JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2024
- Hard
- Numerical
A series \(L C R\) circuit with \(L=\frac{100}{\pi} \mathrm{mH}, C=\frac{10^{-3}}{\pi} \mathrm{F}\) and \(R=10 \Omega\), is connected across an ac source of \(220 \mathrm{~V}, 50 \mathrm{~Hz}\) supply. The power factor of the circuit would be___________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\( \mathrm{X}_{\mathrm{c}}=\frac{1}{\omega \mathrm{C}}=\frac{\pi}{2 \pi \times 50 \times 10^{-3}}=10\ \Omega \)
\( \mathrm{x}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3} \)
\( =10\ \Omega \)
\( \because \mathrm{x}_{\mathrm{C}}=\mathrm{x}_{\mathrm{L}}, \text { Hence, circuit is in resonance } \)
\( \therefore \text { power factor }=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{R}}=1\)
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