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So I'm starting with the Nilsson/Riedel electronics text and got confused early on. They show a simple loop circuit (figure 2.15 for those who have it) with 1 ideal voltage source and 3 resistors and, across one of the resistors, the current flows opposite to the rest of the circuit.

enter image description here

I understand that, in the Passive Sign Convention, positive current flows from + to - across a load; what does this mean physically, though? It was my understanding that electric current only moved in one direction; I've apparently forgotten something from my undergrad electronics class.

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  • \$\begingroup\$ A hyperlink to the document you specify is really needed. Also note that current flows through a load and not across it. \$\endgroup\$ – Andy aka Oct 2 '15 at 23:24
  • \$\begingroup\$ I actually lucked out, it is available online: google.com/url?sa=t&source=web&rct=j&url=https://… \$\endgroup\$ – PoGaMi Oct 2 '15 at 23:26
  • \$\begingroup\$ It's 32 pages. Copy the page you refer to and paste it into your question. \$\endgroup\$ – Andy aka Oct 2 '15 at 23:27
  • \$\begingroup\$ I'm not quite sure how to paste images in the phone app, I'll have to try it from my laptop later tonight. \$\endgroup\$ – PoGaMi Oct 2 '15 at 23:29
  • \$\begingroup\$ Wikipedia has a pretty decent article on the passive sign convention. It doesn't seem that the circuit you put up here is much way relevant to your generic question. I could even say it's a red herring. \$\endgroup\$ – Fizz Oct 3 '15 at 4:08
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I can choose the reference polarity of my resistor however I want, for example:

schematic

simulate this circuit – Schematic created using CircuitLab

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

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

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  • \$\begingroup\$ Ahh ok, think I get it. So in their convention, the + and - nodes are just dependent on how you choose to test the system. They (at least prior to a reading) have no bearing on which direction the current is actually traveling. \$\endgroup\$ – PoGaMi Oct 3 '15 at 19:07

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