Determine the equivalent impedances circuit referred to the primary side of a 20- kVA, 8000/240 V, 60 Hz transformer. The open circuit test was performed on the secondary side and the short circuit test was performed on the primary side. (on secondary) Open-circuit test Short-circuit test (on primary) Voc = 240 V VSC = 489 V loc = 7.133 A Isc = 2.5 A Voc = 400 W Psc = 240 W

Determine the equivalent impedances circuit referred to the primary side of a 20- kVA, 8000/240 V, 60 Hz transformer. The open circuit test was performed on the secondary side and the short circuit test was performed on the primary side. (on secondary) Open-circuit test Short-circuit test (on primary) Voc = 240 V VSC = 489 V loc = 7.133 A Isc = 2.5 A Voc = 400 W Psc = 240 W

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 3.10P: A single-phase step-down transformer is rated 13MVA,66kV/11.5kV. With the 11.5 kV winding...

See similar textbooks

Similar questions

The open circuit and short circuit tests were performed on the primary side of the 25 kVA, 7620/240, 60 Hz transformer and the following data were obtained, open circuit test; Eoc = 7820 V, loc = 0.30 A and Poc = 500 W. Short circuit test Esc = 450 V, Isc = 3.5 A and Psc = 220 W. Determine the shunt and series impedances refer to the primary of the transformer." A.116K + j26K and 17.97 + j130 B.124K + jl4K and 2l.85 + j134 C. 112K + j22K and 15.95 + j128 D. 120K + jl0K and 19.95 + j132
Q3/A/ The equivalent circuit impedances of a 20-kVA, 8000/240 V, 60-Hz transformer are to be determined. The open-circuit test was performed on the secondary side of the transformer and the short circuit test was performed on the primary side of the transformer .The following data was taken: Open-circuit Short-circuit test test (on secondary) Voc 240 v (on primary) Vsc 489 v Isc 2.5 A Psc 240 W loc 7.133 A Poc 400 W
Subjcct Electrical machine
The step down DC chopper uses DC source voltage V(dc) = 200 V and resistive load impedance of 10 ohm. The duty cycle is 0.4 the voltage drop across the switch may be taken as zero, Calculate : (a) Average and RMS value of output voltage. (b) Average and RMS value of output current.
dc tansformd
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.
An ac resistance is higher than a dc resistance. True False
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 single-phase, 50-kVA,2400/240-V,60-Hz distribution transformer has the following parameters: Resistance of the 2400-V winding: R1=0.75 Resistance of the 240-V winding: R2=0.0075 Leakage reactance of the 2400-V winding: X1=1.0 Leakage reactance of the 240-V winding: X2=0.01 Exciting admittance on the 240-V side =0.003j0.02S (a) Draw the equivalent circuit referred to the high-voltage side of the transformer. (b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.
Single phase non-ideal transformer. Turn Ratio= 3000/600. R₁ = 0.9Q R2= 36 mQ Ro= 3000 Q X₁₁ = 0.60 X12= 0.0024 02 X= 6000 Input voltage= 1440 V (also known as rms voltage) Load impedance of secondary winding is Z₂= 0.12 + j0.3 a) Copper losses on primary and secondary windings b) Core loss c) Transformer efficiency V₁ Ī₁ Ro ww m Xu R₁ R'2 ww m ww I₂ X12 m
do it by yourself
Ql-phase step down transformer 1250 KVA, 1 1000/2200 V, 50 Hz .The resistance /phase and leakage reactance phase of the primary winding are 0.78 N and 4.82 and the corresponding values for the secondary winding are 0.012 and 0.000 respectively if the primary winding is connected in delta and secondary in star calculate: )The voltage to be applied at normal frequency to the primary to circulate full-load through it when the secondary is short-circuited Ii) The full load copper loss