Initial condition across capacitor - RLC

The switch has been in the original position for a long time and is flipped at t=0 (apologies for my drawing skills)

I am trying to find the initial condition across the capacitor. I am confused because the capacitor will resemble an open circuit and the inductor will resemble a short circuit, right? So there is no current flowing through the capacitor or R2. It was easy enough to get the initial condition for the inductor, i_L(0-), but I am confused on how to get the initial condition of the capacitor v_C(0-) without current.

I thought I could get the voltage across the inductor and then use that to do V_R2+V_C=V_L but then I got even more confused. How do I go about this?

• With DC current the inductor looks like a short circuit. What is the voltage across a short circuit? In that case what is the voltage across R2 and C. So what is the voltage across C?
– RoyC
Commented Feb 20 at 11:44
• Oh, I was overcomplicating, yes the voltage across a short circuit is zero. Thanks Commented Feb 20 at 11:50

In the very long time between $$\t=-\infty\$$ and $$\t=0^-\$$, the current in the inductor $$\L\$$ will stop changing. Because of $$\V_L=L\cdot \frac{\text{d}}{\text{d}t}I_L\$$, this means $$\V_L=0\:\text{V}\$$ and all of the voltage drop available from $$\V_s\$$ will appear across $$\R_1\$$. This also means, regardless of what happened much earlier, prior to $$\t=0^-\$$, there will be $$\0\:\text{V}\$$ across the series branch of $$\R_2\$$+$$\C\$$. If $$\C\$$ ever did have a voltage across it prior to $$\t=0^-\$$, it most certainly cannot have one at $$\t=0^-\$$. Any charge would have discharged via $$\R_2\$$ and $$\L\$$ by $$\t=0^-\$$.
At the moment of switching, the state of $$\C\$$ should be $$\0\:\text{V}\$$ and the state of $$\L\$$ should be $$\I_L=\frac{V_s}{R_1}\$$. At this point, $$\t=0^+\$$, KCL tells you that the current into the top node will be $$\I_s\$$, the current leaving the node via $$\L\$$ will be $$\I_L=\frac{V_s}{R_1}\$$ so any difference, $$\I_s-\frac{V_s}{R_1}\$$, is the downward-pointed current into $$\R_2\$$. From this, and the fact that $$\V_C=0\:\text{V}\$$, you can compute the initial node voltage at $$\t=0^+\$$. That initial node voltage will provide a rate of change in the current in $$\L\$$ to start altering $$\I_L\$$ and will also provide a current into $$\C\$$ that will start altering $$\V_C\$$.