A battery with an emf of 12.00 V has an internal resistance r. When connected to a resistor R, the terminal voltage is 11.70 V and the current is 0.11 A. I have solved that the external resistor R is 106 ohm. I have solved that the internal resistance r of the battery is 2.73 ohm. What is the energy dissipated in the battery's internal resistance in 3.5 minutes? When a second identical battery is added in series and the external resistor is R = 22 Ohms what is the resulting current?