- #1
psparky
Gold Member
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I was working on trying to find the short circuit current rating of a 120 volt panel fed from a 7.5KVA transformer.
Using P=IV you get 7.5KVA=120 * I
I = 62.5 Amps.
Then I believe you divide that current by the impedance of the transformer. .03 is the impedance in this case in the transformer specs. So I do the math and I get about 2,000 amp short circuit rating.
Great. My question is this. If impedance is R + JX, how does .03 relate to that?
Is it possible the R+jx add up to be .03? Or is that nonsense? I believe they even state the .03 as 3% as well.
Using P=IV you get 7.5KVA=120 * I
I = 62.5 Amps.
Then I believe you divide that current by the impedance of the transformer. .03 is the impedance in this case in the transformer specs. So I do the math and I get about 2,000 amp short circuit rating.
Great. My question is this. If impedance is R + JX, how does .03 relate to that?
Is it possible the R+jx add up to be .03? Or is that nonsense? I believe they even state the .03 as 3% as well.
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