The following data are obtained when open-circuit and short-circuit tests are performed on a single-phase, 50 kVA, 2400/240 volt, 60 Hz distribution transformer. VOLTAGE CURRENT POWER (volts) (amperes) (watts) Measurements on low-voltage side with high-voltage winding open Measurements on high-voltage side with low-voltage winding shorted 240 4.85 173 52.0 20.8 650 a) Neglecting the series impedance, determine the exciting admittance referred to the high voltage side. b) Neglecting the exciting admittance, determine the equivalent series impedance referred to the high voltage side.

The following data are obtained when open-circuit and short-circuit tests are performed on a single-phase, 50 kVA, 2400/240 volt, 60 Hz distribution transformer. VOLTAGE CURRENT POWER (volts) (amperes) (watts) Measurements on low-voltage side with high-voltage winding open Measurements on high-voltage side with low-voltage winding shorted 240 4.85 173 52.0 20.8 650 a) Neglecting the series impedance, determine the exciting admittance referred to the high voltage side. b) Neglecting the exciting admittance, determine the equivalent series impedance referred to the high voltage side.

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 3.35P: Consider a bank of this single-phase two-winding transformers whose high-voltage terminals are...

See similar textbooks

Related questions

Q: An open circuit is occurred when: Select one: a. All of the answers. b. a pair of terminals not…

A: We need to select correct option for open circuit.

Q: a) A sinewave is having a frequency of 50 Hz and instantaneous value of 150 Volts at 1.4 milli…

Q: Briefly explain three(3) side effects of the addition of power generation units and their remedies.

A: Nowadays Electricity is a major part of any country's growth. The demand for power supply is…

Q: What is a power system in the context of electrical engineering? Explain the key components and…

A: A power system, in the context of electrical engineering, is a network of electrical components and…

Q: Define a power system in the context of electrical engineering and briefly explain its components.

A: A power system in electrical engineering refers to a network of electrical components and devices…

Q: Select the correct answer for each of the following statements Book Source: Electricity 3 | 10th…

A: Generator, A generator is a machine that is used for the conversion of mechanical energy into useful…

Q: 1.List the three steps of nuclear fuel fabrication processes. Also sketch a schematic diagram and…

A: Different energy sources are required in day to day life. These energy sources can be renewable or…

Q: 1. What is the goal of circuit theory? 2. Why would we want to use circuit theory and when can we…

A: Circuit: The electrical circuit is made up of a resistor, inductor, capacitor, voltage source,…

Q: The small range wind turbine have direct drive generators, direct current output, aero elastic…

Q: Explain with reasons why hydro power generation is sometimes disadvantageous.

A: Hydro-Power generation- Hydropower generation is a process which is used to hydraulic turbine…

Q: )Explain any four basic components of hydro electric power plant.

A: Hydroelectric power plant has various components in it . I have explqined the 4 basic components as…

Q: uppose the three branch currents in this circuit are I₁ = -3 A. 12 = -18 A, and 13 = -15 A. The…

A: Introduction The passive sign convention is used when describing the power dissipated by passive…

Q: Complete the table below. General Classifications According to 1 Power Source Robot Classifications…

A: Here is your answer! I hope this will help you.

Q: basic concepts of direct current and alternating current, and the main applications that each one…

A: Current The flow of electrons through a conductor is termed current (Electric current). Because this…

Q: How to calculate the impedance of a coil?

Q: Draw the phasor diagram for this machine to scale (use graph paper) i) Draw the generated voltage V…

A: It is given that:

Q: In the circuit shown, four devices are connected together. The values for some of the voltages and…

A: We are authorized to answer three subparts at a time, since you have not mentioned which part you…

Q: Briefly explain the Vertical integration in electricity regulation

A: The traditional definition of a "vertically integrated" in electricity regulation is that which owns…

Q: A load can be considered to be capacitive, resistive or inductive in nature. a) Discuss the…

Concept explainers

Compensation Techniques

The compensation techniques are the methods that are employed in the electrical control systems to enhance the performance of the system. The performance are generally affected by the factors like lag in the compensating devices, slow response of the system of the components like sensors used in the control system. The basic compensation technique involves lag compensation, lead compensation or the combination of the both.

Question

PLEASE SOLVE THIS POWER SYSTEM QUESTION

The following data are obtained when open-circuit and short-circuit tests are
performed on a single-phase, 50 kVA, 2400/240 volt, 60 Hz distribution
transformer.
VOLTAGE
CURRENT
POWER
(volts)
(amperes)
(watts)
Measurements on low-voltage side with
high-voltage winding open
Measurements on high-voltage side
with low-voltage winding shorted
240
4.85
173
52.0
20.8
650
a) Neglecting the series impedance, determine the exciting admittance
referred to the high voltage side.
b) Neglecting the exciting admittance, determine the equivalent series
impedance referred to the high voltage side.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Compensation Techniques in Transmission Line

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.

Similar questions

Write in Briefly about circuit breaker, current transformer and voltage transformer
solve with explanation.
18. If the current flowing in an operating circuit must be approximately measured without interrupting the current, which of the following through-type current transformers would you recommend? A. Bar-type B. Plain-through-type C.Split-core-type D. Bushing-type
- Transformers and Special Locations 7.a. Dry-type transformers are available in what configuration(s)?A. Single-phase or auto-phaseB. Poly-phase onlyC. Single- or poly-phaseD. Single-phase only7.b. How many separate and distinct electrical systems does an essential electrical system consist of?A. FiveB. ThreeC. FourD. Six
b) A fault occurs at bus 3 of the network shown in Figure Q4. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, transformer and load are given in Figure Q4. V₁ = 120° p.u. V₂ = 120° p.u. V₂ = 1/0° p.u. V₂= 120° p.u. jXj0.1 p.u. JX2) 0.1 p.u. jX0j0.15 p.u. jXn-j0.2 p.u. 1 JX(2)-j0.2 p.u. 2 jX)=j0.25 p.u. JX20-10.15 p.u. jXa(z)-j0.2 p.u. 4 jX2(0)=j0.2 p.u. jXT(1) j0.1 p.u. jXT(2)=j0.15 p.u. jXT(0)=j0.1 p.u. Figure Q4. Circuit for problem 4b). = jXj0.1 p.u. j0.1 p.u. - JX(2) JXL(0) 10.1 p.u. = (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 3. (ii) Determine the positive sequence fault current for the case when a three- phase-to-ground fault occurs at bus 3 of the network. (iii) Determine the short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus…
h.1. In an alternating-current system, the maximum size for a copper grounding electrode conductor isA. 250 kcmil.B. 3/0.C. 500 kcmil.D. 4/0. h.2. The figure shows a fault condition in an ungrounded electrical system. Which of the following will occurbecause of the fault in this system?A. Only the overcurrent protector at Point A will trip open.B. The ground fault will heavily damage the service.C. Nothing will happen until a second ground fault occurs.D. The overcurrent protection will trip, shutting off all the line side power.
Belonging to a phase measured from a 220/48 V phase transformer, no-load (no load) operation and short-circuit test. values are given in the table below. Low voltage by applying rated voltage to the high voltage (primary) winding. voltage (secondary) winding ends were left unloaded and the results of the no-load operation test were obtained. low voltage By applying nominal current to the windings, the high voltage windings are short-circuited and short-circuit test results are obtained. has been done. In addition, the transformer with 5 V direct voltage applied to the high voltage winding is 2.2 A direct current. attracts. Calculate all equivalent circuit elements moved to the high voltage side. Boşta çalışma testi Gerilim 220 V Güç 80 W Akım 0,44 A Kısa devre testi Gerilim 12 V Güç Akım 65 W 16,1 A
4. A single-phase, 1000-kVA, 2400/8000-V, 60HZ distribution transformer has the following parameters: Resistance of the 2400-V winding: R1 = 2.75 2 Resistance of the 8000-V winding: R2 = 29.4 2 Leakage reactance of the 2400-V winding: X1= 1.8 Q Leakage reactance of the 8000-V winding: X2 = 20 2 Exciting admittance on the 2400-V side: 0.002 - j0.008 S %3| Draw the equivalent circuit referred to the high voltage side of the transformer. a. b. Draw the equivalent circuit referred to the low voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits. 5. The transformer in Problem 4. is supplying a rated load of 1000-kVA at a rated secondary voltage of 8000 V and at 0.8 power factor lagging. Neglecting the transformer exciting current, determine the input terminal voltage of the transformer on the low-voltage side.
A fault occurs at point F of the system in the figure. Calculate the sub-transient current in phase c of machine 1, assuming that the currents before the fault are zero. Both machines have a nominal capacity of 1200 kVA, 480 V, with reactances X”=X2= 11% and X0 = 3.5%. The nominal capacity of each three-phase transformer is 1200 kVA, 480 V Δ/ 2400 V and with reactances X1=X2=X0= 6%. Transmission line reactances are X1=X2=22% and X0 = 35% on a 1200 kVA, 2400 V basis. Neutral grounding reactor reactances are 8% on a kVA and kV basis. of the machines. a) Calculate the fault current if a ground fault occurs in phase a, b) The fault current if it is a line-to-line fault c) the voltage in phase b and c for the line-to-ground fault in phase a .
2. What type of potential transformer would you use for a 362-kV grounded-neutral metering application on an interconnection system? A. Bar-type B. Cascade-type C. Window-type D. Phase-shift-type
The nameplate on a 25-MVA, 60-Hz single-phase transformer indicates that it has a voltage rating of 8.0-kV:78-kV. A short-circuit test from the high-voltage side (low-voltage winding shortcircuited) gives readings of 4.53 kV, 321 A, and 77.5 kW. An open-circuit test is conducted fromthe low-voltage side and the corresponding instrument readings are 8.0 kV, 39.6 A, and 86.2 kW. (a) Calculate the equivalent series impedance of the transformer as referred to the high-voltageterminals.(b) Calculate the equivalent series impedance of the transformer as referred to the low-voltageterminals.(c) Making appropriate approximations, draw a T equivalent circuit for the transformer.(d) Determine the efficiency of the transformer if it is supplying rated load (unity power factor)at rated voltage at its low-voltage terminal
The nameplate on a 25-MVA, 60-Hz single-phase transformer indicates that it has a voltage rating of 8.0-kV:78-kV. A short-circuit test from the high-voltage side (low-voltage winding shortcircuited) gives readings of 4.53 kV, 321 A, and 77.5 kW. An open-circuit test is conducted fromthe low-voltage side and the corresponding instrument readings are 8.0 kV, 39.6 A, and 86.2 kW. (a) Calculate the equivalent series impedance of the transformer as referred to the high-voltageterminals.(b) Calculate the equivalent series impedance of the transformer as referred to the low-voltageterminals.(c) Making appropriate approximations, draw a T equivalent circuit for the transformer.(d) Determine the efficiency of the transformer if it is supplying rated load (unity power factor)at rated voltage at its low-voltage terminals. Use the open-circuit and short-circuit measurements for the computation of the total power losses