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MHT CET 2024 (9 May, Shift 2) · Power, Power Factor & Wattless Current

Power, Power Factor & Wattless Current question from MHT CET 2024 (9 May, Shift 2)

Q37MHT CET 2024 · 9 MAY · SHIFT 2Medium

Power, Power Factor & Wattless Current

An e.m.f. $E = E_0 \cos\omega t$ is applied to circuit containing L and R in series. If $X_L = 2R$ , then the power dissipated in the circuit is

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Answer: (b)

$X_L = 2R$ , so $Z^2 = R^2 + 4R^2 = 5R^2$ .
$P = \frac{E_{rms}^2R}{Z^2} = \frac{E_0^2}{2} \cdot \frac{R}{5R^2}$ .
$P = \frac{E_0^2}{10R}$ .
So the answer is (b) $\frac{E_0^2}{10R}$ .

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