This is an activity that was created by a harness and cable engineer in the aerospace industry. He is in charge of designing all interconnects between all the various systems / components on a satellite. This is an actual design problem that he made during the course of his job. The only thing he changed was the length of the wire from the ICB300 to the LAE ,because he wanted the 24AWG wire to result in a voltage drop greater than one, in order to test whether the students would see that and move to a larger wire.
The solutions are on the second page of the pdf. These solutions were authored by the engineer who wrote the problem.
I have a simplified version of this problem in my Satellite Design Teams activity.
I would make sure that the students know this is an actual design problem for a satellite. It is not a simplified representation of a problem someone might do, rather it is a real problem that someone must do, in order for the satellite to achieve mission success.
Looking at the equation again: V = L * R * A. V is the voltage drop, which is what they are solving for. L is the lenght of the wire, which is given in the problem. R is resistance, which they get from the table (they first choose a wire size, then look at the table for its resistance). A is amps, which is given to be 0.8 in the first bullet point of the problem.
So basically, A is constant, and the students are inputing some value of L and R, in order to find V. Then they are adding up all their V’s, and seeing if the sum is less than 1.
The design tradeoff between weight / voltage drop is key here. Large wires have very low voltage drop, and since we cannot have a voltage drop greater than 1V, we are tempted to just use very large wires. But large wires are also heavy, and satellites need to be as light as possible. Thus the tradeoff between weight and voltage drop. We must select the smallest possible wire, that still has a voltage drop less than 1V.