Δ & Y and Superposition Theorem?

gneill

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

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([itex]\frac{1}{R2}[/itex] + [itex]\frac{1}{R3}[/itex]) - V_A([itex]\frac{1}{R2}[/itex]) = 0

V_C = V_A([itex]\frac{R3}{R2 + R3}[/itex])

Equation 2
V_A([itex]\frac{1}{R1}[/itex] + [itex]\frac{1}{R2}[/itex]) - V_C([itex]\frac{1}{R2}[/itex]) = I₁

Eq 1 + Eq 2
V_A([itex]\frac{1}{R1}[/itex] + [itex]\frac{1}{R2}[/itex]) - V_A([itex]\frac{R3}{R2(R2+R3)}[/itex]) = I₁

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

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

Equation 3
V_A([itex]\frac{1}{R1}[/itex] + [itex]\frac{1}{R2}[/itex]) - V_C([itex]\frac{1}{R2}[/itex]) = I₂

V_A = (I₂ + [itex]\frac{Vc}{R2}[/itex])([itex]\frac{R1*R2}{R1 + R2}[/itex])

Equation 4
V_A([itex]\frac{1}{R2}[/itex]) - V_C([itex]\frac{1}{R2}[/itex] + [itex]\frac{1}{R3}[/itex]) = I₂

Eq 3 + Eq 4
(I₂ + [itex]\frac{Vc}{R2}[/itex])([itex]\frac{R1}{R1 + R2}[/itex]) - V_C([itex]\frac{1}{R2}[/itex] + [itex]\frac{1}{R3}[/itex]) = I₂

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

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

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

Equation 5
V_A([itex]\frac{1}{R1}[/itex] + [itex]\frac{1}{R2}[/itex]) - V_C([itex]\frac{1}{R2}[/itex]) = 0

V_A = V_C([itex]\frac{R1}{R1 + R2}[/itex])

Equation 6
V_C([itex]\frac{1}{R2}[/itex] + [itex]\frac{3}{R2}[/itex]) - V_A([itex]\frac{1}{R2}[/itex]) = I₃

Eq 5 + Eq 6
V_C([itex]\frac{1}{R2}[/itex] + [itex]\frac{1}{R3}[/itex]) - V_C([itex]\frac{R1}{R2(R1+R2)}[/itex]) = I₃

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

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

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)([itex]\frac{1}{R1}[/itex]) = I₁

Equation 8
(V_A - V_O)([itex]\frac{1}{R1}[/itex]) - (V_O)([itex]\frac{1}{R2}[/itex]) = 0

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

Equation 9
(V_A - V_O)([itex]\frac{1}{R1}[/itex]) = I₂

Equation 10
(V_O - V_C)([itex]\frac{1}{R3}[/itex]) = I₂

Equation 11
(V_O)([itex]\frac{1}{R2}[/itex]) + (V_O - V_A)([itex]\frac{1}{R3}[/itex]) + (V_O + V_C)([itex]\frac{1}{R3}[/itex]) = 0

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

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

Equation 12
(V_C - V_O)([itex]\frac{1}{R3}[/itex]) = I₃

Equation 13
(V_C - V_O)([itex]\frac{1}{R3}[/itex]) - (V_O)([itex]\frac{1}{R2}[/itex]) = 0

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

gneill

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



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.

Attached Thumbnails

   

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 thats 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

That's often a good indicator, and generally leads to tidier equations.
Yup. None of the current sources are candidates for transformation.
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

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

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

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.