How do I proceed to solve this problem. If I write a KVL equation, I get
$$4 (4 Vc) +1/10^-3 integral i dt – 10=0$$
What to do next?
Electrical – Solving a source free series RLC circuit
capacitorcircuit analysisinductorresistors
Related Solutions
I don't think I can give you all the details, but this may help you get started. Fortunately, there are lots of information on the Web for this. A typical way of reverse transforming an expression like yours is to look up a table of Laplace transforms, rewrite the expression such that the components match one or more of the forms in the table. With your expression, it is going to fit into these two forms (copied from Wikipedia):
Now, rewrite your expression such that \$I(s) = AP(s) + BQ(s), A,B\$ are constants.
The first step of rewriting would probably be rewriting the denominator by "completing the square". And you would be factoring out constants during the rewrite.
After you get P(s) and Q(s) to match the forms in the table exactly, then, \$i(t) = Ap(t) + Bq(t)\$ by using the original functions given by the table.
New edit: I was curious, so finished the reverse transform as below
$$ I=\frac{8 \times 10^{-5}s + 0.4}{4 \times 10^{-3} s^2 + 32s +10^5} =\frac{0.02s + 100}{s^2 + 8000s + 25000000} $$ $$ =0.02 (\frac{s + 5000}{s^2 + 8000s + 25000000}) =0.02 (\frac{s + 5000}{(s+4000)^2 + 3000^2}) $$ \$\alpha=4000\$, \$\omega=3000\$ These match the roots you calculated for the poles. $$ I = 0.02 (\frac{s + 4000 - 4000 + 5000}{(s+4000)^2 + 3000^2}) = 0.02 (\frac{s + 4000}{(s+4000)^2 + 3000^2}+\frac{1000}{(s+4000)^2 + 3000^2}) = 0.02 (\frac{s + 4000}{(s+4000)^2 + 3000^2}+\frac13\frac{3000}{(s+4000)^2 + 3000^2})$$ $$ i(t) = 0.02(e^{-4000t}cos(3000t) + \frac13e^{-4000t}sin(3000t))$$
What got me curious was I tried to get the reverse transform from WolframAlpha also, and got some really complicated answer in real form. My guess is that it uses a mechanical method that produces an answer too complicated for it to reduce. So if one just plug numbers into a computer to get answers, sometimes simpler relationships may stay hidden.
You're not meant to have 2 independent sources working at the same time in the circuit, so the best way to deal with this is to use source transformation which will produce a voltage source making the circuit a combination of 2 voltage sources. I'm not sure it can be solved otherwise. Besides before you solve for neper frequency the RLC circuit has to be either a completely parallel or series circuit. Your circuit above is neither (series-parallel), to find 'alpha' you have to use source transform to make it a completely series circuit.
Best Answer
All that matter here are the capacitor and the dependent current source. Regardless of the rest of the circuit, the current source adjusts its voltage to guarantee that the current is \$3v_c\$.
Note that, with the polarities given, the current source discharges the capacitor from an initial \$10V\$, and thus we may write:
$$v_c=10-\frac{1}{C}\int 3v_c\:dt\:=10-300\int v_c\:dt$$
Where the minus sign indicates that the current flow is discharging \$C\$.
Differentiating this equation:
$$\frac{dv_c}{dt}=-300v_c$$
This is a first order differential equation and, clearly, the solution is an exponential of the form:
$$v_c=Ae^{\alpha t}$$
By inspection, \$\alpha =-300\$, and using the initial condition: \$t=0, \:v_c=10\$, gives \$A=10\$. Hence: $$v_c=10e^{-300t}$$
From the circuit, we see that \$i_L=-3v_c\$, therefore:
$$i_L=-30e^{-300t}$$