Using a variable-frequency a.c. voltage source, the maximum current measured in the given LCR circuit is 50 mA for V =5 sin (100 t ) . The values of L and R are shown in the figure. The capacitance of the capacitor ( C ) used is \ \ \ \ μ F .

Physics · Alternating Current

JEE Main 2026 — 23 January, Morning Shift — Question 46

Using a variable-frequency a.c. voltage source, the maximum current measured in the given LCR circuit is 50 mA for $\mathrm{V}=5 \sin (100 \mathrm{t})$ . The values of L and R are shown in the figure. The capacitance of the capacitor ( C ) used is $\_\_\_\_$ $\mu \mathrm{F}$ .

Answer: 50

Numerical answer — enter this value.

Step-by-step solution

Current is maximum, so resonance and $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\mathrm{C}=\frac{1}{\omega^{2} \mathrm{~L}}=\frac{1}{2 \times 10^{4}}$ $=50 \times 10^{-6}=50 \mu \mathrm{~F}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: Alternating Current
Topic: Series LCR Circuit and Power Factor