JEE MAIN - Physics (2024 - 27th January Evening Shift - No. 22)
A series LCR circuit with $$\mathrm{L}=\frac{100}{\pi} \mathrm{mH}, \mathrm{C}=\frac{10^{-3}}{\pi} \mathrm{F}$$ and $$\mathrm{R}=10 \Omega$$, is connected across an ac source of $$220 \mathrm{~V}, 50 \mathrm{~Hz}$$ supply. The power factor of the circuit would be ________.
Answer
1
Explanation
$$\begin{aligned}
& \mathrm{X}_{\mathrm{c}}=\frac{1}{\omega \mathrm{C}}=\frac{\pi}{2 \pi \times 50 \times 10^{-3}}=10 \Omega \\
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3} \\
& =10 \Omega \\
& \because \mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}, \text { Hence, circuit is in resonance } \\
& \therefore \text { power factor }=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{R}}=1
\end{aligned}$$
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