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[29] SECTION B: TRANSFORMERS QUESTION 2 2.1 A 240:48-volt, 50-F power factor of 0.8 lagging. If the no load current and iron
2.2.2 2.2.3 2.2.4 225 EMM2602/101/3 The voltage that must be applied to the H.V. side when the L.V. side is short circuited,
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941 sol) 240148 it-s0H2) 1 – 10 (A) , pf = 0.8 logging &- Cont(0.8)= -36.85 No wad current INL 0.12 (A) 12(W) =P Iron loss ruIp- 24-3618607 Gold 2-73° ape 2.14 <-39,55ºCA) 2.2) 301) 3=250 (KVA), 50, (H2), 1200/230 (0) Z1 = (0.0u tjonin ng 22. 6-001422.2.3) 12.00 Vbase Tharpe 25. 76CM base 208.33 Zer (p.4) 0.243 = 0.0422 P. 4) 5.16 Percentage impedance - 4022 s. (en) 2.2.4)1.6310 x0.85 + 3.89 t. sin (31.08) 1.R - 3. 434 R.M) = 0.03 43 R.). R voltage drop from source to load R X 230 Cul = 0.0343x2

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[29] SECTION B: TRANSFORMERS QUESTION 2 2.1 A 240:48-volt, 50-F power factor of 0.8 lagging. If...
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