Chapter 4 Homework solution: P4.2-2, 7 P4.3-2, 3, 6, 9 P4.4-2, 5, 8, 18 P4.5-2, 4, 5 P4.6-2, 4, 8 P4.7-2, 4, 9, 15 P4.8-2
|
|
- Diane Laurel Richardson
- 5 years ago
- Views:
Transcription
1 Chpter 4 Homework solution: P4.2-2, 7 P4.3-2, 3, 6, 9 P4.4-2, 5, 8, 18 P4.5-2, 4, 5 P4.6-2, 4, 8 P4.7-2, 4, 9, 15 P4.8-2 P P Determine the node voltges for the circuit of Figure Answer: v 1 = 2 V, v 2 = 30 V, nd v 3 = 24 V Figure P KCL t node 1: KCL t node 2: KCL t node 3: v v v = 0 5 v v = v v v v = v + 3 v 2 v = v v v = 3 v + 5 v = Solving gives v 1 = 2 V, v 2 = 30 V nd v 3 = 24 V. P The node voltges in the circuit shown in Figure P re v = 7 V nd v b = 10 V. Determine vlues of the current source current, i s, nd the resistnce, R. Figure P 4.2-7
2 Solution Apply KCL t node to get Apply KCL t node b to get v v v vb = + + = + + = + R = 4 Ω R 4 2 R 4 2 R 4 v vb vb vb is + = + = is + = + is = 4 A P The voltges v, v b, v c, nd v d in Figure P re the node voltges corresponding to nodes, b, c, nd d. The current i is the current in short circuit connected between nodes b nd c. Determine the vlues of v, v b, v c, nd v d nd of i. Answer: v = 12 V, v b = v c = 4 V, v d = 4 V, i = 2 ma Figure P Express the brnch voltge of ech voltge source in terms of its node voltges to get: KCL t node b: v = 12 V, v = v = v + 8 b c d ( 12) vb v vb = i = i vb + 12 = i KCL t the supernode corresponding to the 8 V source:
3 so ( ) vd = + i 4 = vd i 4000 v + 4= 4 v v = 4 v v = 4 V b d d d d Consequently 4 vd vb = vc = vd + 8 = 4 V nd i = = 2 ma 4000 Figure P4.3-3 P Determine the vlues of the power supplied by ech of the sources in the circuit shown in Figure P First, lbel the node voltges. Next, express the resistor currents in terms of the node voltges. Identify the supernode corresponding to the 24 V source
4 Apply KCL to the supernode to get The 12 V source supplies The 24 V source supplies ( ) 12 v 24 v 24 v = = 6v v = 32 V ( v ) ( ) = 12 = 4.8 W v = = 4.8 W The current source supplies ( ) 0.6v = = 19.2 W P The voltmeter in the circuit of Figure P mesures node voltge. The vlue of tht node voltge depends on the vlue of the resistnce R. () (b) Determine the vlue of the resistnce R tht will cuse the voltge mesured by the voltmeter to be 4 V. Determine the voltge mesured by the voltmeter when R = 1.2 kω = 1200 Ω. Answer: () 6 kω (b) 2 V Figure P 4.3-6
5 Lbel the voltge mesured by the meter. Notice tht this is node voltge. Write node eqution t the node t which the node voltge is mesured. 12 vm vm vm = R 3000 Tht is v m = 16 R = R 16 3 vm () The voltge mesured by the meter will be 4 volts when R = 6 kω. (b) The voltge mesured by the meter will be 2 volts when R = 1.2 kω. P Determine the vlues of the node voltges of the circuit shown in Figure P Figure P Express the voltge source voltges s functions of the node voltges to get v v = 5 nd v = Apply KCL to the supernode corresponding to the 5 V source to get
6 v1 v3 v = + = = 5v + 2v 5v Apply KCL t node 3 to get v1 v3 v3 v3 15 = + 15v1+ 28v3 = Solving, e.g. using MATLAB, gives So the node voltges re: v 1 5 v v = 80 v = v v v = 22.4 V, v = 27.4 V, v = 17.4 V, nd v = P Find i b for the circuit shown in Figure P Answer: i b = 12 ma Figure P Write nd solve node eqution: v 6 v v 4v + + = 0 v = 12 V i b v 4v = = 12 ma 3000 (checked using LNAP 8/13/02)
7 P Determine the vlue of the current i x in the circuit of Figure P Answer: i x = 2.4 A Figure P First, express the controlling current of the CCVS in v 2 terms of the node voltges: i x = 2 Next, express the controlled voltge in terms of the node voltges: v v2 = 3ix = 3 v2 = V 2 5 so i x = 12/5 A = 2.4 A. P Determine the vlue of the power supplied by the dependent source in Figure P Figure P 4.4-8
8 Lbel the node voltges. First, v 2 = 10 V due to the independent voltge source. Next, express the controlling current of the dependent source in terms of the node voltges: i v3 v2 v3 10 = = Now the controlled voltge of the dependent source cn be expressed s v v1 v3 = 8 i = 8 v1 = v Apply KCL to the supernode corresponding to the dependent source to get v1 v2 v1 v3 v2 v = Multiplying by 48 nd using v 2 = 10 V gives 16v + 9v = Substituting the erlier expression for v V 2 v + v = v = Then v 1 = V nd i = A. Applying KCL t node 2 gives v1 10 v1 = ib + 12 ib = 3+ 4 v1 = So i b = A. Finlly, the power supplied by the dependent source is ( ) b ( ) ( ) p= 8 i i = = W ( )
9 P The voltges v2, v3 nd v4for the circuit shown in Figure P re: Determine the vlues of the following: () The gin, A, of the VCVS (b) The resistnce R 5 (c) The currents i b nd i c (d) The power received by resistor R 4 v = 16 V, v = 8 V nd v = 6 V Figure P Given the node voltges v2 = 16 V, v3 = 8 V nd v4 = 6 V Av 16 8 V A = = = 4 v 8 6 V ( ) v3 v R5 = v4 R5 = = 45 Ω, i b = = 2 A nd ic = = A v p 4 = = = W 15 15
10 P The vlues of the mesh currents in the circuit shown in Figure P re i 1 = 2 A, i 2 = 3 A, nd i 3 = 4 A. Determine the vlues of the resistnce R nd of the voltges v 1 nd v 2 of the voltge sources. Answers: R = 12 Ω, v 1 = 4 V, nd v 2 = 28 V The mesh equtions re: Top mesh: 4 (2 3) + R(2) + 10(2 4) = 0 so R = 12 Ω. Figure P Bottom, right mesh: 8 (4 3) + 10 (4 2) + v 2 = 0 so v 2 = 28 V. Bottom left mesh v1 + 4 (3 2) + 8 (3 4) = 0 so v 1 = 4 V. (checked using LNAP 8/14/02)
11 P Determine the mesh currents, i nd i b, in the circuit shown in Figure P Figure P KVL loop 1: 25 i i + 75 i ( i i ) = 0 b 450 i 100 i = 2 b KVL loop 2: 100( i i ) i i i = 0 b b b b 100 i i = 4 Solving these equtions: i = 6.5 ma, i = 9.3 ma b b (checked using LNAP 8/14/02)
12 P Find the current i for the circuit of Figure P Hint: A short circuit cn be treted s 0-V voltge source. Mesh Equtions: Figure P mesh 1 : 2 i + 2 ( i i ) + 10 = mesh 2 : 2( i i ) + 4 ( i i ) = mesh 3 : ( i3 i2) + 6 i3 = 0 Solving: 5 i = i2 i = = A 17 (checked using LNAP 8/14/02) P Find v c for the circuit shown in Figure P Answer: v c = 15 V Mesh currents: mesh : i = 0.25 A mesh b: ib = 0.4 A Ohm s Lw: v = 100( i i ) = 100(0.15) =15 V c b Figure P 4.6-2
13 P Figure P Find v c for the circuit shown in Figure P Express the current source current in terms of the mesh currents: i b = i 0.02 Apply KVL to the supermesh: 250 i+ 100 ( i 0.02) + 9 = 0 i =.02 A = 20 ma v = 100( i 0.02) = 4 V c (checked using LNAP 8/14/02)
14 P Determine vlues of the mesh currents, i 1, i 2, nd i 3, in the circuit shown in Figure P Figure P Use units of V, ma nd kω. Express the currents to the supermesh to get Apply KVL to the supermesh to get Apply KVL to mesh 2 to get Solving, e.g. using MATLAB, gives i 1 i3 = 2 ( ) ( ) ( )( ) 4 i i + 1 i 3+ 1 i i = 0 i 5i + 5i = ( ) ( )( ) ( ) 2i + 4 i i + 1 i i = 0 1 i + 7i 4i = i 1 2 i i = 3 i = i 3 0 i 3 1 (checked: LNAP 6/21/04)
15 P4.7-2 Determine the vlues of the power supplied by the voltge source nd by the CCCS in the circuit shown Figure P4.7-2 Figure P4.7-2 First, lbel the mesh currents, tking dvntge of the current sources. Next, express the resistor currents in terms of the mesh currents: Apply KVL to the left mesh: ( ) 3 The 2 A voltge source supplies i ( ) i i 2 = 0 i = = ma 8 2 = = 0.25 mw = = = i i mw The CCCS supplies ( ) ( )( ) ( )( ) 2 P Determine the mesh current i for the circuit shown in Figure P Answer: i = 24 ma Figure P Express the controlling voltge of the dependent source s function of the mesh current: Apply KVL to the right mesh: v b = 100 (.006 i ) [ ] 100 (.006 i ) (.006 i ) i = 0 i = 24 ma (checked using LNAP 8/14/02)
16 P Determine the vlue of the resistnce R in the circuit shown in Figure P Figure P Notice tht i b nd 0.5 ma re the mesh currents. Apply KCL t the top node of the dependent source to get 3 1 ib = 4 ib ib = ma 6 Apply KVL to the supermesh corresponding to the dependent source to get b 3 ( R)( ) 3 3 ( R)( ) 5000 i = = R = = kω (checked: LNAP 6/21/04) P Determine the vlues of the mesh currents i 1 nd i 2 for the circuit shown in Figure P Figure P Expressing the dependent source currents in terms of the mesh currents we get: Apply KVL to mesh 2 to get ( ) i = 4i = 4 i + 1 4= i 4i ( ) ( ) 2i + 2 i i i = 0 2= 2i + 6i
17 Solving these equtions using MATLAB we get i 1 = 8 A nd i 2 = 3 A P The circuit shown in Figure P hs two inputs, v s nd i s, nd one output v o. The output is relted to the inputs by the eqution v o = i s + bv s where nd b re constnts to be determined. Determine the vlues nd b by () writing nd solving mesh equtions nd (b) writing nd solving node equtions. () Apply KVL to meshes 1 nd 2: Figure P So = 24 nd b = (b) ( ) ( ) 32i v + 96 i i = 0 1 s 1 s v + 30i i i = 0 s 2 2 s 150i2 =+ 120is vs 4 vs i2 = is v = 30i = 24i v 5 o 2 s s Apply KCL to the supernode corresponding to the voltge source to get
18 So Then ( ) v vs + vo v vo vs + vo vo + = i s vs + vo vo vs vo = + = v = 24i v 5 o s s So = 24 nd b = (checked: LNAP 5/24/04)
Fundamentals of Electrical Circuits - Chapter 3
Fundmentls of Electricl Circuits Chpter 3 1S. For the circuits shown elow, ) identify the resistors connected in prllel ) Simplify the circuit y replcing prllel connect resistors with equivlent resistor.
More informationChapter 5 Solution P5.2-2, 3, 6 P5.3-3, 5, 8, 15 P5.4-3, 6, 8, 16 P5.5-2, 4, 6, 11 P5.6-2, 4, 9
Chapter 5 Solution P5.2-2, 3, 6 P5.3-3, 5, 8, 15 P5.4-3, 6, 8, 16 P5.5-2, 4, 6, 11 P5.6-2, 4, 9 P 5.2-2 Consider the circuit of Figure P 5.2-2. Find i a by simplifying the circuit (using source transformations)
More informationElectric Circuits I. Nodal Analysis. Dr. Firas Obeidat
Electric Circuits I Nodal Analysis Dr. Firas Obeidat 1 Nodal Analysis Without Voltage Source Nodal analysis, which is based on a systematic application of Kirchhoff s current law (KCL). A node is defined
More information332:221 Principles of Electrical Engineering I Fall Hourly Exam 2 November 6, 2006
2:221 Principles of Electricl Engineering I Fll 2006 Nme of the student nd ID numer: Hourly Exm 2 Novemer 6, 2006 This is closed-ook closed-notes exm. Do ll your work on these sheets. If more spce is required,
More informationChapter 4: Methods of Analysis
Chapter 4: Methods of Analysis When SCT are not applicable, it s because the circuit is neither in series or parallel. There exist extremely powerful mathematical methods that use KVL & KCL as its basis
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 informationCURRENT SOURCES EXAMPLE 1 Find the source voltage Vs and the current I1 for the circuit shown below SOURCE CONVERSIONS
CURRENT SOURCES EXAMPLE 1 Find the source voltage Vs and the current I1 for the circuit shown below EXAMPLE 2 Find the source voltage Vs and the current I1 for the circuit shown below SOURCE CONVERSIONS
More informationEE292: Fundamentals of ECE
EE292: Fundamentals of ECE Fall 2012 TTh 10:00-11:15 SEB 1242 Lecture 4 120906 http://www.ee.unlv.edu/~b1morris/ee292/ 2 Outline Review Voltage Divider Current Divider Node-Voltage Analysis 3 Network Analysis
More informationModule 2. DC Circuit. Version 2 EE IIT, Kharagpur
Module 2 DC Circuit Lesson 5 Node-voltage 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 informationEE-201 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
EE-201, Review Probs Test 1 page-1 Spring 98 EE-201 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 informationSeries & Parallel Resistors 3/17/2015 1
Series & Parallel Resistors 3/17/2015 1 Series Resistors & Voltage Division Consider the single-loop circuit as shown in figure. The two resistors are in series, since the same current i flows in both
More informationDIRECT CURRENT CIRCUITS
DRECT CURRENT CUTS ELECTRC POWER Consider the circuit shown in the Figure where bttery is connected to resistor R. A positive chrge dq will gin potentil energy s it moves from point to point b through
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 informationElectrical Technology (EE-101-F)
Electrical Technology (EE-101-F) Contents Series & Parallel Combinations KVL & KCL Introduction to Loop & Mesh Analysis Frequently Asked Questions NPTEL Link Series-Parallel esistances 1 V 3 2 There are
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 informationResistors. Consider a uniform cylinder of material with mediocre to poor to pathetic conductivity ( )
10/25/2005 Resistors.doc 1/7 Resistors Consider uniform cylinder of mteril with mediocre to poor to r. pthetic conductivity ( ) ˆ This cylinder is centered on the -xis, nd hs length. The surfce re of the
More information48520 Electronics & Circuits: Web Tutor
852 Electronics & Circuits: Web Tutor Topic : Resistive Circuits 2 Help for Exercise.: Nodal Analysis, circuits with I, R and controlled sources. The purpose of this exercise is to further extend Nodal
More informationV x 4 V x. 2k = 5
Review Problem: d) Dependent sources R3 V V R Vx - R2 Vx V2 ) Determine the voltage V5 when VV Need to find voltage Vx then multiply by dependent source multiplier () Node analysis 2 V x V x R R 2 V x
More informationLecture 7 notes Nodal Analysis
Lecture 7 notes Nodl Anlysis Generl Network Anlysis In mny cses you hve multiple unknowns in circuit, sy the voltges cross multiple resistors. Network nlysis is systemtic wy to generte multiple equtions
More informationChapter 2 Analysis Methods
Chapter Analysis Methods. Nodal Analysis Problem.. Two current sources with equal internal resistances feed a load as shown in Fig... I a ¼ 00 A; I b ¼ 00 A; R ¼ 00 X; R L ¼ 00 X: (a) Find the current
More informationBasic Electrical Circuits Analysis ECE 221
Basic Electrical Circuits Analysis ECE 221 PhD. Khodr Saaifan http://trsys.faculty.jacobs-university.de k.saaifan@jacobs-university.de 1 2 Reference: Electric Circuits, 8th Edition James W. Nilsson, and
More informationMAE140 - Linear Circuits - Winter 09 Midterm, February 5
Instructions MAE40 - Linear ircuits - Winter 09 Midterm, February 5 (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 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 informationDesigning Information Devices and Systems I Fall 2016 Babak Ayazifar, Vladimir Stojanovic Homework 6. This homework is due October 11, 2016, at Noon.
EECS 16A Designing Informtion Devices nd Systems I Fll 2016 Bk Ayzifr, Vldimir Stojnovic Homework 6 This homework is due Octoer 11, 2016, t Noon. 1. Homework process nd study group Who else did you work
More informationES250: Electrical Science. HW1: Electric Circuit Variables, Elements and Kirchhoff s Laws
ES250: Electrical Science HW1: Electric Circuit Variables, Elements and Kirchhoff s Laws Introduction Engineers use electric circuits to solve problems that are important to modern society, such as: 1.
More informationKirchhoff's Laws and Circuit Analysis (EC 2)
Kirchhoff's Laws and Circuit Analysis (EC ) Circuit analysis: solving for I and V at each element Linear circuits: involve resistors, capacitors, inductors Initial analysis uses only resistors Power sources,
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 informationVoltage Dividers, Nodal, and Mesh Analysis
Engr228 Lab #2 Voltage Dividers, Nodal, and Mesh Analysis Name Partner(s) Grade /10 Introduction This lab exercise is designed to further your understanding of the use of the lab equipment and to verify
More information196 Circuit Analysis with PSpice: A Simplified Approach
196 Circuit Anlysis with PSpice: A Simplified Approch i, A v L t, min i SRC 5 μf v C FIGURE P7.3 () the energy stored in the inductor, nd (c) the instntneous power input to the inductor. (Dul of Prolem
More informationSYMBOLIC ANALYSIS OF LINEAR ELECTRIC CIRCUITS
SYMBOLIC ANALYSIS OF LINEAR ELECTRIC CIRCUITS I. Tomčíová Technical university in Košice, Slovaia Abstract In present days there exist lots of programs such as PSPICE, TINA, which enable to solve circuits
More informationOverview. Before beginning this module, you should be able to: After completing this module, you should be able to:
Module.: Differentil Equtions for First Order Electricl Circuits evision: My 26, 2007 Produced in coopertion with www.digilentinc.com Overview This module provides brief review of time domin nlysis of
More informationECE 2100 Circuit Analysis
ECE 2100 Circuit Analysis Lesson 3 Chapter 2 Ohm s Law Network Topology: nodes, branches, and loops Daniel M. Litynski, Ph.D. http://homepages.wmich.edu/~dlitynsk/ esistance ESISTANCE = Physical property
More informationDesigning Information Devices and Systems I Spring 2018 Homework 7
EECS 16A Designing Informtion Devices nd Systems I Spring 2018 omework 7 This homework is due Mrch 12, 2018, t 23:59. Self-grdes re due Mrch 15, 2018, t 23:59. Sumission Formt Your homework sumission should
More informationI1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
2 The Prllel Circuit Electric Circuits: Figure 2- elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is
More informationChapter 3 Methods of Analysis: 1) Nodal Analysis
Chapter 3 Methods of Analysis: 1) Nodal Analysis Dr. Waleed Al-Hanafy waleed alhanafy@yahoo.com Faculty of Electronic Engineering, Menoufia Univ., Egypt MSA Summer Course: Electric Circuit Analysis I (ESE
More informationmywbut.com Mesh Analysis
Mesh Analysis 1 Objectives Meaning of circuit analysis; distinguish between the terms mesh and loop. To provide more general and powerful circuit analysis tool based on Kirchhoff s voltage law (KVL) only.
More informationChapter 4: Techniques of Circuit Analysis. Chapter 4: Techniques of Circuit Analysis
Chpter 4: Techniques of Circuit Anlysis Terminology Node-Voltge Method Introduction Dependent Sources Specil Cses Mesh-Current Method Introduction Dependent Sources Specil Cses Comprison of Methods Source
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 informationChapter 10: Symmetrical Components and Unbalanced Faults, Part II
Chpter : Symmetricl Components nd Unblnced Fults, Prt.4 Sequence Networks o Loded Genertor n the igure to the right is genertor supplying threephse lod with neutrl connected through impednce n to ground.
More informationDesigning Information Devices and Systems I Anant Sahai, Ali Niknejad. This homework is due October 19, 2015, at Noon.
EECS 16A Designing Informtion Devices nd Systems I Fll 2015 Annt Shi, Ali Niknejd Homework 7 This homework is due Octoer 19, 2015, t Noon. 1. Circuits with cpcitors nd resistors () Find the voltges cross
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 informationUnit 3: Direct current and electric resistance Electric current and movement of charges. Intensity of current and drift speed. Density of current in
Unit 3: Direct current nd electric resistnce Electric current nd movement of chrges. ntensity of current nd drift speed. Density of current in homogeneous currents. Ohm s lw. esistnce of homogeneous conductor
More informationChapter E - Problems
Chpter E - Prolems Blinn College - Physics 2426 - Terry Honn Prolem E.1 A wire with dimeter d feeds current to cpcitor. The chrge on the cpcitor vries with time s QHtL = Q 0 sin w t. Wht re the current
More informationI 3 2 = I I 4 = 2A
ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo 2.13 We re ske to use KCL to fin urrents 1 4. The key point in pplying KCL in this prolem is to strt with noe where only one of the urrents
More informationINTRODUCTION TO ELECTRONICS
INTRODUCTION TO ELECTRONICS Basic Quantities Voltage (symbol V) is the measure of electrical potential difference. It is measured in units of Volts, abbreviated V. The example below shows several ways
More informationChapter 10 Sinusoidal Steady State Analysis Chapter Objectives:
Chapter 10 Sinusoidal Steady State Analysis Chapter Objectives: Apply previously learn circuit techniques to sinusoidal steady-state analysis. Learn how to apply nodal and mesh analysis in the frequency
More informationChapter 7. Chapter 7
Chapter 7 Combination circuits Most practical circuits have combinations of series and parallel components. You can frequently simplify analysis by combining series and parallel components. An important
More informationSystematic Circuit Analysis (T&R Chap 3)
Systematic Circuit Analysis (T&R Chap 3) Nodevoltage analysis Using the voltages of the each node relative to a ground node, write down a set of consistent linear equations for these voltages Solve this
More informationVer 6186 E1.1 Analysis of Circuits (2015) E1.1 Circuit Analysis. Problem Sheet 2 - Solutions
Ver 8 E. Analysis of Circuits (0) E. Circuit Analysis Problem Sheet - Solutions Note: In many of the solutions below I have written the voltage at node X as the variable X instead of V X in order to save
More informationExam 2 Solutions ECE 221 Electric Circuits
Nme: PSU Student ID Numer: Exm 2 Solutions ECE 221 Electric Circuits Novemer 12, 2008 Dr. Jmes McNmes Keep your exm flt during the entire exm If you hve to leve the exm temporrily, close the exm nd leve
More informationSirindhorn International Institute of Technology Thammasat University at Rangsit
Sirindhorn International Institute of Technology Thammasat University at Rangsit School of Information, Computer and Communication Technology COURSE : ECS 304 Basic Electrical Engineering Lab INSTRUCTOR
More informationCOOKBOOK KVL AND KCL A COMPLETE GUIDE
1250 COOKBOOK KVL AND KCL A COMPLETE GUIDE Example circuit: 1) Label all source and component values with a voltage drop measurement (+,- ) and a current flow measurement (arrow): By the passive sign convention,
More informationModule 2. DC Circuit. Version 2 EE IIT, Kharagpur
Module DC Circuit Lesson 4 Loop Analysis of resistive circuit in the context of dc voltages and currents Objectives Meaning of circuit analysis; distinguish between the terms mesh and loop. To provide
More informationDesigning Information Devices and Systems II Fall 2016 Murat Arcak and Michel Maharbiz Homework 0. This homework is due August 29th, 2016, at Noon.
EECS 16B Designing Information Devices and Systems II Fall 2016 Murat Arcak and Michel Maharbiz Homework 0 This homework is due August 29th, 2016, at Noon. 1. Homework process and study group (a) Who else
More informationECE 1311: Electric Circuits. Chapter 2: Basic laws
ECE 1311: Electric Circuits Chapter 2: Basic laws Basic Law Overview Ideal sources series and parallel Ohm s law Definitions open circuits, short circuits, conductance, nodes, branches, loops Kirchhoff's
More informationPreamble. Circuit Analysis II. Mesh Analysis. When circuits get really complex methods learned so far will still work,
Preamble Circuit Analysis II Physics, 8 th Edition Custom Edition Cutnell & Johnson When circuits get really complex methods learned so far will still work, but they can take a long time to do. A particularly
More informationExperiment #6. Thevenin Equivalent Circuits and Power Transfer
Experiment #6 Thevenin Equivalent Circuits and Power Transfer Objective: In this lab you will confirm the equivalence between a complicated resistor circuit and its Thevenin equivalent. You will also learn
More informationENGG 225. David Ng. Winter January 9, Circuits, Currents, and Voltages... 5
ENGG 225 David Ng Winter 2017 Contents 1 January 9, 2017 5 1.1 Circuits, Currents, and Voltages.................... 5 2 January 11, 2017 6 2.1 Ideal Basic Circuit Elements....................... 6 3 January
More informationSolved Problems. Electric Circuits & Components. 1-1 Write the KVL equation for the circuit shown.
Solved Problems Electric Circuits & Components 1-1 Write the KVL equation for the circuit shown. 1-2 Write the KCL equation for the principal node shown. 1-2A In the DC circuit given in Fig. 1, find (i)
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 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 informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : ND_EE_NW_Analog Electronics_05088 Delhi Noida Bhopal Hyderabad Jaipur Lucknow ndore Pune Bhubaneswar Kolkata Patna Web: E-mail: info@madeeasy.in Ph: 0-4546 CLASS TEST 08-9 ELECTCAL ENGNEENG Subject
More informationHOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016
HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M
More informationIntroduction to Electronic Circuits. DC Circuit Analysis: Transient Response of RC Circuits
Introduction to Electronic ircuits D ircuit Anlysis: Trnsient esponse of ircuits Up until this point, we hve een looking t the Stedy Stte response of D circuits. StedyStte implies tht nothing hs chnged
More informationDC STEADY STATE CIRCUIT ANALYSIS
DC STEADY STATE CIRCUIT ANALYSIS 1. Introduction The basic quantities in electric circuits are current, voltage and resistance. They are related with Ohm s law. For a passive branch the current is: I=
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 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 informationSinusoidal Steady State Analysis (AC Analysis) Part II
Sinusoidal Steady State Analysis (AC Analysis) Part II Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/
More informationThe equivalent model of a certain op amp is shown in the figure given below, where R 1 = 2.8 MΩ, R 2 = 39 Ω, and A =
The equivalent model of a certain op amp is shown in the figure given below, where R 1 = 2.8 MΩ, R 2 = 39 Ω, and A = 10 10 4. Section Break Difficulty: Easy Learning Objective: Understand how real operational
More informationDesigning Information Devices and Systems I Discussion 8B
Lst Updted: 2018-10-17 19:40 1 EECS 16A Fll 2018 Designing Informtion Devices nd Systems I Discussion 8B 1. Why Bother With Thévenin Anywy? () Find Thévenin eqiuvlent for the circuit shown elow. 2kΩ 5V
More informationEE40. Lec 3. Basic Circuit Analysis. Prof. Nathan Cheung. Reading: Hambley Chapter 2
EE40 Lec 3 Basic Circuit Analysis Prof. Nathan Cheung 09/03/009 eading: Hambley Chapter Slide Outline Chapter esistors in Series oltage Divider Conductances in Parallel Current Divider Node-oltage Analysis
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 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. 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 informationMethods of Analysis and Selected Topics (dc)
Methods of Anlysis nd Selected Topics (dc) 8 Ojectives Become fmilir with the terminl chrcteristics of current source nd how to solve for the voltges nd currents of network using current sources nd/or
More informationmeas (1) calc calc I meas 100% (2) Diff I meas
Lab Experiment No. Ohm s Law I. Introduction In this lab exercise, you will learn how to connect the to network elements, how to generate a VI plot, the verification of Ohm s law, and the calculation of
More information6. MESH ANALYSIS 6.1 INTRODUCTION
6. MESH ANALYSIS INTRODUCTION PASSIVE SIGN CONVENTION PLANAR CIRCUITS FORMATION OF MESHES ANALYSIS OF A SIMPLE CIRCUIT DETERMINANT OF A MATRIX CRAMER S RULE GAUSSIAN ELIMINATION METHOD EXAMPLES FOR MESH
More informationResistive Network Analysis
C H A P T E R 3 Resistive Network Anlysis his chpter will illustrte the fundmentl techniques for the nlysis of resistive circuits. The methods introduced re sed on the circuit lws presented in Chpter 2:
More informationFigure Circuit for Question 1. Figure Circuit for Question 2
Exercises 10.7 Exercises Multiple Choice 1. For the circuit of Figure 10.44 the time constant is A. 0.5 ms 71.43 µs 2, 000 s D. 0.2 ms 4 Ω 2 Ω 12 Ω 1 mh 12u 0 () t V Figure 10.44. Circuit for Question
More informationE E 2320 Circuit Analysis. Calculating Resistance
E E 30 Circuit Analysis Lecture 03 Simple esistive Circuits it and Applications Calculating esistance l A 6 1.67 10 cm cu 6 al.7010 Area, A When conductor has uniform crosssection cm l 1 Temperature Coefficient
More informationECEN 325 Electronics
ECEN 325 Electronics Introduction Dr. Aydın İlker Karşılayan Texas A&M University Department of Electrical and Computer Engineering Ohm s Law i R i R v 1 v v 2 v v 1 v 2 v = v 1 v 2 v = v 1 v 2 v = ir
More informationHomework 1 solutions
Electric Circuits 1 Homework 1 solutions (Due date: 2014/3/3) This assignment covers Ch1 and Ch2 of the textbook. The full credit is 100 points. For each question, detailed derivation processes and accurate
More informationUNIVERSITY OF UTAH ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT. 10k. 3mH. 10k. Only one current in the branch:
UNIVERSITY OF UTAH ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT ECE 1270 HOMEWORK #6 Solution Summer 2009 1. After being closed a long time, the switch opens at t = 0. Find i(t) 1 for t > 0. t = 0 10kΩ
More informationCross-section section of DC motor. How does a DC Motor work? 2 Commutator Bars N X. DC Motors 26.1
DC Motors 26.1 How does DC Motor work? Crosssection section of DC motor Mgnetic field vector, B oft Iron Core (otor) Wire length vector, dl Force vector, df Current, i Permnent Mgnet (ttor) Crosssection
More informationName Class Date. Match each phrase with the correct term or terms. Terms may be used more than once.
Exercises 341 Flow of Chrge (pge 681) potentil difference 1 Chrge flows when there is between the ends of conductor 2 Explin wht would hppen if Vn de Grff genertor chrged to high potentil ws connected
More informationReview of Circuit Analysis
Review of Circuit Analysis Fundamental elements Wire Resistor Voltage Source Current Source Kirchhoff s Voltage and Current Laws Resistors in Series Voltage Division EE 42 Lecture 2 1 Voltage and Current
More informationKirchhoff's Laws and Maximum Power Transfer
German Jordanian University (GJU) Electrical Circuits Laboratory Section Experiment Kirchhoff's Laws and Maximum Power Transfer Post lab Report Mahmood Hisham Shubbak / / 8 Objectives: To learn KVL and
More informationCIRCUIT ANALYSIS TECHNIQUES
APPENDI B CIRCUIT ANALSIS TECHNIQUES The following methods can be used to combine impedances to simplify the topology of an electric circuit. Also, formulae are given for voltage and current division across/through
More informationENGR 2405 Class No Electric Circuits I
ENGR 2405 Class No. 48056 Electric Circuits I Dr. R. Williams Ph.D. rube.williams@hccs.edu Electric Circuit An electric circuit is an interconnec9on of electrical elements Charge Charge is an electrical
More informationPotential Changes Around a Circuit. You must be able to calculate potential changes around a closed loop.
Tody s gend: Potentil Chnges Around Circuit. You must e le to clculte potentil chnges round closed loop. Electromotive force (EMF), Terminl Voltge, nd Internl Resistnce. You must e le to incorporte ll
More informationChapter 2 Resistive Circuits
1. Sole circuits (i.e., find currents and oltages of interest) by combining resistances in series and parallel. 2. Apply the oltage-diision and current-diision principles. 3. Sole circuits by the node-oltage
More informationKirchhoff s Rules. Kirchhoff s rules are statements used to solve for currents and voltages in complicated circuits. The rules are
Kirchhoff s Rules Kirchhoff s rules are statements used to solve for currents and voltages in complicated circuits. The rules are Rule. Sum of currents into any junction is zero. i = 0 i 1 2 = 12 Why?
More information10/25/2005 Section 5_2 Conductors empty.doc 1/ Conductors. We have been studying the electrostatics of freespace (i.e., a vacuum).
10/25/2005 Section 5_2 Conductors empty.doc 1/3 5-2 Conductors Reding Assignment: pp. 122-132 We hve been studying the electrosttics of freespce (i.e., vcuum). But, the universe is full of stuff! Q: Does
More informationIn this lecture, we will consider how to analyse an electrical circuit by applying KVL and KCL. As a result, we can predict the voltages and currents
In this lecture, we will consider how to analyse an electrical circuit by applying KVL and KCL. As a result, we can predict the voltages and currents around an electrical circuit. This is a short lecture,
More informationPEP 2017 Assignment 12
of the filament?.16.. Aductile metal wire has resistance. What will be the resistance of this wire in terms of if it is stretched to three times its original length, assuming that the density and resistivity
More informationDesigning Information Devices and Systems I Spring 2018 Homework 8
EECS 16A Designing Informtion Devices nd Systems I Spring 2018 Homework 8 This homework is due Mrch 19, 2018, t 23:59. Self-grdes re due Mrch 22, 2018, t 23:59. Sumission Formt Your homework sumission
More informationELE B7 Power Systems Engineering. Power System Components Modeling
Power Systems Engineering Power System Components Modeling Section III : Trnsformer Model Power Trnsformers- CONSTRUCTION Primry windings, connected to the lternting voltge source; Secondry windings, connected
More informationHomework 3 Solution. Due Friday (5pm), Feb. 14, 2013
University of California, Berkeley Spring 2013 EE 42/100 Prof. K. Pister Homework 3 Solution Due Friday (5pm), Feb. 14, 2013 Please turn the homework in to the drop box located next to 125 Cory Hall (labeled
More informationBasics of Network Theory (Part-I)
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 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 information