In an A.C. circuit, the potential difference ' $V$ ' and current 'I' are given respectively by…
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 $\rightarrow \cos \frac{\pi}{3}=\frac{1}{2}$ ]
$10^4 \mathrm{~W}$
10 W
2.5 W
5 W
Solution
$\begin{aligned}
& V=100 \sin (100 t) V \\
& I=100 \sin \left(100 t+\frac{\pi}{3}\right) \mathrm{mA}
\end{aligned}$
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 $\phi=\frac{\pi}{3}$
$\begin{aligned} \therefore \quad \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi & =\frac{\mathrm{V}_0}{\sqrt{2}} \times \frac{\mathrm{I}_0}{\sqrt{2}} \cos \frac{\pi}{3} \\ & =\frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \frac{1}{2}=2.5 \mathrm{~W}\end{aligned}$