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JEE Main 2021 (20 Jul, Shift 2) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2021 (20 Jul, Shift 2)

Q87JEE Main 2021 · 20 JUL · SHIFT 2Easy

Resonance, Bandwidth & Q Factor

A series LCR circuit of R = 5 $\Omega$ , L = 20 mH and C = 0.5 $\mu$ F is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ______________ $\times$ 102 W.

Numerical answer type. Work it out, then open the answer.

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Answer: 125

At resonance $Z = R$ and power factor is 1: $P = \dfrac{V_{rms}^2}{R}$
$P = \dfrac{250^2}{5} = \dfrac{62500}{5} = 12500\text{ W} = 125\times10^{2}\text{ W}$
So the answer is 125 (the '102' in the question means $10^2$ ).

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