I am a science teacher who usually works with match classes or computer courses. Next semester I will be teaching a high school physics course and while I knew how to do this "once upon a time" when I took physics, I no longer remember the details. I remember most of it, but I'm missing some important detail.

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Taking this problem as an example I cretae three equations which are based on the junction rule at A and the loop rule at B and C.

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After solving this system of equations I arrive at the solution I1=1A, I2=3/8A, and I3=5/8A. So now I can work out the voltage drop across each element.

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Okay, here is the problem. I expected (perhaps incorrectly) that the total voltage drops across the resistors would equal the sum of the voltage sources. I intened that this be a check on the answers to ensure they were wrong. In this case (an many others) I am getting a mismatch. 5v+15v+25v=45v which is not equal to 20v+10v=30v. Sometimes this works, but most of the time I don't get equal answers.

It has been so long since I did this, I'm hoping someone will know what I have forgotten. Is there a pattern of behavior for the voltages which I should expect to see? Is there some test which the students can do to check their work?

  • $\begingroup$ Anedar's answer deals with the conceptual stuff. I'll throw in a couple of trivial practical points that save time when you solve these problems.… Call your currents $x,\ y$ … rather than $I_1,\ I_2$ … (2) Work with the minimum number of currents, so in your example put ($x-y$) straight into your voltage equations instead of $I_3$. You'll see that I've just used Kirchhoff's current equation at the 'red dot' junction. $\endgroup$ – Philip Wood Dec 6 '17 at 17:53
  • $\begingroup$ @PhilipWood Right. I actually teach this with a refresher in solving linear equations of X, Y, and Z variables. That is one option, but I wanted to stay with I so they are clear on this being CURRENT. I do show them how to use two currents (one per loop) to solve these. My point there being that you get one variable (current) per equation (loop) so you are guaranteed that the system is solvable. And then I go through multiple solving techniques (matrices, etc.) Thanks for the suggestions. $\endgroup$ – BSD Dec 6 '17 at 17:57

In short: No, not all voltages sum to zero, unless you have a closed path

Take an easy counterexample and put a number $n$ resistors in parallel to a voltage source. The voltage drop over all resistors is just the voltage of the source. If you sum them up, you get something like $(n+1)V_0$. Only voltages in a closed circle sum to 0, which are your equations B and C. As check you might use e.g. $ 40 I_3 - 40 I_2 = 10$, which is just the outer circle.

When thinking about voltages i like to remind myself that theres an electrical potential behind. Every point in the circuit has some potential, like every point on a landscape has a height. Voltages are nothing than differences in potential (or differences in height). If you walk around the circuit you walk sometimes up, sometimes down, but wherever you walk, once you get to the point where you started you are at the same height again. And adding all voltages you passed (watch the sign!) sums to 0. If you end up at a different point however, it can sum to anything.

So as tests qualify all paths that return to the starting point. E.g. also walking from A via 10V soucre, lower resistor, 20V source, middle resistor, upper resistor, lower resistor, 10V source, middle resistor, 20V source, lower resistor and 10V source back to A. Totally overcomplicated, but it adds to 0 if you got the signs right. The easier one is the outer circle which i used in the first paragraph.

The second possible test is like your equation A, that on each point the sum of all currents has to be 0 (again, watch the sign). You have a second point at the right side, even though thats the same equation just with opposite signs.

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