Maybe I shouldn't have started this blog now, not with everything that's been going on.

It’s Monday again and I’m talking about Friday for the third week in a row, but including some of Thursday.

We answered an example of how to solve circuits that are shaped like houses and such, with diagonals, and how to “rewire” them to look like regular series and parallel circuits. In retrospect maybe I’ll give them more examples of those later.

I also answered the two Kirchhoff’s rules applications I gave them last time. Or at least I tried to, for the second one, the one that I described before as being the squares in an L shape.

The first one, being a standard three branch loop with three EMF’s, they were also able to answer. The second one though, had three equations just for the junction rule alone, and had assigned currents that all went in one node, which is already contrary to what I had taught them before. I probably should have made it clear that it was already understood that at least one of those currents would have a negative value in computation. Two of the equations involved three currents each, while one, at the bend of the L, would cater to four currents.

Then there were six loop equations added. Two used only two currents. The rest used three currents each. So it was difficult to find an equation that would yield only one unknown when substituted. What’s worse was that when we tried to reduce the number of unknowns (particularly involving any two of the six loop equations) then we would just get an expression for another loop equation.

In the end we had to stop trying because it was already time, and I just solved it afterwards. I used only three of the loop equations: the ones that I knew would not give me the other equations. And I made sure that I was switching between using the loop and the node equations for variety.

I was able to come up with the answer, although my main test, finding out which of the three currents going into one junction was negative, failed. They all had the same sign.

But I looked for the students anyway and told them to photocopy the solution, which I would not discuss in class anymore, but they had to study on their own.

It was after that when I discovered where I had gone wrong with the computation, forgetting to copy a negative sign in one simplification. I asked Melissa, the president of the Young Educators’ Society (and a Business Administration major, not Engineering), to check my solution, and she found the same error, but went further: solving the values of the five currents. This time the checking procedure I used earlier gave me a negative current out of the three going into one node, and they all “added” up.

I’m planning not to tell the students about the error, and let them discover it for themselves. Then I’ll see who has studied my solution. After all, the process is still correct even if the numeric values are wrong.

There’s the bell. Next time I’ll talk about returning the last quiz to one of my two Trigonometry classes last Friday, and making them compute the minimum percentage in the last exam and in the Finals to get a one-point-zero in the course. Class dismissed.