Some power devices have bidirectional current capability Select one: O a. True O b. False The voltage regulation of the transformer will depend on the load Select one: O a. For mainly inductive or resistive load, the voltage regulation is negative O b. For mainly capacitive, inductive or resistive load, the regulation is positive O. For mainly capacitive load, voltage regulation is positive and other loads it is negative. O d. For mainly capacitive load, voltage regulation is negative.

Some power devices have bidirectional current capability Select one: O a. True O b. False The voltage regulation of the transformer will depend on the load Select one: O a. For mainly inductive or resistive load, the voltage regulation is negative O b. For mainly capacitive, inductive or resistive load, the regulation is positive O. For mainly capacitive load, voltage regulation is positive and other loads it is negative. O d. For mainly capacitive load, voltage regulation is negative.

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 3.9P

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Similar questions

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 symbol shown is a(n) a. iron core transformer. b. auto transformer. c. current transformer. d. air core transformer.
An ideal transformer has no real or reactive power loss. (a) True (b) False
For an ideal transformer, the efficiency is (a) 0 (b) 100 (c) 50
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.
Rework Problem 3.14 if the transformer is delivering rated load at rated secondary voltage and at (a) unity power factor, (b) 0.8 power factor leading. Compare the results with those of Problem 3.14. -
Consider the three single-phase two-winding transformers shown in Figure 3.37. The high-voltage windings are connected in Y. (a) For the low-voltage side, connect the windings in , place the polarity marks, and label the terminals a, b, and c in accordance with the American standard. (b) Relabel the terminals a, b, and c such that VAN is 90 out of phase with Va for positive sequence.
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
The ideal transformer windings are eliminated from the per-unit equivalent circuit of a transformer. (a) True (b) False
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.
For the transformer in Problem 3.10. The open-circuit test with 11.5 kV applied results in a power input of 65 kW and a current of 30 A. Compute the values for GcandBm in siemens referred to the high-voltage winding. Compute the efficiency of the transformer for a load of 10 MW at 0.8 p.f. lagging at rated voltage.
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