Have completed a half a dozen attempts at this problem and the power delivery/absorption always comes out uneven and I can't figure out where at along the way I'm messing up. The teacher taught us the mesh method in class (which wasn't included in the text book) with one example, but from what I'm reading the problem I came up in pspice is a supermesh problem and I'm still not understanding where I'm going wrong or what the process is exactly for a supermesh. I'm using clockwise circles to calculate all of the equations. Top left is Ia, top middle is Ib, top right is Ic, elongated bottom touching all the other meshes is Id, and the bottom middle is Ie. V1 = 5V V2 = 10V V3 = 15V V4 = 20V V5 = 25V R1 = 5Ω R2 = 10Ω R3 = 15Ω R4 = 20Ω R5 = 25Ω R6 = 30Ω R7 = 35Ω R8 = 40Ω The equations I've come up with match the values for all the resistors and power sources but I believe I must have a mistake with a negative sign somewhere when computing all the current values for the voltages and resistors to calculate the power delivered/absorbed by them. EQ1: -5 = 25Ia - 20Id EQ2:35 = 40Ib - 30Ic EQ3:-20 = -30Ib + 80Ic - 35Id EQ4:-10 = -20Ia -35Ic +80Id EQ5:-15 = 40Ie Using matrices I get the values (values aren't precise but close enough, I computed the fractions out by hand as well but these should suffice): Ia ~ -0.3921 Ib ~ 0.8476 Ic ~ -0.0366 Id ~ -0.2406 Ie ~ -0.3766 I assume when calculating the power delivered/absorbed I'll be taking the absolute value of whatever difference (I believe that's what we did in class) there is for resistors/voltage sources bordering two meshes. Current through items (Amps): V1 = abs(-0.3921) = 0.3921 V2 = abs(0.8476--0.3921) = 1.2397 V3 = abs(-0.2406--0.3766) = 0.136 V4 = abs(-0.0366) = 0.0366 V5 = abs(-0.2406-0.8476) = 1.0882 R1 = abs(-0.3921) = 0.3921 R2 = abs(0.8476) = 0.8476 R3 = abs(-0.0366) = 0.0366 R4 = abs(-0.2406--0.3921) = 0.1515 R5 = abs(-0.2406) = 0.2406 R6 = abs(-0.0366-0.8476) = 0.8842 R7 = abs(-0.2406--0.0366) = 0.204 R8 = abs(-0.3766) = 0.3766 Power delivered by voltage sources = Ʃ(Δv*I) ~ 44.33 Ex for V1: 5V*0.3921A = 1.9605 Watts Power absorbed by resistors = Ʃ(Δv*I) = Ʃ(R*I^2) ~ 40.46? Ex for R1: (5Ω*0.3921A)*(0.3921A) = 0.7687 Watts It's weird because the problem we did in class also had a voltage source bordering two meshes and I did it once and the powers were equivalent. Even using precise fractions my powers are off by roughly 4 watts.