Values at t=0- and t=0+ when closing switch; Circuit Analysis

Question

I have a circuit as shown in the schematic and need to find out the values at t=0- and t=0+ but I don't understand it correctly. The condition is that the switch was open for a long time and closes at t=0.

^{simulate this circuit – Schematic created using CircuitLab}

I need to name the values shown in the circuit at t=0- (shortly before SW1 closes):

- i(0-)   = 0A (there is no current because C1 is blocking all DC Current after fully charged)
- V_C(0-) = 10V
- V_L(0-) = 0V (Complete Voltage at C1)
- V_2(0-) = 0V (Complete Voltage at C1)

Now I need to name the values shown in the circuit at t=0+ (shortly after SW1 closes); This is where I struggle:

- i(0+)   = ?
- V_C(0+) = 10V (Voltage cant jump at a capacitor)
- V_L(0+) = ?
- V_2(0+) = ?

I'm missing something really trivial here and would be thankful for some help.

Answer 1

If the switch has been open for a very long time, then it's true that \$V_L=0\text{V}, \: I_C = 0\text{A}\$. This will result in the following circuit.

^{simulate this circuit – Schematic created using CircuitLab}

When you close the switch, you short out the voltage source and leave the energy storing components to discharge. The capacitor voltage and inductor current can't change instantaneously, so you know: $$V_C = 10\text{V}, \: I_L=0\text{A} $$ But one important thing to notice is that you also connect one end of the capacitor directly to ground. So now you have the following circuit.

^{simulate this circuit}

You should be able to see that a thing regarding the voltage across the inductor has changed. I think this should be able to push you through the problem.