JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 24)
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
100
Explanation
$\because$ For maximum amplitude of current, circuit should be at resonance.
$$ \begin{aligned} & \therefore X_{L}=X_{C} \\\\ & \omega L=\frac{1}{\omega C} \\\\ & L=\frac{1}{\omega^{2} C} \\\\ & =\frac{1}{\left(2 \pi \times 2 \times 10^{3}\right)^{2} \times 62.5 \times 10^{-9}} \\\\ & =100 ~\mathrm{mH} \end{aligned} $$
$$ \begin{aligned} & \therefore X_{L}=X_{C} \\\\ & \omega L=\frac{1}{\omega C} \\\\ & L=\frac{1}{\omega^{2} C} \\\\ & =\frac{1}{\left(2 \pi \times 2 \times 10^{3}\right)^{2} \times 62.5 \times 10^{-9}} \\\\ & =100 ~\mathrm{mH} \end{aligned} $$
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