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`.

Explanation

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)]`