Obtain the equation of the bandwidth for an L-C-R series AC circuit and deduce the equation of Q factor.

Explanation

When `omega_(1) = omega_(1) + Delta omega`, the amplitude of current is `I_(m) = ( I_(max))/( sqrt(2))` hence amplitude of current at `omega_(1)`,
`I_(m) = ( V_(m))/( Z) = ( V_(m))/( sqrt( R^(2)) + ( omega_(0) L - ( 1)/( omega_(0)C))^(2))`
`(I_(m)^(max))/(sqrt(2)) = ( V_(m))/( R sqrt( 2))`
or `sqrt( R^(2) + ( omega_(1) L - ( 1)/( omega_(1) C ))^(2)) = Rsqrt(2)`
`:. R^(2) + ( omega_(1) L - (1)/( omega_(1) C ))^(2) = 2R^(2)`
`:. ( omega_(1) L - ( 1)/( omega_(1) C ))^(2) = R^(2)`
`:. omega_(1) L - ( 1)/( omega_(1) C ) = R`
but puttting `omega_(1) = omega_(0) + Delta omega`
`( omega _(0) + Delta omega) L - ( 1)/(( omega _(0) + Delta omega)C ) = R `
`:. omega_(0) L + Delta omega L - ( 1)/( omega_(0) C + Delta omega C ) = R `
`:. omega_(0) L ( 1+ ( Delta omega )/( omega_(0))) - ( 1)/( omega_(0) C ( 1+ ( Delta omega)/( omega_(0)))) = R `
`:. omega_(0) L ( 1+ ( Delta omega )/( omega_(0))) - ( omega _(0) L)/((1+ (Deltaomega)/omega_(0)))= R`
where `omega_(0) C = ( 1)/( omega_(0) L )`
`:. omega _(0) L ( 1+ ( Delta omega)/( omega_(0))) - omega_(0) L ( 1+ ( Delta omega)/( omega_(0)))^(-1)`
since `( Delta omega )/( omega_(0)) lt lt 1 ( 1+ ( Delta omega )/( omega_(0)))^(-1) = ( 1- ( Delta omega )/( omega ))`
`:. omega_(0) L ( 1 + (Delta omega ) /( omega_(0))) - omega_(0) L ( 1 - ( Delta omega)/( omega_(0)))`
`:. omega_(0) L[ 1+ ( Delta omega )/( omega_(0)) - 1+ ( Delta omega ) /( omega_(0))]`
`:. omega _(0) L xx ( 2 Delta omega )/( omega _(0)) = R :. Delta omega = ( R )/( 2L )`