# New NCEES #524



## indy-engineer (Mar 3, 2009)

Can anyone point me in the right direction as to why the increase from 50% to 100% of nameplate KVA has a 4X multiplier? I tried to do it linearly starting with 460W at point 0. Any guidance or formula references would be appreciated. Thanks!


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## Flyer_PE (Mar 3, 2009)

The 460W is core loss and will remain relatively constant for any given power level.

The reason you get 4x the power loss when doubling the current is that the loss is primarily I2R. When you double the current, the power loss due to I2R quadruples.


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## indy-engineer (Mar 3, 2009)

You're brilliant!! Thanks so much!!!


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## cabby (Mar 4, 2009)

Since R remains constant, is it set equal to 1?



Flyer_PE said:


> The 460W is core loss and will remain relatively constant for any given power level.
> The reason you get 4x the power loss when doubling the current is that the loss is primarily I2R. When you double the current, the power loss due to I2R quadruples.


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## Flyer_PE (Mar 4, 2009)

^ No. The actual value of "R" in the problem is irrelevant. In the problem, they give you power loss for a given percentage of transformer loading. They then double the loading and ask what the power loss would be. Since the power loss through the transformer increases with the square of the current, a 2x increase in current results in a 4x increase in power loss through the transformer. If they had tripled the current, the loss would have gone up by a factor of 9.


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## JillSuzanne (Apr 2, 2013)

Where are they doubling the current? I see them doubling kVA? 100% kVA = 2 x 50% kVA?


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## DK PE (Apr 2, 2013)

In short yes, since voltage is assumed constant, the only way one could go from 25% kVA to 50% of some kVA is to double current. Same as if you go from 50 % kVA to 100% kVA current needs to double.

If you have a transformer rated 25kVA and is is operating at 50% load and at a later time it is operating at 100% rated load, the current must have doubled.


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