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MHT CET 2026 (12 May, Shift 1) · Power, Power Factor & Wattless Current

Power, Power Factor & Wattless Current question from MHT CET 2026 (12 May, Shift 1)

Q8MHT CET 2026 · 12 MAY · SHIFT 1Medium

Power, Power Factor & Wattless Current

An e.m.f. $e = E_0 \cos\omega t$ is applied to a circuit containing L, C and R in series where $X_L = 4R$ and $X_C = R$ . The average power dissipated in the circuit is

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Answer: (a)

$X_L - X_C = 4R - R = 3R$ , so $Z = \sqrt{R^2 + 9R^2} = \sqrt{10}R$ .
$P = \frac{E_{rms}^2R}{Z^2}$ with $E_{rms}^2 = \frac{E_0^2}{2}$ .
$P = \frac{E_0^2}{2} \cdot \frac{R}{10R^2} = \frac{E_0^2}{20R}$ .
So the answer is (a) $\frac{E_0^2}{20R}$ .

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