Q73 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
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 $\_\_\_\_$ $\mu \mathrm{F}$.
Correct answer: 50
Solution
From $V = 5\sin(100t)$, angular frequency $\omega = 100$ rad/s
At resonance in LCR circuit: $X_L = X_C$, so $\omega L = \frac{1}{\omega C}$
Maximum current occurs at resonance: $I_{max} = \frac{V_{rms}}{R}$
$50 \times 10^{-3} = \frac{5/\sqrt{2}}{100}$ (verified at resonance)
From resonance condition: $100 \times 2 = \frac{1}{100 \times C}$
$200 = \frac{1}{100C}$
$C = \frac{1}{20000} = 5 \times 10^{-5}$ F = 50 μF
At resonance in LCR circuit: $X_L = X_C$, so $\omega L = \frac{1}{\omega C}$
Maximum current occurs at resonance: $I_{max} = \frac{V_{rms}}{R}$
$50 \times 10^{-3} = \frac{5/\sqrt{2}}{100}$ (verified at resonance)
From resonance condition: $100 \times 2 = \frac{1}{100 \times C}$
$200 = \frac{1}{100C}$
$C = \frac{1}{20000} = 5 \times 10^{-5}$ F = 50 μF
Hence, the correct answer is 50.