DOCSLIB.ORG

Chapter 5 Solution-1

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Chapter 5 Solution-1 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 ia by simplifying the circuit (using source transformations) to a single-loop circuit so that you need to write only one KVL equation to find ia. Figure P 5.2-2 Solution: 16 Finally, apply KVL: −+10 3ii + 4 − = 0 ∴= i 2.19 A aa3 a (checked using LNAP 8/15/02) P 5.2-3 Find vo using source transformations if i = 5/2 A in the circuit shown in Figure P 5.2-3. Hint: Reduce the circuit to a single mesh that contains the voltage source labeled vo. Answer: vo = 28 V Figure P 5.2-3 Solution: Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors: Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top: Source transformation at left; series resistors at right: Parallel resistors, then source transformation at left: Finally, apply KVL to loop −+6 iv (9 + 19) − 36 −o = 0 iv=5/ 2 ⇒=−+o 42 28 (5/ 2) = 28 V (checked using LNAP 8/15/02) P 5.2-6 Use source transformations to find the value of the voltage va in Figure P 5.2-6. Answer: va = 7 V Figure P 5.2-6 Solution: A source transformation on the right side of the circuit, followed by replacing series resistors with an equivalent resistor: Source transformations on both the right side and the left side of the circuit: Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources with an equivalent current source: 50( 100) 100 Finally, v = (0.21) =( 0.21) = 7 V a 50+ 100 3 (checked using LNAP 8/15/02) P5.3-3 The circuit shown in Figure P5.3-3 has two inputs, vs and is, and one output io. The output is related to the inputs by the equation ioss= ai + bv Given the following two facts: The output is io = 0.45 A when the inputs are is = 0.25 A and vs = 15 V. and The output is io = 0.30 A when the inputs are is = 0.50 A and vs = 0 V. Determine the values of the constants a and b and the values of the resistances are R 1 and R 2. Answers: a = 0.6 A/A, b = 0.02 A/V, R 1 = 30 Ω and R 2 = 20 Ω. Figure P5.3-3 Solution: From the 1st fact: 0.45=ab( 0.25) + ( 15) From the 2nd fact: 0.30 0.30=ab( 0.50) +( 0) ⇒= a =0.60 0.50 0.45− ( 0.60)( 0.25) Substituting gives 0.45=( 0.60)( 0.25) +bb( 15) ⇒= =0.02 15 Next, consider the circuit: R1 aiso= io1 = i = is vs =0 + RR12 R1 so 0.60 = ⇒=2RR12 3 RR12+ vs and bvso= io2 = i = is =0 RR12+ 11 so 0.02 = ⇒RR12 += =Ω50 RR12+ 0.02 Solving these equations gives R 1 = 30 Ω and R 2 = 20 Ω. Figure P5.3-5 P5.3-5 Determine v(t), the voltage across the vertical resistor in the circuit in Figure P5.3-5. Solution; We’ll use superposition. Let v1(t) the be the part of v(t) due to the voltage source acting alone. Similarly, let v2(t) the be the part of v(t) due to the voltage source acting alone. We can use these circuits to calculate v1(t) and v2(t). Notice that v1(t) is the voltage across parallel resistors. Using equivalent resistance, we calculate 40||10 = 8 Ω. Next, using voltage division we calculate 8 vt( ) =(12) = 2 V 1 8+ 40 Similarly v2(t) is the voltage across parallel resistors Using equivalent resistance we first determine 40||40 = 20 Ω and then calculate 20 vt( ) = (12cos( 5t)) = 8cos( 5 t) V 2 10+ 20 Using superposition vt( ) =+=+ v12( t) v( t) 2 8cos( 5t) V P 5.3-8 Use superposition to find the value of the current ix in Figure P 5.3-8. Answer: i = 3.5 mA Figure P5.3-8 Solution: Consider 8 V source only (open the 2 A source) Let i1 be the part of ix due to the 8 V voltage source. Apply KVL to the supermesh: 6(iii111) ++−= 3( ) 3( ) 80 82 i = = A 1 12 3 Consider 2 A source only (short the 8 V source) Let i2 be the part of ix due to the 2 A current source. Apply KVL to the supermesh: 6(ii22) + 3( ++ 23) i 2 = 0 −61 i = = − A 2 12 2 21 1 Finally, i=+=−= ii A x 12326 P 5.3-15 The circuit shown in Figure P 5.3-15 has three inputs: v1, i2, and v3. The output of the circuit is the current io. The output of the circuit is related to the inputs by i1 = avo + bv2 + ci3 where a, b, and c are constants. Determine the values of a, b, and c. Figure P 5.3-15 Solution: 10 v1 10 20 ii=− +− o2 10+ 40 20 ++ 12( 40� 10) 10+ 40 20++ 12( 40� 10) 20+ 12 v3 +− 40++( 20 12) 10++ 40�( 20 12) 111 io=− vi 12 +− +− v 3 200 10 62.5 So ab=−=−=−0.05, 0.1 and c0.016 (checked: LNAP 6/19/04) P 5.4-3 The circuit shown in Figure P 5.4-3b is the Thévenin equivalent circuit of the circuit shown in Figure P 5.4-3a. Find the value of the open-circuit voltage, voc, and Thévenin resistance, Rt. Answer: voc = 2 V and Rt = 4 Ω Figure P 5.4-3 Solution: The circuit from Figure P5.4-3a can be reduced to its Thevenin equivalent circuit in five steps: (a) (c) (b) (d) (e) Comparing (e) to Figure P5.4-3b shows that the Thevenin resistance is Rt = 4 Ω and the open circuit voltage, voc = 2 V. (checked using LNAP 8/15/02) P 5.4-6 Find the Thévenin equivalent circuit for the circuit shown in Figure P 5.4-6. Figure P 5.4-6 Solution: Find voc: 2 vv− v Apply KCL at the top, middle node: aa= a ++3 0 ⇒v = 18 V 36 a The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V Find isc: 2 vv− v v Apply KCL at the top, middle node: aa= a ++3 a ⇒v =−18 V 363 a v −18 Apply Ohm’s law to the right-hand 3 Ω resistor : i =a = = −6 V sc 33 voc 18 Finally: Rt = = =−Ω3 i sc −6 (checked using LNAP 8/15/02) P 5.4-8 A resistor, R, was connected to a circuit box as shown in Figure P 5.4-8. The voltage, v, was measured. The resistance was changed, and the voltage was measured again. The results are shown in the table. Determine the Thévenin equivalent of the circuit within the box and predict the voltage, v, when R = 8 kΩ. Figure P 5.4-8 Solution: From the given data: 2000 6 = v + oc Rt 2000 voc =1.2 V ⇒ 4000 R =−Ω1600 2 = v t + oc Rt 4000 When R = 8000 Ω, R vv= 8000 RR+ oc v = (1.2) = 1.5 V t −+1600 8000 P 5.4-16 An ideal voltmeter is modeled as an open circuit. A more realistic model of a voltmeter is a large resistance. Figure P 5.4- 16a shows a circuit with a voltmeter that measures the voltage vm. In Figure P 5.4-16b the voltmeter is replaced by the model of an ideal voltmeter, an open circuit. The voltmeter measures vmi, the ideal value of vm. As Rm → ∞, the voltmeter becomes an ideal voltmeter and vm → vmi. When Rm < ∞, the voltmeter is not ideal and vm > vmi. The difference between vm and vmi is a measurement error caused by the fact that the voltmeter is not ideal. (a) Determine the value of vmi. (b) Express the measurement error that occurs when Rm = 1000 Ω as a percentage of vmi. (c) Determine the minimum value of Rm required to ensure that the measurement error is smaller than 2 percent of vmi. Figure P 5.4-16 Solution: Replace the circuit by its Thevenin equivalent circuit: R v = m 5 m + Rm 50 (a) vvmi =limm = 5 V Rm →∞ (b) When Rvmm=1000 Ω=, 4.763 V so 5− 4.762 % error = ×=100 4.76% 5 R 55− m + Rm 50 Rm (c) 0.02 ≥ ⇒ ≥0.98 ⇒ Rm ≥Ω2450 5 Rm + 50 (checked: LNAP 6/16/04) P 5.5-2 Two black boxes are shown in Figure P 5.5-2. Box A contains the Thévenin equivalent of some linear circuit, and box B contains the Norton equivalent of the same circuit. With access to just the outsides of the boxes and their terminals, how can you determine which is which, using only one Figure P 5.5-2 shorting wire? Solution: When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt.
Load more

Recommended publications
  • EE2003 Circuit Theory
    Chapter 10: Sinusoidal Steady-State Analysis 10.1 Basic Approach 10.2 Nodal Analysis 10.3 Mesh Analysis 10.4 Superposition Theorem 10.5 Source Transformation 10.6 Thevenin & Norton Equivalent Circuits 10.7 Op Amp AC Circuits 10.8 Applications 10.9 Summary 1 10.1 Basic Approach • 3 Steps to Analyze AC Circuits: 1. Transform the circuit to the phasor or frequency domain. 2. Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc.). 3. Transform the resulting phasor to the time domain. Phasor Phasor Laplace xform Inv. Laplace xform Fourier xform Fourier xform Solve variables Time to Freq Freq to Time in Freq • Sinusoidal Steady-State Analysis: Frequency domain analysis of AC circuit via phasors is much easier than analysis of the circuit in the time domain. 2 10.2 Nodal Analysis The basic of Nodal Analysis is KCL. Example: Using nodal analysis, find v1 and v2 in the figure. 3 10.3 Mesh Analysis The basic of Mesh Analysis is KVL. Example: Find Io in the following figure using mesh analysis. 4 5 10.4 Superposition Theorem When a circuit has sources operating at different frequencies, • The separate phasor circuit for each frequency must be solved independently, and • The total response is the sum of time-domain responses of all the individual phasor circuits. Example: Calculate vo in the circuit using the superposition theorem. 6 4.3 Superposition Theorem (1) - Superposition states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (or currents through) that element due to EACH independent source acting alone.
    [Show full text]
  • Chung-Ang University School of Electrical and Electronics Engineering
    Lecture 04 Chung-Ang University School of Electrical and Electronics Engineering Prof. Kwee-Bo SIM, Michael Circuit Theory Chapter 04 : Circuit Theorems Your success as an engineer will be directly proportional to your ability to communicate! - Charles K. Alexander ☞ Learning Objectives 2 Circuit Theory Chapter 04 : Circuit Theorems Your success as an engineer will be directly proportional to your ability to communicate! - Charles K. Alexander 4.1 Introduction A major advantage of analyzing circuits using Kirchhoff’s laws as we did in Chapter 3 is that we can analyze a circuit without tampering with its original configuration. A major disadvantage of this approach is that, for a large, complex circuit, tedious computation is involved. The growth in areas of application of electric circuits has led to an evolution from simple to complex circuits. To handle the complexity, engineers over the years have developed some theorems to simplify circuit analysis. Such theorems include Thevenin’s and Norton’s theorems. Since these theorems are applicable to linear circuits, we first discuss the concept of circuit linearity. In addition to circuit theorems, we discuss the concepts of superposition, source transformation, and maximum power transfer in this chapter. The concepts we develop are applied in the last section to source modeling and resistance measurement. 3 Circuit Theory Chapter 04 : Circuit Theorems Your success as an engineer will be directly proportional to your ability to communicate! - Charles K. Alexander 4.2 Linear Property (1/2) Linearity is the property of an element describing a linear relationship between cause and effect. Although the property applies to many circuit elements, we shall limit its applicability to resistors in this chapter.
    [Show full text]
  • Chapter 5: Circuit Theorems
    Chapter 5: Circuit Theorems 1. Motivation 2. Source Transformation 3. Superposition (2.1 Linearity Property) 4. Thevenin’s Theorem 5. Norton’s Theorem 6. Maximum Power Transfer 7. Summary 1 5.1 Motivation If you are given the following circuit, are there any other alternative(s) to determine the voltage across 2Ω resistor? In Chapter 4, a circuit is analyzed without tampering with its original configuration. What are they? And how? Can we work it out by inspection? In Chapter 5, some theorems have been developed to simplify circuit analysis such as Thevenin’s and Norton’s theorems. The theorems are applicable to linear circuits. Discussion: Source Transformation, Linearity, & superposition. 2 5.2 Source Transformation (1) ‐ Like series‐parallel combination and wye‐delta transformation, source transformation is another tool for simplifying circuits. ‐ An equivalent circuit is one whose v-i characteristics are identical with the original circuit. ‐ A source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, and vice versa. • Transformation of independent sources The arrow of the current + + source is directed toward the positive terminal of the voltage source. ‐ ‐ • Transformation of dependent sources The source transformation + + is not possible when R = 0 for voltage source and R = ∞ for current source. 3 ‐ ‐ 5.2 Source Transformation (1) A voltage source vs connected in series with a resistor Rs and a current source is is connected in parallel with a resistor Rp are equivalent circuits provided that Rp Rs & vs Rsis 4 5.2 Source Transformation (3) Example: Find vo in the circuit using source transformation.
    [Show full text]
  • Current Source & Source Transformation Notes
    EE301 – CURRENT SOURCES / SOURCE CONVERSION Learning Objectives a. Analyze a circuit consisting of a current source, voltage source and resistors b. Convert a current source and a resister into an equivalent circuit consisting of a voltage source and a resistor c. Evaluate a circuit that contains several current sources in parallel Ideal sources An ideal source is an active element that provides a specified voltage or current that is completely independent of other circuit elements. DC Voltage DC Current Source Source Constant Current Sources The voltage across the current source (Vs) is dependent on how other components are connected to it. Additionally, the current source voltage polarity does not have to follow the current source’s arrow! 1 Example: Determine VS in the circuit shown below. Solution: 2 Example: Determine VS in the circuit shown above, but with R2 replaced by a 6 k resistor. Solution: 1 8/31/2016 EE301 – CURRENT SOURCES / SOURCE CONVERSION 3 Example: Determine I1 and I2 in the circuit shown below. Solution: 4 Example: Determine I1 and VS in the circuit shown below. Solution: Practical voltage sources A real or practical source supplies its rated voltage when its terminals are not connected to a load (open- circuited) but its voltage drops off as the current it supplies increases. We can model a practical voltage source using an ideal source Vs in series with an internal resistance Rs. Practical current source A practical current source supplies its rated current when its terminals are short-circuited but its current drops off as the load resistance increases. We can model a practical current source using an ideal current source in parallel with an internal resistance Rs.
    [Show full text]
  • Source Transformation
    1 HW Chapter 10: 14, 20, 26, 44, 52, 64, 74, 92. Henry Selvaraj 2 Source Transformation • Source transformation in frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance. • Or vice versa: Vs VZIsss I s Zs Henry Selvaraj 3 Example: using the source transformation method. We transform the voltage source to a current source and obtain the circuit. Convert the current source to a voltage source Henry Selvaraj 4 Thevenin and Norton Equivalency • Both Thevenin and Norton’s theorems are applied to AC circuits the same way as DC. • The only difference is the fact that the calculated values will be complex. VZIZZTh N N Th N Henry Selvaraj 5 Example • Find the Thevenin and Norton equivalent circuits at terminals a-b for each of the circuit. Henry Selvaraj 6 Op Amp AC Circuits • As long as the op amp is working in the linear range, frequency domain analysis can proceed just as it does for other circuits. • It is important to keep in mind the two qualities of an ideal op amp: – No current enters either input terminals. – The voltage across its input terminals is zero with negative feedback. Henry Selvaraj 7 Example if vs =3 cos 1000t V. frequency-domain equivalent Henry Selvaraj Example contd… 8 Applying KCL at node 1, Henry Selvaraj 9 Example For the differentiator shown in Fig. obtainVo/Vs. Find vo(t) when vs (t) = Vm sin ωt and ω = 1/RC. Henry Selvaraj 10 Example Henry Selvaraj 11 Application • The op amp circuit shown here is known as a capacitance multiplier.
    [Show full text]
  • Network Theory
    NETWORK THEORY For ELECTRICAL ENGINEERING INSTRUMENTATION ENGINEERING ELECTRONICS & COMMUNICATION ENGINEERING NETWORK THEORY SYLLABUS Network graph, KCL, KVL, Node and Mesh analysis, Transient response of dc and ac networks, Sinusoidal steady‐state analysis, Resonance, Passive filters, Ideal current and voltage sources, Thevenin’s theorem, Norton’s theorem, Superposition theorem, Maximum power transfer theorem, Two‐port networks, Three phase circuits, Power and power factor in ac circuits. ANALYSIS OF GATE PAPERS ELECTRONICS ELECTRICAL INSTRUMENTATION 1 Mark 2 Mark 1 Mark 2 Mark 1 Mark 2 Mark Exam Year Ques. Ques. Total Ques. Ques. Total Ques. Ques. Total 2003 4 7 18 3 6 15 ‐ 1 2 2004 5 5 15 1 7 15 ‐ ‐ ‐ 2005 5 6 17 4 7 18 ‐ 2 4 2006 6 ‐ 6 2 6 14 1 4 9 2007 2 4 10 ‐ 7 14 2 4 10 2008 2 7 16 2 6 14 2 7 16 2009 3 4 11 2 6 14 ‐ 2 4 2010 2 3 8 3 4 11 2 2 6 2011 3 3 9 3 5 13 2 3 8 2012 4 4 12 5 6 17 4 6 16 2013 3 6 15 2 3 8 3 5 13 2014 Set‐1 2 4 10 2 2 6 3 4 11 2014 Set‐2 2 4 10 3 2 7 ‐ ‐ ‐ 2014 Set‐3 2 4 10 3 3 9 ‐ ‐ ‐ 2014 Set‐4 2 4 10 ‐ ‐ ‐ ‐ ‐ ‐ 2015 Set‐1 4 3 10 4 3 10 3 5 13 2015 Set‐2 3 3 9 3 3 9 ‐ ‐ ‐ 2015 Set‐3 3 2 7 ‐ ‐ ‐ ‐ ‐ ‐ 2016 Set‐1 1 2 5 4 5 14 3 3 9 2016 Set‐2 4 2 8 5 4 13 ‐ ‐ ‐ 2016 Set‐3 1 3 7 ‐ ‐ ‐ ‐ ‐ ‐ 2017 Set‐1 2 3 8 1 2 5 4 4 12 2017 Set‐1 0 4 8 2 2 6 ‐ ‐ ‐ 2018 2 3 8 2 3 8 3 3 9 © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission CONTENTS Topics Page No 1.
    [Show full text]
  • Concept Inventory Assessment Instruments for Circuits Courses
    AC 2011-1962: CONCEPT INVENTORY ASSESSMENT INSTRUMENTS FOR CIRCUITS COURSES Tokunbo Ogunfunmi, Santa Clara University TOKUNBO OGUNFUNMI, Ph.D., P.E. is the Associate Dean for Research and Faculty Development in the School of Engineering at Santa Clara University, Santa Clara, California. He is also an Associate Professor of Electrical Engineering and Director of the Signal Processing Research Lab. (SPRL). He earned his BSEE (First Class Honors) from Obafemi Awolowo University (formerly University of Ife), Nigeria, his MSEE and PhDEE from Stanford University, Stanford, California. His teaching and research interests span the areas of Circuits and Systems, Digital Signal Processing (theory, applications and implementations), Adaptive Systems, VLSI/ASIC Design and Multimedia Signal Processing. He is a Senior Member of the IEEE, Member of Sigma Xi, AAAS and ASEE. Mahmudur Rahman, Santa Clara University Dr. Mahmudur Rahman received M.S. Engg. and Dr. Engg. from Tokyo Institute of Technology, and then worked as a research scientist in NEC Corporation at Tamagawa, Tokyo, Japan during 1981 -1985. He ac- tively co-organized 1st through 5th International Conference on Silicon Carbide and Related Materials in various capacities including Conference Chair and Editor of Conference Proceedings during 1987-1993. Presently he is an Associate Professor of Electrical Engineering and Director of the Electron Devices Laboratory at Santa Clara University. His current research interests include: i) Stable High Current Den- sity Carbon Nanotube Cold Field Emission Electron Source Technology ii) Wavelet Approach to Sys- tematic Quantitative Characterization of Surface Nanostructures of Thin-Films based on Scanning Probe Microscopy, iv) Statistical and Temperature Map Dependent Electromigration Modeling and Simulation of Deep-Submicron Processes Applied to Design for Manufacturing, v) Low Cost Efficient Thin-film Organic Solar Cells by Optimization of Pentacene/Fullerene Interface.
    [Show full text]
  • Fundamentals of Electrical Circuits
    Fundamentals of Electrical Circuits Printed September 2016 First printed in 2007 Background and Acknowledgements This material is intended for the first course sequence in Electrical Engineering focused on Electrical Circuit Analysis and Design. The content is derived from the author’s educational, engineering and management career, and teaching experience. Additionally, the following resources have informed the development of content and format: Katz, R. Contemporary Logic Design . (2005) Pearson. Wakerly, I. Digital Design . (2001) Prentice Hall. Sandige, R. Digital Design Essentials . (2002) Prentice Hall. Nilsson, J. Electrical Circuits . (2004) Pearson. MathWorks. MATLAB Reference Material Version R2000a . (2007) MathWorks I would like to extend special thanks to the many students and colleagues for their contributions in making this material a more effective learning tool. Further, I would invite the reader to forward corrections, additional topics, examples and problems to me for future updates. Thanks, Izad Khormaee www.EngrCS.com © 2009 Izad Khormaee, All Rights Reserved. Fundamentals of Electrical Circuits, V3.6 Page 2 Table of Contents CHAPTER 1. INTRODUCTION .............................................................................................................................. 6 1.1. OVERVIEW OF ELECTRICAL ENGINEERING .......................................................................................................... 7 1.2. PROBLEM SOLVING .............................................................................................................................................
    [Show full text]
  • Voltage/Current Sources & Source Transformations
    ELL 100 - Introduction to Electrical Engineering LECTURE 3: ANALYSIS OF DC CIRCUITS VOLTAGE/CURRENT SOURCES & SOURCE TRANSFORMATIONS Outline Introduction Voltage and Current Sources Passive Sign Convention Series and Parallel Connection of Sources Node, Branches and Loops Series and Parallel Elements Source Transformation Exercise/Numerical Analysis Unsolved Problems References 2 INTRODUCTION Electrical Circuits seem to be everywhere! 3 INTRODUCTION Analysis of Circuits implies: • Finding voltage across any element of the circuit. • Finding current through any element of the circuit. 4 INTRODUCTION Analysis of Circuits Modern trains are powered by electric Knowledge of circuit analysis enable oneself to motors. Their electrical systems are best use circuit simulator developed for students to analyzed using AC or phasor analysis learn analog, digital, and power electronics. techniques. 5 INTRODUCTION Circuit representation of a Physical Transformer Knowledge of circuit analysis helps to analyse any electrical device with their equivalent circuit representation. 6 INTRODUCTION DC Shunt Motor Lf Ra Circuit representation of a DC Shunt Motor Vdc A Rf AA 7 INTRODUCTION iPad Charger with internal Circuitry Internal Circuitry 8 INTRODUCTION Single Phase Inverter Internal Circuitry AC Load IAC IDC t t 9 INTRODUCTION Single phase diode bridge rectifier Equivalent Circuit 10 INTRODUCTION DC-DC step down converter Equivalent Circuit Regulated DC Output 11 INTRODUCTION Lighting System Parallel Connection Break in wire 220 or 110 volt AC Supply Rest of the bulbs are ON except the parallel path with break in circuit. 12 INTRODUCTION Lighting System Series Connection 220 or 110 volt Break AC Supply in wire Rest of the bulbs are OFF after breaking point. Correct? 13 INTRODUCTION Lighting System Series Connection 220 or 110 volt Break AC Supply in wire X No! TheseRest 2 bulbsof the will bulbs also be are OFF.
    [Show full text]
  • Laboratory Experiment 6 Pre-Reading Circuit Theorems ©2008 by Professor Mohamad H
    ECE 3300 Electrical Circuits: Laboratory Experiment 6 Pre-Reading Circuit Theorems ©2008 by Professor Mohamad H. Hassoun In this experiment, the student will verify two important circuit theorems: Thevenin’s Theorem and the Maximum Power Transfer Theorem. Learning Objectives By the end of this unit, the student should be able to build simple resistive circuits, powered by independent sources, and employ laboratory measurements to: Experimentally obtain the Thevenin equivalent circuit. Employ source transformation to the Thevenin circuit to obtain the Norton equivalent circuit. Experimentally verify the Maximum Power Transfer Theorem. Thevenin’s Theorem Back in 1883, a French engineer by the name M. L. Thevenin had developed a theorem that allows a portion of a complex resistive circuit to be reduced to a simple circuit consisting of a voltage source in series with a resistor. This simple circuit is called the Thevenin equivalent circuit. The following is a formal statement of Thevenin’s Theorem: For any circuit of resistive elements and power sources (current and/or voltage) with an identified terminal pair a-b, the circuit can be replaced by a series combination of an ideal voltage source vt and a resistor Rt. Here, the Thevenin voltage vt is the voltage across a-b (or the open-circuit voltage, voc) and Rt (the Thevenin resistance) is the ratio of the open-circuit voltage to the short-circuit current (isc) at the terminal pair a-b. Once the terminal pair a-b has been identified, the Thevenin equivalent circuit can be easily obtained, experimentally, by making one voltage measurement and one current measurement.
    [Show full text]
  • Response of First Order RL and RC Circuits
    Response of First Order RL and RC Circuits First Order Circuits: Overview In this chapter we will study circuits that have dc sources, resistors, and either inductors or capacitors (but not both). Such circuits are described by first order differential equations. They will include one or more switches that open or close at a specific point in time, causing the inductor or capacitor to see a new circuit configuration. This in turn will cause a time-dependent change in voltages and currents. We will find that the equations describing the voltages and currents in these circuits (i.e., the circuit responses) are exponential in time, and characterized by a single time constant. In other words, we will have responses of the form / ( ) ( ) − For the natural response, and ∝ / ( ) ( ) (1 ) − for the step response, where τ is the time constant ∝. These− are single time constant circuits. Natural response occurs when a capacitor or an inductor is connected, via a switching event, to a circuit that contains only an equivalent resistance (i.e., no independent sources). In that case, if the capacitor is initially charged with a voltage, or the inductor is initially carrying a current, the capacitor or inductor will release its energy to the resistance. The circuit below shows an inductor that was initially connected to a current source, which establishes a current in the inductor. The switching event at t = t0 results in the inductor being connected to only a resistance. In this case, the inductor current iL(t) will decrease exponentially in time. t = t0 R i (t) S L L R iS We next consider a capacitor initially connected to a voltage source, which establishes a voltage across the capacitor.
    [Show full text]
  • Src Transformation & Thevenin
    2/19/20 Overview • Review concepts of • Voltage Equivalent Circuits: • Req Thevenin Theorem • Thevenin Equivalent Circuit • Equivalent Vs-with-series-R • “Equivalent” V-I-R Behavior to an actual power supply or EGR 220, Chapter 4.5 - 4.11 ‘driving’ circuit. (not section 4.9) February 20, 2020 2 1 2 Equivalent Resistance •Equivalent circuits concept • Equivalent resistance and voltage are terminal- dependent a) IndistinguishaBle from each other, • Ohm’s Law tells us that V = I*R so… b) …In terms of the V – I – Req characteristics, V • R = /I c) …At the specified terminals (nodes) • Electrical resistance is the ratio of: • The (open circuit) voltage across a pair of nodes to • The (short circuit) through the pair of nodes 3 4 3 4 1 2/19/20 Source Transformation & Equivalents Source Transformation • Voltage Vab and Req can Be measured across any nodes of any device or circuit. • We are interested in this measured orcalculated V – I – R behavior Isc v v = i R!!!!!!!!!!!!! ⇔ !!!!!!!!!!!!!i = s s s s R Req • Caution: maintain polarity of sources (“sc” = short circuit; “oc” = open circuit) 5 6 Your Mission Will your computer melt/do nothing? • Find two circuits with equivalent behavior with 1) 1 current source + 1 resistor 2) 1 voltage source + 1 resistor • If you design a power source with output of • I = ∞� à Your computer will melt Find Voc and Isc • V = 0� à Your computer will Be a paper weight 7 8 7 8 2 2/19/20 Will your computer melt/do nothing? Later: Can these be equivalent? Find Voc and Isc 9 10 9 10 Later: Can these be equivalent? Later: Can these be equivalent? 11 12 11 12 3 2/19/20 • Find i in the circuit below Source Transformation • Find i in the circuit below 13 14 Thevenin Equivalent Thevenin Equivalent – Process 1) Find RTh: ⏚ 2) Find VTh: 15 16 15 16 4 2/19/20 Thevenin Equivalent – Process Thevenin Equivalent – Process 1) Find RTh: 1) Find RTh: a) Remove the load resistor (if there is one) a) Remove the load resistor (if there is one) b) Set all (independent) sources equal to zero.
    [Show full text]