A series LCR circuit is connected to an AC voltage source. When the capacitor is removed, the phase difference between current and voltage is $\pi /3$. When the inductor is removed, the phase difference is again $\pi /3$. The power factor of the original circuit is:

Solution

1. When the capacitor is removed, the circuit is inductive, so $$\begin{array}{rl}\tan (\varphi )& =X_L/R\\ & =\tan (\pi /3)\\ & =\sqrt{3}\end{array}$$.
2. When the inductor is removed, the circuit is capacitive, so $$\begin{array}{rl}\tan (\varphi )& =X_C/R\\ & =\tan (\pi /3)\\ & =\sqrt{3}\end{array}$$.
3. Equating the two expressions gives $X_L/R=X_C/R$, which implies $X_L=X_C$.
4. Since $X_L=X_C$, the circuit is at resonance, meaning the net reactance is zero and the phase angle is $0$.
5. The power factor is $\cos (0)=1$.
6. Hence the answer is (B).