Question:

A series LR circuit is connected to an ac source of frequency \(\omega\) and the inductive reactance is equal to 2R. A capacitance of capacitive reactance equal to R is added in series with L and R. The ratio of the new power factor to the old one is :

Show Hint

Power factor \(= R/Z\). Compute Z before and after adding the capacitor.
Updated On: Nov 4, 2026
  • \(\sqrt{\frac{2}{3}}\)
  • \(\sqrt{\frac{2}{5}}\)
  • \(\sqrt{\frac{3}{2}}\)
  • \(\sqrt{\frac{5}{2}}\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

Old circuit: \(X_L = 2R\), \(Z_1 = \sqrt{R^2 + 4R^2} = \sqrt5 R\), so \(\cos\phi_1 = \dfrac{R}{Z_1} = \dfrac{1}{\sqrt5}\).
New circuit: \(X_L - X_C = 2R - R = R\), \(Z_2 = \sqrt{R^2 + R^2} = \sqrt2 R\), so \(\cos\phi_2 = \dfrac{1}{\sqrt2}\).
Ratio \(= \dfrac{\cos\phi_2}{\cos\phi_1} = \dfrac{\sqrt5}{\sqrt2} = \sqrt{\dfrac{5}{2}}\).
Answer: (4) \(\sqrt{\dfrac{5}{2}}\)
Was this answer helpful?
0
0

Top JEE Main Physics Questions

View More Questions

Top JEE Main Alternating current Questions

View More Questions