The same 12.0 V battery with 0.100 ohm internal resistance now drives a 0.500 ohm load. What are the current and the terminal voltage?

1. Internal resistance is in series with the load, so the total resistance is 0.500 ohm + 0.100 ohm = 0.600 ohm. 2. Current follows from I = emf / (Rload + r) = 12.0 V / 0.600 ohm = 20.0 A. 3. The drop inside the battery is Ir = (20.0 A)(0.100 ohm) = 2.00 V. 4. Terminal voltage is V = emf minus Ir = 12.0 V minus 2.00 V = 10.0 V. 5. Losing 2.00 V out of 12.0 V marks 0.500 ohm as a heavy load for this battery. 6. The pair 1.19 A and 11.9 V belongs to the 10.0 ohm light load, where the current is far smaller. 7. 24.0 A comes from dividing 12.0 V by the load alone, which leaves the internal resistance out of the loop. 8. Holding the terminal voltage at 12.0 V while 20.0 A flows would need an internal resistance of zero. _Source: OpenStax College Physics (CC BY 4.0), Ch 21 "Circuits and DC Instruments", section 21.2 Electromotive Force: Terminal Voltage_