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

• A. zero
• B. 0.5
• C. 1.0
• D. -1.0

Answer 1

Answer: (C)

When L is removed,
$tan\varphi =\frac{|X_C|}{R}$
$\Rightarrow tan\frac{\pi }{3}=\frac{X_c}{R}\dots \left(i\right)$
When C is removed,
$tan\varphi =\frac{|X_L|}{R}$
$\Rightarrow tan\frac{\pi }{3}=\frac{X_L}{R}\dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
Since $X_L=X_C$, the circuit is in resonance
$Z=R$
Power factor $=cos\varphi =\frac{R}{Z}=1$