- #1

rbrayana123

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## Homework Statement

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?

## Homework Equations

V = IR

Superposition Theorem

## 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.

R

_{ab,y}: 10 + 20 = 30 (Series Circuits)

1/R

_{ab,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?...?

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