A series resonant L C R circuit has a quality factor (Q-factor) 0.4 . If R=2 k Ω, C=0.1 μ F , then the value of inductance is

Correct answer: (B) 0.064 H. Q=1/R √(L/C) or L/C=(Q R)^2 ∴ L= (0.4 × 2 × 10^3 )^2 × 0.1 × 10^-6=0.064 H

Q129 / 180 · AIIMS 2016 · Physics

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

A
$0.1 \mathrm{H}$
Correct answer
B
$0.064 \mathrm{H}$
C
$2 \mathrm{H}$
D
$5 \mathrm{H}$

Solution

$\begin{aligned} & \text { } Q=\frac{1}{R} \sqrt{\frac{L}{C}} \text { or } \frac{L}{C}=(Q R)^2 \\ & \therefore \quad L=\left(0.4 \times 2 \times 10^3\right)^2 \times 0.1 \times 10^{-6}=0.064 \mathrm{H}\end{aligned}$

Hence, the correct answer is option: 'B'.

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