In a series `LCR` circuit, at the frequencies `f_(1)` and `f_(2)` of `AC` source, the current amplitude falls to `(1)/(sqrt(2))` of the current amplitude at resonance. Then the value of `f_(2) - f_(1)` is

Options

A. `(R )/(2 pi L)`
B. `(R )/(L)`
C. `(R )/(pi L)`
D. `(R )/(pi^(2) L)`

Answer

A

Explanation

`i = (i_(r))/(sqrt(2))` (say at angular frequencies `omega_(1)` and `omega_(2)`)
`(V_(0))/(sqrt(R^(2) + (omega L + (1)/(omega C)))^(2)) = (V_(0))/(sqrt(2) R)`
solving `omega_(1) L = (1)/(omega_(1) C) = - R` ….(i)
`omega_(2) L - (1)/(omega_(2) C) = + R` ....(ii)
Adding, `omega_(1) omega_(2) = (1)/(LC)` .....(iii)
Substracting and using (iii) `omega_(2) -omega_(2) = (R )/(LC)`
`:. f_(2) - f_(1) = (R )/(2 pi L)`