Alternating Current Question 5
Question 5 - 2024 (27 Jan Shift 2)
A series LCR circuit with $L=\frac{100}{\pi} mH, C=\frac{10^{-3}}{\pi} F$ and $R=10 \Omega$, is connected across an ac source of $220 V, 50 Hz$ supply. The power factor of the circuit would be
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Answer: (1)
Solution:
$X _c=\frac{1}{\omega C}=\frac{\pi \text { nathon }}{2 \pi \times 50 \times 10^{-3}}=10 \Omega$
$X _L=\omega L=2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3}$
$=10 \Omega$
$\because X _C=X _L$, Hence, circuit is in resonance
$\therefore$ power factor $=\frac{R}{Z}=\frac{R}{R}=1$
