Question

The instantaneous values of alternating current and voltages in a circuit are given as,
$i=\frac{1}{\sqrt{2}}\sin \left(100\pi t\right)\text{A}$ and
$E=\frac{1}{\sqrt{2}}\sin (100\pi t+\frac{\pi }{3})\text{V}$
The average power (in watts) consumed in the circuit is,

• A. $\frac{1}{8}$
• B. $\frac{\sqrt{3}}{4}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{8}$

Given: $i=\frac{1}{\sqrt{2}}\sin \left(100\pi t\right)\text{A}$

Comparing it with $i=i_0\sin \left(\omega t\right)$, we get, $i_0=\frac{1}{\sqrt{2}}\text{A}$

Also, $E=\frac{1}{\sqrt{2}}\sin (100\pi t+\frac{\pi }{3})\text{volt}$

Comparing it with, $E=E_0\sin \left(\omega t\right)$, we get,
$E_0=\frac{1}{\sqrt{2}}\text{V},\varphi =\frac{\pi }{3}$

$\therefore i_{rms}=\frac{i_0}{\sqrt{2}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}\text{A}$

$E_{rms}=\frac{E_0}{\sqrt{2}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}\text{V}$

Average power consumed in the circuit,
$P=i_{rms}{E}_{rms}\cos \varphi$

$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\cos \frac{\pi }{3}$

$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{8}\text{W}$

Hence, $\left(D\right)$ is the correct answer.