1. The problem statement, all variables and given/known data The battery in the figure below has negligible internal resistance. Find (a) the current in each resistor: (b) the power delivered by the battery: 2. Relevant equations V = IR Resistor Connected in series Req = R1 + R2 + R3 Resistor Connected in Parallel 1/Req = 1/R1 + 1/R1 + 1/R3 3. The attempt at a solution I think that the resistors 3omega and vertical-2omega are connected in series, and the other 2 are connected in parallel. So the ones in Parallel.. 1/Req1 = 1/2 + 1/4 1/Req1 = 3/4 Req1 = 1.333 The ones in Series Req2 = 3 + 2 = 5 so... Req = Req1 +Req2 =1.333 + 5 = 6.333 But to find the Current, I you just use Ohm's law I = V/r so the 3.0-omega resistor would be I = 6/3 = 2 ...but thats wrong and Im at a loss to know why?? thanks for the help!