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A coil with active resistance `R` and inductance `L` was connected at the moment `t=0` to a source of voltage `V=V_(m)cos omegat.` Find the current in the coil as a function of time `t`. |
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Answer» The equatio of the circuit is `(I` is the current `)` `L(dI)/(dt)=RI=V_(m) cos omegat ` From the theory of different equations. `I=I_(p)+I_(C)` where `I_(p)` is a particular integral and `I_(C)` is the complementary function `(` Solution of the differential equation with the RHS `=0). N`ow `I_(C)=I_(CO)e^(-tR//L)` and for `I_(p)` we write `I_(p)=I_(m) cos ( omegat-varphi)` Substituting we get `I_(m)=(V_(m))/( sqrt(R^(2)+omega^(2)L^(2))), varphi=tan ^(-1)(omegaL)/(R)` Then `I_(m)=(V_(m))/( sqrt(R^(2)+omega^(2)L^(2)))cos ( omegat- varphi)+I_(CO) e^(-tR//L` Now in an inductive circuit `I=0` at `t=0` because a current cannot change suddently. Thus `I_(CO)=-(V_(m))/(sqrt(R^(2)=omega^(2)L^(2))) cos varphi` and so ` I=(V_(m))/(sqrt(R^(2)+ omega^(2)L^(2)))[cos ( omegat - varphi)- cos varphie^(-tR//L)]` |
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