1. The problem statement, all variables and given/known data http://imageshack.us/a/img685/602/homeworkprobsg29.jpg [Broken] The bridge circuit provides a good illustration of the use of Thevenin's theorem. Find the current through the bridge resistor (50Ω). 2. Relevant equations V = IR source transformations, current division, voltage division, KCL, KVL 3. The attempt at a solution Some trouble understanding this whole procedure, but I think you cut out the 100V so it becomes short circuited and then cut out the bridge resistor (50Ω) out of the picture and then the wire with the bridge resistor becomes open-circuited? (With the bridge resistor acting as a "load resistor"?) Then the rest of the bridge resistors, after adding in series, become 1000Ω and 3000Ω in parallel. Which is then R = 750Ω equivalent. I think it is also supposed to be the Thevenin resistance too So RTh = 750Ω then I think you have 750Ω and the original 50Ω resistor back in series, I'm stuck with what to substitute back as the original source (unless I already did something else wrong too). I think I'm supposed to use current division / voltage division later on to find the current through the 50Ω.