A 12.0 V battery ages until its internal resistance reaches 0.500 ohm, and it still drives a 0.500 ohm load. What power does the load now receive?

1. The internal resistance is now as large as the load, so the total is 0.500 ohm + 0.500 ohm = 1.00 ohm. 2. Current is I = 12.0 V / 1.00 ohm = 12.0 A, down from the 20.0 A a fresh battery supplied. 3. Power in the load is P = I squared times R = (12.0 A)(12.0 A)(0.500 ohm). 4. (12.0 A)(12.0 A) = 144 A squared, and 144 times 0.500 ohm gives 72.0 W. 5. Terminal voltage has fallen to 12.0 V minus (12.0 A)(0.500 ohm) = 6.00 V, half the emf. 6. Checking with P = V squared divided by R gives (6.00 V)(6.00 V) / 0.500 ohm = 72.0 W, the same answer. 7. 200 W was the load power while the internal resistance was still 0.100 ohm, so it is the fresh-battery figure. 8. 144 W comes from putting the full 12.0 V across the load instead of the 6.00 V that actually reaches it. 9. 36.0 W comes from dividing by the total 1.00 ohm rather than by the 0.500 ohm load. _Source: OpenStax College Physics (CC BY 4.0), Ch 21 "Circuits and DC Instruments", section 21.2 Electromotive Force: Terminal Voltage_