In LCR series circuit, when ' L ' is removed from the circuit, the phase difference between voltage and current in the circuit is π/3. If ' C ' is removed from the circuit instead of L then phase difference is again π/3. The power factor of the circuit is (tan 60° = √(3)) (a) √(3)/2 (b) √(2) (c) 1/√(2) (d) 1

Answer: (d). L removed: XC = √(3)R; C removed: XL = √(3)R. XL = XC ⇒ resonance, Z = R. Power factor = R/Z = 1. So the answer is (d) 1.

In LCR series circuit, when ' L ' is removed from the circuit, the… | MHT CET 2025 (25 Apr, Shift 1)

Q20MHT CET 2025 · 25 APR · SHIFT 1Medium

Series LCR Circuit

In LCR series circuit, when ' $L$ ' is removed from the circuit, the phase difference between voltage and current in the circuit is $\frac{\pi}{3}$ . If ' $C$ ' is removed from the circuit instead of $L$ then phase difference is again $\frac{\pi}{3}$ . The power factor of the circuit is $\left(\tan 60^\circ = \sqrt{3}\right)$

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Answer: (d)

L removed: $X_C = \sqrt{3}R$ ; C removed: $X_L = \sqrt{3}R$ .
$X_L = X_C \Rightarrow$ resonance, $Z = R$ .
Power factor $= \dfrac{R}{Z} = 1$ . So the answer is (d) 1.

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Series LCR Circuit question from MHT CET 2025 (25 Apr, Shift 1)

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