Question

Obtain expression for an a.c. circuit:
$P_{av}=V_{rms}\times I_{rms}\cos \varphi$

Answer 1

Let the potential difference and current of A.C. circuit is given by:
$V=V_0\sin \omega t$
$I=I_0\sin (\omega t-\varphi )$
Where $\theta$ is phase difference between e.m.f. and current. Now instantaneous power $P=VI$:
$P=V_0\sin \omega t\cdot I_0\sin (\omega t-\varphi )$
$P=V_0{I}_0\sin \omega t\cdot \sin (\omega t-\varphi )$
$P=\frac{1}{2}{V}_0{I}_{0}2\sin \omega t\sin (\omega t-\varphi )$
$P=\frac{1}{2}{V}_0{I}_0[\cos \{\omega t-(\omega t-\varphi )\}-\cos \{\omega t+(\omega t-\varphi )\}]$
$P=\frac{V_0}{\sqrt{2}}\cdot \frac{I_0}{\sqrt{2}}[\cos \varphi -\cos (2\omega t-\varphi )]$
where $2\sin A\sin B=\cos (A-B)-\cos (A+B)$
$P=V_{rms}{I}_{rms}[\cos \varphi -0]$
where $\cos (2\omega t-\varphi )$ average value for a complete cycle is zero.
$P_{av}=\frac{1}{2}{V}_0{I}_0\cos \varphi$
$\therefore$ Average power for a complete cycle is:
$P_{av}=V_{ams}\times I_{rms}\cos \varphi$
where $\frac{I_0}{\sqrt{2}}=I_{rms}$, $\frac{V_0}{\sqrt{2}}=V_{rms}$, are average AC e.m.f. and current.