An AC voltage source is applied across an R-C circuit. Angular frequency of the source is `omega`, resistance is R and capacitance is C. The current registered is I. If now the frequency of source is changed to `omega/2` (but maintaining the same voltage), the current in the circuit is found to be two third. calculate the ratio of reactance to resistance at the original frequency `omega` .

Explanation

At angular frequency w, the current in R-C circuit is given by
`i^(rms)=(epsilon^(rms)/(sqrt((R^(2))+((1//omega^(2)C^(2))^(2)`(1)
when frequency is changed to w/3, the current is halved.Thus
`i_(rms)=epsilon^(rms)/(sqrt((R^(2))+(1//(omega//3)^(2)C^(2))))=epsilon_(rms)/sqrt(R^(2)+(9//omega^(2)C^(2))`
From equation (i) and (ii), we have
`(1)/sqrt(R^(2)+(1//omega^(2)(C^(2))))=(2)/(sqrt(R^(2)+(9//omega^(2)C^(2))`
Solving this equation, we get` 3R^(2) =(5)/omega^(2)C^(2)`
Hence, the ratio of reactance to resistance is`((1//omegaC))/(R)=sqrt((3)/(5)`