# Delta and star transformation of AC circuits

• Engineering
js3
So do i need to combine two deltas using product over the sum rule in (a) and use that answer to make a Y transformation in(b)?
Hi, I'm just after a bit of help. I understand that gneill confirmed that using the product over sum is the correct technique to combine the two delta impedances, but in my text book it quotes this as a Π to T transformation. Would anybody be able to shed a bit more light on this? Thanks

gneill
Mentor
Hi, I'm just after a bit of help. I understand that gneill confirmed that using the product over sum is the correct technique to combine the two delta impedances, but in my text book it quotes this as a Π to T transformation. Would anybody be able to shed a bit more light on this? Thanks
The original circuit had both Δ and a Y loads (Some call a Δ configuration a π configuration). The idea was to transform one of the loads to the same form as the other so that they could then be easily "summed" to a single load. That combined load could then be transformed as required.

Converting the Δ to a Y first leaves you with two Δ loads. They can be easily combined since the corresponding impedances are in parallel (hence the product over sum rule to combine them). The resulting single Δ can then be transformed to its Y equivalent as required.

js3
Thank you for your help on that, much clearer now. I have another question, I've been watching some tutorials and is converting a star impedance to it's equivalent delta impedance really as easy as multiplying by 3?

gneill
Mentor
Thank you for your help on that, much clearer now. I have another question, I've been watching some tutorials and is converting a star impedance to it's equivalent delta impedance really as easy as multiplying by 3?
Yes, if all the impedances are identical.

Hi, regarding question d. This is how I expressed my working out to get my answer (looking at the previous posts I now realise my answer is wrong but would like to know if my train of thought is correct)
Total impedance = 45+j45
Voltage = 415v
I =v/r* = 415(45-j45)/45+j45(45-j45)
= 18675-j18675/2025-j2025
total current = 9.2-j9.2 amps
P=vi*
P=415x(9.2-j9.2)
Real power = 3818watts. think I am supposed to times this by 3 but still not sure why(even after reading the posts!)

gneill
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Hi, regarding question d. This is how I expressed my working out to get my answer (looking at the previous posts I now realise my answer is wrong but would like to know if my train of thought is correct)
Total impedance = 45+j45
In what way is that the "total impedance"? Total impedance of what?

For part C would the circuit look like this (see picture) before I make it into 1 single Y load.

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gneill
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For part C would the circuit look like this (see picture) before I make it into 1 single Y load.
Part C also specifies: "...and with the star-points of the two Y-sub-circuits connected together," What do you suppose is meant by that?

Would it be like this?

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gneill
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Would it be like this?
That would be my interpretation, yes.

ok brilliant i will give it a go. Thank you for your help gneill

So i have worked out that each of the impedance's would be 7.5+j7.5
Is this correct?

gneill
Mentor
So i have worked out that each of the impedance's would be 7.5+j7.5
Is this correct?
Yes. But you could have verified this by checking the earlier parts of this thread.

Yes. But you could have verified this by checking the earlier parts of this thread.
Hi Gneill,
Reading through the thread, I can't quite figure out why part (c) is the same as part (b). I have labelled up his drawing to reflect these calcs:
So Zy1=45+j45Ω
Zy2=ZΔ / 3 = 15+j15

Would each Y1 impedance not be in parallel with a Y2 impedance hence the calculation for part (c) would then be:
Zy combined = ((45+j45)(15+j15)) / ((45+j45)+(15+j15)) = j1350 / (60+j60)

j1350 / (60+j60) multiplied by conjugate = (81000+j81000) / 7200 = 11.25+j11.25 Ω

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gneill
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Reading through the thread, I can't quite figure out why part (c) is the same as part (b). I have labelled up his drawing to reflect these calcs:
So Zy1=45+j45Ω
Zy2=ZΔ / 3 = 15+j15
Zy1 is not 45+j45Ω. It's the original given value of (15 + j15)Ω

Zy1 is not 45+j45Ω. It's the original given value of (15 + j15)Ω
Sorry got a bit lost there, I follow you now and have gotten 7.5Ω.

One other query regarding part (d), I have calculated out the same procedure as mentioned earlier in this thread using P = 3VI for total power consumed, which gives 11481.667-j11481.667. Could I assume that the 'imaginary' number here is effectively Reactive power Q? So my final answer should be 11481.667W or 11.48kW?

I know you mentioned that real power is just P = VI but should that not be S = VI, hence using the two numbers I get for S = VI would then be reactive & real power rather than using P = VI which would be the wrong formula technically?

gneill
Mentor
One other query regarding part (d), I have calculated out the same procedure as mentioned earlier in this thread using P = 3VI for total power consumed, which gives 11481.667-j11481.667. Could I assume that the 'imaginary' number here is effectively Reactive power Q? So my final answer should be 11481.667W or 11.48kW?
The imaginary part is the reactive power, yes. I think you might have forgotten to take the complex conjugate of the current when you calculated ##p = V I##, as the sign of the imaginary component is suspect (to me).

The final real power consumed is 11.48 kW as you've stated.

I know you mentioned that real power is just P = VI but should that not be S = VI, hence using the two numbers I get for S = VI would then be reactive & real power rather than using P = VI which would be the wrong formula technically?
I'm not sure where this question comes from as it's been a long time since the thread has been active and I haven't re-read through all the posts.

You can calculate the complex power by ##S = V I^*##, where ##I^*## is the complex conjugate of the current ##I##. ##P = VI## wouldn't make sense if V and I are both complex numbers, as ##P## must be strictly real.

You can calculate the complex power by S=VI∗S=VI∗S = V I^*, where I∗I∗I^* is the complex conjugate of the current III. P=VIP=VIP = VI wouldn't make sense if V and I are both complex numbers, as PPP must be strictly real.
I see, that does answer why it should be S = VI* thanks. Why is it you multiply by the conjugate? Would the negative Q value not mean it was leading?

gneill
Mentor
Why is it you multiply by the conjugate? Would the negative Q value not mean it was leading?
It's related to the fact for finding the power you want the phase difference between the current and voltage. Say that the voltage phasor ##V## has some phase angle ##\phi_V## and the current phasor ##I## some phase angle ##\phi_I##. Multiplying these two phasors would yield ##|V||I|## with a phase angle of ##\phi_V + \phi_I##. But you want the phase difference, ##\phi_V - \phi_I##. Taking the complex conjugate of the current gives you this.

You should be able to find the mathematical derivation of the complex power in a text book, or via google.

Jason-Li