JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 13)
In an a.c. circuit, voltage and current are given by:
$$V=100 \sin (100 t) V$$ and $$I=100 \sin \left(100 t+\frac{\pi}{3}\right) \mathrm{mA}$$ respectively.
The average power dissipated in one cycle is:
5 W
25 W
2.5 W
10 W
Explanation
$$\begin{aligned}
& P_{\text {avg }}=V_{\text {rms }} I_{r m s} \cos (\Delta \phi) \\
& =\frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \cos \left(\frac{\pi}{3}\right) \\
& =\frac{10^4}{2} \times \frac{1}{2} \times 10^{-3} \\
& =\frac{10}{4}=2.5 \mathrm{~W}
\end{aligned}$$
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