Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

In an electrical circuit R, L, C, and an AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is \pi /3. If instead, C is removed from the circuit, the phase difference is again \pi /3. The power factor of the circuit is:

1/2

1/√2

√3/2

Step-by-Step Solution

In a series AC circuit, the phase difference $\phi$ is determined by the reactance and resistance.

Case 1 (L removed): The circuit acts as an RC circuit. The phase angle is given by $\tan \phi = X_C / R$ . With $\phi = \pi/3$ ( $60^\circ$ ), we have $X_C = R \tan(60^\circ) = R\sqrt{3}$ .
Case 2 (C removed): The circuit acts as an RL circuit. The phase angle is given by $\tan \phi = X_L / R$ . With $\phi = \pi/3$ , we have $X_L = R \tan(60^\circ) = R\sqrt{3}$ .
Combined LCR Circuit: Since $X_L = X_C = R\sqrt{3}$ , the net reactance $X = X_L - X_C$ is zero. This is the condition for electrical resonance .
Power Factor: At resonance, the circuit is purely resistive, the phase difference $\phi$ is $0$ , and the power factor is $\cos \phi = \cos(0) = 1$ .

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