So...in the case of the DC battery showing 12.8 volts, can I assume that the battery is capable of providing up to 31.5 amps? I=E/R = 12.8/.4 = 31.5 amps.
Correct -- Although as you need a dead short across the terminals to achieve that much current, you couldn't put it to any productive use. There's also the point that such a small battery wouldn't supply this much power (12.8V x 31.5A = 403.2W) for more than a very brief period. The battery would be fully exhausted very quickly and the EMF (and therefore the current also) would quickly taper off.
Do I work from the full, unladen voltage to calculate amps and ohms, or do I have to let the amp load bring the voltage down before I'll have good numbers to work with?
Ohm's Law applies equally to the entire circuit, or to any portion of the circuit. You just have to make sure you match up the values correctly.
Let's take the last part of your post and apply it to the example in hand:
It would be easy to make the mistake of working off the full voltage divided by the amps drawn, which would be 11.8 volts/2 amps = 5.9 ohms.
What you've done there is to calculate the resistance of the
load. We know that 2A is flowing through the whole circuit, and 11.8V is the voltage which appears across the load. That's applying Ohm's Law to the load portion of the circuit.
Let's extend it to the whole circuit. We know from the previous measurement that the actual EMF of the battery is 12.6V. Now, applying Ohm's Law to the entire circuit we get:
R = E / I = 12.6 / 2 = 6.3 ohms.
That's the resistance of the complete circuit,
including the internal resistance of the battery.
We've already calculated the internal resistance of the battery as 0.4 ohm. Add that to the 5.9 ohms of the load and you get the same result for total circuit resistance of 6.3 ohms. It checks out.
So it looks like you're saying that we're not working off the full EMF when we calculate the resistance through a battery, but instead, we work off the voltage drop and the full amp reading.
That's right. It's appying Ohm's Law to the appropriate part of the circuit:
1. To calculate the resistance of the entire circuit (battery+load), you need the full EMF which is powering that circuit - 12.6V.
2. To calculate the resistance of the load, you need the voltage which appears across the load, i.e. 11.8V.
3. To calculate the internal resistance of the battery, you use the voltage which is dropped internally by the battery's resistance, i.e. 0.8V.
