In a series LR circuit, \( X_L=3R \). Now a capacitor with \( X_C=R \) is added in series. The ratio of the new power factor to the old power factor is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sqrt{2} \)
In a series ac circuit the power factor is \( \cos\phi=R/Z \), where the impedance is \( Z=\sqrt{R^{2}+(X_C-X_L)^{2}} \). Adding the capacitor reduces the net reactance, so the impedance falls and the power factor rises.
For the LR circuit (no capacitor):
\( \begin{aligned} Z &= \sqrt{R^{2}+X_L^{2}} \\ &= \sqrt{R^{2}+9R^{2}} \\ &= \sqrt{10}\,R \end{aligned} \)
\( \begin{aligned} \cos\phi &= \frac{R}{Z} \\ &= \frac{1}{\sqrt{10}} \end{aligned} \)
With the capacitor added in series:
\( \begin{aligned} Z' &= \sqrt{R^{2}+(X_C-X_L)^{2}} \\ &= \sqrt{R^{2}+(R-3R)^{2}} \\ &= \sqrt{5}\,R \end{aligned} \)
\( \begin{aligned} \cos\phi' &= \frac{R}{Z'} \\ &= \frac{1}{\sqrt{5}} \end{aligned} \)
The ratio of the new power factor to the old one is:
\( \begin{aligned} \frac{\cos\phi'}{\cos\phi} &= \frac{1/\sqrt{5}}{1/\sqrt{10}} \\ &= \sqrt{\frac{10}{5}} \\ &= \sqrt{2} \end{aligned} \)