JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

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.

Your answer

Solution

Current is maximum at resonance, where $ω = \frac{1}{L C}$ .

From the source, $ω = 100 rad/s$ . Also, at resonance the impedance equals R, so $I_{m a x} = V_{0} / R = 5/100 = 50 mA$ ✓ — confirming resonance.

Solve for C:

$C = \frac{1}{ω ^{2} L} = \frac{1}{( 100 ) ^{2} \times 2} = \frac{1}{2 \times 1 0 ^{4}} F = 50 \times 1 0^{- 6} F = 50 μ F$

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