2 series resistors are in parallel with another resistor.. 1. The problem statement, all variables and given/known data Resistors R1, R2, R3 have resistances of 15ohms, 9, 8.. R1 and R2 are in series and the combination is in parallel with R3 to form the load across a 12 volt battery. a)diagram b)total current c)Current across R3 branch? d)voltage/potential drop across R2? 2. Relevant equations !!!!!!Series and Parallel and Combo circuit equations.. 3. The attempt at a solution 1. The problem statement, all variables and given/known data I drew it.. 1st thing I did was figure total resistance.. to do that, I combined R1 and R2 serially by adding..15+9 is 24.. Then, now that I have one and one parallel (R[1,2] and R3), I do 1/Rt=1/R1+1/R2... And got R(t) to be 6 ohms.. With R(t) calculated, and the given V(t), I can figure B, total current I(t)=V(t)/R(t)... ding.. 2 A or 2 amps.. ------ Now, I'm kind of lost for the rest of the problem.. What is the current in R3 branch? Erm..To find individual current of a parallel resistor, I need to know the voltage and the resistance.. I would think that the voltage of R3 be 12V.. but my teacher said it was 6V..and I had to compact the whole circuit, then repackage it to get 12V..and I'm lost -- Well, assuming that for some reason, V3 is 6V.. I did I3=V3/R3... 6V/8 ohms=0.75A...that would be C Then.. for D..I did I(t)=I(3)+I(1,2).. I(1,2)=1.25A.. So.. V2 would be I2R2.. If I(1,2)=1.25A, then I1 and I2 also equal 1.25A? Then.. 1.25A*9 ohms.. 11.25V equals V2? --- meh.. ok.. Can anyone confirm whether 2A, total current, and I3 of 0.75 A is right? And I'm pretty sure the rest is wrong.. PS, any freeware circuit drawers?