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NETA III Test April 2019

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  1. rofo42 is offline Member Pro Subscriber
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    Quote Originally Posted by Kalbi_Rob View Post
    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.
    Click image for larger version. 

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    Quote Originally Posted by rofo42 View Post
    That is how I see it also. This question was also on my level II test.

    This is how I visualize it.
    Click image for larger version. 

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    Thank you

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    clipboardwarrior is offline Junior Member Pro Subscriber
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    Quote Originally Posted by SecondGen View Post
    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.

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  7. jgraves is offline Junior Member Pro Subscriber
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    Quote Originally Posted by clipboardwarrior View Post
    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

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  9. clipboardwarrior's Avatar
    clipboardwarrior is offline Junior Member Pro Subscriber
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    Quote Originally Posted by jgraves View Post
    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.

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