1. The problem statement, all variables and given/known data An ammeter whose internal resistance is 63 ohm reads 5.25 mA when connected in a circuit comtaining a battery and two resistors in series whose values are 750 ohm and 480 ohm. What is the actual current when the ammeter is absent? 2. Relevant equations V = IR Rseries = R1 + R2 3. The attempt at a solution First I found the voltage using the internal resistance of the ammeter and the current: V = IR V = 5.25 x 10-3 (63) = 0.33 V Then I added the two resistances in series to find the total resistance R = 750 + 480 = 1230 ohm Then I tried to apply V = IR with the new resistance to find the current 0.33 V = I (1230 ohm) And get a value of 2.68 x 10-4A, which is not correct (correct answer: 5.52 x 10-3) I'm not very good at these problems, so I have a feeling I'm doing something silly wrong, thanks for the input!