Question 304: AC fundamentals

Question

In LCR series circuit, an alternating e.m.f. 'e' and current ' $i$ ' are given by equations $\mathrm{e}=160 \sin (100 \mathrm{t})$ Volt and $\mathrm{i}=250 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}$.
The average power dissipated in the circuit is

Accepted Answer

$e=160 \sin (100 t)$ volt and
$\mathrm{i}=250 \sin \left(100 \mathrm{t}+\frac{\pi}{3}ight) \mathrm{mA}$
Comparing given equations with the standard forms,
$e=e_0 \sin \omega t$ and $\mathrm{i}=\mathrm{i}_0 \sin (\omega \mathrm{t}+\phi)$ we get,
$$\begin{aligned}
\mathrm{e}_0= & 160 \mathrm{~V}, \mathrm{I}_0=250 \mathrm{~mA} \
	herefore \quad 	ext { Power } & =\frac{\mathrm{e}_0}{\sqrt{2}} \cdot \frac{\mathrm{I}_0}{\sqrt{2}} \cos \phi \
& =\frac{160 	imes 250 	imes 10^{-3}}{2} 	imes \cos \left(\frac{\pi}{3}ight) \
& =\frac{40}{2} 	imes \frac{1}{2} \
& =10 \mathrm{~W}
\end{aligned}$$

Answer

2.5 W

Answer

4.0 W

Answer

10 W

Answer

100 W

AC fundamentals

In LCR series circuit, an alternating e.m.f. 'e' and current ' i ' are given by equations \mathrm{e}=160 \sin (100 \mathrm{t}) Volt and \mathrm{i}=250 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}. The average power dissipated in the circuit is

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