Question

In the circuit (see attachment) the switch is thrown to position 2 at t=0.Assuming that steady-state conditions exist prior to t=0,solve for i(t) for t>0.

Answer 1

Here are the possible answers for the exercise.

Answer 2

At t=0⁻, steady-state involve that inductance tension is null and the current is constant.
\[L.di_0(t)/dt=E-R.i_0(t)=0=E-R.I_0=>I_0=E/R=10/10=1A\]
At t=0⁺
\[L.di(t)/dt+R.i(t)+q(t)/C=0\]
And
\[i(t)=dq(t)/dt\]
\[L.d^2q(t)/dt^2+R.dq(t)/dt+q(t)/C=0\]
\[d^2q(t)/dt^2+(R/L).dq(t)/dt+q(t)/(LC)=0\]
Differential equation :
\[r^2 +2*(R/(2.L)).r+1/LC=0 \]
\[\Delta'=[R/(2.L)]^2 - 1/(L.C)\]
\[r_{1,2}=- R/(2.L)\pm \sqrt{\Delta'}\]
Solution :
\[q(t)=\alpha.e^{r_1.t}+\beta.e^{r_2.t}=>i(t)=dq(t)/dt=\alpha.r_1.e^{r_1.t}+\beta.r_2.e^{r_2.t}\]
You know the current at t=0, i(t=0)=1A and q(t=0)=0
Good luck !

Answer 3

The form of the solution is an exemple, whether
\[(R/2.L)^2 - 1/(LC) = 0; (R/2.L)^2 - 1/(LC) > 0; (R/2.L)^2 - 1/(LC) < 0\]
the solution hasn't the same form (see aperiodic, periodic or critical regime).

Answer 4

Here, the form of the solution is (because (R/2.L)²-1/(L.C)<0:
\[q(t)=e^{−R.t/(2.L)}.f(wt)=e^{−t/2}.[α.\cos(w.t)+β.\sin(w.t)]\]
with
\[w=\sqrt{1/(LC)-R^2/(2.L)^2}=\sqrt{3}/2≈0,866\]
So I think the answer is (5). (I don't calculate i(t) but with w and the exponent I've choiced (5))

Answer 5

Thank you !!!!!!!!