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Archived Transformer physics help

  • Thread starter graygoh
  • Start date
1. Homework Statement
A 10-kVA, 200:400 V transformer gave these test results:
open circuit (LV winding supplied): 200V, 3.2A, 450W
short circuit (HV winding supplied) : 38 V, 25A, 600W
calculate the efficiency when the transformer delivers its rated kVA at 0.85 power factor lagging

2. Homework Equations

efficiency = output power / input power

3. The Attempt at a Solution
I totally don't know how to start. help pls. thanks


Homework Helper
Gold Member
open circuit (LV winding supplied): 200V, 3.2A, 450W
This gives the core loss of the transformer.
short circuit (HV winding supplied) : 38 V, 25A, 600W
This gives the copper loss of the transformer at full load.
Iron loss is fixed loss and doesn't depend on the load. Copper loss varies proportional to the square of the current. In general, copper loss=x2Pcu, where x is the fraction of full load with which the transformer is loaded. Since the transformer is operated at full load, x=1 and the losses will be,
Efficiency η= Output power/input power
=Output kW/(Output kW+losses)
∴ η=10×0.85/(10×0.85+1.050)
∴ η=0.89
The transformer is 89% efficient.
Last edited:

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