Question 352: AC fundamentals

Question

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 $\rightarrow \cos \frac{\pi}{3}=\frac{1}{2}$]

Accepted Answer

$$\begin{aligned}
& V=100 \sin (100 \mathrm{t}) \mathrm{V} \
& \mathrm{I}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}ight) \mathrm{mA}
\end{aligned}$$
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,
$$\mathrm{V}_0=100 \mathrm{~V}, \mathrm{I}_0=100 	imes 10^{-3} \mathrm{~A} 	ext { and } \phi=\frac{\pi}{3}$$
$$\begin{aligned}
	herefore \quad \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi & =\frac{\mathrm{V}_0}{\sqrt{2}} 	imes \frac{\mathrm{I}_0}{\sqrt{2}} \cos \frac{\pi}{3} \
& =\frac{100}{\sqrt{2}} 	imes \frac{100 	imes 10^{-3}}{\sqrt{2}} 	imes \frac{1}{2}=2.5 \mathrm{~W}
\end{aligned}$$

Answer

$10^4 \mathrm{~W}$

Answer

10 W

Answer

2.5 W

Answer

5 W

AC fundamentals

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