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: (a) 0.5 (b) 1.0 (c) - 1.0 (d) zero

Answer: (b). L removed: XC = √(3)R; C removed: XL = √(3)R. XL = XC ⇒ resonance, Z = R. Power factor = R/Z = 1.0. So the answer is (b) 1.0.

A series LCR circuit is connected to an ac voltage source. When L is… | NEET 2020 (Phase 1)

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Series LCR Circuit

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 $\frac{\pi}{3}$ . If instead C is removed from the circuit, the phase difference is again $\frac{\pi}{3}$ between current and voltage. The power factor of the circuit is:

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Answer: (b)

L removed: $X_C = \sqrt{3}R$ ; C removed: $X_L = \sqrt{3}R$ .
$X_L = X_C \Rightarrow$ resonance, $Z = R$ .
Power factor $= \dfrac{R}{Z} = 1.0$ . So the answer is (b) 1.0.

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