Recent content by ErnieChicken

1) The no-load primary current is 5% less than the full-load primary current. (You need to establish full load impedance by way of calculations, Which was what i was given today!) Primary Full Load Current = 3.33A Impedance Primary = 2700V / 3.33A = 810.8 Ohms Impedance Secondary = 450v / 20A =...

From what he was saying. "No Load current is less than 5% of the full current load". Whether or not the interpretation here is that it is the 0.028A calculated earlier or it is actually more than than because if so 5% of I1 at full load --> 3.33a = 3.165a Have i actually calculated an impedance...

Hi Nascant, I can see the confusion. Let me take it back to the beginning and explain the question much better. A 6/1 step down transformer has a full load secondary current of 20A and is rated at 9kVA. The copper loss at full load is 110W. The primary winding has a resistance of 0.35Ω. a)...

From Reusing the KvA formula using P = 1000 x 3kVA x 0.75 = 6750W. And approximately 30kW is where it needs to be at full load. (75 x 400 = 30kW)

So this would be what i am looking --> Rc (Iron Loss) + Resistance Primary( Np/Ns)^2 which is on the secondary side. I am truly sorry for being a pain. Thanks

Sorry for the confusion, This is what i am trying to work out. Part B)

Thanks, Now i know that those figures have been correctly worked out. I have my 4W which is what the loss is at the primary side under full load. At non load, I can work out the resistance using the same method as before to give me a resistance then by using the current i can give myself a...

This has been specified as a guideline to work to i believe. It just states that it needs to be "no-load primary current is ≈ 5% less than the full-load primary current." But if this is between 0% - 5% i guess this is ok since the high voltage and the low losses. Yes, i understand this. my...

Hi Guys, I am currently looking for some book resources on Both Electrical and Mechanical engineering (Maths Related) . I know we have a full raft of information available on the internet but its nothing like having a good book infront of you. Any recommendations? I do work in Industry but...

Thanks Nascant, That has finally clicked in and i was wondering where i was going wrong. When it stated "The Non Load Current must be 5% of the Full Load Current" i was thinking of it being 5% less of the 3.33A but clearly i had misread it. Thanks for that clarification. So i have the 0.0277A...

Thanks man I appreciate it. It must seem like I am a pain as I am struggling to get my head around it. So for the iron loss (6x450^2) /75w = 92.2kOhms --> 2700 / 97.2k = 0.02777A Subtract this from my current I calculated earlier 3.33A - 0.2777A = 3.3A which is within the 5% My question would...

I am sorry to have wasted your time. I am going through a rough time right now and my brain has turned to mush which isn't an excuse. I would like to thank you guys for your help

Quoting From Hersch, "Iron losses = 75W - 4W = 71W. Riron ≈ (6*V2)2 / 71W = 103kΩ" This is what Hersch Used, would that be the correct application ?

Ok i have the copper resistance. 4W for Primary, 106w for Secondary with the currents i had posted. If i am correct, this is for the circuit on load since there is a completed circuit of both sides but i need to find the unloaded values for both current and copper resistance.

Since i have only got a Resistance for both Windings i am not sure what i need to calculate.