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# NETA III Test April 2019

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1.  Member Pro Subscriber
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Apr 2018
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Reputation   Originally Posted by Kalbi_Rob You just put all the resistances in parallel between the end 15 ohm resistances. They are in series on the parallel bus bars.

R= 15+(1/((1/(5+10+5)+(1/(5+10+5)))+15 = 15+(1/((1/20)+(1/20)))+15 = 15+10+15 = 40

Don't quote me on this being the right answer, but that's how I see it.
That is how I see it also. This question was also on my level II test.

This is how I visualize it.   Reply With Quote

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Mar 2015
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Reputation  Originally Posted by rofo42 That is how I see it also. This question was also on my level II test.

This is how I visualize it. Thank you  Reply With Quote

3. Originally Posted by SecondGen Easy way to calculate:
A2 + B2 = C2

750,000 = C2
750,000 x 0.95 = 712,500 = A2
750,000 - 712,500 = 37,500 = B2
sqrt(37,500) = 193.64
I agree with most of this. Until you came up with B2.

sqrt(750,000sq - 712,500sq) = 234.42 = B2

Thoughts? I know this is an old post, but, for those of us studying, it's still relevant.  Reply With Quote

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Reputation  Originally Posted by clipboardwarrior I agree with most of this. Until you came up with B2.

sqrt(750,000sq - 712,500sq) = 234.42 = B2

Thoughts? I know this is an old post, but, for those of us studying, it's still relevant.
I don't think A^2+B^=C^2 method is the right way to go about the problem. Kw loading will stay the same, but changing the kvar will cause the kva to change.

Page 4 of (https://www.eaton.com/ecm/groups/pub...a02607001e.pdf) has a good visual of the change.

The OP's method seems the quickest to answer the problem:
1. Solve for kw. (pf x KVAinitial =kw) --- 0.75 x 750000 = 562500 Kw
2. Find the angle for .95 pf. (cos-1 [Pfdesired]= angle) --- cos-1 (0.95) = 18.19 degrees
3. Solve for Kvar. (tan[angle] x kw = kvar) --- tan(18.19) x 562500 = 184831 Kvar

For fun you can then use A^2+B^=C^2 to solve for the new Kvar = 592088 Kvar  Reply With Quote

5. Originally Posted by jgraves I don't think A^2+B^=C^2 method is the right way to go about the problem. Kw loading will stay the same, but changing the kvar will cause the kva to change.

Page 4 of (https://www.eaton.com/ecm/groups/pub...a02607001e.pdf) has a good visual of the change.

The OP's method seems the quickest to answer the problem:
1. Solve for kw. (pf x KVAinitial =kw) --- 0.75 x 750000 = 562500 Kw
2. Find the angle for .95 pf. (cos-1 [Pfdesired]= angle) --- cos-1 (0.95) = 18.19 degrees
3. Solve for Kvar. (tan[angle] x kw = kvar) --- tan(18.19) x 562500 = 184831 Kvar

For fun you can then use A^2+B^=C^2 to solve for the new Kvar = 592088 Kvar
You're absolutely right. Between the time I read the OP and then read the follow up comment, I spaced on the VA changing! Yoinks! Hopefully, I don't pull any dumb stunts like that next week when I take my test.   Reply With Quote

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