A series LCR circuit has L = 0.01 H, R = 10 Ω and C = 1 and it is connected to ac voltage of amplitude (Vm) 50 V. At frequency 60% lower than resonant frequency, the amplitude of current will be approximately: (a) 466 mA (b) 312 mA (c) 238 mA (d) 196 mA

Answer: (c). ω₀ = (1)/(√(LC)) = 1√0.01×10⁻⁶ = 10⁴ rad/s 60% lower: ω = 0.4ω₀ = 4000 rad/s. XL = 4000×0.01 = 40 Ω, XC = 14000×10⁻⁶ = 250 Ω Z = √(10² + 210²) ≈ 210.2 Ω, I₀ = (50)/(210.2) ≈ 0.238 A So the answer is (c) 238 mA.

A series LCR circuit has L = 0.01 H, R = 10 Ω and C = 1 and it is… | JEE Main 2022 (27 Jul, Shift 2)

Q79JEE Main 2022 · 27 JUL · SHIFT 2Medium

Resonance, Bandwidth & Q Factor

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

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Answer: (c)

$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.01\times10^{-6}}} = 10^4\text{ rad/s}$
60% lower: $\omega = 0.4\omega_0 = 4000\text{ rad/s}$ .
$X_L = 4000\times0.01 = 40\ \Omega$ , $X_C = \frac{1}{4000\times10^{-6}} = 250\ \Omega$
$Z = \sqrt{10^2 + 210^2} \approx 210.2\ \Omega$ , $I_0 = \frac{50}{210.2} \approx 0.238\text{ A}$
So the answer is (c) 238 mA.

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