A 5KVA, 240/120 V single phase transformer has the following parameters:R1 = 0.06 QR2 = 0.015 QX1 = 0.18 QX2 = 0.045 QRc =2000 0Xm = 400 0Transformer secondary winding feeds an inductive load of 5KVA power factor0.9 at 120V rated voltage. Using the simplest equivalent circuit given in thefigure, find the primary voltage, primary current and apparent power factorof the transformer. What is the efficiency of the transformer?X,+X2V1=.Vul1 =AV,V2cose =n =%uuwmm

5kVA, 240/120V Single phase transformer has given parameters. Find the primary current, voltage…

A 5KVA, 240/120 V single phase transformer has the following parameters R1 = 0.06 Q R2 = 0.015 Q X1 = 0.18 Q X2 = 0.045 Q Rc =2000 0 Xm = 400 0 Transformer secondary winding feeds an inductive load of 5KVA power fact 0.9 at 120V rated voltage. Using the simplest equivalent circuit given in the figure, find the primary voltage, primary current and apparent power facto of the transformer. What is the efficiency of the transformer? X,+X2 V1=. V u l1 = A V, V2 cose = n = % a

A 5KVA, 240/120 V single phase transformer has the following parameters R1 = 0.06 Q R2 = 0.015 Q X1 = 0.18 Q X2 = 0.045 Q Rc =2000 0 Xm = 400 0 Transformer secondary winding feeds an inductive load of 5KVA power fact 0.9 at 120V rated voltage. Using the simplest equivalent circuit given in the figure, find the primary voltage, primary current and apparent power facto of the transformer. What is the efficiency of the transformer? X,+X2 V1=. V u l1 = A V, V2 cose = n = % a

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 3.16P: A single-phase, 50-kVA,2400/240-V,60-Hz distribution transformer has the following parameters:...

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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.
Consider an ideal transformer with N1=3000andN2=1000 turns. Let winding 1 be connected to a source whose voltage is e1(t)=100(1| t |)volts for 1t1ande1(t)=0 for | t |1 second. A2- farad capacitor is connected across winding 2. Sketch e1(t),e2(t),i1(t),andi2(t) versus time t.
The ratings of a three-phase three-winding transformer are Primary(1): Y connected 66kV,15MVA Secondary (2): Y connected, 13.2kV,10MVA Tertiary (3): A connected, 2.3kV,5MVA Neglecting winding resistances and exciting current, the per-unit leakage reactances are X12=0.08 on a 15-MVA,66-kV base X13=0.10 on a 15-MVA,66-kV base X23=0.09 on a 10-MVA,13.2-kV base (a) Determine the per-unit reactances X1,X2,X3 of the equivalent circuit on a 15-MVA,66-kV base at the primary terminals. (b) Purely resistive loads of 7.5 MW at 13.2 kV and 5 MW at 2.3kV are connected to the secondary and tertiary sides of the transformer, respectively. Draw the per- unit impedance diagram, showing the per-unit impedances on a 15-MVA,66-kV base at the primary terminals.
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