Solution:
When \(L\) is removed, \(\tan \phi=\frac{\left|X_{C}\right|}{R}\) \(\Rightarrow \tan \frac{\pi}{3}=\frac{X_{c}}{R} \ldots(i)\)
When \(C\) is removed, \(\tan \phi=\frac{\left|X_{L}\right|}{R}\) \(\Rightarrow \tan \frac{\pi}{3}=\frac{X_{L}}{R} \ldots\)
\(\Rightarrow \tan \frac{\pi}{3}=\frac{X_{L}}{R} \ldots\) From \((i)\) and \((i i)\),
Since \(X_{L}=X_{C}\), the circuit is in resonance \(Z=R\)
Power factor \(=\cos \phi=\frac{R}{Z}=1\)
