Q81 / 120 · NEET 2012 (Mains) · Physics
The instantaneous values of alternating current and voltages in a circuit are given as
$$
\begin{aligned}
& i=\frac{1}{\sqrt{2}} \sin (100 \pi t) \text { ampere } \\
& e=\frac{1}{\sqrt{2}} \sin (100 \pi t+\pi / 3) \text { volt }
\end{aligned}
$$
The average power in Watts consumed in the circuit is
- A$\frac{1}{4}$
- B$\frac{\sqrt{3}}{4}$
- C$\frac{1}{2}$
- Correct answerD$\frac{1}{8}$
Solution
Given equations
$$
i=\frac{1}{\sqrt{2}} \sin (100 \pi t)
$$
and $e=\frac{1}{\sqrt{2}} \sin (100 \pi t+\pi / 3)$
$\therefore \quad i_o=\frac{1}{\sqrt{2}}$ and $V_o=\frac{1}{\sqrt{2}}$
We know that average power
$$
\begin{aligned}
& P_{\mathrm{av}}=V_{\mathrm{rms}} \times i_{\mathrm{rms}} \cos \phi \\
&=\frac{1}{2} \times \frac{1}{2} \times \cos 60^{\circ} \\
& \qquad\left[\because i_{\mathrm{rms}}=\frac{i_o}{\sqrt{2}} \text { and } V_{\mathrm{rms}}=\frac{V_o}{\sqrt{2}}\right] \\
&=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\
&=\frac{1}{8} \mathrm{~W}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.