Delta and star transformation of AC circuits

In summary: Yes, the single Delta load from part (a) represents the entire load. After all, you took the time to change the Y load to its Delta form and then combine it with the existing Delta, resulting in a "single Delta load", right?
  • #36
ok brilliant i will give it a go. Thank you for your help gneill
 
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  • #37
So i have worked out that each of the impedance's would be 7.5+j7.5
Is this correct?
 
  • #38
Spongecake said:
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
gneill said:
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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  • #40
Jason-Li said:
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
gneill said:
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
Jason-Li said:
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.

Jason-Li said:
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
gneill said:
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
Jason-Li said:
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 textbook, or via google.
 
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