Δ & Y and Superposition Theorem?

[rbrayana123]

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?...?

 

[rbrayana123]

I'm still stuck on this question. Any help?

 

[gneill]

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]

I'm familiar with a voltage supply (battery) which I'm using in this case. Why won't that work?

 

[gneill]

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]

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]

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]

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]

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]

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]

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]

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.

 

[rbrayana123]

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]

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]

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]

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]

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]

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.

 

[rbrayana123]

Now what's confusing me is the non-existence of a voltage source. Is it not necessary to have one for a current to run through?

 

[gneill]

Now what's confusing me is the non-existence of a voltage source. Is it not necessary to have one for a current to run through?

Nope. There are two types of sources: Voltage sources and Current sources. Either are common in circuits.

A voltage source will produce any amount of current required to maintain its specified voltage (regardless of the load). Think of it as a really good battery.

A current source will produce any amount of voltage required to maintain its specified current (regardless of the load).

 

[rbrayana123]

Thanks for the help! There appears to be a lot my book (Purcell's EM) hasn't covered about circuits such as Nodal Analysis & Current Sources. I'll work through the Math and get back to you.

 

[rbrayana123]

My question is in the next post. I think this is accurate but somethings wrong with my Y configuration.
My question is in the next post. I think this is accurate but somethings wrong with my Y configuration.
My question is in the next post. I think this is accurate but somethings wrong with my Y configuration.

This is just my work for the Δ configuration.

The last two statements before each dashed line are the voltages differences of V_A & V_C with V_B. Subtracting the two should get me the voltage difference between V_A & V_C.

I'll work through Y and post it next. No questions as of yet.

Let R₁ = 34, R₂ = 85 and R₃ = 170
===========================================================================
Suppressing I₂ and I₃:

Equation 1
V_C($\frac{1}{R2}$ + $\frac{1}{R3}$) - V_A($\frac{1}{R2}$) = 0

V_C = V_A($\frac{R3}{R2 + R3}$)

Equation 2
V_A($\frac{1}{R1}$ + $\frac{1}{R2}$) - V_C($\frac{1}{R2}$) = I₁

Eq 1 + Eq 2
V_A($\frac{1}{R1}$ + $\frac{1}{R2}$) - V_A($\frac{R3}{R2(R2+R3)}$) = I₁

V_A = I₁/($\frac{1}{R1}$ + $\frac{1}{R2}$ - $\frac{R3}{R2(R2+R3)}$)

Back to Eq 1
V_C = I₁($\frac{R3}{R2 + R3}$)/($\frac{1}{R1}$ + $\frac{1}{R2}$ - $\frac{R3}{R2(R2+R3)}$)
===========================================================================
Suppressing I₁ and I₃:

Equation 3
V_A($\frac{1}{R1}$ + $\frac{1}{R2}$) - V_C($\frac{1}{R2}$) = I₂

V_A = (I₂ + $\frac{Vc}{R2}$)($\frac{R1*R2}{R1 + R2}$)

Equation 4
V_A($\frac{1}{R2}$) - V_C($\frac{1}{R2}$ + $\frac{1}{R3}$) = I₂

Eq 3 + Eq 4
(I₂ + $\frac{Vc}{R2}$)($\frac{R1}{R1 + R2}$) - V_C($\frac{1}{R2}$ + $\frac{1}{R3}$) = I₂

V_C($\frac{R1}{R2(R1 + R2)}$ - $\frac{1}{R2}$ - $\frac{1}{R3}$) = I₂($\frac{R2}{R1 + R2}$)

V_C = I₂($\frac{R2}{R1 + R2}$)/($\frac{R1}{R2(R1 + R2)}$ - $\frac{1}{R2}$ - $\frac{1}{R3}$)

Back to Eq 3
V_A = I₂($\frac{R1*R2}{R1 + R2}$)(1 + $\frac{\frac{1}{R1 + R2}}{\frac{R1}{R2(R1 + R2)} - \frac{1}{R2} - \frac{1}{R3}}$)
===========================================================================
Suppressing I₁ and I₂:

Equation 5
V_A($\frac{1}{R1}$ + $\frac{1}{R2}$) - V_C($\frac{1}{R2}$) = 0

V_A = V_C($\frac{R1}{R1 + R2}$)

Equation 6
V_C($\frac{1}{R2}$ + $\frac{3}{R2}$) - V_A($\frac{1}{R2}$) = I₃

Eq 5 + Eq 6
V_C($\frac{1}{R2}$ + $\frac{1}{R3}$) - V_C($\frac{R1}{R2(R1+R2)}$) = I₃

V_C = I₃/($\frac{1}{R2}$ + $\frac{1}{R3}$ - $\frac{R1}{R2(R1+R2)}$)

Back to Eq 5
V_A = I₃($\frac{R1}{R1 + R2}$)/($\frac{1}{R2}$ + $\frac{1}{R3}$ - $\frac{R1}{R2(R1+R2)}$)
===========================================================================

 

[rbrayana123]

Voltage Differences from Δ data:
V_a,b = 30I₁ + 10I₂ + 20I₃
V_c,b = 20I₁ - 50I₂ + 70I₃
V_a,c = 10I₁ + 60I₂ - 50I₃
Question:
1) Why do these calculations give a voltage drop that occurs towards the reference node and not away?
2) When I do voltage difference calculations for Y, I'm missing the 20I₃ term for V_a,b & 20I₁ term for V_c,b. Both seem to depend on the top-most resistor but I can't seem to figure out how to bring in the extraneous term.

I've tried a source transformation; however, it just seems strange on the wye-configuration.

Let R₁ = 10, R₂ = 20 and R₃ = 50
===========================================================================
Suppressing I₂ and I₃:

Equation 7
(V_A - V_O)($\frac{1}{R1}$) = I₁

Equation 8
(V_A - V_O)($\frac{1}{R1}$) - (V_O)($\frac{1}{R2}$) = 0

Eq 7 + Eq 8
(V_O) = I₁R₂
(V_A) = I₁(R₁ + R₂)
===========================================================================
Suppressing I₁ and I₃:

Equation 9
(V_A - V_O)($\frac{1}{R1}$) = I₂

Equation 10
(V_O - V_C)($\frac{1}{R3}$) = I₂

Equation 11
(V_O)($\frac{1}{R2}$) + (V_O - V_A)($\frac{1}{R3}$) + (V_O + V_C)($\frac{1}{R3}$) = 0

Eq 9 & 10 + 11
(V_O)($\frac{1}{R2}$) - I₂ + I₂ = 0
V_O = 0
V_A = I₂R₁
V_C = -I₂R₃

===========================================================================
Suppressing I₁ and I₂:

Equation 12
(V_C - V_O)($\frac{1}{R3}$) = I₃

Equation 13
(V_C - V_O)($\frac{1}{R3}$) - (V_O)($\frac{1}{R2}$) = 0

Eq 12 + Eq 13
(V_O) = I₃R₂
(V_C) = I₃(R₂ + R₃)
===========================================================================

 

[gneill]

Voltage Differences from Δ data:
V_a,b = 30I₁ + 10I₂ + 20I₃
V_c,b = 20I₁ - 50I₂ + 70I₃
V_a,c = 10I₁ + 60I₂ - 50I₃
Question:
1) Why do these calculations give a voltage drop that occurs towards the reference node and not away?

It depends upon the chosen directions of the currents. Switch the currents and the signs of the node potentials will change.

2) When I do voltage difference calculations for Y, I'm missing the 20I₃ term for V_a,b & 20I₁ term for V_c,b. Both seem to depend on the top-most resistor but I can't seem to figure out how to bring in the extraneous term.

When you suppress two current sources in this Y configuration, you should note that it leaves a "floating" node. For example, suppressing i2 and i3 leaves node c unconnected. While no current will flow through R3 in this situation, nevertheless node c will have a potential! It will have the same potential as Vo where the other end of R3 is connected. You need to include these potentials in the superposition reckoning.

I've tried a source transformation; however, it just seems strange on the wye-configuration.

What sort of source transformation were you looking at? No obviously helpful transformations jump out at me when I look at the circuit.

If I may make a suggestion about the Y circuit analysis, the central node you've labelled "o" in your diagram is entirely internal to the "black box", so you don't need its value when comparing the two configurations. Why not take advantage of this by moving the reference node from b to o?

It also has the advantage that the potentials of open nodes will be zero, so there's nothing to "carry along" for them for the superposition sum.

 

[rbrayana123]

I see. The lack of current across the wire connecting o and c means no voltage potential drop; however that doesn't mean no voltage potential. In my diagrams, I took to the habit of completely dropping open circuits but now I see why that's incorrect.

As for the reference node, o is a better choice because it connects to more wires I suppose.

Out of desperation, I tried to transform the current source and R2 resister into a voltage source and parallel resistor but there was always a resistor in the way. I'm sure this is incorrect.

You mentioned the direction of the current decides the direction of voltage drop; however I2 seems arbitrary. I'm assuming the sign would've worked out either way to give the right direction. Does this mean that only I1 & I3 are determine the direction of voltage drop?

 

[gneill]

I see. The lack of current across the wire connecting o and c means no voltage potential drop; however that doesn't mean no voltage potential. In my diagrams, I took to the habit of completely dropping open circuits but now I see why that's incorrect.

As for the reference node, o is a better choice because it connects to more wires I suppose.

That's often a good indicator, and generally leads to tidier equations.

Out of desperation, I tried to transform the current source and R2 resister into a voltage source and parallel resistor but there was always a resistor in the way. I'm sure this is incorrect.

Yup. None of the current sources are candidates for transformation.

You mentioned the direction of the current decides the direction of voltage drop; however I2 seems arbitrary. I'm assuming the sign would've worked out either way to give the right direction. Does this mean that only I1 & I3 are determine the direction of voltage drop?

All three current sources contribute to the potentials (via superposition). Since as variables they can have arbitrary values (including negative values), their assumed directions are really only a book keeping measure to keep the equations self consistent.

 

[rbrayana123]

All three current sources contribute to the potentials (via superposition). Since as variables they can have arbitrary values (including negative values), their assumed directions are really only a book keeping measure to keep the equations self consistent.

But if I₁ and I₃ were positioned downward, would I still come up with the same voltage drop? By same voltage drop, I mean the signs in front of I₁ and I₃ change as a result of the sign change. (i.e. +2(+1) changes to -2(+1) where each +1 is in reference to a different direction so the second +1 is really -1 with respect to the co-ordinate system of the first). In this case, wouldn't it be so that the voltage drop is always towards the reference node regardless of assigned directions. If so, why? If not, then what's the behavior of assigned current directions and voltage potentials and why? I understand the computation behind nodal analysis but this detail is bothering me because it appears to be fundamental.

 

[gneill]

If you change the directions of one or more of the currents you'll change the signs of one or more coefficients of the resulting voltage equations (Vab, Vcb,...) This is why it's important to have the same currents in the same locations and orientations for both circuits if you wish to compare them.

When you calculate a potential between nodes it's up to you to choose the direction that you want to consider "positive". It's like which node you place the negative lead of meter on, and which the positive lead; switch the leads around and the sign of measured voltage changes while the magnitude remains the same. So, for example, Vac = -Vca.

For a given setup of the current sources, you should arrive at the same set of equations for the oriented potentials Vab, Vcb, etc., regardless of which node you choose for the reference node.

 

[rbrayana123]

If you change the directions of one or more of the currents you'll change the signs of one or more coefficients of the resulting voltage equations (Vab, Vcb,...) This is why it's important to have the same currents in the same locations and orientations for both circuits if you wish to compare them.

When you calculate a potential between nodes it's up to you to choose the direction that you want to consider "positive". It's like which node you place the negative lead of meter on, and which the positive lead; switch the leads around and the sign of measured voltage changes while the magnitude remains the same. So, for example, Vac = -Vca.

For a given setup of the current sources, you should arrive at the same set of equations for the oriented potentials Vab, Vcb, etc., regardless of which node you choose for the reference node.

Alrighty. I think I understand now. The only thing that's certain are potential differences, however the potentials themselves may differ, as in mechanics when it comes to gravitational potential.

Thank you so much sir for answering all my questions and helping me understand the nitty-gritty details of this problem. Most of the stuff about circuit analysis (Nodal Analysis, Mesh Analysis, Superposition Theorem, etc.) wasn't in my book and I either learned it here from you or heard about it so I knew what to further investigate.

 

[gneill]

Alrighty. I think I understand now. The only thing that's certain are potential differences, however the potentials themselves may differ, as in mechanics when it comes to gravitational potential.

Thank you so much sir for answering all my questions and helping me understand the nitty-gritty details of this problem. Most of the stuff about circuit analysis (Nodal Analysis, Mesh Analysis, Superposition Theorem, etc.) wasn't in my book and I either learned it here from you or heard about it so I knew what to further investigate.

I'm glad I could help. I was impressed by your dedication to solving a general case, rather than just the specific case with the given resistor values.