# Delta - Wye transform help

Discussion in 'Electronics Homework Help' started by kimbokasteniv, Jan 26, 2012.

1. ### kimbokasteniv

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Jan 26, 2012
Completing the actual transform isn't the problem, I am just unable to see how I can use it to my advantage in the following problem: Circuit (A) is the circuit given to me in the book. Circuit (C) is as far as I have been able to take the circuit doing what I thought was the appropriate delta-wye transform. I am attempting to find the current at the location of probe (1) in circuit (A). Of course multisim has given me the answer of 259mA.

I guess my question breaks down into two parts:
Is circuit (C) what I need to solve the problem, or at least headed in the right direction?
And, if the circuit (C) is what I need, how do I apply KCL at the node above probe (4) in circuit (C)?
I am having trouble applying KCL or KVL to the circuit since it seems different from any previous circuit I have dealt with.

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
2,788
Jan 21, 2010
This looks like a trick question.

Once you see the trick you'll almost be able to do it in your head.

hint: What is the difference in current through each battery? So where does (or doesn't) this difference current flow?

3. ### kimbokasteniv

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Jan 26, 2012
Oh I see, there is no current flow across R3 in circuit (A). Yup, ignoring that branch results in a 259mA current.

Thanks for the help.

But I don't think I can fully justify to myself why there is no current flow. Current flows when there is a difference in charge between two points. Correct? So then, current does not flow to the node between the voltage sources because that node is neutral.
Is that correct?

4. ### Merlin3189

250
69
Aug 4, 2011
Yes.
Intuitively, the circuit is symmetric, so the potential at junction R1-R2 is always the same as at junction V1-V2. If the potential at both ends of R3 is the same, there is no PD so no current. You could therefore replace it with any passive circuit, but an open circuit is the most useful, because it just disappears.
You could use mesh analysis to show that the mesh currents in {V1-R3-R2-R4} and in {V2-R5-R1-R3} are equal and opposite in R3, so net current in R3 is zero and it can be removed. But this is just a slightly obscure mathematical trick to justify what is obvious anyway.

5. ### kimbokasteniv

6
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Jan 26, 2012
I appreciate the response Merlin. I have since learned mesh analysis, and I see what you mean.

In case anyone is curious, I have also seen the solution provided by the author, and after the delta-wye transform, the author uses source transformation which results in a simple series circuit.

6. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
2,788
Jan 21, 2010
In these cases it would seem easier to have removed the redundant resistor and then employed simple calculations of resistors in parallel.

If R1 and R2 were different values, this wouldn't work, and I suspect that the easiest method may be the one in the official answer.

But I wonder at why each of the examples had such a simple shortcut.  