AWOL:
What makes the voltage fall depending on whether I'm using 270s or 10ks?
The lower the resistance, the higher the current.
The higher the current, the faster the battery drains.
I wasn't clear, sorry...
Unloaded battery voltage: 5.8V
Loaded battery voltage measured with 10k resistors: 5.6V
Loaded battery voltage measured with 270 resistors: 3.5V
10k current, measured: 0.28mA, and calculated: 0.28mA
270 current, measured: 6.48mA, and calculated: 6.48mA
I understand that a higher current will run the battery down faster, meaning that it will eventually deliver 0V at any load, and we'll get there sooner with Rt=540 than with Rt=20k. But what I meant by my question was, why does it deliver lower voltage with a higher load right now?
I feel like I'm missing some important terms to express myself, I hope you can decode what I'm trying to say.