Again, I refer the reader to the HO:ME blog for further discussion of same. They also refer to some "math" that we will get to in a bit, although probably not at this blog post.
This first is just to document the voltage from the single AA battery.
I then wired the circuit below, and found that as I was wiring the circuit I made a mistake in the way I wired it up. I had everything connected, below, except for the two dotted lines. I then had two choices, to connect via path 1 below, or path 2. I honestly wasn't sure which, so I just did both and measured the resistance:
This is path 1, current of 162 milliAmps:
This is path 2, current of 320 milliAmps:
So right at twice the flow. Which I thought was wrong. What happened to the other resistor? I had two resistors in the circuit, but it seemed that when I ran the path 1 circuit, one of the resistors was effectively cut out.
For comparison, the current just running through a single resistor is 162, same as the path 1 wiring.
So I drew this out with the path 1 section of the circuit tipped over a bit, just to help me visualize, and came up with this funny-looking triangle circuit. It occurred to me that this triangle loop was a bit deceptive, and what I really have is two resistors in parallel. One is the actual resistor on the up-and-back part of the triangle, and the other is just a wire.
The resistor itself is 10K ohms. The wire, 0.4 ohms. This, in my mind, indicated that there probably is current flowing through the resistor, but it's, oh, 5 or so orders of magnitude lower than that flowing through the wire. No wonder it looked like I had simply cut that resistor out of the circuit.
This, then, is the diagram for path 2, which is the proper way to hook up the resistors in parallel.
A very simple circuit, which still provided ample opportunity for learning (that is, confusion that required resolution via reading and further experimentation).