An `AC` source of variable frequency is applied across a series `L-C-R` circuit. At a requency double the resonace frequency. The impedance is `sqrt(10)` times the minimum impedance. . The inductive reactance is

Options

A. `R`
B. `2 R`
C. `3 R`
D. `4 R`

Answer

4

Explanation

`Z^(2) = R^(2) + (omega L - 1//omega C)^(2)`
`10 R^(2) = R^(2) + (2 omega_(0)L - 1// 2 omega_(0)C)^(2)`
minimum impedance `Z` min `= R`
`omega_(0)^(2) LC = 1` -------(1)
`2 omega_(0) L - (1)/(2 omega_(0)C) = 3R` -------(2)
from (1) `(1)/(2 omega_(0)C) = R" ":. X_(C ) = R`
from (2) `X_(C ) = 2 omega_(0) L = 3 R + R = 4 R`