For the circuit in the figure:

**Answer 1**

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.