In an L - R circuit, the inductive reactance is equal to √(3) times the resistance ' R ' of the circuit. An e.m.f. E = E₀ sin (ω t) is applied to the circuit. The power consumed in the circuit is (a) (E₀²)/(4R) (b) (E₀²)/(6R) (c) (E₀²)/(8R) (d) (E₀²)/(12R)

Answer: (c). XL = √(3)R, so Z² = R² + 3R² = 4R². P = Erms²RZ² = (E₀²)/(2) · R/4R². P = (E₀²)/(8R). So the answer is (c) (E₀²)/(8R).

In an L - R circuit, the inductive reactance is equal to √(3) times… | MHT CET 2025 (19 Apr, Shift 2)

Q24MHT CET 2025 · 19 APR · SHIFT 2Medium

Power, Power Factor & Wattless Current

In an $L - R$ circuit, the inductive reactance is equal to $\sqrt{3}$ times the resistance ' $R$ ' of the circuit. An e.m.f. $E = E_0 \sin(\omega t)$ is applied to the circuit. The power consumed in the circuit is

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Answer: (c)

$X_L = \sqrt{3}R$ , so $Z^2 = R^2 + 3R^2 = 4R^2$ .
$P = \frac{E_{rms}^2R}{Z^2} = \frac{E_0^2}{2} \cdot \frac{R}{4R^2}$ .
$P = \frac{E_0^2}{8R}$ .
So the answer is (c) $\frac{E_0^2}{8R}$ .

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Power, Power Factor & Wattless Current question from MHT CET 2025 (19 Apr, Shift 2)

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