The series combination of R`(Omega)` and capacitor C(F) is connected to an A.C. source of V volts and angular frequency `omega`. If the angular frequency is reduced to `omega/3`, the current is found to be reduced to one-half without changing the value of the voltage. Determine the ratio of the capacitive reactance and the resistance.

Explanation

First case : (Here, r.m.s. value of I and V as I and V for the sake of convenience)
`I=V/sqrt(R^2 + 1/(omega^2C^2)) " " therefore I^2=V^2/(R^2+X_C^2)`...(i)
Second case :
`I/2=V/sqrt(R^2+9/(omega^2C^2)) " " therefore I^2/4 = V^2/(R^2+9X_C^2)`...(ii)
Dividing equ. (i) by (ii), we have
`therefore 4=(R^2+9X_C^2)/(R^2+X_C^2)`
`therefore 4R^2+4X_C^2=R^2+9X_C^2`
`therefore 5X_C^2=3R^2`
`therefore X_C/R=sqrt(3/5)`