Transformer engineering problem

[neraketam]

Homework problem:
A transformer on a pole near a factory steps the voltage down from 3900V to 220V. The transformer is to deliver 1180kW to the factory at 91% efficiency. Find the power delivered to the primary in kW.

What I've done:
Well, I thought that power in equals power out. So I worked all the math to prove that and used P=IsVs, solved for Is then used that to find Ip (Ip=VsIs/Vp). Then found P=IpVp. I got 1180kW! I then calculated in the efficiency--not correct. And then in the follow up question, I found out that my Ip isn't even correct. What's wrong here?

 

[Doc Al]

Originally posted by neraketam
Well, I thought that power in equals power out. So I worked all the math to prove that and used P=IsVs, solved for Is then used that to find Ip (Ip=VsIs/Vp). Then found P=IpVp. I got 1180kW!

Sounds like you are going in a circle.

True, power in = power out for an ideal transformer. A nonideal one will waste some of that power. If I read the question correctly, the first part just asks for what input power is needed to get the output power given, given the efficiency:

P_in X 0.91 = P_out

 

[neraketam]

transformers

I did that but my silly internet homework submission page is saying it's not correct. That is not the first time it's said that so...at least I know I'm doing the problem right and thanks for getting back to me so quickly!

 

[Averagesupernova]

You don't need to know anything about voltage and current to do this. All you need to know is the power out and the efficiency. It is supposed to be 1180 Kwatts out. So, figure the efficiency from that. Can't help you with your internet problem though. Sorry.

P.S. Why did I just post this when Doc Al did it so nicely with just a formula?