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NEET PHYSICSALTERNATING CURRENTMedium

Question

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:

A

1/2

B

1/√2

C

1

D

√3/2

Step-by-Step Solution

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

  1. Case 1 (L removed): The circuit acts as an RC circuit. The phase angle is given by tan⁡ϕ=XC/R\tan \phi = X_C / R. With ϕ=π/3\phi = \pi/3 (60∘60^\circ), we have XC=Rtan⁡(60∘)=R3X_C = R \tan(60^\circ) = R\sqrt{3}.
  2. Case 2 (C removed): The circuit acts as an RL circuit. The phase angle is given by tan⁡ϕ=XL/R\tan \phi = X_L / R. With ϕ=π/3\phi = \pi/3, we have XL=Rtan⁡(60∘)=R3X_L = R \tan(60^\circ) = R\sqrt{3}.
  3. Combined LCR Circuit: Since XL=XC=R3X_L = X_C = R\sqrt{3}, the net reactance X=XL−XCX = X_L - X_C is zero. This is the condition for electrical resonance .
  4. Power Factor: At resonance, the circuit is purely resistive, the phase difference ϕ\phi is 00, and the power factor is cos⁡ϕ=cos⁡(0)=1\cos \phi = \cos(0) = 1 .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ALTERNATING CURRENT. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSALTERNATING CURRENTelectricalcircuitvoltagesourceconnected

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