Obtain an anaalytical solution for the relation of phase between instantaneous current and voltage for an LCR series AC circuit.

Explanation

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) omega cos ( 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)/( omega C ) 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) sin omegat `
`:. 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.