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Question

A current transformer has a single turn primary and 400 turns on secondary winding. The resistance annd reactance of the secondary circuit are 2Ω and 3Ω respectively including transformer winding. When 6 A current is flowing in the secondary winding, the magnetizing mmf is 100AT and iron loss is 2W. The value of ratio error will be

A
3.42%
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B
4.14%
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C
2.24%
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D
1.62%
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Solution

The correct option is B 4.14%
Primary winding turns,
NP=1

Secondary winding turns,
Ns=400

Turn ratio,
n=NSNP=400

Seconday circuit burden impedance
=(2)2+(3)2=3.606Ω

For secondary winding circuit.
cosδ=23.606=0.554
and, sinδ=33.606=0.832

Secondary induced voltage
ES=6×3.606
=21.64V

Primary inuced voltage
EP=ESn=21.64400=0.054V

Loss component of current referred to primary winding
Ic=Iron lossE=20.054=37.04A

Magnetizing current,
Im=Magnetizing mmfPrimary winding turns=1001=100A

Acutal ratio,
R=n+IC cosδ+Im sin δIS

=400+37.04cosδ+100sinδIs

=400+103.726=400+17.28

=417.28

Percentage ratio error

=KnRR×100; Where Ka= nominal ratio

=400417.28417.28×100=4.14%

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