In a series LR ac circuit, the inductive reactance \( (X_{L}) \) is \( 5\ R \). Now the inductor \( L \) is replaced by a capacitor \( C \) with capacitive reactance \( (X_{C}) \) of \( 3\ R \). Ratio of new power factor to old power factor is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \sqrt{\displaystyle\frac{13}{5}} \)
The power factor of a series circuit is \( \cos\phi = R/Z \), where the impedance of a resistor with one reactive element is \( Z = \sqrt{R^{2} + X^{2}} \). It does not matter whether the reactance is inductive or capacitive.
Old circuit (LR, \( X_{L} = 5R \)):
\( \displaystyle \cos\phi_{1} = \frac{R}{\sqrt{R^{2} + 25R^{2}}} = \frac{1}{\sqrt{26}} \)
New circuit (RC, \( X_{C} = 3R \)):
\( \displaystyle \cos\phi_{2} = \frac{R}{\sqrt{R^{2} + 9R^{2}}} = \frac{1}{\sqrt{10}} \)
\( \displaystyle \frac{\cos\phi_{2}}{\cos\phi_{1}} = \frac{\sqrt{26}}{\sqrt{10}} = \sqrt{\frac{13}{5}} \)