Practice portal › Alternating Current › Resonance in Series LCR

Using a variable-frequency source, the maximum current in the LCR circuit (see figure, L=2 H, R=100 Ω) is 50 mA for V=5 sin(100t) V. The capacitance C used is ______ μF.

Asked in JEE Main 23rd Jan 1st Shift 2026 · Resonance condition and tuning

Figure: Resonance condition and tuning
Answer: 50

Step-by-step solution

Maximum current occurs at resonance, so the source frequency equals ω₀: ω₀=100 rad/s.

(Check: at resonance Iₚₑₐₖ=(V₀)/R=5/(100)=50 mA.)

ω₀=1/(√LC)⇒ C=1/(ω₀² L)=1/(100²×2)=5×10⁻⁵ F=50 μF.

More Resonance in Series LCR questionsAll Resonance in Series LCR questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer