Reversed Current in Passive Loads?
I can choose the reference polarity of my resistor however I want, for example:







simulate this circuit &ndash; Schematic created using CircuitLab

Now if I find \$V_{R1}&gt;0\$ that means that the left side is at higher potential than the right side. If I find \$V_{R1}&lt;0\$ it means the right side is at higher potential than the left side.

Similarly if I find \$I_{R1}&gt;0\$ it means current is flowing in from the left (because of the passive sign convention), and if I find \$I_{R1}&lt;0\$ it means current is flowing in from the right.

If you initially chose to put the "+" node on the right and the "-" node on the left, then when you solved your circuit you'd just find that \$V_{R1}\$ and \$I_{R1}\$ with the same magnitude and opposite sign from if you chose to assign the nodes the way I did.

Assuming you're talking about Figure 2.15 in the file you linked, you're going to find that \$i_1 = -i_2\$, for example.