Question

Prove that for A.C. circuit:
$P_{av}=V_{r.m.s.}\times I_{r.m.s.}cos\varphi$

Answer 1

Let the potential difference and current A.C. circuit is given by

$V=V_{0}sin\omega t$

$I=I_{0}sin(\omega t-\varphi )$

Where $\varphi$ is phase difference between e.m.f. and current. Now instantaneous power P = VI.

$P=V_{0}sin\omega t.I_{0}sin(\omega t-\varphi )$

$P=V_0{I}_{0}sin\omega t.sin(\omega t-\varphi )$

$P=\frac{1}{2}{V}_0{I}_{0}2sin\omega tsin(\omega t-\varphi )$

$P=\frac{1}{2}{V}_0{I}_0\left[cos\right\{\omega t-(\omega t-\varphi )\}-cos\{\omega t+(\omega t-\varphi )\left\}\right]$

$P=\frac{V_0}{\sqrt{2}}.\frac{I_0}{\sqrt{2}}[cos\varphi -cos(2\omega t-\varphi \left)\right]$

Where $2sinAsinB=cos(A-B)-cos(A+B)$

$P=V_{rms}{I}_{rms}[cos\varphi -0]$

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

$P_{av}=V_{rms}\times I_{rms}cos\varphi$

Where, $\frac{I_0}{\sqrt{2}}=I_{rms},\frac{V_0}{\sqrt{2}}=V_{rms}$, are average AC e.m.f. and current.