Physics - EMI/AC Question with Solution | TestHub

The average power consumed in an AC circuit is given by

$\mathrm{P}=\frac{{\mathrm{V}}_0{\mathrm{I}}_0}{2}\cos \mathrm{\varphi }$

Where$\mathrm{\varphi }$is phase angle and${\mathrm{V}}_0$and${\mathrm{I}}_0$are the peak values of voltage and current respectively.

Given,${\mathrm{V}}_0=200 \mathrm{V},\mathrm{ }{\mathrm{I}}_0=2\mathrm{ }\mathrm{A},\mathrm{ }\mathrm{\varphi }=\frac{\mathrm{\pi }}{3}$.

$\mathrm{P}=\frac{200\times 2}{2}\cos \frac{\mathrm{\pi }}{3}$

$=\frac{200\times 2}{2}\times \frac{1}{2}=100 \mathrm{W}$