Single phase non-ideal transformer. Turn Ratio= 3000/600.R₁ = 0.9QR2= 36 mQRo= 3000 QX₁₁ = 0.60X12= 0.0024 02X= 6000Input voltage= 1440 V (also known as rms voltage)Load impedance of secondary winding is Z₂= 0.12 + j0.3a) Copper losses on primary and secondary windingsb) Core lossc) Transformer efficiencyV₁Ī₁RowwmXuR₁R'2wwmwwI₂X12m

Answered: Single phase non-ideal transformer.…

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.28MCQ

See similar textbooks

Similar questions

An ideal transformer has no real or reactive power loss. (a) True (b) False
With the American Standard notation, in either a Y-or-Y transformer, positive- sequen quantities on the high-voltage side shall lead their corresponding quantities on the low-voltage side by 30. (a) True (b) False
A single-phase step-down transformer is rated 13MVA,66kV/11.5kV. With the 11.5 kV winding short-circuited, rated current flows when the voltage applied to the primary is 5.5 kV. The power input is read as 100 kW. Determine Req1andXeq1 in ohms referred to the high-voltage winding.
For an ideal 2-winding transformer, an impedance Z2 connected across winding 2 (secondary) is referred to winding 1 (primary) by multiplying Z2 by (a) The turns ratio (N1/N2) (b) The square of the turns ratio (N1/N2)2 (c) The cubed turns ratio (N1/N2)3
The direct electrical connection of the windings allows transient over voltages to pass through the auto transfonner more easily, and that is an important disadvantage of the autotransformer. (a) True (b) False
A 250/125 V, five kVA single-phase transformer has a primary resistance of 0.2 ohms and reactance of 0.75 ohms. The secondary resistance is 0.05 ohm and reactance of 0.2 ohms.a) Determine its regulation while supplying full load on 0.8 leading p.f.b) The secondary terminal voltage on full load and 0.8 leading p.f.Ans.: -0.4.88%; 131.1
Q7/ Complete the blank The copper loss of a transformer at 4/5 of the full load is 4000 W, full load copper loss will be..........................................?
A single-phase transformer has a voltage ratio on open circuit of 7900/660 V. The primary and secondary resistances are 0.55 ohm and 0.035 ohm respectively. The corresponding leakage reactances being 4.22 ohm and 0.18 ohm. The load is equivalent to a coil of resistance 3.85 ohm and inductive reactance of 4.2 ohm. Determine the terminal voltage of the transformer and the output in kW when supplying a given load using approximate equivalent circuit when the applied voltage is 7912 V. Your answer
10
With base MVA of 150 and base voltage of 33. If the reactance of the transformer is 0.23 pu. The parameter of the reactance of the transformer in the Neplan software is O 1.66 2.30 23 0.23 O Non of them
In a transformer that has 10 times more turns on the secondary side (side 2) than on the primary side (side 1), the magnitude of the voltage/EMF across the windings on the secondary side, E₂ will be... O a.... 10 times lower than the voltage/emf on the primary side, E₁. O b.... the same as the voltage/emf on the primary side. O c.... always zero regardless of the voltage on the primary side, E₁. O d.... 10 times higher than the voltage/emf on the primary side, E₁.
Seconday winding of transformer is made-up of material. O insulated CRGO conductor O insulated copper conductor O Cold Rolled Grain Oriented (CRGO) conductor Cold Rolled Grain Oriented (CRGO) Silicon Steel

SEE MORE QUESTIONS

Recommended textbooks for you

Power System Analysis and Design (MindTap Course …

Electrical Engineering

ISBN:

9781305632134

Author:

J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:

Cengage Learning