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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`. |
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Answer» 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)` |
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