The instantaneous values of alternating current and voltages in a circuit are given as\ $i=\frac{1}

Question

The instantaneous values of alternating current and voltages in a circuit are given as\ $i=\frac{1}{\sqrt{2}} \sin (100 \pi t)$ ampere $e=\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}ight)$ volt\ The average power in watts consumed in the circuit is\

Professor PC Thomas Classes · Answer

(d) : Given : $i=\frac{1}{\sqrt{2}} \sin (100 \pi t)$ ampere Compare it with $i=i_{0} \sin (\omega t)$, we get $$ i_{0}=\frac{1}{\sqrt{2}} \mathrm{~A} $$ Given: $e=\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}ight)$ volt\ Compare it with $e=e_{0} \sin (\omega t+\phi)$, we get $$ e_{0}=\frac{1}{\sqrt{2}} \mathrm{~V}, \phi=\frac{\pi}{3} $$ $	herefore \quad i_{\mathrm{rms}}=\frac{i_{0}}{\sqrt{2}}=\frac{1}{2} \mathrm{~A}$ and $e_{\mathrm{rms}}=\frac{e_{0}}{\sqrt{2}}=\frac{1}{2} \mathrm{~V}$\ Average power consumed in the circuit,\ $P=i_{	ext {rms }} e_{	ext {rms }} \cos \phi$\ $=\left(\frac{1}{2}ight)\left(\frac{1}{2}ight) \cos \frac{\pi}{3}=\left(\frac{1}{2}ight)\left(\frac{1}{2}ight)\left(\frac{1}{2}ight)=\frac{1}{8} \mathrm{~W}$\