Resistor of 2 Ω, inductor of 100 and capacitor of 400 pF are connected in series across a source of erms = 0.1 Volt. At resonance, voltage drop across inductor is (a) 25 V (b) 2.5 V (c) 250 V (d) 20 V

Answer: (a). At resonance Z = R: I = 0.1/2 = 0.05 A ω₀ = (1)/(√(LC)) = 1√10⁻⁴ × 4×10⁻¹⁰ = 12×10⁻⁷ = 5×10⁶ rad/s XL = ω₀ L = 5×10⁶ × 10⁻⁴ = 500 Ω VL = I XL = 0.05 × 500 = 25 V So the answer is (a) 25 V.

Resistor of 2 Ω, inductor of 100 and capacitor of 400 pF are… | MHT CET 2023 (12 May, Shift 2)

Q57MHT CET 2023 · 12 MAY · SHIFT 2Easy

Resonance, Bandwidth & Q Factor

Resistor of $2\ \Omega$ , inductor of $100\ \mu\text{H}$ and capacitor of $400\text{ pF}$ are connected in series across a source of $e_{\text{rms}} = 0.1$ Volt. At resonance, voltage drop across inductor is

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Answer: (a)

At resonance $Z = R$ : $I = \dfrac{0.1}{2} = 0.05\text{ A}$
$\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{10^{-4} \times 4\times10^{-10}}} = \dfrac{1}{2\times10^{-7}} = 5\times10^{6}\text{ rad/s}$
$X_L = \omega_0 L = 5\times10^6 \times 10^{-4} = 500\ \Omega$
$V_L = I X_L = 0.05 \times 500 = 25\text{ V}$
So the answer is (a) 25 V.

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Resonance, Bandwidth & Q Factor question from MHT CET 2023 (12 May, Shift 2)

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