Alternating Current Question 5
Question 5 - 2024 (27 Jan Shift 2)
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
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Answer: (1)
Solution:
$\mathrm{X}_{\mathrm{c}}=\frac{1}{\omega \mathrm{C}}=\frac{\pi \text { nathon }}{2 \pi \times 50 \times 10^{-3}}=10 \Omega$
$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3}$
$=10 \Omega$
$\because \mathrm{X}{\mathrm{C}}=\mathrm{X}{\mathrm{L}}$, Hence, circuit is in resonance
$\therefore$ power factor $=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{R}}=1$
