Question 212: AC fundamentals

Question

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 $$\pi^2=10$$)

Accepted Answer

$\because$ For maximum amplitude of current, circuit should be at resonance.
$$
\begin{aligned}
& 	herefore X_{L}=X_{C} \\
& \omega L=\frac{1}{\omega C} \\
& L=\frac{1}{\omega^{2} C} \\
& =\frac{1}{\left(2 \pi 	imes 2 	imes 10^{3}ight)^{2} 	imes 62.5 	imes 10^{-9}} \\
& =100 ~\mathrm{mH}
\end{aligned}
$$

AC fundamentals

An LCR series circuit of capacitance 62.5 nF and resistance of 50 $\Omega, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is __________ mH. (Take \pi^2=10$)

Integer / Numeric Answer Type