Physics - EMI/AC Question with Solution | TestHub

The potential difference is $V=5\cos (\omega t-\pi /6)$ V.

The current is $I=10\sin (\omega t)$ A, which can be written as $I=10\cos (\omega t-\pi /2)$ A.

The peak voltage is $V_0=5$ V and peak current is $I_0=10$ A.

The phase difference is $\varphi =(\omega t-\pi /6)-(\omega t-\pi /2)=\pi /2-\pi /6=3\pi /6-\pi /6=2\pi /6=\pi /3$.

The average power is $P_{avg}=\frac{V_0{I}_0}{2}\cos \varphi$.

$P_{avg}=\frac{5\times 10}{2}\cos (\pi /3)=25\times \frac{1}{2}=12.5$ W.