In LCR series circuit, an alternating e.m.f. 'e' and current ' $i$ ' are given by equations $\mathrm{e}=160…
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
$\quad 2.5 \mathrm{~W}$
4.0 W
10 W
100 W
Solution
$e=1.60 \sin (100 t)$ volt and
$\mathrm{i}=250 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \quad \mathrm{mA}$
Comparing given equations with the standard forms, $e=e_0 \sin \omega t$ and $i=i_0 \sin (\omega t+\phi)$ we get,
$\begin{aligned}
& \mathrm{e}_0=160 \mathrm{~V}, \mathrm{I}_0=250 \mathrm{~mA} \\
& \therefore \quad \text { Power }=\frac{\mathrm{e}_0}{\sqrt{2}} \cdot \frac{\mathrm{I}_0}{\sqrt{2}} \cos \phi \\
&=\frac{160 \times 250 \times 10^{-3}}{2} \times \cos \left(\frac{\pi}{3}\right) \\
&=\frac{40}{2} \times \frac{1}{2} \\
&=10 \mathrm{~W}
\end{aligned}$