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Physics
A series resonant LCR circuit has a quality factor (Q-factor) 0.4. If R = 2 kΩ, C = 0.1 μ F, then the value of inductance is
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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
JIPMER
JIPMER 2010
Alternating Current
A
0.1 H
9%
B
0.064 H
70%
C
2 H
13%
D
5 H
7%
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$