Two Batteries In Series--Total EMF and Internal Resistance 1. The problem statement, all variables and given/known data Two 1.50-V batteries—with their positive terminals in the same direction—are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.260 Ω, the other an internal resistance of 0.180 Ω. When the switch is closed, a current of 600 mA occurs in the lamp. What is the bulb's resistance? What fraction of the chemical energy transformed appears as internal energy in the batteries? 2. Relevant equations 3. The attempt at a solution I have already developed the proper solution to this problem; however, I did so on a presumption (or intuition--that sounds a lot better), and would like help clarifying a concept. I understand that the emf is the actual voltage applied by the battery, or the intended voltage, but is there a reason why I can sum the voltage of the two batteries, to get that emf? Likewise, is there a reason I can just sum the resistance of each battery, which constitute the internal resistance?