JEE Main 2022 · 27 Jul · Shift 2 · Physics Q44

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PhysicsSingle correct+4 / -1Question 44 of 90

A series LCR circuit has L = 0.01 H, R = 10 Ω and C = 1 μF and it is connected to ac voltage of amplitude $(V_m)$ 50 V. At frequency 60% lower than resonant frequency, the amplitude of current will be approximately :

466 mA

312 mA

238 mA

196 mA

Answer and worked solution

Official answer: C

Resonance at $\omega_0=10^4$ rad/s; 60% lower is $\omega=4000$ rad/s.

Reactances. $X_L=40$ Ω and $X_C=250$ Ω, so $Z=\sqrt{10^2+210^2}\approx210$ Ω and $I=\frac{50}{210}\approx238$ mA.

(C)

Topics:LCR circuits and resonance Reactance and impedance

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JEE Main 2022 · 27 Jul · Shift 2, Physics question 44