ok, so the problem states: the current in a loop circuit that has a resistance of R1 is 3A. The current is reduced to 1.6A when an additional resistor R2=2ohms is added in series with R1. What is the value of R1? I drew two diagrams. My logic was that the voltage would be the same, b/c the voltage is supplied by the battery which does not change. Thus you have V=I1R1 for the first scenario; and you have V=I2(R1+R2). Since V is the same, you can set the two equations equal to each other, and solve for R1. Is the logic right so far? If so, maybe I just have an algebratic error... (3A)R1=(1.6A)(R1+2ohms) (3/1.6)=(R+2)/R 1.875=1+R/2 .875=R/2 R=1.75 ohms???