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 - Physics MCQ NEET 2020 E1 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

When L is removed, it is an RC circuit: tan ϕ = \frac{X _{C}}{R} = tan (π /3) = 3 \Rightarrow X_{C} = R 3 .When C is removed, it is an RL circuit: tan ϕ = \frac{X _{L}}{R} = tan (π /3) = 3 \Rightarrow X_{L} = R 3 .This implies X_{L} = X_{C}, which is the condition for resonance.In resonance, the phase difference ϕ = 0 .Power factor cos ϕ = cos 0 = 1.0 .

Try it yourself before checking the explanation above.

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:

zero

0.5

1.0

- 1.0

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