1. The problem statement, all variables and given/known data A black box with three terminals, a, b and c, contains nothing but three resistors and connecting wire. Show that no external measurement can distinguish between these two possible set-ups, known as Delta and Y: http://i45.tinypic.com/mcd9gp.png Is there any other possibility? 2. Relevant equations V = IR Superposition Theorem 3. The attempt at a solution First, I connected a wire with a voltage source between two terminals and showed used parallel/series to show they're equivalent. Rab,y: 10 + 20 = 30 (Series Circuits) 1/Rab,Δ: 1/(34) + 1/(85 + 170) = 1/30 (Parallel Circuits; One Parallel has Two in Series) This applies to the rest of the circuits. The only measurable quantities outside are total resistance, total current and total voltage of entire circuit; all are equivalent. Next, I turned my attention to proving that these are the only possibilities. For Y, algebra provides a unique solution, For Δ, I'm still in the process but I'm confident I can prove whether it's unique or not through some heavy algebra.... Once I'm done with that, I'll get started on a general case. My actual questions: 1) I've tried re-arranging three terminals and three resistors in a way that combines series and parallel but the only forms appear to be Δ & Y. Is this absolutely true? I feel like there's a limit on geometry that should convince me on this but I'm overlooking it. In Purcell, I often run into a lot of problem based on geometric test cases and sound reasoning. I really like these problems but is there a formal study of mathematics that may help me out a tad bit? 2) There's one last test case left: all three terminals are connected to one another. The attempt is to prove the external currents on all three wires are equivalent in both the Δ & Y cases. However, I'm running into trouble applying the superposition theorem. Here's my set-up but I'm unsure because I'm always doubting how my current splits up in complex circuits (any tips) http://i45.tinypic.com/2ujkyeb.png V = I2R2 + I6R6 V = I4R4 + I6R6 I1 = I2 + I3 I3 = I4 + I5 I6 = I2 + I4 I1 = I5 + I6 However, the sixth equation follows directly from equations 3 - 5 so it seems like I don't have enough information to solve for I1, I3 & I5. (5 equations, 6 unknowns). Should I include two additional currents going from R2 to R4 and one from R4 to R2?...?