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 . If instead C is removed from the circuit, the phase difference is again between current and voltage. The

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Solution

Q.
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 $π / 3$ . If instead C is removed from the circuit, the phase difference is again $π / 3$ between current and voltage. The power factor of the circuit is [2020]

1 zero

2 0.5

3 1.0

4 - 1.0

Ans.
(3)

When $L$ is removed

$\tan ϕ = \frac{| X_{C} |}{R} \Rightarrow \tan \frac{π}{3} = \frac{X_{C}}{R}$ ...(i)

When C is removed,

$\tan ϕ = \frac{| X_{L} |}{R} \Rightarrow \tan \frac{π}{3} = \frac{X_{L}}{R}$ ...(ii)

From (i) and (ii), $X_{C} = X_{L}$

Since, $X_{L} = X_{C},$ the circuit is in resonance.

$Z = R$

Power factor, $\cos ϕ = \frac{Z}{R} = \frac{R}{R} = 1$

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