An a.c. source is applied to a series LR circuit with XL = 3R and power factor is X₁. Now a capacitor with Xc = R is added in series to LR circuit and the power factor is X₂. The ratio X₁ to X₂ is (a) 1:2 (b) 2:1 (c) 1:√(2) (d) √(2):1

Answer: (c). LR: Z₁ = √(R² + 9R²) = √(10)R, X₁ = (1)/(√(10)). LCR: XL - XC = 2R, Z₂ = √(5)R, X₂ = 1/√(5). (X₁)/(X₂) = (√(5))/(√(10)) = 1/√(2). So the answer is (c) 1:√(2).

An a.c. source is applied to a series LR circuit with XL = 3R and… | MHT CET 2025 (19 Apr, Shift 2)

Q28MHT CET 2025 · 19 APR · SHIFT 2Medium

Power, Power Factor & Wattless Current

An a.c. source is applied to a series LR circuit with $X_L = 3R$ and power factor is $X_1$ . Now a capacitor with $X_c = R$ is added in series to LR circuit and the power factor is $X_2$ . The ratio $X_1$ to $X_2$ is

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Answer: (c)

LR: $Z_1 = \sqrt{R^2 + 9R^2} = \sqrt{10}R$ , $X_1 = \frac{1}{\sqrt{10}}$ .
LCR: $X_L - X_C = 2R$ , $Z_2 = \sqrt{5}R$ , $X_2 = \frac{1}{\sqrt{5}}$ .
$\frac{X_1}{X_2} = \frac{\sqrt{5}}{\sqrt{10}} = \frac{1}{\sqrt{2}}$ .
So the answer is (c) $1:\sqrt{2}$ .

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

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