In series LCR circuit, at resonance the peak value of current will be [ E₀ is peak emf, R is resistance, ω L is inductive reactance and ω C is capacitive] (a) E₀/R (b) (E₀)/(√(2)R) (c) (E₀)/(√(R² + (ω L - (1)/(ω C))²)) (d) (E₀)/(√(2)√(R² + (ω L - (1)/(ω C))²))

Answer: (a). I₀ = (E₀)/(√(R² + (ω L - (1)/(ω C))²)) At resonance ω L = (1)/(ω C), so Z = R. So the answer is (a) E₀/R.

In series LCR circuit, at resonance the peak value of current will be… | MHT CET 2021 (24 Sep, Shift 2)

Q85MHT CET 2021 · 24 SEP · SHIFT 2Easy

Resonance, Bandwidth & Q Factor

In series LCR circuit, at resonance the peak value of current will be [ $E_0$ is peak emf, R is resistance, $\omega L$ is inductive reactance and $\omega C$ is capacitive]

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Answer: (a)

$I_0 = \dfrac{E_0}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}$
At resonance $\omega L = \dfrac{1}{\omega C}$ , so $Z = R$ .
So the answer is (a) $\dfrac{E_0}{R}$ .

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Resonance, Bandwidth & Q Factor question from MHT CET 2021 (24 Sep, Shift 2)

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