An LCR series circuit of capacitance 62.5 nF and resistance of 50 $\Omega$, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is __________ mH.
(Take $\pi^2=10$)
Answer (integer)
100
Solution
$\because$ For maximum amplitude of current, circuit should be at resonance.
<br/><br/>
$$
\begin{aligned}
& \therefore X_{L}=X_{C} \\\\
& \omega L=\frac{1}{\omega C} \\\\
& L=\frac{1}{\omega^{2} C} \\\\
& =\frac{1}{\left(2 \pi \times 2 \times 10^{3}\right)^{2} \times 62.5 \times 10^{-9}} \\\\
& =100 ~\mathrm{mH}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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