In a L-R circuit the inductive reactance is equal to the resistance R in the circuit. An emf E = E₀ cos ω t is applied to the circuit. The power consumed in the circuit is (a) (E₀²)/(√(2)R) (b) (E₀²)/(4R) (c) (E₀²)/(2R) (d) (E₀²)/(8R)

Answer: (b). XL = R, so Z² = 2R². P = Erms²RZ² = (E₀²)/(2) · R/2R². P = (E₀²)/(4R). So the answer is (b) (E₀²)/(4R).

In a L-R circuit the inductive reactance is equal to the resistance R… | MHT CET 2023 (13 May, Shift 2)

Q46MHT CET 2023 · 13 MAY · SHIFT 2Medium

Power, Power Factor & Wattless Current

In a $L\text{-}R$ circuit the inductive reactance is equal to the resistance $R$ in the circuit. An emf $E = E_0 \cos\omega t$ is applied to the circuit. The power consumed in the circuit is

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Answer: (b)

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

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