A circuit consisting of a capacitor with capacitance `C` and a resistance `R` connected in series was connected at the moment `t=0` to a source of ac voltage `V=V_(m) cos omegat` . Find the current in the circuit as a function of time `t`.

Explanation

Here the equation is `(Q` is charge on the capacitor `)`
`(Q)/(C)+R(dQ)/( dt)=V_(m) cos omegat`
A solution subject to `Q=0 ` at `t=0` is of the form `(` as in the previous problem `)`
`Q=Q_(m)[cos ( omegat-bar(varphi))-cos bar(varphi)e ^(-t//RC)]`
Subtituting back
`(Q_(m))/(C) cos ( omegat-bar(varphi))-omegaRQ_(m) sin ( omegat- bar( varphi))`
`=V_( m) cos omega t `
`=V_(m){ cos bar(varphi)( omegat-bar(varphi))-sin bar(varphi) sin ( omegat-bar(varphi))}`
so `Q_(m)=CV_(m) cos bar(varphi)`
`omega RQ_(m)=V_(m) sin bar ( varphi)`
This leads to
`Q_(m)=(CV_(m))/( sqrt(1+ ( omegaRC)^(2))), tan bar(varphi)= omega RC`
Hence `I=(dQ)/( dt)=(V_(m))/(sqrt(R^(2)+((1)/( omegaC))^(2)))[- sin omegat-bar(varphi)+( cos ^(2) bar(varphi))/( sin bar(varphi))e^(-t//RC)]`
The solution given in the book satisfies `I=0` at `t=0`. Then `Q=0` at `t=0` but this will not satisfy the equation at `t=0`. Thus `I=0`, ( Equation will be satisfies with `I=0` only if `Q=0` at ` t=0)`
With out `I, I(t=0) =(V_(m))/( R)`