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Question: A series resonant LCR circuit has a quality factor (Q-factor)=0.4.If R=2\(k\Omega\), C= 0.1\(\mu F\)...

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

A

0.1H

B

0.064H

C

2 H

D

5 H

Answer

0.064H

Explanation

Solution

: Quality factor ,Q=1RLCQ = \frac{1}{R}\sqrt{\frac{L}{C}} or LC=(QR)2\frac{L}{C} = (QR)^{2}

Here, , Q=0.4,R=2kΩ=2×103ΩQ = 0.4,R = 2k\Omega = 2 \times 10^{3}\Omega

C=0.1μF=0.1×10−6FC = 0.1\mu F = 0.1 \times 10^{- 6}F

∴\therefore L=(QR)2CL = (QR)^{2}C

∴\therefore L=(0.4×2×103)2×0.1×10−6=0.064HL = (0.4 \times 2 \times 10^{3})^{2} \times 0.1 \times 10^{- 6} = 0.064H