- #36
Spongecake
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ok brilliant i will give it a go. Thank you for your help gneill
Yes. But you could have verified this by checking the earlier parts of this thread.Spongecake said:So i have worked out that each of the impedance's would be 7.5+j7.5
Is this correct?
gneill said:Yes. But you could have verified this by checking the earlier parts of this thread.
Zy1 is not 45+j45Ω. It's the original given value of (15 + j15)ΩJason-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
gneill said:Zy1 is not 45+j45Ω. It's the original given value of (15 + j15)Ω
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).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?
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.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?
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.
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.Jason-Li said:Why is it you multiply by the conjugate? Would the negative Q value not mean it was leading?