Help with Series RLC circuit

Hi wiz0r,
Presumably the initial conditions that you're looking for is the state of the circuit at the time t=0⁺ when the switch has just opened.

Immediately before the switch opens, and assuming that the switch had been closed for "a long time", then your statement that the capacitor will look like an open circuit is correct -- no current will flow into or out of it. Similarly, the inductor will look like a short circuit.

Can you determine what voltage should be across the capacitor at t=0^-?

 

That looks fine. So when the switch opens the capacitor will have that voltage on it as an initial condition. And the current will still be zero because of the inductor.

To find i(t) you'll have to either write and solve the differential equation for the new circuit configuration, or use a "canned" solution for the RLC circuit.

 

It's looking good!

 

The equation should yield current values, and Vc(0) is a voltage.

At time t = 0 the current is zero, so K1 + K2 = 0. That is, K1 = -K2. You'll need to find another way to solve for the values. Perhaps if you could determine the initial rate of change of the current...

 

Right

s1K1 + s2K2 won't be zero. It'll equal the initial rate of change of the current. Go back to the circuit diagram and see if you can't determine what that might be. You've an inductor that has suddenly had a voltage Vc impressed upon it... how will it react?

 

The equation of interest for the inductor is its defining equation: V = L dI/dt. In this case there's an impressed voltage Vc across inductor L=2.5H. What's dI/dt?

 

Right. At t=0 dI/dt = Vc/L = 10.286 A/s. So, given K1 = -K2,...

 

Looks good! Bravo!