So here is the circuit schematic:
Known values: \$E_6=200V\$, \$E=35V\$, \$R_2=50\Omega\$, \$R_3=100\Omega\$, \$R_4=140\Omega\$, \$R_5=120\Omega\$, \$R_6=100\Omega\$, \$C=10\mu F\$.
When \$P\$ is opened, capacitor load \$Q=1mC\$. When \$P\$ is closed, we know that the voltage across the capacitor \$C\$ is ten times lower than the voltage in case of opened \$P\$.
The task is to find \$I_{g1}\$ and \$R\$.
What I started with are the voltages across the capacitor.
In the first case (\$P\$ opened): \$U_{C1}=\frac{Q}{C}=\frac{1mC}{10\mu F}=100V\$
In the second case (\$P\$ closed): \$U_{C2}=\frac{U_{C1}}{10}=10V\$
So, \$Q\$ when P is closed is: \$Q=C\cdot U_{C2}=0.1mC\$
Could you give me a hint on what to do next? Just give me an idea and I think I can work it through.
Thank you for your time.
Edit:
Solution:
1) First, I solved the circuit when the switch is opened. In that case, I knew the voltage across the capacitor, which is: \$U_{C_1}=100V\$. Using superposition, I calculated the voltage \$U_{C_1}'\$ with respect only to \$I_{g1}\$ and then calculated \$U_{C_1}''\$ with respect to \$E_6\$.
\$U_{C_1}'=10I_{g1}\$ and \$U_{C_1}''=20V\$. Since I knew that \$U_{C_1}=U_{C_1}'+U_{C_1}''\$ I got \$I_{g1}=8A\$.
2) Now that I knew \$I_{g1}\$, I had to find \$R\$ and since I know the voltage across the capacitor when the switch is closed \$U_{C_2}=\frac{U_{C_1}}{10}=10V\$, the voltage across the resistor is then: \$U_R=U_{C_2}+E=45V\$. Then, I used mesh current analysis to get the current through \$R\$, and in the end I got: \$I=\frac{135}{R+100}\$. Then, using Ohm's law, \$U_R=R\cdot I\$ I got \$R=50\Omega\$.
