Promotion Test

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Re: Promotion Test

If you answered NO they showed you the door, because your probably lying. I believe most people have lied at least once in their life.

Do I look good in this dress?
Do you think I'm getting fat?

Have you answered this question truthfully?
 
Re: Promotion Test

SPSnyder;
Sorry I didnt answer sooner, had a family reunion to visit.
I will confirm what Charlie b has said, a leading power factor is Capacitive and conversely a lagging pf would be Inductive.
I thought it interesting all the different approaches to one problem.

WOC
 
Re: Promotion Test

I would like to thank everyone who responded.
When I said there was only one question I should have been more clear, only one math problem.
But thanks again
Tom
 
Re: Promotion Test

This whole problem can be done graphically. To go one step further than Ed carried it, this can be laid out on a piece of graph and a protractor and a compass.
The Given: 25kva Transformer Real power is 1200 watts and the pf of .6 lagging.
By marking off 25 kw on the horizontal (X) axis which is the Real power, set the compass to this length, and strike an arc from the origin and 25 kw point to the vertical axis (Y) this will give the limits of the transformer. The power triangle will now fit within this arc. Whenever the hypoteneuse( of the power triangle) is equal to the radius of the arc you will be at full load or 25kva/kw of the transformer(depending on the powerfactor) as long as you know the power factor or an assumed power factor you can plot any parameter of that transformer.
And as long as the S side (kva) is not equal the the distance from the origin to the 25(or whatever size is used) kva arc you will be able to add to the current load on the transformer.
The answer in question will be as accurate as how accurate you lay out the graph, in other words accurate to 1/2 the least count.

WOC
 
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