Lucas Organ
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- NL CANADA
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- Industrial Electrician
wouldnt this be correct 600/120=5*112.5=562.5V 250/5=50*3=150A 562.5*150*1.732*.9=131 523.75/1000= 131.52kw
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Looking for someone to help me with a code question and a power question as I prepare to write my IP!
- Thread starter Lucas Organ
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Lucas Organ
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- Location
- NL CANADA
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- Industrial Electrician
correct but we are not dealing with 100V it is 562 VI assume you meant 300 watts not amps
Eddy_Current
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- Location
- Canada
The PT and CT ratio question is great practice, but I highly doubt you will get a question like that in the exam.
Lucas Organ
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- Location
- NL CANADA
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- Industrial Electrician
I certainly hope your right, but what do you think of the calculations do they look right to you?The PT and CT ratio question is great practice, but I highly doubt you will get a question like that in the exam.
?? WYE or Delta, P(kW) = (VL * I * 1.732 * PF) / 1000. In a WYE you would multiple voltage by 1.732, and in a Delta you would multiple current by 1.732.That would be if the PT's were connected in delta or open delta to measure the L-L voltages. If the PT's were connected as wye-wye, then the power delivered would be 1.732 times higher.
Yes.I assume you meant 300 watts not amps
wouldnt this be correct 600/120=5*112.5=562.5V 250/5=50*3=150A 562.5*150*1.732*.9=131 523.75/1000= 131.52kw
is it 600:120? Your original post said 600:20. but yes your above calculation is correct.
Lucas Organ
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- NL CANADA
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- Industrial Electrician
Very sorry about that, yes I did mean to say 120is it 600:120? Your original post said 600:20. but yes your above calculation is correct.
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