Q134 / 200 · NEET 2012 (Screening) · Physics
In an electrical circuit $R, L, C$ and an $\mathrm{AC}$ voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\pi / 3$. If instead, $C$ is removed from the circuit, the phase difference is again $\pi / 3$. The power factor of the circuit is
- A$1 / 2$
- B$1 / \sqrt{2}$
- Correct answerC1
- D$\sqrt{3} / 2$
Solution
Here, phase difference
$\begin{aligned}
\tan \phi & =\frac{X_L-X_C}{R} \\
\tan \frac{\pi}{3} & =\frac{X_L-X_C}{R}
\end{aligned}$
When, $L$ is removed
$\begin{aligned}
& \sqrt{3}=\frac{X_C}{R} \\
& X_C=\sqrt{3} R
\end{aligned}$
When $C$ is removed
$\begin{gathered}
\tan \frac{\pi}{3}=\sqrt{3}=\frac{X_L}{R} \\
X_L=R \sqrt{3}
\end{gathered}$
Hence, in resonant circuit
$\begin{aligned}
\tan \phi & =\frac{\sqrt{3} R-\sqrt{3} R}{R}=0 \\
\phi & =0
\end{aligned}$
$\therefore$ Power factor $\cos \phi=1$
It is the condition of resonance therefore phase difference between voltage and current is zero and power factor is $\cos \phi=1$.
Hence, the correct answer is option: 'C'.