I think you will find these much easier than first appearance. For number 3, just be careful with your signs. I have included the solutions for the currents below for a random sample problem.

Notes on problem 3:

Many of you have asked about the negative current. Here you go.

So, about the currents…as I explained it is something of a guessing game. I think that I misconstrued a question in class today about negative currents. Is it OK to have a negative current? Yes! As long as your algebra proves the validity of your solution. A negative indicates that you chose the wrong direction for one or two branches of your loop rules. If your algebra does NOT work out, then you have to go back and redo your direction in your loops.

So, to my example problem. First, there are a huge range of values available on that problem so yours may be VERY different. In fact, I just recycled the problem twice on my teacher view (I can do that with my version of the assignment) and the values of battery 1 and 2 flip flopped…something that would change the way I did my loops. Before I get into details here is something to remember: You have two parallel branches with batteries and one with a resistor. If it helps you model it against my work, you can rotate your circuit by 180 degrees out of the plane of the paper until your larger voltage is on the left like mine was. As long as you keep the right resistors attached to the right batteries in the right order you have the same circuit!

1) I used the inside loops rather than the outside since that would give the the current through the central 4 ohm resistor in both equations…easier algebra.

2) I choose the direction of loop 1 to be consistent with my 5V battery

3) I choose the direction of loop 2 to be consistent with the larger voltage, in this case the 5V battery. Basically when 5V meets 3V in the opposite direction, you net 2V in the direction of the 5V source. Essentially the bigger battery wins, right? (underlying big picture question: why? More unbalanced charge at the terminals of course). So, you are correct that loops appear to conflict in the center 4 ohm resistor but…

4) Look at equation 2–notice how the signs of the voltage losses and gains compliment the direction established in equation 1. Consistency is the key! The middle term in equation 1 (-I1(4)) indicates a loss through the center 4 ohm resistor. Look at equation 2. The middle term is positive reflecting a “backflow” of current in the opposite direction that I established in equation 1. Again, it doesn’t matter which is correct. Consistency wins. Your math will tell you the direction (positive or negative).

5) Equation 3 represents the node rule for the junction at the top of the circuit, again using the larger battery for the positive direction. Again, consistency.

I trust you can follow the algebra? Let me know if you need assistance with that. It is no bother. I should ask you to check me as I actually did this problem last year and just re-posted the work today, so by all means scrutinize my algebra. It is very easy to switch a sign when you are dealing with systems of equations, as you surely appreciate.

Now, back to my original point. Is it OK that I got a negative value for I2? You bet! Why? When I check it against equation 3 it returns a true solution. What does it mean? It means that the direction of current flow I established in I2 (to the right) is wrong. Turns out this current should be flowing to the left. Does it matter? No. The value of the current is correct. Remember also that I2 represents the current in the 5 ohm resistor. Why should it be to the left? Well, the math really answers the question but lets go back to the problem. Set aside the need for consistency of signs in your algebra and just study the figure. We have two batteries working “against each other” in the outside loop, but complicated by the fact that both are connected by the central branch with the 4 ohm resistor. Both batteries are similar in voltage (one is not way bigger than the other), so imagine a positive charge (conventional current) in the wires. If you are by the first 4 ohm resistor you dont want to go toward that battery. If you are by the 5 ohm resistor you don’t want to go towards that battery. Imagine both batteries pushing out positive charge…where does it go? Towards the middle and down the central branch, getting sucked away to the negative terminals on both sides. Now does the negative answer for I2 make sense?

So why didn’t I do that in the first place? Consistency. I made a guess, that turned out to be wrong, but because I was consistent in establishing direction on all of my work, it still returned an algebraically true solution that also makes good physical, conceptual sense for the circuit in the problem.

Below please find the key to the back side of the recent quiz (that I handed back today). I did not grade this problem since you were pressed for time. Please be sure to review prior to your exam.