1. The problem statement, all variables and given/known data http://imgur.com/6Cds4YF [Broken] What will the ammeter read when the switch is closed? 2. Relevant equations V = IR 3. The attempt at a solution I already calculated the emf of the battery one the left to be about 27 V. It seems that the answer to the question is as simple as using V = IR where V is from the battery. So I = V/R = 25/50 = 0.5 A. I don't understand why the other battery to the left and the other resistors don't affect the current through the ammeter though. Wouldn't using the numbers for calculating the current, assume that the only item in the circuit is a 25 V battery and one 50 ohm resistor?