A series LCR circuit is connected to an ac source having voltage `v=v_(m) sin omega t`. Derive the expression for the instantaneous current I and its phase relationship to the applied voltage.
(i) Maximum and (ii) minimum.

Explanation

`v=v_(m)sin omegat`
Let the current in the circuit be lead the applied voltage by an angle `phi`.
`i=i_(m) sin (omegat+phi)`
The Kirchhoff's voltage law gives `L=(di)/(dt)+Ri+q/C=v`
It is given that `v=v_(m)sin omegat`(applied voltage)
`L(d^(2)q)/(dt^(2))+Ri(dq)/(dt)+q/C=v_(m)sin omegat`..(1)
On solving the equation , we obtain `q=q_(m)sin (omegat+theta)`
`(dq)/(dt)=q_(m)omega cos (omegat+theta)`
`(d^(2)q)/(dt^(2))=-q_(m)omega^(2) sin (omegat+theta)`
On substituting these values in equation (1), we obtain
`q_(m)omega[R cos (omegat+theta)+(X_(C)-X_(L)) sin (omegat+theta)]=v_(m) sin omegat` `X_(C)=1/(omegaC),X_(1)=omegaL`
`Z=sqrt(R^(2)+(X_(C)-X_(L))^(2))`
`q_(m)omegaZ[R/Zcos(omegat+R)+((X_(C)-X_(L)))/Zsin (omegat+theta)]=v_(m)sin omegat`....(2)
Let `cos phi=R/Z` and `((X_(C)-X_(L)))/Z=sin phi`
This gives `tan phi=((X_(C)-X_(L)))/R`
On substituting this in equation (2), we obtain
`q_(m)omegaZcos (omegat+theta-phi)=v_(m)sin omegat`
On comparing the two sides , we obtain
`v_(m)=q_(m)omegaZ=i_(m)Z`
`i_(m)=q_(m)omega`
and `theta=-pi/2+phi`
`i=(dq)/(dt)=q_(m)omegacos(omegat+theta)=i_(m) cos (omegat+theta)`
or `i=i_(m)sin (omegat+theta)`
Where `i_(m)=v_(m)/Z=v_(m)/sqrt(R^(2)+(X_(C)-X_(L)^(2))`
And `phi=tan^(-1) ((X_(C)+X_(L))/R)`
The condition for resonance to occur
`i_(m)=v_(m)/sqrt(R^(2)+(X_(C)-X_(L)^(2))`
For resonance to occur, the value of `i_(m)` has to be the maximum.
The value of `i_(m)` will be the maximum when
`X_(C)=X_(L)`
`1/(omegaC)=omegaL`
`omega^(2)=1/sqrt(LC)`
`2pif=1/(2pisqrt(LC))`
Power factor `=cos phi`
Where, `cosphi=R/Z=R/sqrt(R^(2)+(X_(C)-X_(L))^(2))`
(i)Conditions for maximum power factor (i.e., `cos phi=1`)
`* X_(C)=X_(L)`
Or
`*R=0`
(ii)Conditions for minimum power factor
`*` When the circuit is purely inductive
`*` When the circuit is purely capacitve