Question 39: AC fundamentals

Question

In an electrical circuit $$R, L, C$$ and $$\mathrm{AC}$$ 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

Accepted Answer

Here, phase difference
$$	an \phi=\frac{X_L-X_C}{R} \Rightarrow 	an \frac{\pi}{3}=\frac{X_L-X_C}{R}$$
When $$L$$ is removed, $$	an \frac{\pi}{3}=\frac{X_C}{R}=\sqrt{3}$$
$$	herefore \quad X_C=\sqrt{3} R$$ .... (i)
Similarly, when $$C$$ is removed
$$\begin{aligned}
& 	an \frac{\pi}{3}=\frac{X_L}{R}=\sqrt{3} \
\Rightarrow \quad X_L & =\sqrt{3} R \
\Rightarrow \quad X_C & =X_L \quad 	ext{...(ii) [from Eq. (i)]}
\end{aligned}$$
Now, $$	an \phi=0$$
$$\Rightarrow \quad \phi=0^{\circ}$$
$$	herefore$$ Power factor, $$\cos \phi=\cos 0^{\circ}=1$$

Answer

$$1 / 2$$

Answer

$$1 \sqrt{2}$$

Answer

1

Answer

$$\frac{\sqrt{3}}{2}$$

AC fundamentals

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