I've the following circuit:
simulate this circuit – Schematic created using CircuitLab
The capacitor is charged up to 10 V according to the following formula (when the switch is closed):
$$\text{V}_\text{C}\left(t\right)=\hat{\text{u}}\cdot\left(1-\exp\left(-\frac{t}{\text{CR}_1}\right)\right)\tag1$$
Now, when I open the switch the capacitor will discharge its energy in the resistors that are standing in series with the capacitor.
To find the current that runs in the circuit when I open the switch is:
$$\text{V}_\text{C}\left(t\right)+\text{V}_{\text{R}_1}\left(t\right)+\text{V}_{\text{R}_2}\left(t\right)=0\tag2$$
Writing that in terms of the the current (using Ohm's law and \$\text{I}_\text{C}\left(t\right)=\text{C}\cdot\text{V}'_\text{C}\left(t\right)\$) in the circuit gives:
$$\text{I}_\text{C}\left(t\right)\cdot\frac{1}{\text{C}}+\text{I}_{\text{R}_1}'\left(t\right)\cdot\text{R}_2+\text{I}_{\text{R}_2}'\left(t\right)\cdot\text{R}_2=0\tag3$$
Now, if I define a new time \$t=0\$ when I open the switch, the initial current in the capacitor is equal to \$0\$ amps. So \$\text{I}_\text{C}\left(0\right)=0\$. Now it is a series circuit (when the switch is open) so I can rewrite:
$$\text{I}\left(t\right)\cdot\frac{1}{\text{C}}+\text{I}'\left(t\right)\cdot\text{R}_2+\text{I}'\left(t\right)\cdot\text{R}_2=0\tag4$$
Now, when I use the initial condition (to solve the DE given in equation \$(4)\$), I get the current in the circuit (after the switch is opened) equals \$0\$, but that is not possible. What is mistake here?
