For a series `RLC` circuit `R=X_(L)=2X_(C)`. The impedence of the current and phase different (between) `V` and `i` will be

Options

A. `(sqrt5R)/(2), tan^(-1)(2)`
B. `(sqrt5R)/(2), tan^(-1)(1/2)`
C. `sqrt5X_(C),tan^(-1)(2)`
D. `sqrt5R, tan^(-1)(1/2)`

Answer

B

Explanation

`R = X_(L) = 2X_(C)`
`Z = sqrt(R^(2) + (X_(L) - X_(C))^(2)) = sqrt((2X_(C)^(2) + (2X_(C) - X_(C) )^(2)`
`= sqrt((4X_(c)^(2)) + X_(C)^(2))`
`= sqrt((4X_(c)^(2)) + X_(C)^(2))`
` = sqrt5X_(C) = (sqrt5R)/(2)`
`tan phi = (X_(L) - X_(C))/(R) = (2X_(C) - X_(C))/(2X_(C))`
`tan phi = 1/2, phi = tan ^(-1) ((1)/(2))`