A series combination of resistor 'R' and capacitor 'C' is connected to an a.c. source of angular frequency ' '. Keeping the voltage same, if the frequency is changed to ( 3 ) the current becomes half of the original current. Then the ratio of capacitive reactance and resistance is

Step-by-step solution

Initial current I₁ = V R^2 + X_C^2 where X_C = 1 C When frequency is changed to 3 , the new capacitive reactance becomes X_C' = 1 ( /3)C = 3X_C New current I₂ = V R^2 + (3X_C)^2 = V R^2 + 9X_C^2 Given I₂ = I₁ 2 V R^2 + 9X_C^2 = 1 2 V R^2 + X_C^2 Squaring both sides: R^2 + 9X_C^2 = 4(R^2 + X_C^2) R^2 + 9X_C^2 = 4R^2 + 4X_C^2 5X_C^2 = 3R^2 X_C^2 R^2 = 3 5 = 0.6 X_C R = 0.6 Answer: 0.6