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NEET PhysicsGeneralHard

Question

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 π3\frac{\pi}{3}. If instead C is removed from the circuit, the phase difference is again π3\frac{\pi}{3} between current and voltage. The power factor of the circuit is :

1

zero

2

0.5

3

1.0

4

-1.0

Step-by-Step Solution

When L is removed, tanϕ=XCRtanπ3=XCR\tan \phi = \frac{|X_C|}{R} \Rightarrow \tan \frac{\pi}{3} = \frac{X_C}{R}. When C is removed, tanϕ=XLRtanπ3=XLR\tan \phi = \frac{|X_L|}{R} \Rightarrow \tan \frac{\pi}{3} = \frac{X_L}{R}. From (i) and (ii), XL=XCX_L = X_C. Since XL=XCX_L = X_C, the circuit is in resonance. Z=RZ = R. Power factor = cosϕ=RZ=1\cos \phi = \frac{R}{Z} = 1

Exam Context & Concepts Covered

This question aligns with the NEET Physics syllabus, specifically targeting concepts from General. 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.

Physicsseriescircuitconnectedvoltagesource

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