A 2300/208-V 500-kVA 60-Hz single-phase transformer was tested by means of the open-circuit test (on the low-voltage winding) and short-circuit test (on the high-voltage winding). The test data obtained are: Open-circuit test: Voc = Short-circuit test: Vsc = 95 V, 208 V, Ioc = 52.5 A, Isc = 217.4 A, Poc = 3800 W Psc = 6200 W %3D %3D OC %3D %3D SC Calculate a. the efficiency of the transformer at rated kVA and half-rated kVA, when the power factor is 0.866 lagging; and

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter3: Power Transformers
Section: Chapter Questions
Problem 3.11MCQ: For a short-circuit test on a 2-winding transformer, with one winding shorted, can you apply the...
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A 2300/208-V 500-kVA 60-Hz single-phase transformer was tested by means of
the open-circuit test (on the low-voltage winding) and short-circuit test (on the
high-voltage winding). The test data obtained are:
Open-circuit test: Voc = 208 V, Ioc = 52.5 A,
Short-circuit test: Vsc = 95 V,
Poc = 3800 W
Psc = 6200 W
ос
%3D
ос
Isc
= 217.4 A,
%3D
%3D
Calculate
a. the efficiency of the transformer at rated kVA and half-rated kVA, when the
power factor is 0.866 lagging; and
Transcribed Image Text:A 2300/208-V 500-kVA 60-Hz single-phase transformer was tested by means of the open-circuit test (on the low-voltage winding) and short-circuit test (on the high-voltage winding). The test data obtained are: Open-circuit test: Voc = 208 V, Ioc = 52.5 A, Short-circuit test: Vsc = 95 V, Poc = 3800 W Psc = 6200 W ос %3D ос Isc = 217.4 A, %3D %3D Calculate a. the efficiency of the transformer at rated kVA and half-rated kVA, when the power factor is 0.866 lagging; and
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