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Question

The maximum efficiency of a 100 kVA, 1 - phase transformer is 98% and occurs at 80% of full load at 0.8 pf lagging. If leakage reactance of transformer is 5%. The voltage regulation at full load will be (Take power factor = 0.8)

A
2.125%
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B
4.275%
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C
5.673%
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D
3.753%
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Solution

The correct option is D 3.753%
Given maximum efficiency of transformer is 98% at 80% of full load at 0.8 p.f lagging.

At maximum efficiency,

x2Pcu=Pi

x=0.8

So, (0.8)2Pcu=Pi

0.98=0.8×0.80.8×0.8+Pi+(0.8)2Pcu

2Pi=0.64(10.981)

2Pi=0.01306 p.u

Pi=(0.8)2Pcu

=0.00653p.u

(0.8)2Pcu=0.00653

Pcu=0.0102p.u.

Pcu=I22R2=0.0102p.u.

%ofR2=1.02%

and leakage reactance,

X2=5%or0.05p.u.

Voltage regulation=I2R2cosϕ+I2X2sinϕV2

=0.0102×0.8+0.05×0.61=0.03816(or)3.816%

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