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Home Q & A Board Science-Related Homework Help High School Level Science
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angie19 angie19
wrote...
7 years ago
a 100 MW power station generates energy at 26.0 kV. The power lines must carry only 200A of current to be efficient enough. To what voltage must the potential be increased? show your work ( formula,substitution, and round correctly and with the appropriate unit
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wrote...
#1
7 years ago
In single-phase:

10*10^7/(2*10^2) = 5*10^5 volt (500 kV)

In three-phase:

10*10^7/(2*10^2*√3) = 5*10^5*√3/3 volt (288.7 kV)
angie19 Author
wrote...
#2
7 years ago
Can you please explain this formula?
Post Merge: 7 years ago

What's this single and three phase ?
wrote...
#3
Educator
7 years ago
I'm confused by this too

100,000,000 divided by 200 = 500 kV

10*10^7 Rightwards Arrow 100,000,000 (got that)

How did you get the rest?
angie19 Author
wrote...
#4
7 years ago
is the answer just 500 kV?Neutral Face
angie19 Author
wrote...
#5
7 years ago
@ bio man looking at the question its worth 4 marks so I feel like theres more to the question than just the one step
looking at my formulas I have theres: 3 formulas to work with

#1) Vp/ Vs = Is/ Ip    #2) Np/Ns = Vp/Vs    or #3) Np/Ns = Is/Ip

Looking at the question I see two Voltages and a current so applying the  first formula: Vp/Vs = Is/Ip
** however the question is asking for to what the voltage must be increased ... there is where I am stuck..     
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