A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is (π)/(3). If instead C is removed from the circuit, the phase difference is again (π)/(3) between current and voltage. The power factor of the circuit is :

(C) 1.0. L removed: (π)/(3) = (X_C)/(R). C removed: (π)/(3) = (X_L)/(R). So X_L = X_C: with all three present the circuit is at resonance, Z = R and φ = 0. φ = 1

Alternating Current question for NEET (NEET 2020), with solution

Q 12-07-013NEETNEET 2020Top questionMedium

A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is $\dfrac{\pi}{3}$. If instead C is removed from the circuit, the phase difference is again $\dfrac{\pi}{3}$ between current and voltage. The power factor of the circuit is :

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Alternating Current question for NEET (NEET 2020), with solution

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