A series LCR circuit | Alternating Current | JEE MAIN_2026

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JEE MAIN_2026

05-04-26_S2

Question

A series LCR circuit with R $=20 \Omega, \mathrm{~L}=1.6 \mathrm{H}$ and $\mathrm{C}=40 \mu \mathrm{~F}$ is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is $\_\_\_\_$ $\Omega$.

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Solution

For resonance $$ \begin{aligned} & \omega=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{1.6 \times 40 \times 10^{-6}}}=\frac{10^3}{8}=125 \\ & X_L=\omega L=125 \times 1.6=200 \Omega \end{aligned} $$

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