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Δ & Y and Superposition Theorem?

by rbrayana123
Tags: Δ, superposition, theorem
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rbrayana123
#1
Jan6-13, 02:40 AM
P: 45
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?...?
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rbrayana123
#2
Jan9-13, 11:41 PM
P: 45
I'm still stuck on this question. Any help?
gneill
#3
Jan10-13, 10:00 AM
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P: 11,829
If you drive each port (ab, ac, cb) in turn with a current supply, leaving the other ports open, you can write expressions for the resulting voltage at each port for each case. By superposition you can then write expressions for the voltages at the ports if they were being driven by three separate current supplies simultaneously. If the three voltage expressions are identical for the two circuits, they must be indistinguishable.

(alternatively you can attach three current supplies at once and solve the circuit for the voltage expressions all at once)

rbrayana123
#4
Jan10-13, 04:01 PM
P: 45
Δ & Y and Superposition Theorem?

I'm familiar with a voltage supply (battery) which I'm using in this case. Why won't that work?
gneill
#5
Jan10-13, 04:31 PM
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Quote Quote by rbrayana123 View Post
I'm familiar with a voltage supply (battery) which I'm using in this case. Why won't that work?
You can use voltage supplies if you wish. Just be sure to take into account that the sum of the potential changes around the outside path must be zero (a constraint equation). Solve for the currents in each supply.
rbrayana123
#6
Jan10-13, 04:43 PM
P: 45
Quote Quote by rbrayana123 View Post

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?...?
I end up not having enough equations in that case. I feel it's because I'm not interpreting the flow of current correctly.
gneill
#7
Jan10-13, 04:46 PM
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P: 11,829
The problem is, you've created a short circuit for the battery around the outer loop. This is not a physically viable circuit and will defy sensible analysis.
rbrayana123
#8
Jan10-13, 04:57 PM
P: 45
Let me know if this is correct:

The outer rectangle loop shouldn't be connected. In actuality, there are two wires coming from each node connecting to each of the others. Of course, between each wire is a battery. But I can consider the other two short-circuited and superimpose to calculate current flow.

This is probably what offset my equations?
gneill
#9
Jan10-13, 05:21 PM
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Quote Quote by rbrayana123 View Post
Let me know if this is correct:

The outer rectangle loop shouldn't be connected. In actuality, there are two wires coming from each node connecting to each of the others. Of course, between each wire is a battery. But I can consider the other two short-circuited and superimpose to calculate current flow.

This is probably what offset my equations?
I'm having trouble picturing what you describe. So I can't answer your question with any confidence.

However, if you mean that you're connecting voltage sources to each port and suppressing them two at a time (by "replacing" them with short circuits), then if the suppression leads to an impossible situation like a short circuit across the remaining source, then I'd say that yes, this is going to muck up your analysis.

If I may suggest, you might find that using current sources and nodal analysis will be more straightforward. You can connect them all at the same time and determine the voltage at each port as a function of the three currents, essentially handling all the superpositions in one go. Remember that to suppress a current source you remove it rather than short it; much easier on this circuit topology.
rbrayana123
#10
Jan10-13, 05:32 PM
P: 45
Quote Quote by gneill View Post
I'm having trouble picturing what you describe. So I can't answer your question with any confidence.

However, if you mean that you're connecting voltage sources to each port and suppressing them two at a time (by "replacing" them with short circuits), then if the suppression leads to an impossible situation like a short circuit across the remaining source, then I'd say that yes, this is going to muck up your analysis.

I'm more than willing to true current sources. However, I'm having a difficult time seeing why the voltage sources provide an impossible scenario. Wouldn't there be a closed loop?
gneill
#11
Jan10-13, 05:45 PM
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Quote Quote by rbrayana123 View Post
I'm more than willing to true current sources. However, I'm having a difficult time seeing why the voltage sources provide an impossible scenario. Wouldn't there be a closed loop?
If you suppress two at a time (in order to determine the effect of the remaining source), then you create an impossible situation with the remaining source being shorted.
gneill
#12
Jan10-13, 06:20 PM
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Here's your Δ configuration redrawn with current supplies in place.



I chose node b as the reference node for convenience. You could perform superposition analysis, suppressing two sources at a time and working out the resulting node voltages, or just do nodal analysis for the whole circuit to find the potentials at nodes a and c (with respect to the reference node b); The math then handles the superposition for you.

With Va and Vc in hand, you can then take differences to determine Vab, Vac, etc.

If you do a similar analysis for the Y configuration and arrive at the same set of equations for Vab, Vac, Vbc, then the circuits are externally indistinguishable.
Attached Thumbnails
Fig1.gif  
rbrayana123
#13
Jan11-13, 09:41 PM
P: 45
I'm really not following the placement of the voltage source. I guess I'm having trouble with why you made the choice to place it there and can't seem to pinpoint where it would be located on an actual real-world black box. Also, would I not have to suppress the voltage source at b?
gneill
#14
Jan11-13, 09:46 PM
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Quote Quote by rbrayana123 View Post
I'm really not following the placement of the voltage source. I guess I'm having trouble with why you made the choice to place it there and can't seem to pinpoint where it would be located on an actual real-world black box.
Which voltage source would that be? Which circuit are you referring to?
rbrayana123
#15
Jan11-13, 09:50 PM
P: 45
Sorry for lack of clarity. I'm talking about the delta circuit you labeled and I'm confused about the nature of the reference node you placed at B.
gneill
#16
Jan11-13, 09:59 PM
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Quote Quote by rbrayana123 View Post
Sorry for lack of clarity. I'm talking about the delta circuit you labeled and I'm confused about the nature of the reference node you placed at B.
Ah. The reference node, indicated with the ground symbol, is just there to facilitate nodal analysis (if you choose to use it). Otherwise the circuit is just your Delta circuit with current sources "bolted on".

When doing nodal analysis, the procedure is to pick a reference node and arbitrarily assign it as the "zero" reference for other potentials in the circuit (as though you were to attach your voltmeter's negative probe there and poke around the rest of the circuit with the positive probe). Then the potentials at the other nodes are obtained by writing nodal equations with that node as a reference.
rbrayana123
#17
Jan11-13, 10:04 PM
P: 45
I think I'm understanding a little bit more now. I was unfamiliar with the reference node symbol and interpreted it as a voltage source, which made the topology seem strange. As long as I'm consistent with my definitions of reference node and the directionality of the current sources, equality between the delta and wye should be provable. Should I choose to change the reference node or direction of the current sources, my voltages would differ I'm assuming.
gneill
#18
Jan11-13, 10:11 PM
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P: 11,829
Quote Quote by rbrayana123 View Post
I think I'm understanding a little bit more now. I was unfamiliar with the reference node symbol and interpreted it as a voltage source, which made the topology seem strange. As long as I'm consistent with my definitions of reference node and the directionality of the current sources, equality between the delta and wye should be provable. Should I choose to change the reference node or direction of the current sources, my voltages would differ I'm assuming.
You are free to choose the current source directions any way you wish, but stick with your choice throughout; It would make sense to keep consistent current source directions since you want to stimulate both arrangements in identical fashion.

You can choose any node you wish as the reference node in either circuit. The particular potentials at given nodes with respect to an arbitrary reference point are not as important as the potential differences between pairs of nodes -- the Vab, Vac, Vbc potentials.


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