0 \sin(\omega t - \phi)$. Using the trigonometric identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$, the average power over a cycle is determined. The term containing $2\omega t$ averages to zero over a complete cycle, leaving only the constant term. Thus, the average power is $P = \frac{1}{2} E_0 I_0 \cos \phi$ .

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