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I'm trying to upskill by teaching myself electrical engineering. I already have a degree in theoretical physics and pure mathematics so I don't mind math-heavy answers. While I don't find the maths of the electronics problems I've encountered so far to be too bad, some of the theory is a bit confusing to me.

I'm trying to understand how to calculate the total impedance of a 3-phase overhead transmission line.

I know that usually transmission lines are connected in delta and I know how to calculate the impedance in each phase.

In the particular problem I'm working on we have 7 Ω of resistance from a resistor and 10j Ω from some conductor in each phase.

So using Pythagoras' Theorem on the phasor diagram that means there is \$Z_p={\sqrt {7^2+10^2}}={\sqrt {149}}\$ in each phase.

In polar notation there is \$Z_p= {\sqrt{149}} \angle 55\$ impedance in each phase.

The next bit is where I get confused.

I would have thought to calculate the total impedance, as a delta connection is connected in parallel, we would use:

$$\tfrac{1}{Z_{total}}=\tfrac{1}{Z_1}+\tfrac{1}{Z_2}+\tfrac{1}{Z_3}$$

Where \$Z_1=Z_2=Z_3=Z_p\$ as the delta connection is balanced.

This would give:

$$\tfrac{1}{Z_{total}}=\tfrac{Z_2Z_3+Z_1Z_3+Z_1Z_2}{Z_1Z_2Z_3}=\tfrac{3}{\sqrt{149}} \Rightarrow Z_{total}=\tfrac{\sqrt{149}}{3}$$

Then the polar form of this would be \${\sqrt{149}} \angle 55\$.

$${BUT!!!}$$

according to page 6 of this powerpoint

$$Z_1=Z_2=Z_3=Z_{total}$$

So \$Z_{total}\$ in polar form is just \$Z_p= {\sqrt{149}} \angle 55\$.

Could someone please explain why this is the case?

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  • \$\begingroup\$ The word "total" does not appear on page 6. It says Za = Zb = Zc = ZΔ. It also says Z1 = Z2 = Z3 = Zy. You seem to feel you have uncovered an inconsistency but I am not seeing it. Also, it is not clear to me what Ztotal should mean, so it is hard for me to figure how I would want to calculate it. \$\endgroup\$
    – user57037
    Mar 15, 2022 at 3:44
  • \$\begingroup\$ @mkeith I had thought that $$Z_{\Delta}$$ meant the total impedance experienced by the circuit. By Z total I had meant as well the total impedance experienced by the circuit \$\endgroup\$
    – hoff
    Mar 15, 2022 at 3:49
  • \$\begingroup\$ @mkeith the line Impedance is what I thought both had meant \$\endgroup\$
    – hoff
    Mar 15, 2022 at 3:53
  • \$\begingroup\$ I don't know what the total impedance experienced by the circuit means, either. Why don't we wait a bit and see if someone gets what you are talking about. It is not my intention to be pedantic. It's just that I see no way to answer your question right now so I am worried that it needs some kind of touch up to be answerable. But maybe it is just a deficiency in my understanding. You could possibly explain why you are trying to calculate Rtotal. Maybe that would clear things up. Like are you trying to calculate a power loss or something? \$\endgroup\$
    – user57037
    Mar 15, 2022 at 3:55
  • \$\begingroup\$ @JamieOM as delta connection is connected in parallel why do you think so? \$\endgroup\$
    – across
    Mar 15, 2022 at 4:02

2 Answers 2

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The presentation doesn't explain what it means by \$Z_\Delta\$ or \$Z_Y\$, and I think from context they are referring to the equivalent impedance between a phase node (say \$a\$) and a hypothetical ground, at 0V potential. That ground is shown as node \$n\$ in the wye configuration, but there's no such physical node in the delta setup.

The hypothetical ground (0V) I am speaking of is the potential with respect to which each of the phase nodes oscillates above and below. The mean potential of each of the phases. It's the zero implied in the three phase potentials \$v = sin(\omega t + \phi)\$.

So instead of using the word "total" in the description of \$Z_\Delta\$ or \$Z_Y\$, I would say that these are the hypothetical equivalent impedances that a phase node would "see" when looking into the network towards 0V.

For the wye configuration, by symmetry one can intuit that the common centre node potential is the average of the potential of the three phase nodes (which I admit criminally lacks rigour), which is zero. That means that each phase node sees only its corresponding Z as an impedance to neutral.

That's where I get confused by the document, because the relationship between \$Z_{1|2|3}\$ (from the Y network) and \$Z_{a|b|c}\$ from the delta, is not clear. But, I can propose an explanation for the claim that \$Z_\Delta=3Z_Y\$, if we assume that \$Z_Y\$ at least represents, as I suggest, the equivalent impedance to neutral, as experienced by each phase.

In the Δ-Y transformation, the impedances you need to produce a wye-network which is behaviourally identical to a delta formation of three impedances \$Z_{AB}\$, \$Z_{AC}\$ and \$Z_{AC}\$ are calculated:

$$ \begin{aligned} Z_A = \frac{Z_{AB}Z_{AC}}{Z_{AB}+Z_{AC}+Z_{BC}} \\ \\ Z_B = \frac{Z_{AB}Z_{BC}}{Z_{AB}+Z_{AC}+Z_{BC}} \\ \\ Z_C = \frac{Z_{AC}Z_{BC}}{Z_{AB}+Z_{AC}+Z_{BC}} \\ \\ \end{aligned} $$

In the case where \$Z_{AB}=Z_{AC}=Z_{BC}=Z_\Delta\$, this simplifies to:

$$ \begin{aligned} Z_Y = Z_A = Z_B = Z_C &= \frac{{Z_\Delta}^2}{3Z_\Delta} \\ \\ Z_Y &= \frac{Z_\Delta}{3} \end{aligned} $$

That gives us this equivalency:

schematic

simulate this circuit – Schematic created using CircuitLab

The right "wye" arrangement has the central neutral node, and the impedance to that node from each phase is:

$$ Z_Y = \frac{1}{3}Z_\Delta $$

That's as far as I have gone thinking about this, but maybe it will be helpful to get past any confusion.

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There is a major difference between a parallel and a delta connection. In a delta connection, 3 impedances are connected to form a triangle. However, in a parallel connection, the resistances are connected across the same pair of nodes. There is no sense in writing total impedance for a delta connection as there is no series or parallel connections, unless it is per branch. The \$\sqrt{149}\angle{55}\$ is the impedance in one branch.

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