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JEE Main 2022 (26 Jul, Shift 1) · Power, Power Factor & Wattless Current

Power, Power Factor & Wattless Current question from JEE Main 2022 (26 Jul, Shift 1)

Q53JEE Main 2022 · 26 JUL · SHIFT 1Medium

Power, Power Factor & Wattless Current

In a series $LR$ circuit $X_L = R$ and power factor of the circuit is $P_1$ . When capacitor with capacitance $C$ such that $X_L = X_C$ is put in series, the power factor becomes $P_2$ . The ratio $\frac{P_1}{P_2}$ is:

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Answer: (b)

$P_1 = R/\sqrt{R^2+R^2} = 1/\sqrt{2}$
With $X_C=X_L$ : resonance, $P_2 = 1$ .
$P_1/P_2 = 1/\sqrt{2}$ , so the answer is (b).

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