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Question

# The open circuit test and the short circuit test were performed on the primary side of a 10 kVA, 2000/240 V, 50 Hz transformer and the following data were obtained OC test:VOC=240 V; IOC=0.6 A, POC=80 W SC test:VSC=112 V; ISC=5 A, PSC=160 W Find the approximate equivalent circuit referred to the primary side,

A
Req=6.4 Ω;Xeq=j21.46 Ω;RC=49.96 kΩ;Xm=j33.38 kΩ
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B
Req=0.092Ω;Xeq=j0.31Ω;RC=720kΩ;Xm=j480kΩ
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C
Req=0.092Ω:Xeq=j0.31Ω;RC=49.96kΩ;Xm=j33.34kΩ
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D
Req=7.2 Ω:Xeq=j21.46 Ω;RC=720 Ω;Xm=j480 Ω
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Solution

## The correct option is A Req=6.4 Ω;Xeq=j21.46 Ω;RC=49.96 kΩ;Xm=j33.38 kΩFrom the given data it can be seen that, OC test is performed on the LV side. Thus the equivalent core-loss resistance as referred to the low-voltage side is RC=240280=720Ω The apparent power under no load ; SOC=VOCIOC=240×0.6 =144 VA Thus, the reactive power is QOC=√1442−802 =119.73VAR The magnetizing reactance as referred to the low voltage side is Xm=2402119.73=481.082Ω The core-loss resistance and the magnetizing reactance as referred to HV side are obtained as follows: N1N2=2000240=8.33 RCHV=(8.33)2×720=49.96 kΩ Xm HV=(8.33)2×481.082=33.378 kΩ SC best is performed on HV side, thus Req HV=PscI2sc=16052=6.4 Ω Zeq HV=VscIsc=1125=22.4 Ω Xeq HV=√22.42−6.42=21.46 Ω

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