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2. The secondary winding of a transformer has a terminal voltage of (t)=169.7 sin3141 V. The turn ratio of the transformer is

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Answer:- - 51 w 40 where, Real = 0.152 Rc =50r Xeq = 0.252 Xm = 252 v₂ = 169.7 sin 314 .: . 169.710 I s Y =120169 - 0 Now, UmIz- 14.175 in (3147- 53.13) There 0 = 53.13(kga) Caso = 0.6 =14.14 (53.13° I sind=0.8 s F2=10153.13-0. I2 =101053,130 CSS Scacompare 0 4 0 Is logs the voltage by 5 3.13° wkt 베의 ER I = KI2 = 5 Iz = I1 = 50 A then vez nous pure asuv ise, har Now Iwsian# *100 wkt %2 = (uz I2C050). X(V2 I 20050) + Pi +(x)? Peu x = fraction of load (1,3 11/21 lu etc). Now. %. @ Full load - (120

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