Give analytical solution to a series RLC circuit connected to ac.

Explanation

The voltage equation for the circuit is `L(di)/(dt)+Ri+(q)/(c)=v=v_(m)sinomegat`
The solution to the equation will be
`q=q_(m)sin(omegat+theta)`
`therefore q_(m)omega[Rcos(omegat+theta)+(X_(C)-X_(L))sin(omegat+theta)]=v_(m)sinomegat`
where, `X_(C)=(1)/(Comega)andX_(L)=Lomega`
Also, `q_(m)omegaZ[(R)/(Z)cos(omegat+theta)+(X_(C)-X_(L))/(Z)sin(omegat+theta)]=v_(m)sinomegat`

Put, `(R)/(Z)=cosphiand((X_(C)-X_(L))/(Z))=sinphi" "...(1)`
Hence, `q_(m)omegaZ(cos(omegat+theta-phi))=v_(m)sinomegat`
i.e., `[q_(m)omega(cos(omegat+theta-phi)]Z=v_(m)sinomegat`
If `theta-phi=-pi//2`, then `cos(omegat+theta-phi)=cos(pi//2-omegat)=sin(omegat)`
`therefore v_(m)=q_(m)omegaZ,i_(m)=q_(m)omegaandZ=(v_(m))/(i_(m))`
Since, `i=(dq)/(dt)=(d)/(dt)(q_(m)sin(omegat+theta)=q_(m)omegacos(omegat+theta)`
i.e., `i=i_(m)cos(omegat-(pi)/(2)+phi)=i_(m)cos((pi)/(2)-(omegat+phi))`
i.e., `i=i_(m)sin(omegat+phi)`
By using (1), `tanphi=(X_(C)-X_(L))/(Z)`
graphically we represent it as in fig (i) and fig (ii)
Hence, `v^(2)=v_(R)^(2)+(V_(C)-V_(L))^(2)` using fig (ii)
or `Z=((v_(m))/(i_(m)))=sqrt(R^(2)+(X_(C)-X_(L))^(2))`