For series LCR circuit, the current is observed to lead the applied voltage. An additional capacitor C₁ when joined with capacitor C, makes the power factor of the circuit unity. The capacitor C₁ must be connected in (a) series with C and has magnitude (C)/((ω² LC - 1)) (b) series with C and has magnitude ((1 - ω² LC))/(ω² L) (c) parallel with C and has magnitude ((1 - ω² LC))/(ω² L) (d) parallel with C and has magnitude (C)/((ω² LC - 1))

Answer: (c). Current leads voltage, so XC > XL: the circuit is capacitive. For unity power factor, XC must drop to XL, so the capacitance must increase: connect C₁ in parallel with C. Resonance: C + C₁ = (1)/(ω²L). C₁ = (1)/(ω²L) - C = (1 - ω²LC)/(ω²L). So the answer is (c).

For series LCR circuit, the current is observed to lead the applied… | MHT CET 2026 (13 May, Shift 1)

Q11MHT CET 2026 · 13 MAY · SHIFT 1Medium

Power, Power Factor & Wattless Current

For series LCR circuit, the current is observed to lead the applied voltage. An additional capacitor $C_1$ when joined with capacitor C, makes the power factor of the circuit unity. The capacitor $C_1$ must be connected in

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Answer: (c)

Current leads voltage, so $X_C > X_L$ : the circuit is capacitive.
For unity power factor, $X_C$ must drop to $X_L$ , so the capacitance must increase: connect $C_1$ in parallel with $C$ .
Resonance: $C + C_1 = \frac{1}{\omega^2L}$ .
$C_1 = \frac{1}{\omega^2L} - C = \frac{1 - \omega^2LC}{\omega^2L}$ .
So the answer is (c).

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Power, Power Factor & Wattless Current question from MHT CET 2026 (13 May, Shift 1)

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