An AV 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

Consider the R-C circuit as shown below ,

because `X_(C) = 1/(omegaC) implies Z = sqrt (R^(2) + X_(C)^(2))`
` therefore I = V/Z = V/(sqrt (R^(2) + X_(C)^(2)))`
When frequency is changed to `omega/2` ,

The value of new capacitive reactance is
`X_(C)^(') = 1/((omega/2)C) = 2/omegaC = 2X_(C)`
New impedance , `Z^(') = sqrt (R^(2) + X_(C)^(2)) = sqrt (R^(2) + (2X_(C))^(2))`
` implies I^(') = V/(sqrt (R^(2) + 4X_(C)^(2)))`
It is given that `I^(') = 2/3I`
`implies 2/3 xx V/(sqrt (R^(2) + X_(C)^(2))) = V/(sqrt (R^(2) + 4X_(C)^(2)))`
`implies 4R^(2) + 16X_(C)^(2) = 9R^(2) + 9X_(C)^(2)`
`implies 7X_(c)^(2) = 5R^(2) implies X_(C)/R = sqrt (5/7)`