A series L C R circuit with L=100/π mH , C =frac 10 -3 π F and R =10 Ω , is connected across an ac source of 220 V , 50 Hz supply. The power factor of the circuit would be \ \ \ \ \ \ \ .

Physics · Alternating Current

JEE Main 2024 — 27 January, Shift 2 — Question 59

A series $L C R$ circuit with $L=\frac{100}{\pi} \mathrm{mH}, \mathrm{C}=\frac{10^{-3}}{\pi} \mathrm{~F}$ 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

Numerical answer — enter this value.

Step-by-step solution

$X_{C}=\frac{1}{\omega C}=\frac{\pi}{2 \pi \times 50 \times 10^{-3}}=10 \Omega$

$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3}$

$=10 \Omega$

$\because \mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}$

Hence, circuit is in resonance $\therefore$

power factor $=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{R}}=1$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Physics
Chapter: Alternating Current
Topic: Series LCR Circuit and Power Factor