Homework Help: Star Delta Transformation

The question asks you to find a single Δ-connected load that is equivalent to total of the two loads shown.

 

We won't confirm or deny a guess. You'll have to show how you got there.

 

Can you justify adding them? How do the voltage sources "see" them when you combine them?

 

Ignore the phase angles of the sources for now. How do the individual impedances of the two loads combine as seen by the sources? You have two Δ loads. What connections are in common? Can you see serial or parallel reduction possibilities?

 

Can you elaborate?

 

Okay, that is correct. But it is important that you can see why the individual impedances are in parallel by considering the circuit diagram. If you can't see it, be sure to ask.

 

How are we to interpret your coded messages? A few words of explanation and justification would help.

 

Okay, what leads you conclude that you can combine the Y's in that fashion? I'm not saying it's incorrect for this very special case, but you should be able to justify it. That way you won't inadvertently make the error of trying to do the same thing when the scenario is different and combining them in this fashion would be incorrect.

 

What do you mean by "add 0 into it"?

The diagram alone doesn't explain why you can combine the like-labeled impedances in parallel. Note that none of those pairs shares two nodes... So you need to make a special case deduction about the nature of the central nodes of each Y...

 

If you read the question statement carefully, for (b) you are expected to perform a Δ-Y transformation of the single Δ load obtained in part (a). That should be a straightforward application of the Δ-Y formulas.

It's part (c) that asks you to combine two identical Y loads. A very important fact to note is that they are identical. When driven in parallel by the same sources you can be sure that they will behave identically as well. That lets you say something about their center nodes.

 

With a Δ load you have the voltage across each of the individual load impedances. The voltage and impedance are sufficient to calculate the power.