From the image attached: A. Calculate the current through the 22Ω resistor. B. What is the voltage across ab Here is what I've done: A. ) I labelled the currents and applied the Junction Rule I1 + I2 = I3 -I1*22Ω + 8V + 13*6Ω + 18V - I3*6Ω = 0 -I1*22Ω + 8V + I2*14Ω - 12V = 0 Which I rearrange into: 2.17A - I1*1.83 = I3 and I2 = 0.29A + I1*1.57 Inserting these two equations into I1 + I2 = I3 I1 = (2.17A - I1*1.83) - (0.29A + I1*1.57) I1 = 0.43A I3 = 2.17A - (0.43A)*1.83 I3 = 1.38A I2 = 0.29A + (0.43A)*1.57 I2 = 0.96A Therefore the current through the 22Ω will be 0.43A ? B.) For this question, I really don't know how to tackle this, do I just do V=I3*R + 18v ? For R, do I just add up the resistors (6Ω + 6Ω) in series?