An alternating emf of 0.2 V is applied across an L-C-R series circuit having R = 4 Ω, C = 80 and L = 200 mH. At resonance the voltage drop across the inductor is (a) 10V (b) 2.5V (c) 1V (d) 5V

Answer: (b). At resonance Z = R: I = 0.2/4 = 0.05 A ω₀ = (1)/(√(LC)) = 1√0.2 × 80×10⁻⁶ = 14×10⁻³ = 250 rad/s XL = 250 × 0.2 = 50 Ω VL = I XL = 0.05 × 50 = 2.5 V So the answer is (b) 2.5 V.

An alternating emf of 0.2 V is applied across an L-C-R series circuit… | MHT CET 2020 (16 Oct, Shift 1)

Q101MHT CET 2020 · 16 OCT · SHIFT 1Easy

Resonance, Bandwidth & Q Factor

An alternating emf of $0.2\text{ V}$ is applied across an L-C-R series circuit having $R = 4\ \Omega, C = 80\ \mu\text{F}$ and $L = 200\text{ mH}$ . At resonance the voltage drop across the inductor is

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Answer: (b)

At resonance $Z = R$ : $I = \dfrac{0.2}{4} = 0.05\text{ A}$
$\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{0.2 \times 80\times10^{-6}}} = \dfrac{1}{4\times10^{-3}} = 250\text{ rad/s}$
$X_L = 250 \times 0.2 = 50\ \Omega$
$V_L = I X_L = 0.05 \times 50 = 2.5\text{ V}$
So the answer is (b) 2.5 V.

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Resonance, Bandwidth & Q Factor question from MHT CET 2020 (16 Oct, Shift 1)

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