1. The problem statement, all variables and given/known data A battery of EMF 12.6 V and internal resistance 0.1 Ω is being charged from a DC source of EMF 24.0 V and internal resistance 1.0 Ω using the circuit shown in the figure below. V1 and V2 are high resistance voltmeters and R is a fixed resistor. (a) What is the polarity of terminal A of the source? (b) I the charging current is 5.0 A, determine the resistance of the resistor R. (c) If the resistance of R were changed to 0.9 Ω what would be the reading on each voltmeter? Answers: (b) 1.18 Ω, (c) 18.3 V on V1 and 13.2 V on V2. 2. The attempt at a solution Every part I am in doubt whether it is correct or not. (a) Since the battery has a polarity of + - Battery + - (of we look from the top to the bottom), I would say that the polarity of the terminal A would be negative (-). Since the current flows from the positive sign of the battery to the negative sign A of the DC source. (b) This is only a guess: 24 - 12.6 = 5 (0.1 + 1 + R) → R = 1.18 Ω. Maybe because the voltmeters have large resistance they don't disturb the circuit and so we just calculate the way I did (like it is a simple circuit only with two batteries). (c) No idea.