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. $\frac{\sqrt{3}}{4}$
• 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
$e_0=\frac{1}{\sqrt{2}}V,\varphi =\frac{\pi }{3}$
$\therefore i_{rms}=\frac{i_0}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2}}A=\frac{1}{2}A$
$e_{rms}=\frac{e_0}{\sqrt{2}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}V=\frac{1}{2}V$
Average power consumed in the circuit,
$P=i_{rms}{e}_{rms}\cos \varphi$
$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\cos \frac{\pi }{3}$
$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{8}W$