A series combination of resistor ( R ), capacitor ( C ) is connected to an .C. source of angular frequency `omega`. Keeping the voltage same, if the frequency is charged to `(omega)/(3)`, the current becomes half of the original current. Then the ratio of the capacitive reactance and resistance at the former frequency is

Options

A. `sqrt(0.6)`
B. `sqrt(3)`
C. `sqrt(2)`
D. `sqrt(6)`

Answer

A

Explanation

In the first case,
Capacitive reactance, `X_(C) = (1)/(omega C)`
Impedance of the circuit,
`Z= sqrt(R^(2) + X_(C)^(2))`
Current in the circuit
`I =(V)/(R ) = (V)/(sqrt(R^(2) + X_(C)^(2)))`
In the second case,
`omega'= (omega)/(3)`
`:.` Capacitive reactance,
`X'_(C) = (1)/(omega'C) = (1)/((omega)/(3)C) = (3)/(omegaC) = 3X_(C)`
Impedance of the circuit
`Z' = sqrt(R^(2) + X'_(C)^(2)) = sqrt(R^(2) + 9X_(C)^(2))`
Current in the
`(I)/(2) = (V)/(Z') = (V)/(sqrt(R^(2) + 9X_(C)^(2)))`....(ii)
Divide (ii) by (i) we get
`(1)/(2) = (sqrt(R^(2) + X_(C)^(2)))/(sqrt(R^(2) + 9X_(C)^(2)))`
Squaring both sides , we get
`(1)/(4) = (R^(2) + X_(C)^(2))/(R^(2) + 9X_(C)^(2))`
`R^(2) + 9X_(C)^(2) = 4R^(2) + 4X_(C)^(2)`
`5X_(C)^(2) = 3R^(2) or (X_(C))/(R ) = sqrt((3)/(5)) = sqrt0.6`