In an A.C. circuit, the potential difference ' \(V\) ' and current 'I' are given respectively by \(\mathrm{V}=100 \sin (100 \mathrm{t}) \mathrm{V}, \mathrm{I}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}\)
The power dissipated in the circuit will be [Given \(\rightarrow \cos \frac{\pi}{3}=\frac{1}{2}\) ]

\(\begin{aligned}
& V=100 \sin (100 t) V \\
& I=100 \sin \left(100 t+\frac{\pi}{3}\right) \mathrm{mA}
\end{aligned}\)
Comparing these equations with the standard forms, \(\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}\) and \(\mathrm{I}=\mathrm{I}_0 \sin \omega \mathrm{t}\)
we get, \(\mathrm{V}_0=100 \mathrm{~V}, \mathrm{I}_0=100 \times 10^{-3} \mathrm{~A}\) and \(\phi=\frac{\pi}{3}\)
\(\therefore \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi\) \(=\frac{\mathrm{V}_0}{\sqrt{2}} \times \frac{\mathrm{I}_0}{\sqrt{2}} \cos \frac{\pi}{3} \)
\( =\frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \frac{1}{2}=2.5 \mathrm{~W}\)