A series combination of resistor (R) and capacitor (C) is connected to an AC 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 the capacitive reactance and resistance at the former frequency is

Step-by-step solution

In the first case, capacitive reactance, X_C= 1 C Impedance of the circuit, Z= R^2+X_C^2 Current in the circuit, I= V Z = V R^2+X_C^2 (i) In the second case, ^ = 3 Capacitive reactance, X_C^ = 1 ^ C = 1 3 C = 3 C =3 X_C Impedance of the circuit, Z^ = R^2+X_C^ 2 = R^2+9 X_C^2 Current in the circuit, I 2 = V Z^ = V R^2+9 X_C^2 Divide (ii) by (i), we get aligned & 1 2 = R^2+X_C^2 R^2+9 X_C^2 1 4 = R^2+X_C^2 R^2+9 X_C^2 & 5 X_C^2=3 R^2 or X_C R = 3 5 = 0.6 aligned