1. The problem statement, all variables and given/known data When switch S in the figure below is open, the voltmeter V of the battery reads 3.68 V. When the switch is closed, the voltmeter reading drops to 3.50 V, and the ammeter A reads 1.80 A. Assume that the two meters are ideal, so they don't affect the circuit. Find the emf of the battery. Find the internal resistance of the battery. Find the circuit resistance R. 2. Relevant equations V=IR Resistors in series: Req=R1+R2+R3... 3. The attempt at a solution I found the emf of the battery to be 3.68 V, because it is the only voltage source being read by the voltmeter when the switch is open. I am not sure how to find the resistance. I have come up with this: Req = r + R Vtotal = I*Req ==> Req = Vtotal/I = 3.5/1.8 = 1.944 Ω 1.944 = r + R I am not sure where to go from here.