Q.Derive the expression $P_{avg}=V_{rms}I_{rms}\cos\phi$ for the average power dissipated, over one complete cycle, in a general series AC circuit carrying current at phase angle $\phi$ relative to the applied voltage.

Instantaneous power. For $v=v_0\sin\omega t$ and $i=i_0\sin(\omega t-\phi)$,

Expanding with a product-to-sum identity. Using $\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]$ with $A=\omega t$, $B=\omega t-\phi$:

Averaging over a full cycle. The first term, $\tfrac12v_0i_0\cos\phi$, does not depend on time at all. The second term oscillates at angular frequency $2\omega$ and, being a pure cosine, averages to exactly zero over any whole number of cycles. So

$P_{avg} = \frac{v_0i_0}{2}\cos\phi$ …

Write out $p=vi$, expand using the product-to-sum identity, discard the zero-average double-frequency term, …

• Averaging $v$ and $i$ separately and multiplying the results, instead of correctly averaging their PRODUCT (which is not the same thing when they are out of phase). …