A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase diffe

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 :

Accepted Answer

When L is removed, $	an \phi = \frac{|X_C|}{R} \Rightarrow 	an \frac{\pi}{3} = \frac{X_C}{R}$. When C is removed, $	an \phi = \frac{|X_L|}{R} \Rightarrow 	an \frac{\pi}{3} = \frac{X_L}{R}$. From (i) and (ii), $X_L = X_C$. Since $X_L = X_C$, the circuit is in resonance. $Z = R$. Power factor = $\cos \phi = \frac{R}{Z} = 1$

Step-by-Step Solution

Practice All Physics Questions