Question

The instantaneous values of alternating current and voltages in a circuit are given as $i=\frac{1}{\sqrt2}\sin(100\pi t)$ ampere $e=\frac{1}{\sqrt2}\sin(100\pi t+\frac{\pi}{3})$ 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)

$\frac{1}{8}$

Answer 2

Given : $i=\frac{1}{\sqrt2}\sin(100\pi t)$ ampere
Compare it with $i=i_0 \, \sin(\omega t)$ we get
\hspace20mmi_0=\frac{1}{\sqrt2}A
Given : $e=\frac{1}{\sqrt2}\sin\big(100\pi t+\frac{\pi}{3}\big)$ volt
Compare it with , we get
\hspace20mme_0=\frac{1}{\sqrt2}V,\phi=\frac{\pi}{3}
$\therefore \, \, i_{rms}=\frac{i_0}{\sqrt2}=\frac{\sqrt2}{\sqrt2}A=\frac{1}{2}A$
\hspace20mm e_{rms}=\frac{e_0}{\sqrt2}=\frac{\frac{1}{\sqrt2}}{\sqrt2}V=\frac{1}{2}V
Average power consumed in the circuit,
\hspace20mmP=i_{rms} \, e_{rms} \, cos\phi
\hspace20mm=\big(\frac{1}{2}\big)\big(\frac{1}{2}\big)cos\frac{\pi}{3}=\big(\frac{1}{2}\big)\big(\frac{1}{2}\big)\big(\frac{1}{2}\big)=\frac{1}{8}W