Solved: RC First Order Circuit Voltage Drop after Switch Opens

[raface]

Homework Statement

Find the voltage drop across the capacitor just after the switch is opened v(0⁺)



Most variables are in picture attached
C = 100mF

Homework Equations

V_c(t)= V_c($$\infty$$)+[V_c(t₊)-V_c($$\infty$$))*e^{-(t-t₁/$$\tau$$)}

The Attempt at a Solution

First i assumed that the capacitor would be charge in steady state before the switch is opened. This would mean it is a short circuit. So the voltage in the capacitor would be 12V? Not sure if this is correct.

Secondly i assumed that as t tends to infinity, after the switch is opened that the voltage across the capacitor will become 30V

I then worked put the thevenin resistance of the capacitor circuit after the switch is opened.

R_Th = 9.6154 ohms

Then
$$\tau$$ = R_Th * C

I am not sure how to apply this to the first order equation to find v(0⁺)

 

[CEL]

In steady state, the capacitor is an open circuit, so its voltage is the difference between the voltage drop at the 4 ohm, 6 ohm resistors, caused by the 5A source and the voltage drop in the voltage divider (5 ohm, 20 ohm) caused by the voltage source.
After the switch opens, the capacitor will be connected to ground via the 6 ohm resistor and will discharge. At t = infinity, the voltage in the capacitor will be zero.