A series LCR circuit with resistance (R) 500 ohm is connected to an a.c. source of 250 V. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. The impedance of the circuit is (tan π/3 = √(3)) (a) (500)/(√(3)) Ω (b) 500√(3) Ω (c) 250 Ω (d) 500 Ω

Answer: (d). C removed: XL = Rtan 60° = √(3)R. L removed: XC = Rtan 60° = √(3)R. XL = XC ⇒ resonance, Z = R = 500 Ω. So the answer is (d) 500 Ω.

A series LCR circuit with resistance (R) 500 ohm is connected to an… | MHT CET 2021 (20 Sep, Shift 2)

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Series LCR Circuit

A series LCR circuit with resistance (R) $500\text{ ohm}$ is connected to an a.c. source of $250\text{ V}$ . When only the capacitance is removed, the current lags behind the voltage by $60^\circ$ . When only the inductance is removed, the current leads the voltage by $60^\circ$ . The impedance of the circuit is $\left(\tan\frac{\pi}{3} = \sqrt{3}\right)$

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Answer: (d)

C removed: $X_L = R\tan 60^\circ = \sqrt{3}R$ .
L removed: $X_C = R\tan 60^\circ = \sqrt{3}R$ .
$X_L = X_C \Rightarrow$ resonance, $Z = R = 500\ \Omega$ .
So the answer is (d) $500\ \Omega$ .

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