For the circuit in the figure:
You’ve got a 12 V battery driving a 5 Ohm load, and the battery has an effective internal resistance of 1 Ohm.
You can see that the battery resistance and the load are in series (more or less by definition*), so you can get the total power dissipation in a couple of ways. The first is to first get the current in the circuit using Ohms Law in the form I = V/R.
The equivalent series resistance is Rint + Rload = 6 Ohms
The rate of energy conversion is the power, which is P = IV.
12V/(5+1 Ohms) = 2 Amps
Power = 12 V * 2 Amps = 24 Watts.
This power is all coming from stored chemical energy.
You could have gotten here in one step using P = V^2/R = 144/6 = 24 Watts.
To get the dissipation in the battery, use the other form for Joule power, P = R* I^2, using just the battery internal resistance: 1 Ohm * (2 amps)^2 = 4 Watts.
Use the same equation for the dissipation in the external load: 5 Ohms * (2 amps)^2 = 20 Watts
* If the battery has a slow self-discharge, you can have a resistance in parallel to the external load as well.
Source(s): Any Physics I book.