An a.c. source of angular frequency ' ' is fed across a resistor ' R ' and a capacitor ' C ' in series. The current registered is I. If now the frequency of source is changed to 3 (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency ' ' will be

Step-by-step solution

aligned & Initial current I= V Z , Final current I^ = V Z^ & 1 2 = I^ I = Z Z^ Z Z^ = 1 2 & Z^2 Z^ 2 = 1 4 R^2+X_c^2 R^2+X_c^ 2 = 1 4 & 4 R^2+4 X_c^2=R^2+X_c^ 2 & 3 R^2=X_c^ 2 -4 X_c^2 .(1) & X_c= 1 C , X^ = 3 C X^ =3 X_c & Putting this value of X^ in (1) we get & 3 R^2=9 X_c^2-4 X_c^2=5 X_c^2 & X_c^2 R^2 = 3 5 =0.6 & X_c R = 3 5 aligned