Alternating Current

JEE Main level, single correct, medium

3 of 190

Question 3 of 190

In a series LCR circuit, R=10 Ω,L=100 mH,C=10 μFR=10\,\Omega, L=100\,\text{mH}, C=10\,\mu\text{F}. Quality factor QQ of the circuit is:

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Answer: (C) 1010

Q=1RLC=1100.110−5=110104=10010=10Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{10}\sqrt{\frac{0.1}{10^{-5}}} = \frac{1}{10}\sqrt{10^4} = \frac{100}{10} = 10.

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