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NEET PhysicsHard

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, tan⁡ϕ=∣XC∣R⇒tan⁡π3=XCR\tan \phi = \frac{|X_C|}{R} \Rightarrow \tan \frac{\pi}{3} = \frac{X_C}{R}. When C is removed, tan⁡ϕ=∣XL∣R⇒tan⁡π3=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 = cos⁡ϕ=RZ=1\cos \phi = \frac{R}{Z} = 1

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