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Obtain an anaalytical solution for the relation of phase between instantaneous current and voltage for an LCR series AC circuit. |
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Answer» Solution :The voltage equation for L-C-R series circuit is `L(dI)/(dt) + RI + ( q)/( C ) = V` `L (dI)/( dt) + RI + ( q)/( C ) = V_(m) sin omegat `....(1) `I = ( dq)/( dt) :. (dI)/( dt) - ( d^(2)q)/( dt^(2)) ` HENCE from equation (1) `L (d^(2)q)/( dt^(2)) + R(dq)/( dt) + ( q)/( C ) = V_(m) sin omega t ` `:. ( d^(2) q)/( dt^(2)) + ( R )/(l) ( dq)/( dt) + ( q)/( LC ) = ( V_(m))/( L) sin omega t ` ....(2) This is like the equation for a forced, damped oscillator. Let us ASSUME a solution, `q = q_(m) sin ( omega t + theta ) ` ....(3) `:. ( dq)/( dt) = q_(m) omegacos ( omega t + theta ) ` ...(4) and `(d^(2)q)/( dt^(2)) = - q_(m) omega^(2) sin ( omega t + theta ) `.....(5) `:.` Putting VALUE of equation (3), (4) and (5) in equ. (2), `-q_(m)omega^(2) L sin(omega t + theta ) + Rq_(m) omega COS ( omega t + theta ) + ( q_(m))/( C ) sin (omega t + theta ) = V_(m ) sin omega t ` `q_(m) omega [-L omega sin ( omega t + theta ) + R cos ( omega t + theta ) + ( 1)/( omegaC ) sin ( omega t+ theta ) = V_(m) sin omega t] ` `:. q_(m) omega [ R cos ( omega t + theta ) + ( 1)/( omega C ) sin ( omega t + theta ) - omega L sin ( omega t + theta ) = V_(m) sin omega t ]` `:. q_(m) omega [ R cos ( omega t + theta ) + ( X_(C ) - X_(L)) sin ( omega t + theta ) = V_(m) sin oemga t ` [ Putting `(1)/( omega C ) = X_(C )` and `omega L = X_(L)]` Multiplying and dividing by Z, `:. q_(m) omega Z [ ( R )/( Z) cos ( omega t + theta ) - (( X_(C ) - X_(L))/( Z)) sin ( omega t + theta ) = ( V_(m) Z sin omega t )/( Z )]` `:. q_(m)omega Z [ cos phi cos ( omega t + theta ) - sin phi sin ( omega t + theta ) ]= V_(m) sin omega t `.....( 6) where `( R )/( Z ) = cos phi ` and `( X_( C ) - X_(L))/( Z) = sin phi ` `:. (X_(C ) - X_(L))/( R ) = tan phi ` `:. phi = tan^(-1) ((X_(C)-X_(L))/(R )) `....(7) `:. q_(m) omega Z [ cos (omega t + theta ) cos phi + sin ( omega t + theta ) sin phi ]= V_(m) sinomegat ` `:. q_(m) omega Z [ cos ( omega t + theta- phi]=V_(m) sin oemga t ` `[ :. cos alpha cos beta - sin alpha sin beta = cos ( alpha - beta )]` `:. q_(m) omega Z [ cos { omega t + ( [ ( pi )/( 2)) } ] = V_(m) sin omega t ``[ :. ` Takin `theta - phi = - ( pi )/( 2) ]` `:. q_(m) omega Z sin omega t = V _(m) sin omega t `...(8) Comparing with, `V_(m) = q_(m) omega Z` `:. V_(m) = I_(m) Z ``[ :. I_(m) = q_(m) omega ]` Current in circuit, `I = ( dq)/( dt) = ( d)/( dt) [ q_(m) sin (omega t + theta )]` `= q_(m) omega cos ( omega t + theta )` `= I_(m) cos ( omega t + theta ) ``[ :. q_(m) omega = I_(m)]` or `I = I_(m) sin ( omega t + phi ) ` where `theta = phi - ( pi )/(2)` and `I_(m) = ( V_(m))/( Z ) = ( V_(m))/( sqrt( R^(2) + (X_(C ) -X_(L))^(2)))` and `phi = tan^(-1) ( X_(C ) - X_(L))/(R )` Hence, in L-C-R series circuit, current is`I_(m) cos ( omega t + theta ) ` or `I_(m) sin ( omega t + phi ) ` and amplitude `I_(m) = ( V_(m))/( sqrt(R^(2) + ( X_(C )- X_(L))^(2)))` Hence, solution obtained by both techniques are same. |
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