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Q. A series resonant $LCR$ circuit has a quality factor ($Q$-factor) 0.4. If $R = 2\, k\Omega, C = 0.1 \, \mu F$, then the value of inductance is

JIPMERJIPMER 2010Alternating Current

Solution:

$Q = \frac{1}{R} \sqrt{\frac{L}{C}} $ or $\frac{L}{C} = (QR)^2$
$ \therefore \:\:\: L = (0.4 \times 2 \times 10^3)^2 \times 0.1 \times 10^{-6} = 0.064 \, H$