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MHT CET 2024 (10 May, Shift 1) · Power, Power Factor & Wattless Current

Power, Power Factor & Wattless Current question from MHT CET 2024 (10 May, Shift 1)

Q36MHT CET 2024 · 10 MAY · SHIFT 1Medium

Power, Power Factor & Wattless Current

An e.m.f. $E = E_0 \cos\omega t$ is applied to the $L - R$ circuit. The inductive reactance is equal to the resistance ' $R$ ' of the circuit. The power consumed in the circuit is

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Answer: (c)

$X_L = R$ , so $Z^2 = R^2 + R^2 = 2R^2$ .
$P = \frac{E_{rms}^2R}{Z^2} = \frac{E_0^2}{2} \cdot \frac{R}{2R^2}$ .
$P = \frac{E_0^2}{4R}$ .
So the answer is (c) $\frac{E_0^2}{4R}$ .

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