JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Alternating Current

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

Answer: 50

Solution

The current is maximum at resonance, where the inductive and capacitive reactances cancel and

ω = \frac{1}{L C}

. The source gives

ω = 100

rad/s, and the figure shows

L = 2

H, so

C = \frac{1}{ω ^{2} L} = \frac{1}{( 100 ) ^{2} \times 2} = 5 \times 1 0^{- 5}

F. (Consistently, at resonance the peak current is

\frac{V _{0}}{R} = \frac{5}{100} = 50

mA.) The capacitance is

50 μ

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Integer answer