Sinusoidal Steady State Analysis (AC Analysis) Part II


 Edwin Black
 2 years ago
 Views:
Transcription
1 Sinusoidal Steady State Analysis (AC Analysis) Part II Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University
2 OUTLINE Previously on ELCN102 AC Power Analysis Instantaneous Power Average Power Complex Power Root Mean Square Maximum Average Power Transfer 2
3 Previously on ELCN102 Methods of Solution of AC Circuits To solve a AC circuit you can use one or more of the following methods: Simplification Method Loop Analysis Method Node Analysis Method Superposition Method Thevenin equivalent circuit Norton equivalent circuit
4 Previously on ELCN102 Simplification Method In step by step simplification we can use: Source transformation Combination of active elements Combination of series and parallel elements Stardelta & deltastar transformation
5 Previously on ELCN102 Source Transformation A voltage source V AC with a series impedance Z can be transformed into a current source I AC = V AC /Z and a parallel impedance Z A current source I AC with a parallel impedance Z can be transformed into a voltage source V AC = I AC Z and a series impedance Z
6 Previously on ELCN102 Loop Analysis Method The Loop Analysis Method (Mesh Method) uses KVL to generate a set of simultaneous equations. 1) Convert the independent current sources into equivalent voltage sources 2) Identify the number of independent loop (L) on the circuit 3) Label a loop current on each loop. 4) Write an expression for the KVL around each loop. 5) Solve the simultaneous equations to get the loop currents.
7 Previously on ELCN102 Matrix Form Z 11 Z 12 Z 1N Z 21 Z 22 Z 2N Z N1 Z N2 Z NN I 1 I 2 I N = V 1 V 2 V N Z ii = impedance in loop i Z ij = Common impedance between loops i and j = Z ji V i = voltage sources in loop i V is +ve if it supplies current in the direction of the loop current
8 Previously on ELCN102 Node Analysis Method The Node Analysis Method (Nodal Analysis) uses KCL to generate a set of simultaneous equations. 1) Convert independent voltage sources into equivalent current sources. 2) Identify the number of non simple nodes (N) of the circuit. 3) Write an expression for the KCL at each N 1 Node (exclude the ground node). 4) Solve the resultant simultaneous equations to get the voltages. 8
9 Previously on ELCN102 Matrix Form Y 11 Y 12 Y 1N Y 21 Y 22 Y 2N Y N1 Y N2 Y NN V 1 V 2 V N = I 1 I 2 I N Y ii = admittance of node i Y ij = common admittance between node i and j = Y ji I i = current sources at node i I is +ve if it supply current into the node
10 Previously on ELCN102 Superposition Theorem For a linear circuit containing multiple independent sources, the voltage across (or current through) any of its elements is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone o V 5 0 o A Total I a I b I = I a + I b
11 Previously on ELCN102 Thevenin s Theorem A linear twoterminal circuit, can be replaced by an equivalent circuit consisting of a voltage source V th in series with a impedance Z th.
12 Previously on ELCN102 Thevenin s Theorem Steps 1) Identify the load impedance and introduce two nodes a and b 2) Remove the load impedance between node a and b 3) Calculate the open circuit voltage between nodes a and b. This voltage is V th of the Thevenin equivalent circuit. 4) Set all the independent sources to zero (voltage sources are SC and current sources are OC) and calculate the impedance seen between nodes a and b. This impedance is Z th of the Thevenin equivalent circuit.
13 Previously on ELCN102 Norton s Theorem A linear twoterminal circuit can be replaced by equivalent circuit consisting of a current source I N in parallel with a impedance Z N
14 Previously on ELCN102 Norton s Theorem Steps 1) Identify the load impedance and introduce two nodes a and b 2) Remove the load impedance between node a and b and set all the independent sources to zero (voltage sources are SC and current sources are OC) and calculate the impedance seen between nodes a and b. This resistance is Z N of the Norton equivalent circuit. 3) Replace the load impedance with a short circuit and calculate the short circuit current between nodes a and b. This current is I N of the Norton equivalent circuit.
15 Previously on ELCN102 Thevenin and Norton equivalent circuits Thevenin equivalent circuit must be equivalent to Norton equivalent circuit Z N = Z th, V th = I N Z N, I N = V th Z Z th = V th th I N
16 AC Power Analysis Instantaneous Power Instantaneous power p t is the product of the instantaneous voltage v(t) across the element and the instantaneous current i(t) through it p t = v t i t v t = V m cos ωt + φ v, i t = I m cos ωt + φ i p t = V m cos ωt + φ v I m cos ωt + φ i = V m I m cos ωt + φ v cos ωt + φ
17 AC Power Analysis Instantaneous Power Instantaneous power p t is the product of the instantaneous voltage v(t) across the element and the instantaneous current i(t) through it p t = V m I m cos ωt + φ v cos ωt + φ i Hint: cos A cos B = 1 2 cos A + B + cos A B p t = V m I m 2 cos 2ωt + φ v + φ i + cos φ v φ i
18 AC Power Analysis Instantaneous Power p t = 1 2 V mi m cos 2ωt + φ v + φ i + cos φ v φ i
19 Average Power AC Power Analysis Average Power (P av ) is the average of p t over one period P av = 1 T න 0 T p t dt = 1 T න 0 T V m I m 2 cos 2ωt + φ v + φ i + cos φ v φ i dt = V mi m 2T sin 2ωt + φ v + φ i 2ω T + t cos φ v φ i 0
20 Average Power AC Power Analysis Average Power (P av ) is the average of p t over one period P av = V mi m 2T sin 2ωt + φ v + φ i 2ω T + t cos φ v φ i 0 sin 2ωt + φ v + φ i sin 2ωt + φ v + φ i t=0 t=t sin φ v + φ i sin 2ωT + φ v + φ i ω = 2π T sin 2 2π + φ v + φ i sin φ v + φ i
21 Average Power AC Power Analysis Average Power (P av ) is the average of p t over one period P av = V mi m 2T T cos φ v φ i = 1 2 V mi m cos φ v φ i
22 Average Power AC Power Analysis Average Power (P av ) is the average of p t over one period P av = 1 2 V mi m cos φ v φ i = A cos φ v φ i cos φ v φ i is called the Power factor. A = 1 2 V mi m is referred to as the Apparent power. Hint: cos x is even function cos φ v φ i = cos φ i φ v
23 Average Power AC Power Analysis For an impedance Z P av = 1 2 V mi m cos φ v φ i V = V m φ v, I = I m φ i Z = V I = V m I m φ v φ i = Z Z P av = 1 2 Z I m I m cos φ v φ i = I m 2 2 Z cos Z
24 Average Power AC Power Analysis Z = Z Z = R + jx R = Z cos Z, X = Z sin Z P av = I m 2 2 Z cos Z = I m 2 R 2 For impedance Z, the average power is given by P av = I m 2 R 2 R = Re Z is the resistive part of Z
25 Average Power AC Power Analysis For φ v φ i = 0 P av = 1 2 V mi m cos φ v φ i Z is pure resistive and the average power absorbed by the impedance will be maximum. For φ v φ i = ±90 o Z is pure reactive and the average power absorbed by the impedance will be zero.
26 Example (1) AC Power Analysis For a linear circuit, v t = 120cos(377t + 45 o ), and i(t) = 10cos(377t 10 o ). Calculate the average power.
27 Example (2) AC Power Analysis For the shown circuit, calculate the average power delivered by the source and the average power absorbed by the circuit.
28 Example (3) AC Power Analysis Verify the power balance for the shown circuit
29 AC Power Analysis Root Mean Square Root Mean Square (rms) or Effective value of a varying signal is defined as the value of DC signal that would produce the same power dissipation in a resistive load DC Signal Sinusoidal Signal 2 P av = I eff R 2 I eff = I m 2 2 I eff = I m 2 P av = I m 2 2 R The eff value of a sinusoidal signal = Maximum of the sinusoidal 2
30 AC Power Analysis Root Mean Square Root Mean Square (rms) or Effective value of a varying signal is defined as the value of DC signal that would produce the same power dissipation in a resistive load DC Signal 2 P av = I eff R Any periodic signal P av = 1 T න 0 T i t v t dt I eff = 1 T න 0 T i 2 t dt = R T න 0 T i 2 t dt
31 AC Power Analysis Root Mean Square For any periodic function x t, the eff value is given by X eff = 1 T න 0 T x 2 t dt The effective value is the (square) root of the mean (or average) of the square of the periodic signal. Thus, the effective value is often known as the rootmeansquare value, or rms value for short. X eff = X rms
32 AC Power Analysis Root Mean Square For any periodic function x t, the eff value is given by X rms = X eff = 1 T න 0 T x 2 t dt For a voltage signal v t = V m cos ωt v 2 t = V 2 m cos 2 ωt 1 Tv T න 2 t dt = V m 2 0 T න 0 T cos 2 ωt dt
33 AC Power Analysis Root Mean Square V m 2 Tcos 2 ωt dt = V m 2 T න 0 2T න 0 T 1 + cos 2ωt dt Hint: cos 2 θ = cos 2θ = V m 2 2T t + sin 2ωt 2ω T = V m V rms = 1 T න 0 T v 2 t dt = V m 2
34 AC Power Analysis Root Mean Square For a resistor the average power is given by P av = I m 2 2 R = I m 2 I m 2 R = I2 rms R = V rms 2 R
35 AC Power Analysis Root Mean Square For a resistor the average power is given by P av = 1 2 V mi m cos φ v φ i = V rms I rms cos φ v φ i = A pf A = V rms I rms is called the Apparent power because it similar to the average power of the DC circuit. Apparent power is measured in voltamperes (VA) to distinguish it from the average power, which is measured in watts.
36 AC Power Analysis Root Mean Square For a resistor the average power is given by P av = 1 2 V mi m cos φ v φ i = V rms I rms cos φ v φ i = A pf pf = cos φ v φ i is called the power factor. It is factor by which the apparent power must be multiplied to obtain the average power. For a purely resistive load, pf = 1. For a purely reactive load, pf = 0.
37 AC Power Analysis Root Mean Square For a resistor the average power is given by P av = 1 2 V mi m cos φ v φ i = V rms I rms cos φ v φ i = A pf pf is said to be leading if the current leads the voltage, which implies a capacitive load. pf is said to be lagging if the current lags the voltage, which implies a inductive load.
38 Example (4) AC Power Analysis Determine the rms value of the current waveform shown. If the current is passed through a 2Ω resistor, find the average power absorbed by the resistor.
39 Complex Power AC Power Analysis Complex power is important in power analysis because it contains all the information about the power absorbed by a given load. The complex power S absorbed by an AC load is the product of the voltage and the complex conjugate of the current S = 1 V I 2 V = V m φ v, I = I m φ i S = 1 2 V m φ v I m φ i
40 Complex Power For an impedance Z AC Power Analysis S = 1 2 V mi m φ v φ i = V m I m 2 2 φ v φ i = V rms I rms φ v φ i Z = V I = V m φ v I m φ i = V m φ v φ i I m = V rms φ v φ i I rms 2 S = Z I rms I rms = I rms Z = V 2 rms Z = I2 rms R + jx = I2 rms R + ji2 rms X
41 Complex Power AC Power Analysis S = V rms I rms φ v φ i = V rms I rms cos φ v φ i + jv rms I rms sin φ v φ i 2 2 = I rms R + ji rms X = P + jq P = Re S is called the Active Power and its unit is Watt. Q = Im S is called the Reactive Power and its unit is VAR. A = S = V rms I rms is the Apparent Power and its unit is VA. φ v φ i is the Power Factor Angle.
42 Complex Power AC Power Analysis S = V rms I rms φ v φ i = V rms I rms cos φ v φ i + jv rms I rms sin φ v φ i 2 2 = I rms R + ji rms X = P + jq The real power P is the average power in watts delivered to a load. P is the actual power dissipated by the load. The reactive power Q is a measure of the energy exchange between the source and the reactive part of the load.
43 Complex Power AC Power Analysis S = V rms I rms φ v φ i = V rms I rms cos φ v φ i + jv rms I rms sin φ v φ i 2 2 = I rms R + ji rms X = P + jq For resistive load, φ v φ i = 0 Q = 0. For capacitive load (leading pf), φ v φ i < 0 Q < 0. For inductive load (lagging pf), φ v φ i > 0 Q > 0.
44 Complex Power AC Power Analysis S = V rms I rms φ v φ i = V rms I rms cos φ v φ i + jv rms I rms sin φ v φ i 2 2 = I rms R + ji rms X = P + jq S, P, and Q construct the power triangle.
45 Example (5) AC Power Analysis The voltage across a load is v(t) = 60cos(ωt 10 o ) V and the current through the element in the direction of the voltage drop is i(t) = 1.5cos(ωt + 50) A. Find: (a) The complex and apparent powers. (b) The real and reactive powers. (c) The power factor and the load impedance.
46 Maximum Average Power Transfer Definition For maximum average power transfer, the load impedance Z L must be equal to the complex conjugate of the Thevenin/Norton impedance Z Th. For maximum power P ZL Z L = Z th = Z N
47 Maximum Average Power Transfer Definition Z th = R th + jx th Z L = Z th = R th jx th Z T = Z th + Z L = 2R th I = V th Z T = V th 2R th V th = V m 0 o I = V m 2R th 0 o P ZL = I m 2 2 R th = V m 2R th 2 Rth 2
48 Maximum Average Power Transfer Definition P ZL = V m 2 R th 2 4R th 2 P ZL = V m 2 8R th V rms = V m 2 V 2 rms = V m 2 2 P ZL = V 2 rms 4R th P ZL = I m 2 2 R th = V m 2R th 2 Rth 2
49 Maximum Average Power Transfer Example (6) For the circuit shown: a) Determine the load impedance Z L that maximizes the average power drawn from the circuit. b) What is the maximum average power?
Sinusoidal Steady State Analysis (AC Analysis) Part I
Sinusoidal Steady State Analysis (AC Analysis) Part I Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/
More informationThree Phase Circuits
Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/ OUTLINE Previously on ELCN102 Three Phase Circuits Balanced
More informationLecture 11  AC Power
 AC Power 11/17/2015 Reading: Chapter 11 1 Outline Instantaneous power Complex power Average (real) power Reactive power Apparent power Maximum power transfer Power factor correction 2 Power in AC Circuits
More information11. AC Circuit Power Analysis
. AC Circuit Power Analysis Often an integral part of circuit analysis is the determination of either power delivered or power absorbed (or both). In this chapter First, we begin by considering instantaneous
More informationIntroduction to AC Circuits (Capacitors and Inductors)
Introduction to AC Circuits (Capacitors and Inductors) Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/
More informationChapter 5 SteadyState Sinusoidal Analysis
Chapter 5 SteadyState Sinusoidal Analysis Chapter 5 SteadyState Sinusoidal Analysis 1. Identify the frequency, angular frequency, peak value, rms value, and phase of a sinusoidal signal. 2. Solve steadystate
More information12. Introduction and Chapter Objectives
Real Analog  Circuits 1 Chapter 1: SteadyState Sinusoidal Power 1. Introduction and Chapter Objectives In this chapter we will address the issue of power transmission via sinusoidal or AC) signals. This
More informationRefresher course on Electrical fundamentals (Basics of A.C. Circuits) by B.M.Vyas
Refresher course on Electrical fundamentals (Basics of A.C. Circuits) by B.M.Vyas A specifically designed programme for Da Afghanistan Breshna Sherkat (DABS) Afghanistan 1 Areas Covered Under this Module
More informationSinusoidal Response of RLC Circuits
Sinusoidal Response of RLC Circuits Series RL circuit Series RC circuit Series RLC circuit Parallel RL circuit Parallel RC circuit RL Series Circuit RL Series Circuit RL Series Circuit Instantaneous
More informationEE 3120 Electric Energy Systems Study Guide for Prerequisite Test Wednesday, Jan 18, pm, Room TBA
EE 3120 Electric Energy Systems Study Guide for Prerequisite Test Wednesday, Jan 18, 2006 67 pm, Room TBA First retrieve your EE2110 final and other course papers and notes! The test will be closed book
More informationAC Power Analysis. Chapter Objectives:
AC Power Analysis Chapter Objectives: Know the difference between instantaneous power and average power Learn the AC version of maximum power transfer theorem Learn about the concepts of effective or value
More informationChapter 10: Sinusoidal SteadyState Analysis
Chapter 10: Sinusoidal SteadyState Analysis 1 Objectives : sinusoidal functions Impedance use phasors to determine the forced response of a circuit subjected to sinusoidal excitation Apply techniques
More information4/27 Friday. I have all the old homework if you need to collect them.
4/27 Friday Last HW: do not need to turn it. Solution will be posted on the web. I have all the old homework if you need to collect them. Final exam: 79pm, Monday, 4/30 at Lambert Fieldhouse F101 Calculator
More informationChapter 5. Department of Mechanical Engineering
Source Transformation By KVL: V s =ir s + v By KCL: i s =i + v/r p is=v s /R s R s =R p V s /R s =i + v/r s i s =i + v/r p Two circuits have the same terminal voltage and current Source Transformation
More informationBASIC NETWORK ANALYSIS
SECTION 1 BASIC NETWORK ANALYSIS A. Wayne Galli, Ph.D. Project Engineer Newport News Shipbuilding SeriesParallel dc Network Analysis......................... 1.1 BranchCurrent Analysis of a dc Network......................
More informationChapter 10 Sinusoidal Steady State Analysis Chapter Objectives:
Chapter 10 Sinusoidal Steady State Analysis Chapter Objectives: Apply previously learn circuit techniques to sinusoidal steadystate analysis. Learn how to apply nodal and mesh analysis in the frequency
More informationSinusoidal SteadyState Analysis
Chapter 4 Sinusoidal SteadyState Analysis In this unit, we consider circuits in which the sources are sinusoidal in nature. The review section of this unit covers most of section 9.1 9.9 of the text.
More informationChapter 10 AC Analysis Using Phasors
Chapter 10 AC Analysis Using Phasors 10.1 Introduction We would like to use our linear circuit theorems (Nodal analysis, Mesh analysis, Thevenin and Norton equivalent circuits, Superposition, etc.) to
More informationUNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 1.0 Kirchoff s Law Kirchoff s Current Law (KCL) states at any junction in an electric circuit the total current flowing towards that junction is equal
More informationChapter 10: Sinusoidal SteadyState Analysis
Chapter 10: Sinusoidal SteadyState Analysis 10.1 10.2 10.3 10.4 10.5 10.6 10.9 Basic Approach Nodal Analysis Mesh Analysis Superposition Theorem Source Transformation Thevenin & Norton Equivalent Circuits
More informationSinusoids and Phasors
CHAPTER 9 Sinusoids and Phasors We now begins the analysis of circuits in which the voltage or current sources are timevarying. In this chapter, we are particularly interested in sinusoidally timevarying
More informationThevenin Norton Equivalencies  GATE Study Material in PDF
Thevenin Norton Equivalencies  GATE Study Material in PDF In these GATE 2018 Notes, we explain the Thevenin Norton Equivalencies. Thevenin s and Norton s Theorems are two equally valid methods of reducing
More informationModule 4. Singlephase AC Circuits
Module 4 Singlephase AC Circuits Lesson 14 Solution of Current in RLC Series Circuits In the last lesson, two points were described: 1. How to represent a sinusoidal (ac) quantity, i.e. voltage/current
More informationBFF1303: ELECTRICAL / ELECTRONICS ENGINEERING. Alternating Current Circuits : Basic Law
BFF1303: ELECTRICAL / ELECTRONICS ENGINEERING Alternating Current Circuits : Basic Law Ismail Mohd Khairuddin, Zulkifil Md Yusof Faculty of Manufacturing Engineering Universiti Malaysia Pahang Alternating
More informationConsider a simple RC circuit. We might like to know how much power is being supplied by the source. We probably need to find the current.
AC power Consider a simple RC circuit We might like to know how much power is being supplied by the source We probably need to find the current R 10! R 10! is VS Vmcosωt Vm 10 V f 60 Hz V m 10 V C 150
More informationChapter 9 Objectives
Chapter 9 Engr8 Circuit Analysis Dr Curtis Nelson Chapter 9 Objectives Understand the concept of a phasor; Be able to transform a circuit with a sinusoidal source into the frequency domain using phasor
More informationBASIC PRINCIPLES. Power In SinglePhase AC Circuit
BASIC PRINCIPLES Power In SinglePhase AC Circuit Let instantaneous voltage be v(t)=v m cos(ωt+θ v ) Let instantaneous current be i(t)=i m cos(ωt+θ i ) The instantaneous p(t) delivered to the load is p(t)=v(t)i(t)=v
More informationPower Factor Improvement
Salman bin AbdulazizUniversity College of Engineering Electrical Engineering Department EE 2050Electrical Circuit Laboratory Power Factor Improvement Experiment # 4 Objectives: 1. To introduce the concept
More informationCircuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer
Circuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer J. McNames Portland State University ECE 221 Circuit Theorems Ver. 1.36 1
More informationBasics of Network Theory (PartI)
Basics of Network Theory (PartI). A square waveform as shown in figure is applied across mh ideal inductor. The current through the inductor is a. wave of peak amplitude. V 0 0.5 t (m sec) [Gate 987: Marks]
More informationEE313 Fall 2013 Exam #1 (100 pts) Thursday, September 26, 2013 Name. 1) [6 pts] Convert the following timedomain circuit to the RMS Phasor Domain.
Name If you have any questions ask them. Remember to include all units on your answers (V, A, etc). Clearly indicate your answers. All angles must be in the range 0 to +180 or 0 to 180 degrees. 1) [6 pts]
More informationD C Circuit Analysis and Network Theorems:
UNIT1 D C Circuit Analysis and Network Theorems: Circuit Concepts: Concepts of network, Active and passive elements, voltage and current sources, source transformation, unilateral and bilateral elements,
More informationECE2262 Electric Circuits. Chapter 5: Circuit Theorems
ECE2262 Electric Circuits Chapter 5: Circuit Theorems 1 Equivalence Linearity Superposition Thevenin s and Norton s Theorems Maximum Power Transfer Analysis of Circuits Using Circuit Theorems 2 5. 1 Equivalence
More informationMidterm Exam (closed book/notes) Tuesday, February 23, 2010
University of California, Berkeley Spring 2010 EE 42/100 Prof. A. Niknejad Midterm Exam (closed book/notes) Tuesday, February 23, 2010 Guidelines: Closed book. You may use a calculator. Do not unstaple
More informationECE2262 Electric Circuits
ECE2262 Electric Circuits Equivalence Chapter 5: Circuit Theorems Linearity Superposition Thevenin s and Norton s Theorems Maximum Power Transfer Analysis of Circuits Using Circuit Theorems 1 5. 1 Equivalence
More information15884/484 Electric Power Systems 1: DC and AC Circuits
15884/484 Electric Power Systems 1: DC and AC Circuits J. Zico Kolter October 8, 2013 1 Hydro Estimated U.S. Energy Use in 2010: ~98.0 Quads Lawrence Livermore National Laboratory Solar 0.11 0.01 8.44
More informationECE 421/521 Electric Energy Systems Power Systems Analysis I 2 Basic Principles. Instructor: Kai Sun Fall 2013
ECE 41/51 Electric Energy Systems Power Systems Analysis I Basic Principles Instructor: Kai Sun Fall 013 1 Outline Power in a 1phase AC circuit Complex power Balanced 3phase circuit Single Phase AC System
More informationTransformer. Transformer comprises two or more windings coupled by a common magnetic circuit (M.C.).
. Transformers Transformer Transformer comprises two or more windings coupled by a common magnetic circuit (M.C.). f the primary side is connected to an AC voltage source v (t), an AC flux (t) will be
More informationREACTANCE. By: Enzo Paterno Date: 03/2013
REACTANCE REACTANCE By: Enzo Paterno Date: 03/2013 5/2007 Enzo Paterno 1 RESISTANCE  R i R (t R A resistor for all practical purposes is unaffected by the frequency of the applied sinusoidal voltage or
More informationEE201 Review Exam I. 1. The voltage Vx in the circuit below is: (1) 3V (2) 2V (3) 2V (4) 1V (5) 1V (6) None of above
EE201, Review Probs Test 1 page1 Spring 98 EE201 Review Exam I Multiple Choice (5 points each, no partial credit.) 1. The voltage Vx in the circuit below is: (1) 3V (2) 2V (3) 2V (4) 1V (5) 1V (6)
More informationModule 2. DC Circuit. Version 2 EE IIT, Kharagpur
Module 2 DC Circuit Lesson 5 Nodevoltage analysis of resistive circuit in the context of dc voltages and currents Objectives To provide a powerful but simple circuit analysis tool based on Kirchhoff s
More informationBasics of Electric Circuits
António Dente Célia de Jesus February 2014 1 Alternating Current Circuits 1.1 Using Phasors There are practical and economic reasons justifying that electrical generators produce emf with alternating and
More informationEIT Review. Electrical Circuits DC Circuits. Lecturer: Russ Tatro. Presented by Tau Beta Pi The Engineering Honor Society 10/3/2006 1
EIT Review Electrical Circuits DC Circuits Lecturer: Russ Tatro Presented by Tau Beta Pi The Engineering Honor Society 10/3/2006 1 Session Outline Basic Concepts Basic Laws Methods of Analysis Circuit
More information3.1 Superposition theorem
Many electric circuits are complex, but it is an engineer s goal to reduce their complexity to analyze them easily. In the previous chapters, we have mastered the ability to solve networks containing independent
More informationLecture 05 Power in AC circuit
CA2627 Building Science Lecture 05 Power in AC circuit Instructor: Jiayu Chen Ph.D. Announcement 1. Makeup Midterm 2. Midterm grade Grade 25 20 15 10 5 0 10 15 20 25 30 35 40 Grade Jiayu Chen, Ph.D. 2
More informationEXP. NO. 3 Power on (resistive inductive & capacitive) load Series connection
OBJECT: To examine the power distribution on (R, L, C) series circuit. APPARATUS 1signal function generator 2 Oscilloscope, A.V.O meter 3 Resisters & inductor &capacitor THEORY the following form for
More informationECE 420. Review of Three Phase Circuits. Copyright by Chanan Singh, Panida Jirutitijaroen, and Hangtian Lei, For educational use onlynot for sale.
ECE 40 Review of Three Phase Circuits Outline Phasor Complex power Power factor Balanced 3Ф circuit Read Appendix A Phasors and in steady state are sinusoidal functions with constant frequency 5 0 15 10
More informationElectric Circuits II Sinusoidal Steady State Analysis. Dr. Firas Obeidat
Electric Circuits II Sinusoidal Steady State Analysis Dr. Firas Obeidat 1 Table of Contents 1 2 3 4 5 Nodal Analysis Mesh Analysis Superposition Theorem Source Transformation Thevenin and Norton Equivalent
More informationElectric Circuit Theory
Electric Circuit Theory Nam Ki Min nkmin@korea.ac.kr 01094192320 Chapter 11 Sinusoidal SteadyState Analysis Nam Ki Min nkmin@korea.ac.kr 01094192320 Contents and Objectives 3 Chapter Contents 11.1
More informationFundamentals of Electric Circuits, Second Edition  Alexander/Sadiku
Chapter 3, Problem 9(8). Find V x in the network shown in Fig. 3.78. Figure 3.78 Chapter 3, Solution 9(8). Consider the circuit below. 2 Ω 2 Ω j 8 30 o I j 4 j 4 I 2 j2v For loop, 8 30 = (2 j4)i ji 2
More informationECE Spring 2017 Final Exam
ECE 20100 Spring 2017 Final Exam May 2, 2017 Section (circle below) Qi (12:30) 0001 Tan (10:30) 0004 Hosseini (7:30) 0005 Cui (1:30) 0006 Jung (11:30) 0007 Lin (9:30) 0008 PeleatoInarrea (2:30) 0009 Name
More information04Electric Power. ECEGR 452 Renewable Energy Systems
04Electric Power ECEGR 452 Renewable Energy Systems Overview Review of Electric Circuits Phasor Representation Electrical Power Power Factor Dr. Louie 2 Introduction Majority of the electrical energy
More informationLO 1: Three Phase Circuits
Course: EEL 2043 Principles of Electric Machines Class Instructor: Dr. Haris M. Khalid Email: hkhalid@hct.ac.ae Webpage: www.harismkhalid.com LO 1: Three Phase Circuits Three Phase AC System Three phase
More informationChapter 5 Objectives
Chapter 5 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 5 Objectives State and apply the property of linearity State and apply the property of superposition Investigate source transformations Define
More informationFall 2011 ME 2305 Network Analysis. Sinusoidal Steady State Analysis of RLC Circuits
Fall 2011 ME 2305 Network Analysis Chapter 4 Sinusoidal Steady State Analysis of RLC Circuits Engr. Humera Rafique Assistant Professor humera.rafique@szabist.edu.pk Faculty of Engineering (Mechatronics)
More informationEIT Review 1. FE/EIT Review. Circuits. John A. Camara, Electrical Engineering Reference Manual, 6 th edition, Professional Publications, Inc, 2002.
FE/EIT eview Circuits Instructor: uss Tatro eferences John A. Camara, Electrical Engineering eference Manual, 6 th edition, Professional Publications, Inc, 00. John A. Camara, Practice Problems for the
More informationELECTRIC POWER CIRCUITS BASIC CONCEPTS AND ANALYSIS
Contents ELEC46 Power ystem Analysis Lecture ELECTRC POWER CRCUT BAC CONCEPT AND ANALY. Circuit analysis. Phasors. Power in single phase circuits 4. Three phase () circuits 5. Power in circuits 6. ingle
More informationECE 5260 Microwave Engineering University of Virginia. Some Background: Circuit and Field Quantities and their Relations
ECE 5260 Microwave Engineering University of Virginia Lecture 2 Review of Fundamental Circuit Concepts and Introduction to Transmission Lines Although electromagnetic field theory and Maxwell s equations
More informationQUESTION BANK SUBJECT: NETWORK ANALYSIS (10ES34)
QUESTION BANK SUBJECT: NETWORK ANALYSIS (10ES34) NOTE: FOR NUMERICAL PROBLEMS FOR ALL UNITS EXCEPT UNIT 5 REFER THE EBOOK ENGINEERING CIRCUIT ANALYSIS, 7 th EDITION HAYT AND KIMMERLY. PAGE NUMBERS OF
More informationNotes on Electric Circuits (Dr. Ramakant Srivastava)
Notes on Electric ircuits (Dr. Ramakant Srivastava) Passive Sign onvention (PS) Passive sign convention deals with the designation of the polarity of the voltage and the direction of the current arrow
More informationSinusoidal Steady State Analysis
Sinusoidal Steady State Analysis 9 Assessment Problems AP 9. [a] V = 70/ 40 V [b] 0 sin(000t +20 ) = 0 cos(000t 70 ).. I = 0/ 70 A [c] I =5/36.87 + 0/ 53.3 =4+j3+6 j8 =0 j5 =.8/ 26.57 A [d] sin(20,000πt
More informationHomework 2 SJTU233. Part A. Part B. Problem 2. Part A. Problem 1. Find the impedance Zab in the circuit seen in the figure. Suppose that R = 5 Ω.
Homework 2 SJTU233 Problem 1 Find the impedance Zab in the circuit seen in the figure. Suppose that R = 5 Ω. Express Zab in polar form. Enter your answer using polar notation. Express argument in degrees.
More informationSolution: Based on the slope of q(t): 20 A for 0 t 1 s dt = 0 for 3 t 4 s. 20 A for 4 t 5 s 0 for t 5 s 20 C. t (s) 20 C. i (A) Fig. P1.
Problem 1.24 The plot in Fig. P1.24 displays the cumulative charge q(t) that has entered a certain device up to time t. Sketch a plot of the corresponding current i(t). q 20 C 0 1 2 3 4 5 t (s) 20 C Figure
More informationElectrical Circuit & Network
Electrical Circuit & Network January 1 2017 Website: www.electricaledu.com Electrical Engg.(MCQ) Question and Answer for the students of SSC(JE), PSC(JE), BSNL(JE), WBSEDCL, WBSETCL, WBPDCL, CPWD and State
More informationChapter 10 Objectives
Chapter 10 Engr8 Circuit Analysis Dr Curtis Nelson Chapter 10 Objectives Understand the following AC power concepts: Instantaneous power; Average power; Root Mean Squared (RMS) value; Reactive power; Coplex
More informationBasics of Network Theory (PartI)
Basics of Network Theory (PartI) 1. One coulomb charge is equal to the charge on (a) 6.24 x 10 18 electrons (b) 6.24 x 10 24 electrons (c) 6.24 x 10 18 atoms (d) none of the above 2. The correct relation
More informationEE 212 PASSIVE AC CIRCUITS
EE 212 PASSIVE AC CIRCUITS Condensed Text Prepared by: Rajesh Karki, Ph.D., P.Eng. Dept. of Electrical Engineering University of Saskatchewan About the EE 212 Condensed Text The major topics in the course
More informationSinusoidal Steady State Power Calculations
10 Sinusoidal Steady State Power Calculations Assessment Problems AP 10.1 [a] V = 100/ 45 V, Therefore I = 20/15 A P = 1 (100)(20)cos[ 45 (15)] = 500W, 2 A B Q = 1000sin 60 = 866.03 VAR, B A [b] V = 100/
More informationChapter 10: Sinusoids and Phasors
Chapter 10: Sinusoids and Phasors 1. Motivation 2. Sinusoid Features 3. Phasors 4. Phasor Relationships for Circuit Elements 5. Impedance and Admittance 6. Kirchhoff s Laws in the Frequency Domain 7. Impedance
More informationCHAPTER 4. Circuit Theorems
CHAPTER 4 Circuit Theorems The growth in areas of application of electrical circuits has led to an evolution from simple to complex circuits. To handle such complexity, engineers over the years have developed
More informationPhysics 116A Notes Fall 2004
Physics 116A Notes Fall 2004 David E. Pellett Draft v.0.9 Notes Copyright 2004 David E. Pellett unless stated otherwise. References: Text for course: Fundamentals of Electrical Engineering, second edition,
More informationChapter 5 Solution P5.22, 3, 6 P5.33, 5, 8, 15 P5.43, 6, 8, 16 P5.52, 4, 6, 11 P5.62, 4, 9
Chapter 5 Solution P5.22, 3, 6 P5.33, 5, 8, 15 P5.43, 6, 8, 16 P5.52, 4, 6, 11 P5.62, 4, 9 P 5.22 Consider the circuit of Figure P 5.22. Find i a by simplifying the circuit (using source transformations)
More informationCIRCUIT ANALYSIS II. (AC Circuits)
Will Moore MT & MT CIRCUIT ANALYSIS II (AC Circuits) Syllabus Complex impedance, power factor, frequency response of AC networks including Bode diagrams, secondorder and resonant circuits, damping and
More informationSINUSOIDAL STEADY STATE CIRCUIT ANALYSIS
SINUSOIDAL STEADY STATE CIRCUIT ANALYSIS 1. Introduction A sinusoidal current has the following form: where I m is the amplitude value; ω=2 πf is the angular frequency; φ is the phase shift. i (t )=I m.sin
More informationELEC 250: LINEAR CIRCUITS I COURSE OVERHEADS. These overheads are adapted from the Elec 250 Course Pack developed by Dr. Fayez Guibaly.
Elec 250: Linear Circuits I 5/4/08 ELEC 250: LINEAR CIRCUITS I COURSE OVERHEADS These overheads are adapted from the Elec 250 Course Pack developed by Dr. Fayez Guibaly. S.W. Neville Elec 250: Linear Circuits
More informationSeries & Parallel Resistors 3/17/2015 1
Series & Parallel Resistors 3/17/2015 1 Series Resistors & Voltage Division Consider the singleloop circuit as shown in figure. The two resistors are in series, since the same current i flows in both
More informationPOLYTECHNIC UNIVERSITY Electrical Engineering Department. EE SOPHOMORE LABORATORY Experiment 2 DC circuits and network theorems
POLYTECHNIC UNIVERSITY Electrical Engineering Department EE SOPHOMORE LABORATORY Experiment 2 DC circuits and network theorems Modified for Physics 18, Brooklyn College I. Overview of Experiment In this
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : CH_EE_B_Network Theory_098 Delhi Noida Bhopal Hyderabad Jaipur Lucknow ndore Pune Bhubaneswar Kolkata Patna Web: Email: info@madeeasy.in Ph: 056 CLASS TEST 089 ELECTCAL ENGNEENG Subject : Network
More information= 32.0\cis{38.7} = j Ω. Zab = Homework 2 SJTU233. Part A. Part B. Problem 2. Part A. Problem 1
Homework 2 SJTU233 Problem 1 Find the impedance Zab in the circuit seen in the figure. Suppose that R = 5 Ω. Express Zab in polar form. Enter your answer using polar notation. Express argument in degrees.
More informationToolbox: Electrical Systems Dynamics
Toolbox: Electrical Systems Dynamics Dr. John C. Wright MIT  PSFC 05 OCT 2010 Introduction Outline Outline AC and DC power transmission Basic electric circuits Electricity and the grid Image removed due
More informationAn op amp consisting of a complex arrangement of resistors, transistors, capacitors, and diodes. Here, we ignore the details.
CHAPTER 5 Operational Amplifiers In this chapter, we learn how to use a new circuit element called op amp to build circuits that can perform various kinds of mathematical operations. Op amp is a building
More informationPower and Energy Measurement
Power and Energy Measurement EIE 240 Electrical and Electronic Measurement April 24, 2015 1 Work, Energy and Power Work is an activity of force and movement in the direction of force (Joules) Energy is
More informationECE Spring 2015 Final Exam
ECE 20100 Spring 2015 Final Exam May 7, 2015 Section (circle below) Jung (1:30) 0001 Qi (12:30) 0002 Peleato (9:30) 0004 Allen (10:30) 0005 Zhu (4:30) 0006 Name PUID Instructions 1. DO NOT START UNTIL
More informationNotes for course EE1.1 Circuit Analysis TOPIC 10 2PORT CIRCUITS
Objectives: Introduction Notes for course EE1.1 Circuit Analysis 45 Reexamination of 1port subcircuits Admittance parameters for port circuits TOPIC 1 PORT CIRCUITS Gain and port impedance from port
More informationEE221  Practice for the Midterm Exam
EE1  Practice for the Midterm Exam 1. Consider this circuit and corresponding plot of the inductor current: Determine the values of L, R 1 and R : L = H, R 1 = Ω and R = Ω. Hint: Use the plot to determine
More informationChapter 2. Engr228 Circuit Analysis. Dr Curtis Nelson
Chapter 2 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 2 Objectives Understand symbols and behavior of the following circuit elements: Independent voltage and current sources; Dependent voltage and
More informationLecture #3. Review: Power
Lecture #3 OUTLINE Power calculations Circuit elements Voltage and current sources Electrical resistance (Ohm s law) Kirchhoff s laws Reading Chapter 2 Lecture 3, Slide 1 Review: Power If an element is
More informationLABORATORY MODULE ELECTRIC CIRCUIT
LABORATORY MODULE ELECTRIC CIRCUIT HIGH VOLTAGE AND ELECTRICAL MEASUREMENT LAB ELECTRICAL ENGINEERING DEPARTMENT FACULTY OF ENGINEERING UNIVERSITAS INDONESIA DEPOK 2018 MODULE 1 LABORATORY BRIEFING All
More informationBasic Electrical Circuits Analysis ECE 221
Basic Electrical Circuits Analysis ECE 221 PhD. Khodr Saaifan http://trsys.faculty.jacobsuniversity.de k.saaifan@jacobsuniversity.de 1 2 Reference: Electric Circuits, 8th Edition James W. Nilsson, and
More informationECE 205: Intro Elec & Electr Circuits
ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide Version 1.00 Created by Charles Feng http://www.fenguin.net ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 1 Contents 1 Introductory
More informationMAE140  Linear Circuits  Fall 14 Midterm, November 6
MAE140  Linear Circuits  Fall 14 Midterm, November 6 Instructions (i) This exam is open book. You may use whatever written materials you choose, including your class notes and textbook. You may use a
More informationECE 201 Fall 2009 Final Exam
ECE 01 Fall 009 Final Exam December 16, 009 Division 0101: Tan (11:30am) Division 001: Clark (7:30 am) Division 0301: Elliott (1:30 pm) Instructions 1. DO NOT START UNTIL TOLD TO DO SO.. Write your Name,
More informationELECTRONICS E # 1 FUNDAMENTALS 2/2/2011
FE Review 1 ELECTRONICS E # 1 FUNDAMENTALS Electric Charge 2 In an electric circuit it there is a conservation of charge. The net electric charge is constant. There are positive and negative charges. Like
More informationWork, Energy and Power
1 Work, Energy and Power Work is an activity of force and movement in the direction of force (Joules) Energy is the capacity for doing work (Joules) Power is the rate of using energy (Watt) P = W / t,
More informationPhysics 142 AC Circuits Page 1. AC Circuits. I ve had a perfectly lovely evening but this wasn t it. Groucho Marx
Physics 142 A ircuits Page 1 A ircuits I ve had a perfectly lovely evening but this wasn t it. Groucho Marx Alternating current: generators and values It is relatively easy to devise a source (a generator
More informationPower Systems  Basic Concepts and Applications  Part I
PDHonline Course E104 (1 PDH) Power ystems Basic Concepts and Applications Part I Instructor: hihmin Hsu PhD PE 01 PDH Online PDH Center 57 Meadow Estates Drive Fairfax A 006658 Phone & Fax: 709880088
More informationChapter 33. Alternating Current Circuits
Chapter 33 Alternating Current Circuits 1 Capacitor Resistor + Q = C V = I R R I + + Inductance d I Vab = L dt AC power source The AC power source provides an alternative voltage, Notation  Lower case
More informationSchedule. ECEN 301 Discussion #20 Exam 2 Review 1. Lab Due date. Title Chapters HW Due date. Date Day Class No. 10 Nov Mon 20 Exam Review.
Schedule Date Day lass No. 0 Nov Mon 0 Exam Review Nov Tue Title hapters HW Due date Nov Wed Boolean Algebra 3. 3.3 ab Due date AB 7 Exam EXAM 3 Nov Thu 4 Nov Fri Recitation 5 Nov Sat 6 Nov Sun 7 Nov Mon
More informationReview of Basic Electrical and Magnetic Circuit Concepts EE
Review of Basic Electrical and Magnetic Circuit Concepts EE 442642 Sinusoidal Linear Circuits: Instantaneous voltage, current and power, rms values Average (real) power, reactive power, apparent power,
More informationPower and Energy Measurement
Power and Energy Measurement ENE 240 Electrical and Electronic Measurement Class 11, February 4, 2009 werapon.chi@kmutt.ac.th 1 Work, Energy and Power Work is an activity of force and movement in the direction
More information