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Delta circuit

Star equivalent of that delta circuit

The formula used for star delta transformations are:

$$ R_A = \frac{R_{AB} R_{AC}}{R_{AB}+R_{AC}+R_{BC}} $$

and

$$ R_{AB} = \frac{R_AR_B + R_AR_C + R_BR_C}{R_C} $$

The normal way that these formulae are derived is by equating the expressions for the effective resistances between any corresponding pairs of terminals.

But in that approach, it is assumed for the calculation of the effective resistances, that the third nodes are not connected to any external circuit. That is, if we are calculating the effecting resistances between nodes \$A\$ and \$B\$ in the circuits, then we assume that no current enters or leaves the star or delta formation from node \$C\$ (\$I_2 = 0\$ in my diagram).

But how is this assumption valid? How are these formulae, which were derived with the assumption that \$I_2 = 0\$ valid even when \$I_2 \ne 0\$ ?

Perhaps this was done to simplify the derivation, and then make a rigorous proof that the formulae obtained work in all scenarios? But I haven't found any such proof following the derivation in the sources that I referred.

Or maybe there is some law/theorem that enables this approach? Maybe superposition theorem?

My alternative to do this derivation rigorously without making the open node assumption, would be to equate the voltages instead of effective resistances. Since voltages only depends on end points, they can be calculated by taking just one path from \$A\$ to \$B\$. To calculate the voltage of individual branches in the path, we use \$V = IR\$. The currents here are found using Kirchoff's current law, and account for all possible external connections to a star or delta circuit.

Balu
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  • *But in that approach, it is assumed for the calculation of the effective resistances, that the third nodes are open.* <-- I believe that to be incorrect. – Andy aka Oct 27 '22 at 12:26
  • @Andyaka Yes, I thought the same. But everywhere I check, the same assumption has been made. Here's one example YouTube video: https://youtu.be/OV0qi7yzKAM?t=4m24s – Balu Oct 27 '22 at 12:30
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    That cannot be true else RAB would equal RA+RB and clearly that is not the case. "Open" means open-circuit in my book. – Andy aka Oct 27 '22 at 12:33
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    @Andyaka That is exactly what they did though. The effective resistance in the star circuit was taken as \$R_A + R_B\$. And in the delta circuit, it was \$R_{AB} || (R_{BC }+ R_{AC})\$ – Balu Oct 27 '22 at 12:48
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    That is not what your question states. Nowhere does it state that \$R_{AB} = R_A+R_B\$ <-- that would be incorrect. – Andy aka Oct 27 '22 at 15:15
  • @Andyaka I did not explain the full derivation in my question. Please refer the YouTube video that I sent for that. They make the equation \$ R_A + R_B = 1R_{AB} || (R_{BC }+ R_{AC}) \$ and then apply algebra to derive the formulae. I do not find this correct either, that's the reason I asked the question – Balu Oct 27 '22 at 16:58
  • @Balu I actually don't see the problem. The resistance between any two terminals can be measured in two ways: (1) apply a current source and measure the voltage difference, or (2) apply a voltage difference and measure the current. There is a singular situation approached from either side when the currents and voltages are zero, where the method fails. What may be bothering you is this singular situation. The solution using multiple equations involving ***only*** resistance appears to work even when the voltages and currents are zero. And it does work there. You want it to fail there, though. – jonk Oct 28 '22 at 01:04
  • @jonk I don't want it to fail. I'm saying the formulae obtained this way are not a general solution. There's a possibility it might fail to work when \$I_2 \ne 0\$ I want to find out if there's any law/principle/theorem/proof, that says that the resistance can be calculated generally by taking the voltage or current as 0. – Balu Oct 28 '22 at 03:19
  • @Balu Tell me how you would *generally prove* in the 2-terminal case that$$R_{_\text{A}}+R_{_\text{C}}=R_{_\text{AC}}\,\vert\vert\, \left(R_{_\text{AB}}+ R_{_\text{BC}}\right)$$? Because if you can do that, then you have the rest already. – jonk Oct 28 '22 at 03:33
  • @jonk The objective is to figure out transformation formulae such that when viewed only from the terminals, there is no difference in the circuits whatsoever. So, the effective resistances between A and B of both the circuits should be equal. \$R_A\$ want \$R_B\$ are in series in the star circuit. In the delta circuit, \$R_{AB}\$ and \$R_{BC}\$ are in series, and they are collectively parallel to \$R_{AB}\$. But I don't see how this considerations about series and parallel are valid when there are other branches coming out of the third terminal – Balu Oct 28 '22 at 07:05

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