Does 208V 100A 3ph input approx. equal 480V 30A 3ph output for a step-up transformer?

[EEEC]

True, false, or somewhat true and/or false?

 

[LarryFine]

True, false, or somewhat true and/or false?

Multiply each side's voltage x current and compare the results.

Hint: P = E x I
(W = V x A)


We don't like to simply answer questions like this; we want you to figure it out.

We don't like to answer anything simply here.

 

[Open Neutral]

We don't like to answer anything simply here.

For every complex problem there is an answer that is clear, simple, and wrong.

H. L. Mencken

 

[EEEC]

The sub panel is a 100A 120/208V 4 wire. I am trying to determine how many available amps at 480V I'd have. I did some research and it said that I'd have one-third of the current available on the secondary side. But didn't know how they came up with that figure, so...

P = 208V (E) x 100A (i)
P = 20,800

20,800 = 480V (E) x X (I)
I = 20,800 (P) / 480V (E)
I = 43.333

So I'd have 43.3A?

 

[LarryFine]

So I'd have 43.3A?

Exactamundo! :happyyes:

Of course, we/re talking 100%-efficiency hypothetical amps.

But, yes. "Well done, Grasshopper!"

 

[oldsparky52]

The sub panel is a 100A 120/208V 4 wire. I am trying to determine how many available amps at 480V I'd have. I did some research and it said that I'd have one-third of the current available on the secondary side. But didn't know how they came up with that figure, so...

P = 208V (E) x 100A (i)
P = 20,800

20,800 = 480V (E) x X (I)
I = 20,800 (P) / 480V (E)
I = 43.333

So I'd have 43.3A?

Just an FYI, when dealing with calculating power on a 3-phase system, the voltage is (208*1.732=360) and (480*1.732=831). Since this is a basic equation where the 1.732 is influencing both side of the equation, you come up with the same answer as if you didn't use it.

You could also take 480/208=2.30769; 100(amps at 208) / 2.30769 = 43.33 (amps at 480). If voltage goes up by a % then the the current goes down by the same % if the power is the same.

Like Larry alluded too, you have not allowed for transformer losses (don't know how important this is).

 

[EEEC]

Thanks a lot fellas. Now I know how to calculate it. Thanks again!

 

[LarryFine]

Now I know how to calculate it.

I rest my case.



Added: ​Of course, in practice, we would start with the requirements of the load to be served, and work back toward the source.

 

[Besoeker3]

Multiply each side's voltage x current and compare the results.

Hint: P = E x I
(W = V x A)


We don't like to simply answer questions like this; we want you to figure it out.

We don't like to answer anything simply here.

VA, not W
And all you need to do is multiply the current by the voltage ratio.
That gives you the 43.3A

 

[LarryFine]

VA, not W

One lesson at a time. I used what I believed OP to be familiar with.

 

[Besoeker3]

One lesson at a time. I used what I believed OP to be familiar with.

Actually, you don't need either.