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Obtain the equation of the bandwidth for an L-C-R series AC circuit and deduce the equation of Q factor. |
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Answer» SOLUTION :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 )` |
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