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A 450KVA single-phase transformer has a turns ratio of 15/60. The primary winding is connected to 33000V, 50HZ. supply. Calculate: a. the secondary voltage on no load b. the approximate values of the primary and secondary currents on full load. » Single Phase TransFarmer -> S=ASOkva -> TUIA Ratio n= -> %3D 60 4[60 -> Primary winding VP = 33000v, > Fregvancy f = 50 Hz a) we Know Vs ニ So Vs = 4Vp = 4x33000 = 132000V ã 132KV gining secondary vatage (Vs) = 132 kp b) Primary siole エpニア CUsrent 4 SO10-3 ニ VP = 13.6363A 33, secoundary Side cullent. 4S016-3 132 10-3 エs Is = V = 3.409A

A 450KVA single-phase transformer has a turns ratio of 15/60. The primary winding is connected to 33000V, 50HZ. supply. Calculate: a. the secondary voltage on no load b. the approximate values of the primary and secondary currents on full load. » Single Phase TransFarmer -> S=ASOkva -> TUIA Ratio n= -> %3D 60 4[60 -> Primary winding VP = 33000v, > Fregvancy f = 50 Hz a) we Know Vs ニ So Vs = 4Vp = 4x33000 = 132000V ã 132KV gining secondary vatage (Vs) = 132 kp b) Primary siole エpニア CUsrent 4 SO10-3 ニ VP = 13.6363A 33, secoundary Side cullent. 4S016-3 132 10-3 エs Is = V = 3.409A

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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A 450KVA single-phase transformer has a turns ratio of 15/60. The primary winding is connected to 33000V, 50HZ. supply. Calculate: a. the secondary voltage on no load b. the approximate values of the primary and secondary currents on full load. -» Single Phase Transfaramer -> S= ASOkva -> Tun Ratio n= N2 -> 60 -> Primary winding VP = 33000V, -> Freguancy { = 50 Hz %3D Vp a) we Know Vs So Vs = 4 Vp = 4x33000 = 132000V % 132 Kv %3D giving secondary vortage (Vs) = 132 kv b) Primary sSile Ipニ。 CUsrent 4SO103 = 13.6363A VP 33,03 secondeery Side cullet. 45016-3 132 10-3 Is = = 3.409A
Transcribed Image Text:A 450KVA single-phase transformer has a turns ratio of 15/60. The primary winding is connected to 33000V, 50HZ. supply. Calculate: a. the secondary voltage on no load b. the approximate values of the primary and secondary currents on full load. -» Single Phase Transfaramer -> S= ASOkva -> Tun Ratio n= N2 -> 60 -> Primary winding VP = 33000V, -> Freguancy { = 50 Hz %3D Vp a) we Know Vs So Vs = 4 Vp = 4x33000 = 132000V % 132 Kv %3D giving secondary vortage (Vs) = 132 kv b) Primary sSile Ipニ。 CUsrent 4SO103 = 13.6363A VP 33,03 secondeery Side cullet. 45016-3 132 10-3 Is = = 3.409A
A 15KVA single-phase transformer, for 10000V/440V at no load, has resistances and leakage reactance as follows. Primary winding: resistance 42, reactance 162. Secondary winding: resistance 0.152, reactance 0.45Q. Determine the approximate value of the secondary voltage at full load, 0.85 power factor (lagging), when the primary supply voltage is 10000V. Ignore core resistance and magnetization reactance. -> Singie Phase 1runs Folme -> S='15kva → 10'000 /44OV Is R, Jx, 0.15 JO.45 Vs 2 (440 R, = 100001 x 4 = 7.744, °440 =(42)x 16= 0030976 r l000. 440 V = X 10000 = 440V 10000 If Power Factes Of Load = 085(1aging) Ø = Cos Co85) = 31.7883° ISo00 at Full Load curent |Isd =SI =34.09A 440 So Is = 34.09<-31.788°A giving, Vs = Vp'- (R,+o.15+j(x,+o.45) Iş * 440-(>,74,340.15+jco.03097640.45) 34.094 - 36.56° to.- giving Secondaty Side vartage =425.974 <-1.3}'
Transcribed Image Text:A 15KVA single-phase transformer, for 10000V/440V at no load, has resistances and leakage reactance as follows. Primary winding: resistance 42, reactance 162. Secondary winding: resistance 0.152, reactance 0.45Q. Determine the approximate value of the secondary voltage at full load, 0.85 power factor (lagging), when the primary supply voltage is 10000V. Ignore core resistance and magnetization reactance. -> Singie Phase 1runs Folme -> S='15kva → 10'000 /44OV Is R, Jx, 0.15 JO.45 Vs 2 (440 R, = 100001 x 4 = 7.744, °440 =(42)x 16= 0030976 r l000. 440 V = X 10000 = 440V 10000 If Power Factes Of Load = 085(1aging) Ø = Cos Co85) = 31.7883° ISo00 at Full Load curent |Isd =SI =34.09A 440 So Is = 34.09<-31.788°A giving, Vs = Vp'- (R,+o.15+j(x,+o.45) Iş * 440-(>,74,340.15+jco.03097640.45) 34.094 - 36.56° to.- giving Secondaty Side vartage =425.974 <-1.3}'
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