Electrical – Voltage Difference in Superposition Theorem

Ideal circuits with two voltage sources in parallel lead to contradiction, unless they are equal and can be simply replaced with a single one. Note that potentials \$ \varphi_1 \$ and \$ \varphi_2 \$ in your circuit must be equal, since there is no impedance of any kind between them, nor do ideal voltage sources have any internal resistance:

In your case, luckily, these sources produce the same voltage, so the simplest thing to do is to simply remove one of them from the circuit. If you had two ideal sources of a different voltage in parallel, that would lead to contradiction.

In a real circuit, connecting two sources in parallel would lead to a circuit with a very small, but still non-zero resistance between them, which would result in one of the sources (the one with a slightly lower voltage) actually sinking current, but the current through the 5 ohm resistor would only depend on the voltage of the right source.

If you want to put some actual numbers, you can try something like this:

Note that if the sources are again ideal and have completely equal voltages, there will still be no current flowing through the tiny resistance between them, but you should be able to apply the superposition principle.

For a circuit like this, the mesh current method should provide the simplest solution and show that the current through the resistor only depends on the right source.

I'm not sure what is meant by "passive configuration", but the current supplied by the voltage source is opposite the direction of \$i_s\$. This is because

In a single source circuit, the direction of positive current flows out of positive terminal of a voltage source. However, in a multiple source circuit, this condition may not be valid.

You are using superposition, so for the analysis of \$i_1\$ you have a single source circuit. Since the current from the voltage source is in the opposite direction of \$i_s\$, \$i_s\$ is negative.

As for the case where the voltage source is turned off, the current through the (upper) \$5\Omega\$ resistor is zero because it is short-circuited by the voltage source (\$v_1\$ and \$v_2\$ are the same node). You can view these two paths as a current divider with the short-circuited path as a \$0\Omega\$ resistor -- if the current entering the two paths is \$i\$, then the current through the short-circuited path is

$$\frac{5\Omega}{5\Omega + 0\Omega}i = i$$

Alternatively, the current through the \$5\Omega\$ path is

$$\frac{0\Omega}{5\Omega + 0\Omega}i = 0$$

In other words, all the current flows through the short circuited path and none of it through the \$5\Omega\$ resistor. This means you can ignore the upper \$5\Omega\$ resistor as it has no effect on the circuit (no current passes through it, nor is there any voltage across it). Without this \$5\Omega\$ resistor, you should see that there are only two resistors and the desired current is easily solved with a current divider as you have stated.