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
$e=\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}\right)$ Volt
The average power in watts consumed in the circuit is

• A. $\frac{1}{4}$
• B. $\frac{\sqrt3}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{8}$

Answer 1

Answer: (D)

Given $: i=\frac{1}{\sqrt{2}} \sin (100 \pi t)$ ampere
Compare it with $i=i_{0} \sin (\omega t)$ we get
$i_{0}=\frac{1}{\sqrt{2}} A$
Given $: e=\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}\right)$ volt
Compare it with, we get
$e_{0}=\frac{1}{\sqrt{2}} V, \phi=\frac{\pi}{3} $
$\therefore i_{r m s}=\frac{i_{0}}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2}} A=\frac{1}{2} A$
$e_{r m s}=\frac{e_{0}}{\sqrt{2}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} V=\frac{1}{2} V$
Average power consumed in the circuit,
$P=i_{r m s} e_{r m s} \cos \phi$
$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \cos \frac{\pi}{3} $
$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{8} W$