% the leading currents. I(1,:) = amps.* ( j*0.6); % Lagging I(2,:) = amps.* ( 1.0 ); % Unity I(3,:) = amps.* ( j*0.
|
|
- Raymond Justin Marshall
- 6 years ago
- Views:
Transcription
1 % the leading currents. I(1,:) = amps.* ( 0.8 j*0.6); % Lagging I(,:) = amps.* ( 1.0 ); % Unity I(3,:) = amps.* ( 0.8 j*0.6); % Leading % Calculate VS referred to the primary side % for each current and power factor. avs = V (Req.*I j.*xeq.*i); % Refer the secondary voltages back to the % secondary side using the turns ratio. VS = avs * (00/15); % lot the secondary voltage (in kv!) versus load plot(amps,abs(vs(1,:)/1000),'b','linewidth',.0); hold on; plot(amps,abs(vs(,:)/1000),'k','linewidth',.0); plot(amps,abs(vs(3,:)/1000),'r.','linewidth',.0); title ('\bfsecondary Voltage Versus Load'); xlabel ('\bfload (A)'); ylabel ('\bfsecondary Voltage (kv)'); legend('0.8 F lagging','1.0 F','0.8 F leading'); grid on; hold off; The resulting plot of secondary voltage versus load is shown below: 1. A threephase transformer bank is to handle 600 kva and have a 34.5/13.8kV voltage ratio. Find the rating of each individual transformer in the bank (high voltage, low voltage, turns ratio, and apparent power) if the transformer bank is connected to (a) YY, (b) Y, (c) Y, (d), (e) open, (f) open Y open. SOLUTION For the first four connections, the apparent power rating of each transformer is 1/3 of the total apparent power rating of the threephase transformer. For the open and openy open connections, the apparent power rating is a bit more complicated. The 600 kva must be 86.6% of the total apparent
2 power rating of the two transformers, implying that the apparent power rating of each transformer must be 31 kva. The ratings for each transformer in the bank for each connection are given below: Connection rimary Voltage Secondary Voltage Apparent ower Turns Ratio YY 19.9 kv 7.97 kv 00 kva.50:1 Y 19.9 kv 13.8 kv 00 kva 1.44:1 Y 34.5 kv 7.97 kv 00 kva 4.33: kv 13.8 kv 00 kva.50:1 open 34.5 kv 13.8 kv 346 kva.50:1 openy open 19.9 kv 13.8 kv 346 kva 1.44:1 Note: The openy open answer assumes that the Y is on the highvoltage side; if the Y is on the lowvoltage side, the turns ratio would be 4.33:1, and the apparent power rating would be unchanged.. A 13,800/480 V threephase Y connected transformer bank consists of three identical 100kVA 7967/480V transformers. It is supplied with power directly from a large constantvoltage bus. In the shortcircuit test, the recorded values on the highvoltage side for one of these transformers are V = 560 V I = 1.6 A = 3300 W (a) If this bank delivers a rated load at 0.85 F lagging and rated voltage, what is the linetoline voltage on the primary of the transformer bank? (b) What is the voltage regulation under these conditions? (c) Assume that the primary voltage of this transformer bank is a constant 13.8 kv, and plot the secondary voltage as a function of load current for currents from noload to fullload. Repeat this process for power factors of 0.85 lagging, 1.0, and 0.85 leading. (d) lot the voltage regulation of this transformer as a function of load current for currents from noload to fullload. Repeat this process for power factors of 0.85 lagging, 1.0, and 0.85 leading. SOLUTION From the shortcircuit information, it is possible to determine the perphase impedance of the transformer bank referred to the highvoltage side. The primary side of this transformer is Yconnected, so the shortcircuit phase voltage is V 560 V 33.3 V 3 3 V φ, = = = the shortcircuit phase current is Iφ, = I = 1.6 A and the power per phase is φ, = = 1100 W 3 Thus the perphase impedance is 33.3 V ZEQ = REQ jxeq = = 5.66 Ω 1.6 A W θ = cos = cos = 74.3 VI ( 33.3 V)( 1.6 A) ZEQ = REQ jxeq = Ω= 6.94 j4.7 Ω
3 R EQ = 6.94 Ω XEQ = j4.7 Ω (a) If this Y transformer bank delivers rated kva (300 kva) at 0.85 power factor lagging while the secondary voltage is at rated value, then each transformer delivers 100 kva at a voltage of 480 V and 0.85 F lagging. Referred to the primary side of one of the transformers, the load on each transformer is equivalent to 100 kva at 7967 V and 0.85 F lagging. The equivalent current flowing in the secondary of one transformer referred to the primary side is 100 kva I φ, S = = 1.55 A 7967 V I φ, S = A The voltage on the primary side of a single transformer is thus V φ, = Vφ, S Iφ, S Z EQ, ( )( j ) V φ, = V A Ω = V The linetoline voltage on the primary of the transformer is V LL, V φ, ( ) = 3 = V = 14. kv (b) The voltage regulation of the transformer is VR = 100% = 3.01%
4 3. Three 5kVA 4,000/77V distribution transformers are connected in Y. The opencircuit test was performed on the lowvoltage side of this transformer bank, and the following data were recorded: V = 480 V I line, = 4.10 A 3 φ,= 945 W line, The shortcircuit test was performed on the highvoltage side of this transformer bank, and the following data were recorded: V = 1600 V I line, =.00 A 3 φ,= 1150 W line, (a) Find the perunit equivalent circuit of this transformer bank. (b) Find the voltage regulation of this transformer bank at the rated load and 0.90 F lagging. (c) What is the transformer bank s efficiency under these conditions? SOLUTION (a) The equivalent of this threephase transformer bank can be found just like the equivalent circuit of a singlephase transformer if we work on a perphase bases. The opencircuit test data on the lowvoltage side can be used to find the excitation branch impedances referred to the secondary side of the transformer bank. Since the lowvoltage side of the transformer is Yconnected, the perphase opencircuit quantities are: V φ = 77 V I φ, = 4.10 A φ, = 315 W, The excitation admittance is given by Iφ, 4.10 A YEX = = = S V 77 V φ, The admittance angle is Therefore, 315 W θ = = = 1 φ, cos cos Vφ, Iφ, ( 77 V)( 4.10 A) YEX = GC jb = = j0.014 RC = 1/ GC = 44 Ω X = 1/ B = 70.3 Ω The base impedance for a single transformer referred to the lowvoltage side is Z base, S ( Vφ, S ) ( 77 V) = = = Ω S 5 kva φ so the excitation branch elements can be expressed in perunit as 44 Ω 70.3 Ω = = 79.5 pu X.9 pu Ω = = Ω 4
5 The shortcircuit test data can be used to find the series impedances referred to the highvoltage side, since the shortcircuit test data was taken on the highvoltage side. Note that the highvoltage is connected, so Vφ, = V = 1600 V, I I / A φ, = =, and / W φ, = =. Z EQ V φ, = = = Ω I φ, 1600 V A 1 φ, 383 W θ = cos = cos 1 = 78.0 V I φ, φ, ( 1600 V)( A) Z = R jx = = 88 j1355 Ω EQ EQ EQ The base impedance referred to the highvoltage side is Z base, ( Vφ, ) ( 4,000 V) = = = 3,040 Ω S 5 kva φ The resulting perunit impedances are 88 Ω = = pu 3,040 Ω X EQ 1355 Ω = = pu 3,040 Ω The perunit, perphase equivalent circuit of the transformer bank is shown below: I I S jx EQ j V jx V S 79.5 j.9 (b) If this transformer is operating at rated load and 0.90 F lagging, then current flow will be at an cos 1 0.9, or 5.8. The perunit voltage at the primary side of the transformer will be angle of ( ) ( )( ) V = V S I Z S EQ = j = The voltage regulation of this transformer bank is VR = 100% = 3.8% 1.0 (c) The output power of this transformer bank is ( )( )( ) OUT = V I cos θ = = 0.9 pu The copper losses are S S I R ( ) ( ) CU = S EQ = = pu 5
6 The core losses are core ( 1.038) V = = = pu R 79.5 C Therefore, the total input power to the transformer bank is IN = OUT CU core = = 0.96 and the efficiency of the transformer bank is OUT 0.9 η = 100% = 100% = 97.% 0.96 IN 4. A 0kVA 0,000/480V 60Hz distribution transformer is tested with the following results: Opencircuit test Shortcircuit test (measured from secondary side) (measured from primary side) V = 480 V V = 1130 V I = 1.60 A I = 1.00 A V = 305 W = 60 W (a) Find the perunit equivalent circuit for this transformer at 60 Hz. (b) What would the rating of this transformer be if it were operated on a 50Hz power system? (c) Sketch the equivalent circuit of this transformer referred to the primary side if it is operating at 50 Hz. SOLUTION (a) The base impedance of this transformer referred to the primary side is ( V ) ( 0,000 V) Zbase, = = = 0 kω S 0 kva The base impedance of this transformer referred to the secondary side is ( V ) ( 480 V) S Z base, S = = = 11.5 Ω S 0 kva The open circuit test yields the values for the excitation branch (referred to the secondary side): Iφ, 1.60 A YEX = = = S Vφ, 480 V W θ = cos = cos = 66.6 V I 480 V 1.60 A ( )( ) Y = G jb = = j R X EX C C = 1/ G = 757 Ω C = 1/ B = 37 Ω The excitation branch elements can be expressed in perunit as 757 Ω 37 Ω = = 65.7 pu X 8.4 pu 11.5 Ω = = 11.5 Ω The short circuit test yields the values for the series impedances (referred to the primary side): 6
7 Z EQ V 1130 V = = = 1130 Ω I 1.00 A W θ = cos = cos = 76.7 V I ( 1130 V)( 1.00 A) Z = R jx = = 60 j1100 Ω EQ EQ EQ The resulting perunit impedances are 60 Ω = = pu 0,000 Ω The perunit equivalent circuit is X EQ 1100 Ω = = pu 0,000 Ω I I S jx EQ j0.055 V jx V S 65.7 j8.4 (b) If this transformer were operated at 50 Hz, both the voltage and apparent power would have to be derated by a factor of 50/60, so its ratings would be kva, 16,667/400 V, and 50 Hz. (c) The transformer parameters referred to the primary side at 60 Hz are: RC = Zbase RC,pu = ( 0 kω )( 65.7) = 1.31 Ω X = Zbase X,pu = ( 0 kω )( 8.4) = 568 kω REQ = ZbaseRE Q,pu = ( 0 kω )( 0.013) = 60 Ω XEQ = Zbase X E Q,pu = ( 0 kω )( 0.055) = 1100 Ω At 50 Hz, the resistance will be unaffected but the reactances are reduced in direct proportion to the decrease in frequency. At 50 Hz, the reactances are 50 Hz X = ( 568 kω ) = 473 kω 60 Hz 50 Hz X EQ = ( 1100 Ω ) = 917 Ω 60 Hz 7
8 The resulting equivalent circuit referred to the primary at 50 Hz is shown below: I I S jx EQ 60 Ω j917 Ω V jx V S 1.31 Ω j473 kω 8
Chapter 2: Transformers
Chapter 2: Transformers 2-1. The secondary winding of a transformer has a terminal voltage of v s (t) = 282.8 sin 377t V. The turns ratio of the transformer is 100:200 (a = 0.50). If the secondary current
More informationECG 741 Power Distribution Transformers. Y. Baghzouz Spring 2014
ECG 741 Power Distribution Transformers Y. Baghzouz Spring 2014 Preliminary Considerations A transformer is a device that converts one AC voltage to another AC voltage at the same frequency. The windings
More information86 chapter 2 Transformers
86 chapter 2 Transformers Wb 1.2x10 3 0 1/60 2/60 3/60 4/60 5/60 6/60 t (sec) 1.2x10 3 FIGURE P2.2 2.3 A single-phase transformer has 800 turns on the primary winding and 400 turns on the secondary winding.
More informationEE 340 Power Transformers
EE 340 Power Transformers Preliminary considerations A transformer is a device that converts one AC voltage to another AC voltage at the same frequency. It consists of one or more coil(s) of wire wrapped
More informationIV. Three-Phase Transfomers
I. Three-Phase Transfomers Three-Phase Transfomers The majority of the power generation/distribution systems in the world are 3- phase systems. The transformers for such circuits can be constructed either
More informationChapter 2-1 Transformers
Principles of Electric Machines and Power Electronics Chapter 2-1 Transformers Third Edition P. C. Sen Transformer application 1: power transmission Ideal Transformer Assumptions: 1. Negligible winding
More informationCork Institute of Technology. Autumn 2008 Electrical Energy Systems (Time: 3 Hours)
Cork Institute of Technology Bachelor of Science (Honours) in Electrical Power Systems - Award Instructions Answer FIVE questions. (EELPS_8_Y4) Autumn 2008 Electrical Energy Systems (Time: 3 Hours) Examiners:
More informationPractical Transformer on Load
Practical Transformer on Load We now consider the deviations from the last two ideality conditions : 1. The resistance of its windings is zero. 2. There is no leakage flux. The effects of these deviations
More informationcos sin XqIq cos sin V X Consider a simple case ignoring R a and X l d axis q axis V q I q V d I d Approximately, the second item can be ignored:
Consider a simple case ignoring R a and X l E cos XdId I d E X d cos sin XqIq E E jxdid jxqi q jxdid q jx I d axis q axis I q q d q q I q sin X q d P3 3 I a cos I d I cos abdei cos I sin a q d E Xd Xq
More informationEE2022 Electrical Energy Systems
EE0 Electrical Energy Systems Lecture : Transformer and Per Unit Analysis 7-0-0 Panida Jirutitijaroen Department of Electrical and Computer Engineering /9/0 EE0: Transformer and Per Unit Analysis by P.
More informationModule 7. Transformer. Version 2 EE IIT, Kharagpur
Module 7 Transformer Lesson 28 Problem solving on Transformers Contents 28 Problem solving on Transformer (Lesson-28) 4 28.1 Introduction. 4 28.2 Problems on 2 winding single phase transformers. 4 28.3
More information148 Electric Machines
148 Electric Machines 3.1 The emf per turn for a single-phase 2200/220- V, 50-Hz transformer is approximately 12 V. Calculate (a) the number of primary and secondary turns, and (b) the net cross-sectional
More informationEN Assignment No.1 - TRANSFORMERS
EN-06 - Assignment No.1 - TRANSFORMERS Date : 13 th Jan 01 Q1) A 0kVA 00/0 Volts, 60Hz, single phase transformer is found to have the following equivalent circuit parameter referred to the high potential
More informationPROBLEMS on Transformers
PROBLEMS on Transformers (A) Simple Problems 1. A single-phase, 250-kVA, 11-kV/415-V, 50-Hz transformer has 80 turns on the secondary. Calculate (a) the approximate values of the primary and secondary
More informationSpring 2000 EE361: MIDTERM EXAM 1
NAME: STUDENT NUMBER: Spring 2000 EE361: MIDTERM EXAM 1 This exam is open book and closed notes. Assume f=60 hz and use the constant µ o =4π 10-7 wherever necessary. Be sure to show all work clearly. 1.
More informationCHAPTER 2. Transformers. Dr Gamal Sowilam
CHAPTER Transformers Dr Gamal Sowilam Introduction A transformer is a static machine. It is not an energy conversion device, it is indispensable in many energy conversion systems. A transformer essentially
More informationQuestions Bank of Electrical Circuits
Questions Bank of Electrical Circuits 1. If a 100 resistor and a 60 XL are in series with a 115V applied voltage, what is the circuit impedance? 2. A 50 XC and a 60 resistance are in series across a 110V
More informationElectric Machinery Fundamentals
Instructor s Manual to accompany Electric Machinery Fundamentals Fourth Edition Stephen J. Chapman Giorgio Ninni i Instructor s Manual to accompany Electric Machinery Fundamentals, Fourth Edition Copyright
More informationRarely used, problems with unbalanced loads.
THREE-PHASE TRANSFORMERS Transformers used in three-phase systems may consist of a bank of three single-phase transformers or a single three-phase transformer which is wound on a common magnetic core.
More informationHours / 100 Marks Seat No.
17415 15162 3 Hours / 100 Seat No. Instructions (1) All Questions are Compulsory. (2) Answer each next main Question on a new page. (3) Illustrate your answers with neat sketches wherever necessary. (4)
More informationElectric Machinery Fundamentals
Solutions Manual to accompany Chapman Electric Machinery Fundamentals Fifth Edition Stephen J. Chapman BE Systems ustralia i Solutions Manual to accompany Electric Machinery Fundamentals, Fifth Edition
More informationVALLIAMMAI ENGINEERING COLLEGE
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF ELECTRONICS AND INSTRUMENTATION ENGINEERING QUESTION BANK IV SEMESTER EI6402 ELECTRICAL MACHINES Regulation 2013 Academic
More informationTransmission Line Models Part 1
Transmission Line Models Part 1 Unlike the electric machines studied so far, transmission lines are characterized by their distributed parameters: distributed resistance, inductance, and capacitance. The
More informationTransformer & Induction M/C
UNIT- 2 SINGLE-PHASE TRANSFORMERS 1. Draw equivalent circuit of a single phase transformer referring the primary side quantities to secondary and explain? (July/Aug - 2012) (Dec 2012) (June/July 2014)
More informationStudy of Inductive and Capacitive Reactance and RLC Resonance
Objective Study of Inductive and Capacitive Reactance and RLC Resonance To understand how the reactance of inductors and capacitors change with frequency, and how the two can cancel each other to leave
More informationSYNCHRONOUS MACHINES
SYNCHRONOUS MACHINES The geometry of a synchronous machine is quite similar to that of the induction machine. The stator core and windings of a three-phase synchronous machine are practically identical
More informationUNIVERSITY OF TORONTO FACULTY OF APPLIED SCIENCE AND ENGINEERING. MIDTERM EXAMINATION, February Forth Year Electrical and Computer Engineering
NAME: LAST UNIVERSITY OF TORONTO FACULTY OF APPLIED SCIENCE AND ENINEERIN MIDTERM EXAMINATION, February 017 Forth Year Electrical and Computer Engineering ECE413 Energy Systems and Distribution eneration
More informationCHAPTER 9. Sinusoidal Steady-State Analysis
CHAPTER 9 Sinusoidal Steady-State Analysis 9.1 The Sinusoidal Source A sinusoidal voltage source (independent or dependent) produces a voltage that varies sinusoidally with time. A sinusoidal current source
More informationExperiment 3 Single Phase Transformer (II)
Objectives To determine the polarity of single phase transformer windings. To determine the internal resistance of single phase transformer windings. To determine the efficiency and voltage regulation
More informationCHAPTER 6: ALTERNATING CURRENT
CHAPTER 6: ALTERNATING CURRENT PSPM II 2005/2006 NO. 12(C) 12. (c) An ac generator with rms voltage 240 V is connected to a RC circuit. The rms current in the circuit is 1.5 A and leads the voltage by
More informationAC Power Instructor Notes
Chapter 7: AC Power Instructor Notes Chapter 7 surveys important aspects of electric power. Coverage of Chapter 7 can take place immediately following Chapter 4, or as part of a later course on energy
More informationEl-Hawary, M.E. The Transformer Electrical Energy Systems. Series Ed. Leo Grigsby Boca Raton: CRC Press LLC, 2000
El-Hawary, M.E. The Transformer Electrical Energy Systems. Series Ed. Leo Grigsby Boca Raton: CRC Press LLC, 000 97 Chapter 4 THE TRANSFORMER 4. NTRODUCTON The transformer is a valuable apparatus in electrical
More informationCourse ELEC Introduction to electric power and energy systems. Additional exercises with answers December reactive power compensation
Course ELEC0014 - Introduction to electric power and energy systems Additional exercises with answers December 2017 Exercise A1 Consider the system represented in the figure below. The four transmission
More informationAC reactive circuit calculations
AC reactive circuit calculations This worksheet and all related files are licensed under the Creative Commons Attribution License, version 1.0. To view a copy of this license, visit http://creativecommons.org/licenses/by/1.0/,
More informationEECS 216 Winter 2008 Lab 2: FM Detector Part I: Intro & Pre-lab Assignment
EECS 216 Winter 2008 Lab 2: Part I: Intro & Pre-lab Assignment c Kim Winick 2008 1 Introduction In the first few weeks of EECS 216, you learned how to determine the response of an LTI system by convolving
More informationINSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad - 00 03 ELECTRICAL AND ELECTRONICS ENGINEERING ASSIGNMENT Course Name : ELECRICAL MACHINES - II Course Code : A0 Class : II B.TECH-II
More informationRESONANT TRANSFORMER
RESONANT TRANSFORMER Whenever the requirement of the test voltage is too much high, a single unit transformer can not produce such high voltage very economically, because for high voltage measurement,
More informationSECTION 4 TRANSFORMERS. Yilu (Ellen) Liu. Associate Professor Electrical Engineering Department Virginia Tech University
SECTION 4 TRANSFORMERS Yilu (Ellen) Liu Associate Professor Electrical Engineering Department Virginia Tech University Analysis of Transformer Turns Ratio......................... 4.2 Analysis of a Step-Up
More information, ,54 A
AEB5EN2 Ground fault Example Power line 22 kv has the partial capacity to the ground 4,3.0 F/km. Decide whether ground fault currents compensation is required if the line length is 30 km. We calculate
More informationR10. III B.Tech. II Semester Supplementary Examinations, January POWER SYSTEM ANALYSIS (Electrical and Electronics Engineering) Time: 3 Hours
Code No: R3 R1 Set No: 1 III B.Tech. II Semester Supplementary Examinations, January -14 POWER SYSTEM ANALYSIS (Electrical and Electronics Engineering) Time: 3 Hours Max Marks: 75 Answer any FIVE Questions
More informationWELCOME TO THE LECTURE
WLCOM TO TH LCTUR ON TRNFORMR Single Phase Transformer Three Phase Transformer Transformer transformer is a stationary electric machine which transfers electrical energy (power) from one voltage level
More informationBasic Analog Circuits
Basic Analog Circuits Overview This tutorial is part of the National Instruments Measurement Fundamentals series. Each tutorial in this series, will teach you a specific topic of common measurement applications,
More informationOpen Circuit (OC) and Short Circuit (SC) Tests on Single Phase Transformer
Open Circuit (OC) and Short Circuit (SC) Tests on Single Phase Transformer 1 Aim To obtain the equivalent circuit parameters from OC and SC tests, and to estimate efficiency & regulation at various loads.
More informationEE 740 Transmission Lines
EE 740 Transmission Lines 1 High Voltage Power Lines (overhead) Common voltages in north America: 138, 230, 345, 500, 765 kv Bundled conductors are used in extra-high voltage lines Stranded instead of
More informationCode No: R Set No. 1
Code No: R05310204 Set No. 1 III B.Tech I Semester Regular Examinations, November 2007 ELECTRICAL MACHINES-III (Electrical & Electronic Engineering) Time: 3 hours Max Marks: 80 Answer any FIVE Questions
More informationLevel 6 Graduate Diploma in Engineering Electrical Energy Systems
9210-114 Level 6 Graduate Diploma in Engineering Electrical Energy Systems Sample Paper You should have the following for this examination one answer book non-programmable calculator pen, pencil, ruler,
More informationEngineering Science OUTCOME 4 - TUTORIAL 3 CONTENTS. 1. Transformers
Unit : Unit code: QCF Level: 4 Credit value: 5 SYLLABUS Engineering Science L/60/404 OUTCOME 4 - TUTOIAL 3 Be able to apply single phase AC theory to solve electrical and electronic engineering problems
More informationIn Class Examples (ICE)
In Class Examples (ICE) 1 1. A 3φ 765kV, 60Hz, 300km, completely transposed line has the following positive-sequence impedance and admittance: z = 0.0165 + j0.3306 = 0.3310 87.14 o Ω/km y = j4.67 410-6
More informationINSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad - 00 0 ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK Course Name Course Code Class Branch : ELECRICAL MACHINES - II : A0 :
More informationElectric Circuits I. Simple Resistive Circuit. Dr. Firas Obeidat
Electric Circuits I Simple Resistive Circuit Dr. Firas Obeidat 1 Resistors in Series The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances. It
More informationCHAPTER 5 SYNCHRONOUS GENERATORS
CHAPTER 5 SYNCHRONOUS GENERATORS Summary: 1. Synchronous Generator Construction 2. The Speed of Rotation of a Synchronous Generator 3. The Internal Generated Voltage of a Synchronous Generator 4. The Equivalent
More informationChapter 30 Inductance, Electromagnetic. Copyright 2009 Pearson Education, Inc.
Chapter 30 Inductance, Electromagnetic Oscillations, and AC Circuits 30-7 AC Circuits with AC Source Resistors, capacitors, and inductors have different phase relationships between current and voltage
More informationUniversity of Jordan School of Engineering Electrical Engineering Department. EE 219 Electrical Circuits Lab
University of Jordan School of Engineering Electrical Engineering Department EE 219 Electrical Circuits Lab EXPERIMENT 7 RESONANCE Prepared by: Dr. Mohammed Hawa EXPERIMENT 7 RESONANCE OBJECTIVE This experiment
More informationELECTRICAL ENGINEERING ESE TOPIC WISE OBJECTIVE SOLVED PAPER-II
ELECTRICAL ENGINEERING ESE TOPIC WISE OBJECTIVE SOLVED PAPER-II From (1992 2017) Office : F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-26522064 Mobile : 8130909220, 9711853908
More informationEQUIVALENT CIRCUIT OF A SINGLE-PHASE TRANSFORMER
Electrical Machines Lab Experiment-No. One Date: 15-11-2016 EQUIVALENT CIRCUIT OF A SINGLE-PHASE TRANSFORMER Aim: The determination of electrical equivalent circuit parameters of a single phase power transformer
More informationQuestion Paper Profile
I Scheme Question Paper Profile Program Name : Electrical Engineering Program Group Program Code : EE/EP/EU Semester : Third Course Title : Electrical Circuits Max. Marks : 70 Time: 3 Hrs. Instructions:
More informationEXPERIMENT 4: RC, RL and RD CIRCUITs
EXPERIMENT 4: RC, RL and RD CIRCUITs Equipment List Resistor, one each of o 330 o 1k o 1.5k o 10k o 100k o 1000k 0.F Ceramic Capacitor 4700H Inductor LED and 1N4004 Diode. Introduction We have studied
More informationLecture Week 4. Homework Voltage Divider Equivalent Circuit Observation Exercise
Lecture Week 4 Homework Voltage Divider Equivalent Circuit Observation Exercise Homework: P6 Prove that the equation relating change in potential energy to voltage is dimensionally consistent, using the
More informationCode No: R Set No. 1
Code No: R05220204 Set No. 1 II B.Tech II Semester Supplimentary Examinations, Aug/Sep 2007 ELECTRICAL MACHINES-II (Electrical & Electronic Engineering) Time: 3 hours Max Marks: 80 Answer any FIVE Questions
More informationElectricity & Optics
Physics 24100 Electricity & Optics Lecture 19 Chapter 29 sec. 1,2,5 Fall 2017 Semester Professor Koltick Series and Parallel R and L Resistors and inductors in series: R series = R 1 + R 2 L series = L
More informationEXPERIMENT 4: RC, RL and RD CIRCUITs
EXPERIMENT 4: RC, RL and RD CIRCUITs Equipment List An assortment of resistor, one each of (330, 1k,1.5k, 10k,100k,1000k) Function Generator Oscilloscope 0.F Ceramic Capacitor 100H Inductor LED and 1N4001
More informationTRANSFORMER THEORY. Mutual Induction
Transformers Transformers are used extensively for AC power transmissions and for various control and indication circuits. Knowledge of the basic theory of how these components operate is necessary to
More informationDownloaded from / 1
PURWANCHAL UNIVERSITY II SEMESTER FINAL EXAMINATION-2008 LEVEL : B. E. (Computer/Electronics & Comm.) SUBJECT: BEG123EL, Electrical Engineering-I Full Marks: 80 TIME: 03:00 hrs Pass marks: 32 Candidates
More informationChapter 11. Alternating Current
Unit-2 ECE131 BEEE Chapter 11 Alternating Current Objectives After completing this chapter, you will be able to: Describe how an AC voltage is produced with an AC generator (alternator) Define alternation,
More informationECE 241L Fundamentals of Electrical Engineering. Experiment 8 A-C Transformer, Magnetization & Hysteresis
ECE 241L Fundamentals of Electrical Engineering Experiment 8 A-C Transformer, Magnetization & Hysteresis A. Objectives: I. Measure leakage inductance and resistance loss II. Measure magnetization inductance
More informationINSTITUTE OF AERONAUTICAL ENGINEERING
POWER POINT PRESENTATION ON ELECTRICAL MACHINES - II 016-017 II B. Tech II semester (JNTUH-R15) Mr. K DEVENDER REDDY, Assistant Professor ELECTRICAL AND ELECTRONICS ENGINEERING INSTITUTE OF AERONAUTICAL
More informationINSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad ELECTRICAL AND ELECTRONICS ENGINEERING
Course Name Course Code Class Branch INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad - 500 043 ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK : ELECRICAL MACHINES I : A40212
More informationECE215 Lecture 7 Date:
Lecture 7 Date: 29.08.2016 AC Circuits: Impedance and Admittance, Kirchoff s Laws, Phase Shifter, AC bridge Impedance and Admittance we know: we express Ohm s law in phasor form: where Z is a frequency-dependent
More informationFigure Derive the transient response of RLC series circuit with sinusoidal input. [15]
COURTESY IARE Code No: R09220205 R09 SET-1 B.Tech II Year - II Semester Examinations, December-2011 / January-2012 NETWORK THEORY (ELECTRICAL AND ELECTRONICS ENGINEERING) Time: 3 hours Max. Marks: 80 Answer
More informationCHAPTER 4. Distribution Transformers
CHAPTER 4 Distribution Transformers Introduction A transformer is an electrical device that transfers energy from one circuit to another purely by magnetic coupling. Relative motion of the parts of the
More informationFilter Design, Active Filters & Review. EGR 220, Chapter 14.7, December 14, 2017
Filter Design, Active Filters & Review EGR 220, Chapter 14.7, 14.11 December 14, 2017 Overview ² Passive filters (no op amps) ² Design examples ² Active filters (use op amps) ² Course review 2 Example:
More informationSAMPLE EXAM PROBLEM PROTECTION (6 OF 80 PROBLEMS)
SAMPLE EXAM PROBLEM PROTECTION (6 OF 80 PROBLEMS) SLIDE In this video, we will cover a sample exam problem for the Power PE Exam. This exam problem falls under the topic of Protection, which accounts for
More informationEEE3441 Electrical Machines Department of Electrical Engineering. Lecture. Basic Operating Principles of Transformers
Department of Electrical Engineering Lecture Basic Operating Principles of Transformers In this Lecture Basic operating principles of following transformers are introduced Single-phase Transformers Three-phase
More informationFor the circuit in Fig. 1, determine the current in the neutral line.
Problem 1 For the circuit in Fig. 1, determine the current in the neutral line. Solution Figure 1 Since the neutral line is present, we can solve this problem on a per-phase basis. For phase a, For phase
More informationFault Analysis. EE 340 Spring 2012
Fault Analysis EE 340 Spring 2012 Introduction A fault in a circuit is any failure that interferes with the normal system operation. Lighting strokes cause most faults on highvoltage transmission lines
More informationEE 340 Transmission Lines
EE 340 Transmission Lines Physical Characteristics Overhead lines An overhead transmission line usually consists of three conductors or bundles of conductors containing the three phases of the power system.
More informationECE 2006 University of Minnesota Duluth Lab 11. AC Circuits
1. Objective AC Circuits In this lab, the student will study sinusoidal voltages and currents in order to understand frequency, period, effective value, instantaneous power and average power. Also, the
More informationAligarh College of Engineering & Technology (College Code: 109) Affiliated to UPTU, Approved by AICTE Electrical Engg.
Aligarh College of Engineering & Technology (College Code: 19) Electrical Engg. (EE-11/21) Unit-I DC Network Theory 1. Distinguish the following terms: (a) Active and passive elements (b) Linearity and
More informationOhm s Law. 1 Object. 2 Apparatus. 3 Theory. To study resistors, Ohm s law, linear behavior, and non-linear behavior.
Ohm s Law Object To study resistors, Ohm s law, linear behavior, and non-linear behavior. pparatus esistors, power supply, meters, wires, and alligator clips. Theory resistor is a circuit element which
More informationLab 1: Basic RL and RC DC Circuits
Name- Surname: ID: Department: Lab 1: Basic RL and RC DC Circuits Objective In this exercise, the DC steady state response of simple RL and RC circuits is examined. The transient behavior of RC circuits
More informationHomework Assignment 07
Homework Assignment 07 Question 1 (Short Takes). 2 points each unless otherwise noted. 1. A single-pole op-amp has an open-loop low-frequency gain of A = 10 5 and an open loop, 3-dB frequency of 4 Hz.
More informationINTRODUCTION TO AC FILTERS AND RESONANCE
AC Filters & Resonance 167 Name Date Partners INTRODUCTION TO AC FILTERS AND RESONANCE OBJECTIVES To understand the design of capacitive and inductive filters To understand resonance in circuits driven
More informationPower Factor & Harmonics
Power Factor & Harmonics Andy Angrick 2014 Harmonic Distortion Harmonic problems are becoming more apparent because more equipment that produce harmonics are being applied to power systems Grounding Harmonics
More informationGRADE 12 SEPTEMBER 2012 ELECTRICAL TECHNOLOGY
Province of the EASTERN CAPE EDUCATION NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2012 ELECTRICAL TECHNOLOGY MARKS: 200 TIME: 3 hours This question paper consists of 11 pages and a formula sheet. 2
More informationThree-phase short-circuit current (Isc) calculation at any point within a LV installation using impedance method
Three-phase short-circuit current (Isc) calculation at any point within a LV installation using impedance method Calculation of Isc by the impedance method In a 3-phase installation Isc at any point is
More informationClass: Second Subject: Electrical Circuits 2 Lecturer: Dr. Hamza Mohammed Ridha Al-Khafaji
10.1 Introduction Class: Second Lecture Ten esonance This lecture will introduce the very important resonant (or tuned) circuit, which is fundamental to the operation of a wide variety of electrical and
More informationECE 215 Lecture 8 Date:
ECE 215 Lecture 8 Date: 28.08.2017 Phase Shifter, AC bridge AC Circuits: Steady State Analysis Phase Shifter the circuit current I leads the applied voltage by some phase angle θ, where 0 < θ < 90 ο depending
More informationExperiment 45. Three-Phase Circuits. G 1. a. Using your Power Supply and AC Voltmeter connect the circuit shown OBJECTIVE
Experiment 45 Three-Phase Circuits OBJECTIVE To study the relationship between voltage and current in three-phase circuits. To learn how to make delta and wye connections. To calculate the power in three-phase
More informationSHRI RAMSWAROOP MEMORIAL COLLEGE OF ENGG. & MANAGEMENT B.Tech. [SEM I (EE, EN, EC, CE)] QUIZ TEST-3 (Session: ) Time: 1 Hour ELECTRICAL ENGINEE
SHRI RAMSWAROOP MEMORIAL COLLEGE OF ENGG. & MANAGEMENT B.Tech. [SEM I (EE, EN, EC, CE)] QUIZ TEST-3 (Session: 2014-15) Time: 1 Hour ELECTRICAL ENGINEERING Max. Marks: 30 (NEE-101) Roll No. Academic/26
More informationTransformers 21.1 INTRODUCTION 21.2 MUTUAL INDUCTANCE
21 Transformers 21.1 INTRODUCTION Chapter 12 discussed the self-inductance of a coil. We shall now examine the mutual inductance that exists between coils of the same or different dimensions. Mutual inductance
More informationTUNED AMPLIFIERS 5.1 Introduction: Coil Losses:
TUNED AMPLIFIERS 5.1 Introduction: To amplify the selective range of frequencies, the resistive load R C is replaced by a tuned circuit. The tuned circuit is capable of amplifying a signal over a narrow
More informationMAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
Important Instructions to examiners: 1) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. 2) The model answer and the answer written by candidate
More informationLab 8 - INTRODUCTION TO AC CURRENTS AND VOLTAGES
08-1 Name Date Partners ab 8 - INTRODUCTION TO AC CURRENTS AND VOTAGES OBJECTIVES To understand the meanings of amplitude, frequency, phase, reactance, and impedance in AC circuits. To observe the behavior
More informationPART A. 1. List the types of DC Motors. Give any difference between them. BTL 1 Remembering
UNIT I DC MACHINES Three phase circuits, a review. Construction of DC machines Theory of operation of DC generators Characteristics of DC generators Operating principle of DC motors Types of DC motors
More informationForm B. Connection Impact Assessment Application Form Distribution System
Form B Connection Impact Assessment Application Form Distribution System This Application Form is for Generators applying for Connection Impact Assessment ( CIA ). It is important that the Generator provides
More informationa) Determine the smallest, standard-sized circuit breaker that should be used to protect this branch circuit.
ECET4520 Exam II Sample Exam Problems Instructions: This exam is closed book, except for the reference booklet provided by your instructor and one (8.5 x11 ) sheet of handwritten notes that may not contain
More informationECE 3600 Transformers b
Transformer basics and ratings A Transformer is two coils of wire that are magnetically coupled. Transformers b Transformers are only useful for AC, which is one of the big reasons electrical power is
More informationChapter 31 Alternating Current
Chapter 31 Alternating Current In this chapter we will learn how resistors, inductors, and capacitors behave in circuits with sinusoidally vary voltages and currents. We will define the relationship between
More informationSIMULATION OF D-STATCOM AND DVR IN POWER SYSTEMS
SIMUATION OF D-STATCOM AND DVR IN POWER SYSTEMS S.V Ravi Kumar 1 and S. Siva Nagaraju 1 1 J.N.T.U. College of Engineering, KAKINADA, A.P, India E-mail: ravijntu@gmail.com ABSTRACT A Power quality problem
More informationReview: Lecture 9. Instantaneous and Average Power. Effective or RMS Value. Apparent Power and Power Factor. Complex Power. Conservation of AC Power
Review: Lecture 9 Instantaneous and Average Power p( t) VmI m cos ( v i ) VmI m cos ( t v i ) Maximum Average Power Transfer Z L R L jx Effective or RMS Value I rms I m L R P * TH Apparent Power and Power
More information