Delta and star transformation of AC circuits

  • #26
js3
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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
 
  • #27
gneill
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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.
 
  • #28
js3
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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?
 
  • #29
gneill
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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.
 
  • #30
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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
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!)
 
  • #31
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?
 
  • #32
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For part C would the circuit look like this (see picture) before I make it into 1 single Y load.
 

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  • #33
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?
 
  • #34
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Would it be like this?
 

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  • #35
gneill
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Would it be like this?
That would be my interpretation, yes.
 
  • #36
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ok brilliant i will give it a go. Thank you for your help gneill
 
  • #37
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So i have worked out that each of the impedance's would be 7.5+j7.5
Is this correct?
 
  • #38
gneill
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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.
 
  • #39
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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


upload_2019-1-22_21-55-18.png
upload_2019-1-22_22-1-11.png
 

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  • #40
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
 
  • #41
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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?
 
  • #42
gneill
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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.
 
  • #43
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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?
 
  • #44
gneill
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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.
 
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