JEE PYQ: Alternating Current Question 18
Question 18 - 2021 (26 Feb Shift 1)
In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be (round off to nearest integer):
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Answer: 283
Solution
$Q = \frac{1}{R}\sqrt{\frac{L}{C}}$. $Q’ = \frac{\sqrt{2L}}{(R/2)\sqrt{C}} = 2\sqrt{2}Q = 2\sqrt{2} \times 100 \approx 283$.
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