simulate this circuit – Schematic created using CircuitLab

Let's say we know the values of \$Z_1\$, \$Z_2\$, \$Z_3\$. We know the value of \$I_3(0)\$ - the current in the wire with \$Z_3\$ and \$E_3\$ when the switch \$S\$ is in the position 0. We also know \$I_3(1)\$ - the same current when the switch is in position 1.

The task is: find \$I_3(2)\$ - the current \$I_3\$ when the switch is in position 2. I have tried using the compensation theorem, replacing the wire with \$Z_2\$ with an ideal current generator, but that didn't work out.


To find \$I_3(2)\$ you need to know the voltage at the top node (call it \$V_t\$). Then you know by Ohm's Law that $$I_3(2) = (E_3 - V_t)/Z_3$$

To find \$V_t\$ you can use superposition (assuming \$E_3\$, \$E_1\$, and \$I_g\$ are independent sources). This requires turning off all sources except one and finding the remaining source's contribution to \$V_t\$, then repeating until you have found the contribution of all three sources. \$V_t\$ due to all three sources is then by superposition the sum of the three contributions.

To turn off a voltage source make it a short circuit, and to turn off a current source make it an open circuit.

For example, to find the contribution of \$E_3\$, \$E_1\$ becomes a short and \$I_g\$ becomes an open circuit. That means \$Z_1\$ and \$Z_2\$ are in parallel, and this parallel resistance is in series with \$Z_3\$. \$V_t\$ is then the output of a voltage divider formed by \$Z_3\$ and \$Z_1 \parallel Z_2\$.

\$E_1\$'s contribution is very similar, except \$Z_2\$ and \$Z_3\$ are in parallel and this parallel resistance is in series with \$Z_1\$.

For \$I_g\$'s contribution, with \$E_1\$ and \$E_3\$ off all three impedances are in parallel so \$V_t = I_g \times (Z_1 \parallel Z_2 \parallel Z_3)\$.


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