Electrical – Getting negative voltage on resistor

For part (b), the circuit to the left of the switch is disconnected so you can disregard it for any \$t>0\$.

The key insight for (b) is that the circuit is once again in DC steady state (all transients decay as \$t \rightarrow \infty\$).

Thus, the voltage across the inductor is \$0V\$, i.e., it can be replaced with an ideal wire.

Now, this leaves a simple current divider circuit so the inductor current is given by current division.

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.