An AC source of angular frequency is fed across a resistor R and a capacitor C in series. The current flowing in the circuit found to be I . Now the frequency of the source is changed to 3 , (maintaining the same voltage) the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

Step-by-step solution

At angular frequency , current through resistance R and capacitance C is given by I_ rms = V_ rms R^2+X_C^2 = V_ rms R^2+ ( 1 C )^2 ...(i) When angular frequency is changed to 3 , then the current becomes I_ rms 2 = V_ rms R^2+ ( 1 3 C )^2 = V_ rms R^2+ ( 3 C )^2 ...(ii) Dividing Eq. (i) by Eq. (ii), we get array rlrl & & 2= R^2+ ( 3 C )^2 R^2+ ( 1 C )^2 & & 4 [R^2+ ( 1 C )^2 ] & =R^2+ ( 3 C )^2 & & 3 R^2 & = 5 ^2 C^2 & & 1 _C & = 3 5 & X_C R & = 3 5 array