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Two identical 1 − ϕ transformer of 15kVA, gave the following results, when tested by back to back method.
W1 = 1000 watt (Primary)
W2 = 1500watt( Secondary )
Calculate the percentage efficiency of the transformer at 1/4th load 0.8 P.F. lag
  • a)
    84.55%
  • b)
    84.56%
  • c)
    84.57%
  • d)
    84.58%
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Two identical 1 − ϕ transformer of 15kVA, gave the following results,...
In the back to back test the primary wattmeter gives the iron loss.
Each transformer has iron loss
=W1 / 2 = 1000 / 2 = 500W and secondary wattmeter reading gives the full load copper loss.
Each transformer has full load copper loss
=W2 / 2 = 1500 / 2 = 750W
Full load loss at 1/4th of load is
=
Efficiency η=
η=0.8458×100%
η=84.58%
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Two identical 1 − ϕ transformer of 15kVA, gave the following results, when tested by back to back method.W1 = 1000 watt (Primary)W2 = 1500watt( Secondary )Calculate the percentage efficiency of the transformer at 1/4th load 0.8 P.F. laga)84.55%b)84.56%c)84.57%d)84.58%Correct answer is option 'D'. Can you explain this answer?
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