Today was the day of my second physics test. I was rather apprehensive about the whole thing considering how I was crushed by the first test.

Things started to go wrong on Monday, when I accidentally dropped my backpack in such a way that it smashed the screen on my HP-48G. I didn’t love the HP-48G, but I liked it well enough. As a geek who went to high school in Texas, I learned to use an HP-32SII as part of the UIL Math/Science team, and suffered brain damage as a result. I can no longer use an algebraic (i.e., non-RPN calculator. Since I don’t know anyone who could let me borrow one, this meant I need to find a calculator, soon.

HP-32SII, despite being something like $30 new, are now going for nearly $200 on eBay for a nice. They were discontinued back in 2002. It might be the greatest calculator ever produced, but I had no way of knowing that at the time. I also didn’t know that when HP stopped producing it, they’d stop producing decent calculators in general. For years, there was nothing but the abomination that is the HP-33s available.

I never had my own 32SII, because I was always assigned one that belonged to whichever high school I was attending. In college, I borrowed one for physics exams, and never got around to getting my own. When I finally went to buy one, they were going for $100 on eBay. I should have bought one then. Instead, I opted for the 48G, which is basically the graphing version of the 32SII. This makes it way bigger than it needed to be, but it had the same basic keyboard feel and layout of the 32SII, so I was content. I never used the graphing functionality, as I don’t believe in graphing calculators, but I liked the big screen because it let you see several lines of the stack at once, which is probably the only thing I liked better about it than the 32SII. Unfortunately, the large screen also cracks pretty easily when it’s in a backpack that gets dropped.

I was going to replace it with another 48G from eBay, but even those seem to be going close to $100 for a nice one these days, and that wouldn’t get here in time for the exam.

But then I remembered someone mentioned that HP started producing what might be a decent RPN calculator a few years ago, the HP 35S. The Amazon reviews included a lot of former 32SII users forced to “upgrade”, and were generally favorable. Furthermore, with Amazon Prime, I could have it shipped next-day for only $3.99, and it would get here in time for the exam.

It did, and it’s an okay calculator. It has a seriously annoying display bug wherein it will display something like 1.60245856345×10^-19 as “1.60245856345E-” and then you have to hit the right arrow key to see the rest of it. Which is phenomenally dumb because no one’s going to care about the last few significant digits, but everyone’s going to care about the magnitude! It would have been trivial to make it so that the “E-19” is always displayed, and you could scroll to the right for more digits. Things like that make me think HP may never make another good calculator.

The physics test was relatively easy this time around, except for one problem. I’ve reproduced it here. (Note my mad circuit drawing skills, which I needlessly acquired just for this post.) The only text with the problem was the following:

Find the current through R₂.

No problem. I couldn’t believe the luck that this was one of the two big-points problems. I just needed to find the equivalent resistance between R₂ and the and the 20Ω resistor in parallel with it. Then find the current through the remaining 3 resistors in series, which would give me the voltage at either side of the parallel resistors, and then the current through R₂. That’s when I realized he didn’t tell us what the value of R₂ was!

I stared at it. No wonder it seemed too easy. But how to solve? Maybe he didn’t want a number, but rather an expression for the resistance? It’d be an ugly expression, but I set about writing it. Then I thought maybe there was something fancy I could do with Kirchoff’s Laws, but what? I started writing down loop and junction equations to see if there were was some way to figure it out. It became obvious that was just going to make the problem worse, not better. Why couldn’t I do this? What trick, equation or law was I forgetting that would let me figure this out for an unknown resistor? I did every other problem on the test, then came back to it, and was still stumped.

There were about 15 minutes left on the test, and only a couple people had turned in their tests. I’m pretty slow and methodical, so I was surprised almost the entire class was still in the room. I figured they were probably all struggling with the same problem, so I didn’t feel too bad about it, and settled on writing down everything about the circuit that might be useful in hopes of getting some partial credit. With about 15 minutes remaining, someone went up and asked the professor something.

He walked to the white board and said, “Uhm. Sorry. Sorry. On Problem 1, R₂ is equal to… err. What number would you like? How about 10 ohms? I’m really sorry about that.” There was an immediate groan, and a lot of laughter as we looked around the room at each other, realizing that we were all in the depths of the same completely pointless hell. If only any of us had asked earlier! There was amazingly furious erasing for the next 30 seconds, as I suspect all of us had covered the entire page in nonsense hoping to crack something. Of course, with the missing resistance provided, the problem became nearly trivial. Definitely one of the most amusing test-taking moments I can remember.

There was a fun problem I hope I got right that went something like this: “The winch on my jeep draws 50 amps from a 12V battery. How much power does it draw? If it weighs 2000kg, how long would it take the winch to winch the jeep up to the top of the library tower (30m)?” Cute. The first part is easy. The second part involved dredging up things from Physics-I I’d forgotten a decade ago. I managed to come up with U=mgh, so I knew how much energy (Joules) would need to be expended by the winch, but how long would it take? Where was time in this problem? Then I remembered: current is dq/dt, with units of Coulombs/second and it all fell together. Not a difficult problem, but cute. There might have been a more elegant way to solve it, but I’m pretty sure I got it right.