Problem Set 5 Solutions


 Jeffery Hardy
 2 years ago
 Views:
Transcription
1 University of California, Berkeley Spring 01 EE /0 Prof. A. Niknejad Problem Set 5 Solutions Please note that these are merely suggested solutions. Many of these problems can be approached in different ways. 1. (a) No. v(t) is continuous, as v(0) = (60 0e 0 ) V = 0 V = v(0 ). (b) First we observe that i(0 ), since voltage is constant before the switch is thrown. After t, we have i(t) = C dv(t) = (0. mf)(00e 5t V) = 0e 5t ma Clearly, i(0 ) = 0 ma, so it is not continuous, leading to an instantaneous change. (c) Energy is given by w(0) = 1 Cv (t ) = 1 (0. mf)(0 V) = 0 mj. (d) After a long time, v(t) converges to 60 V. Using the same formula as before, we have w( ) = 1 Cv (t = ) = 1 (0. mf)(60 V).36 J.. Under DC conditions, all capacitors become open circuits. We redraw the circuit, along with some nodal voltages: v 3 0 kω v x v y 30 kω 5 kω v 1 v 15 kω 15 V v kω Since no current flows across the two leftmost resistors, we have v 1 = v = v x. Also, v 3 = v x v 7 and v = v y. To find v x and v y, we can use a simple voltage divider: v x = 15 (15 V) = 1.5 V 5 15 v y = (15 V) = 5 V 5 15 So v 1 = v = 1.5 V, v 3 = 7.5 V, v = 5 V.
2 3. The required energy is E = P t = (1.5 kw)(0 60s) = 3.6 MJ. So we need a capacitance given by C = E (3.6 MJ) = = 180 F. The amount of charge is V (00 V) Q = CV = (180 F)(00 V) = 36 kc. As a parallel plate capacitor, it would have an area of A = Cd 180 ɛ rɛ 0 = 6 = 1.36 (15)( ) 6 m.. At the time that the first switch closes, the capacitor sees the first 15 kω resistor. Hence the differential equation is v C 0 15 k (00 µ) dv C Rearranging the equation into a standard form gives us v C 3 dv C = 0 Hence, we have a solution of the form v C (t) = Ae t/3 B, where the first term is the homogeneous solution and the second is the particular solution. The time constant is simply the coefficient in front of dv C, so τ = RC = 3 s. The initial and final conditions are v(0) and v( ) = 0, so the full solution is v C (t) = 0 0e t/3, 0 t When switch closes, the equivalent resistance seen by the capacitor is now (15 kω) (15 kω) = 7.5 kω. So the time constant is τ = R eq C = 1.5 s. The form of the solution remains the same, but we have different conditions. The initial condition is equal to the voltage at t = from before: v C () = 0 0e /3 = 19.3 V. The final condition is v C ( ) =, as the capacitor is open in steadystate and we have a voltage divider. Hence our solution is v C (t) = 7308e t/1.5, t Graphically, we have a growing exponential to V, followed by a decaying exponential to V v C (t) t
3 5. At the time of the switch action, both capacitors have 0 V across them, as they act as open circuits. They do not change instantaneously, so they remain at 0 V each after the switch opens. Thus, the voltage drop across the resistor remains 0, and no current flows. We would also expect the current to be 0 at steady state at t =, so i(t) for all time. You would get this solution even if you were to solve the governing ODE due to the 0 boundary conditions. 6. The differential equation is v C (t) RC dv C(t) = v(t) = t If we use the standard homogeneous solution along with the suggested particular solution, we get v C (t) = A Bt De t/rc The initial condition is that v C (0), so A D. Now if we plug in the solution into our original ODE: A Bt De t/rc RC (B DRC ) e t/rc = t We now match terms to determine the coefficients values. The only linear term on the LHS is Bt, so we must have B = 1. The constant terms are A RCB = A RC, and this must be equal to 0, since there are no constants on the RHS. Thus, A = RC and D = A = RC. The full solution is v C (t) = RC t RCe t/rc = t RC(1 e t/rc ) While we need a specific value for RC for an accurate plot, we can sketch a general characteristic by assigning a value of, say, 1 to RC v C (t) t Notice that the voltage originally exhibits a delay due to the exponential term. As time passes, the capacitor s voltage becomes linear and follows that of the source almost identically after it gets past the initial inertia presented by the capacitor. 3
4 7. (a) Writing a KCL equation at v(t) gives us v C (t) 500 k ( µ)dv C(t) = i(t) = ( µ)e t v C (t) dv C(t) = 5e t (b) The time constant is τ = RC = 1 s, the coefficient in front of the derivative term. The complementary (homogeneous) solution thus takes the form v 0 (t) = Ae t. (c) If we were to use a particular solution of the form v p (t) = Ke t, we get the same form as that of the complementary solution. But the complementary solution only works for a homogenous ODE (i.e., the forcing function is 0): v C (t) dv C(t) = Ke t Ke t (d) If we were to try v p (t) = Kte t instead and plug it into the ODE, we now obtain Kte t Ke t Kte t = 5e t So the constant is K = 5. Combining it with the homogeneous solution, we have v C (t) = (A 5t)e t. If we assume that the capacitor is initially uncharged, then v C (0) and A. So our full solution is v C (t) = 5te t. 8. In using superposition, we consider one source at a time. Let s deal with v(t) first. Then the current source becomes an open circuit, and the equivalent resistance seen by the capacitor is 1 kω (the two 1 k are in series, and they are in parallel with the k). Hence the time constant is τ = R eq C = (1 kω)(0.5 µf).5 ms. The forcing function (voltage source) is constant, so we seek a solution of the form v O1 (t) = A Be 000t. Using the conditions v O1 (0) and v O1 ( ).5 V (the capacitor becomes an open circuit, and we have a simple voltage divider at v O1 (t)), the full solution is v O1 (t).5(1 e 000t ). Next we have the current source, and v(t) becomes a short. To make this easier to deal with, we can use a source transformation to turn the current source and resistor into a Thévenin equivalent with a voltage source (alternatively, you can write some KCL equations). 1 kω 1 kω v O (t) kω 1 V 0.5 µf Notice that the circuit is nearly identical to the previous one. So the solution is (without resolving it) v O (t).5(1 e 000t ). The full solution is the sum of the two individual components: v O (t) = 1 e 000t.
5 9. The golden rules hold, as the op amp is in negative feedback. Defining the node between the capacitor and resistor in the feedback loop to be v x, we can write KCL equations at v x and at the inverting input. C d(v x 0) v x v o R 0 v i C d(0 v x) R 1 We can integrate the second equation to solve for v x and plug it back into the first: v x (t) = 1 v i (t) v i(t) 1 v i (t) v o(t) R 1 C R 1 R 1 R C R v o (t) = R v i (t) 1 v i (t) R 1 R 1 C. Since the op amp merely serves as a buffer, we have v i (t) on the left side of the capacitor. The voltage on the right side is i o (t)r by Ohm s law. Then KCL gives us i o (t) C d(i o(t)r v i (t)) i o (t) RC di o(t) = C dv i(t) We cannot simplify further since we do not know the form of v i (t). We can infer that i o (t) will have a homogeneous solution Ae t/rc, but the form of the particular solution depends on v i (t). 5
Electric Circuits Fall 2015 Solution #5
RULES: Please try to work on your own. Discussion is permissible, but identical submissions are unacceptable! Please show all intermeate steps: a correct solution without an explanation will get zero cret.
More informationProblem Set 4 Solutions
University of California, Berkeley Spring 212 EE 42/1 Prof. A. Niknejad Problem Set 4 Solutions Please note that these are merely suggested solutions. Many of these problems can be approached in different
More informationEE292: Fundamentals of ECE
EE292: Fundamentals of ECE Fall 2012 TTh 10:0011:15 SEB 1242 Lecture 14 121011 http://www.ee.unlv.edu/~b1morris/ee292/ 2 Outline Review SteadyState Analysis RC Circuits RL Circuits 3 DC SteadyState
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 information0 t < 0 1 t 1. u(t) =
A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 13 p. 22/33 Step Response A unit step function is described by u(t) = ( 0 t < 0 1 t 1 While the waveform has an artificial jump (difficult
More informationECE2262 Electric Circuit
ECE2262 Electric Circuit Chapter 7: FIRST AND SECONDORDER RL AND RC CIRCUITS Response to FirstOrder RL and RC Circuits Response to SecondOrder RL and RC Circuits 1 2 7.1. Introduction 3 4 In dc steady
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 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 informationReal Analog Chapter 7: First Order Circuits. 7 Introduction and Chapter Objectives
1300 Henley Court Pullman, WA 99163 509.334.6306 www.store. digilent.com 7 Introduction and Chapter Objectives First order systems are, by definition, systems whose inputoutput relationship is a first
More informationRC Circuits (32.9) Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring / 1
(32.9) We have only been discussing DC circuits so far. However, using a capacitor we can create an RC circuit. In this example, a capacitor is charged but the switch is open, meaning no current flows.
More informationPHYSICS 171 UNIVERSITY PHYSICS LAB II. Experiment 6. Transient Response of An RC Circuit
PHYSICS 171 UNIVERSITY PHYSICS LAB II Experiment 6 Transient Response of An RC Circuit Equipment: Supplies: Function Generator, Dual Trace Oscilloscope.002 Microfarad, 0.1 Microfarad capacitors; 1 Kilohm,
More informationDesigning Information Devices and Systems I Spring 2018 Lecture Notes Note 20
EECS 16A Designing Information Devices and Systems I Spring 2018 Lecture Notes Note 20 Design Example Continued Continuing our analysis for countdown timer circuit. We know for a capacitor C: I = C dv
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 informationCapacitors. Chapter How capacitors work Inside a capacitor
Chapter 6 Capacitors In every device we have studied so far sources, resistors, diodes and transistors the relationship between voltage and current depends only on the present, independent of the past.
More informationProf. Anyes Taffard. Physics 120/220. Voltage Divider Capacitor RC circuits
Prof. Anyes Taffard Physics 120/220 Voltage Divider Capacitor RC circuits Voltage Divider The figure is called a voltage divider. It s one of the most useful and important circuit elements we will encounter.
More informationEE 40: Introduction to Microelectronic Circuits Spring 2008: Midterm 2
EE 4: Introduction to Microelectronic Circuits Spring 8: Midterm Venkat Anantharam 3/9/8 Total Time Allotted : min Total Points:. This is a closed book exam. However, you are allowed to bring two pages
More informationTo find the step response of an RC circuit
To find the step response of an RC circuit v( t) v( ) [ v( t) v( )] e tt The time constant = RC The final capacitor voltage v() The initial capacitor voltage v(t ) To find the step response of an RL circuit
More informationSolved Problems. Electric Circuits & Components. 11 Write the KVL equation for the circuit shown.
Solved Problems Electric Circuits & Components 11 Write the KVL equation for the circuit shown. 12 Write the KCL equation for the principal node shown. 12A In the DC circuit given in Fig. 1, find (i)
More informationResponse of SecondOrder Systems
Unit 3 Response of SecondOrder Systems In this unit, we consider the natural and step responses of simple series and parallel circuits containing inductors, capacitors and resistors. The equations which
More informationECE2262 Electric Circuits. Chapter 6: Capacitance and Inductance
ECE2262 Electric Circuits Chapter 6: Capacitance and Inductance Capacitors Inductors Capacitor and Inductor Combinations OpAmp Integrator and OpAmp Differentiator 1 CAPACITANCE AND INDUCTANCE Introduces
More informationSeries RC and RL Time Domain Solutions
ECE2205: Circuits and Systems I 6 1 Series RC and RL Time Domain Solutions In the last chapter, we saw that capacitors and inductors had element relations that are differential equations: i c (t) = C d
More informationRC, RL, and LCR Circuits
RC, RL, and LCR Circuits EK307 Lab Note: This is a two week lab. Most students complete part A in week one and part B in week two. Introduction: Inductors and capacitors are energy storage devices. They
More informationElectric Circuit Theory
Electric Circuit Theory Nam Ki Min nkmin@korea.ac.kr 01094192320 Chapter 8 Natural and Step Responses of RLC Circuits Nam Ki Min nkmin@korea.ac.kr 01094192320 8.1 Introduction to the Natural Response
More informationLinear Circuits. Concept Map 9/10/ Resistive Background Circuits. 5 Power. 3 4 Reactive Circuits. Frequency Analysis
Linear Circuits Dr. Bonnie Ferri Professor School of Electrical and Computer Engineering An introduction to linear electric components and a study of circuits containing such devices. School of Electrical
More informationSourceFree RC Circuit
First Order Circuits SourceFree RC Circuit Initial charge on capacitor q = Cv(0) so that voltage at time 0 is v(0). What is v(t)? Prof Carruthers (ECE @ BU) EK307 Notes Summer 2018 150 / 264 First Order
More informationCircuits Practice Websheet 18.1
Circuits Practice Websheet 18.1 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. How much power is being dissipated by one of the 10Ω resistors? a. 24
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 informationEE40: Introduction to µelectronic Circuits Lecture Notes
EE40: Introduction to µelectronic Circuits Lecture Notes Alessandro Pinto University of California at Berkeley 545P Cory Hall, Berkeley, CA 94720 apinto@eecs.berkeley.edu July 0, 2004 Contents First Order
More informationENGR 2405 Chapter 8. Second Order Circuits
ENGR 2405 Chapter 8 Second Order Circuits Overview The previous chapter introduced the concept of first order circuits. This chapter will expand on that with second order circuits: those that need a second
More informationProblem Set 1 Solutions (Rev B, 2/5/2012)
University of California, Berkeley Spring 2012 EE 42/100 Prof. A. Niknejad Problem Set 1 Solutions (Rev B, 2/5/2012) Please note that these are merely suggested solutions. Many of these problems can be
More informationEE292: Fundamentals of ECE
EE292: Fundamentals of ECE Fall 2012 TTh 10:0011:15 SEB 1242 Lecture 20 121101 http://www.ee.unlv.edu/~b1morris/ee292/ 2 Outline Chapters 13 Circuit Analysis Techniques Chapter 10 Diodes Ideal Model
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 informationMidterm Exam 2. Prof. Miloš Popović
Midterm Exam 2 Prof. Miloš Popović 100 min timed, closed book test. Write your name at top of every page (or initials on later pages) Aids: single page (single side) of notes, handheld calculator Work
More informationElectric Circuits. Overview. Hani Mehrpouyan,
Electric Circuits Hani Mehrpouyan, Department of Electrical and Computer Engineering, Lecture 15 (First Order Circuits) Nov 16 th, 2015 Hani Mehrpouyan (hani.mehr@ieee.org) Boise State c 2015 1 1 Overview
More informationAP Physics C. Electric Circuits III.C
AP Physics C Electric Circuits III.C III.C.1 Current, Resistance and Power The direction of conventional current Suppose the crosssectional area of the conductor changes. If a conductor has no current,
More informationE40M Review  Part 1
E40M Review Part 1 Topics in Part 1 (Today): KCL, KVL, Power Devices: V and I sources, R Nodal Analysis. Superposition Devices: Diodes, C, L Time Domain Diode, C, L Circuits Topics in Part 2 (Wed): MOSFETs,
More informationLAPLACE TRANSFORMATION AND APPLICATIONS. Laplace transformation It s a transformation method used for solving differential equation.
LAPLACE TRANSFORMATION AND APPLICATIONS Laplace transformation It s a transformation method used for solving differential equation. Advantages The solution of differential equation using LT, progresses
More informationDesigning Information Devices and Systems I Fall 2018 Lecture Notes Note Introduction: Opamps in Negative Feedback
EECS 16A Designing Information Devices and Systems I Fall 2018 Lecture Notes Note 18 18.1 Introduction: Opamps in Negative Feedback In the last note, we saw that can use an opamp as a comparator. However,
More informationECE2262 Electric Circuits. Chapter 6: Capacitance and Inductance
ECE2262 Electric Circuits Chapter 6: Capacitance and Inductance Capacitors Inductors Capacitor and Inductor Combinations 1 CAPACITANCE AND INDUCTANCE Introduces two passive, energy storing devices: Capacitors
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 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 informationEIT QuickReview Electrical Prof. Frank Merat
CIRCUITS 4 The power supplied by the 0 volt source is (a) 2 watts (b) 0 watts (c) 2 watts (d) 6 watts (e) 6 watts 4Ω 2Ω 0V i i 2 2Ω 20V Call the clockwise loop currents i and i 2 as shown in the drawing
More informationDEPARTMENT OF COMPUTER ENGINEERING UNIVERSITY OF LAHORE
DEPARTMENT OF COMPUTER ENGINEERING UNIVERSITY OF LAHORE NAME. Section 1 2 3 UNIVERSITY OF LAHORE Department of Computer engineering Linear Circuit Analysis Laboratory Manual 2 Compiled by Engr. Ahmad Bilal
More informationEEE105 Teori Litar I Chapter 7 Lecture #3. Dr. Shahrel Azmin Suandi Emel:
EEE105 Teori Litar I Chapter 7 Lecture #3 Dr. Shahrel Azmin Suandi Emel: shahrel@eng.usm.my What we have learnt so far? Chapter 7 introduced us to firstorder circuit From the last lecture, we have learnt
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 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 informationChapter 28. Direct Current Circuits
Chapter 28 Direct Current Circuits Circuit Analysis Simple electric circuits may contain batteries, resistors, and capacitors in various combinations. For some circuits, analysis may consist of combining
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 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 informationDesigning Information Devices and Systems I Discussion 8B
EECS 16A Spring 2018 Designing Information Devices and Systems I Discussion 8B 1. BioMolecule Detector We ve already seen how to build a biomolecule detector where biomolecules change the resistance
More informationChapter 4 Transients. Chapter 4 Transients
Chapter 4 Transients Chapter 4 Transients 1. Solve firstorder RC or RL circuits. 2. Understand the concepts of transient response and steadystate response. 1 3. Relate the transient response of firstorder
More informationPhysics 102: Lecture 7 RC Circuits
Physics 102: Lecture 7 C Circuits Physics 102: Lecture 7, Slide 1 C Circuits Circuits that have both resistors and capacitors: K Na Cl C ε K ε Na ε Cl S With resistance in the circuits, capacitors do not
More informationUniversity of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences. EECS 40 Midterm II
University of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences EECS 40 Midterm II Spring 2001 Prof. Roger T. Howe April 11, 2001 Name: Last, First Student
More informationElectrical Circuits (2)
Electrical Circuits (2) Lecture 7 Transient Analysis Dr.Eng. Basem ElHalawany Extra Reference for this Lecture Chapter 16 Schaum's Outline Of Theory And Problems Of Electric Circuits https://archive.org/details/theoryandproblemsofelectriccircuits
More informationUNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences
UNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences EECS 40 Spring 2000 Introduction to Microelectronic Devices Prof. King MIDTERM EXAMINATION
More informationThe RC Time Constant
The RC Time Constant Objectives When a directcurrent source of emf is suddenly placed in series with a capacitor and a resistor, there is current in the circuit for whatever time it takes to fully charge
More information2005 AP PHYSICS C: ELECTRICITY AND MAGNETISM FREERESPONSE QUESTIONS
2005 AP PHYSICS C: ELECTRICITY AND MAGNETISM In the circuit shown above, resistors 1 and 2 of resistance R 1 and R 2, respectively, and an inductor of inductance L are connected to a battery of emf e and
More informationECE Circuit Theory. Final Examination. December 5, 2008
ECE 212 H1F Pg 1 of 12 ECE 212  Circuit Theory Final Examination December 5, 2008 1. Policy: closed book, calculators allowed. Show all work. 2. Work in the provided space. 3. The exam has 3 problems
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 informationUnit 2: Modeling in the Frequency Domain. Unit 2, Part 4: Modeling Electrical Systems. First Example: Via DE. Resistors, Inductors, and Capacitors
Unit 2: Modeling in the Frequency Domain Part 4: Modeling Electrical Systems Engineering 582: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 20,
More informationDesigning Information Devices and Systems I Spring 2016 Elad Alon, Babak Ayazifar Midterm 2. Exam location: 145 Dwinelle, last SID# 2
EECS 16A Designing Information Devices and Systems I Spring 2016 Elad Alon, Babak Ayazifar Midterm 2 Exam location: 145 Dwinelle, last SID# 2 PRINT your student ID: PRINT AND SIGN your name:, (last) (first)
More informationCircuits with Capacitor and Inductor
Circuits with Capacitor and Inductor We have discussed so far circuits only with resistors. While analyzing it, we came across with the set of algebraic equations. Hereafter we will analyze circuits with
More informationCS 436 HCI Technology Basic Electricity/Electronics Review
CS 436 HCI Technology Basic Electricity/Electronics Review *Copyright 19972008, Perry R. Cook, Princeton University August 27, 2008 1 Basic Quantities and Units 1.1 Charge Number of electrons or units
More informationExperiment 4. RC Circuits. Observe and qualitatively describe the charging and discharging (decay) of the voltage on a capacitor.
Experiment 4 RC Circuits 4.1 Objectives Observe and qualitatively describe the charging and discharging (decay) of the voltage on a capacitor. Graphically determine the time constant τ for the decay. 4.2
More informationLecture 4: DC & Transient Response
Introduction to CMOS VLSI Design Lecture 4: DC & Transient Response David Harris Harvey Mudd College Spring 004 Outline DC Response Logic Levels and Noise Margins Transient Response Delay Estimation Slide
More informationEECE251. Circuit Analysis I. Set 4: Capacitors, Inductors, and FirstOrder Linear Circuits
EECE25 Circuit Analysis I Set 4: Capacitors, Inductors, and FirstOrder Linear Circuits Shahriar Mirabbasi Department of Electrical and Computer Engineering University of British Columbia shahriar@ece.ubc.ca
More informationE40M Capacitors. M. Horowitz, J. Plummer, R. Howe
E40M Capacitors 1 Reading Reader: Chapter 6 Capacitance A & L: 9.1.1, 9.2.1 2 Why Are Capacitors Useful/Important? How do we design circuits that respond to certain frequencies? What determines how fast
More informationBasic RL and RC Circuits RL TRANSIENTS: STORAGE CYCLE. Engineering Collage Electrical Engineering Dep. Dr. Ibrahim Aljubouri
st Class Basic RL and RC Circuits The RL circuit with D.C (steady state) The inductor is short time at Calculate the inductor current for circuits shown below. I L E R A I L E R R 3 R R 3 I L I L R 3 R
More informationCapacitance, Resistance, DC Circuits
This test covers capacitance, electrical current, resistance, emf, electrical power, Ohm s Law, Kirchhoff s Rules, and RC Circuits, with some problems requiring a knowledge of basic calculus. Part I. Multiple
More informationElectronics Capacitors
Electronics Capacitors Wilfrid Laurier University October 9, 2015 Capacitor an electronic device which consists of two conductive plates separated by an insulator Capacitor an electronic device which consists
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 informationEE40 Midterm Review Prof. Nathan Cheung
EE40 Midterm Review Prof. Nathan Cheung 10/29/2009 Slide 1 I feel I know the topics but I cannot solve the problems Now what? Slide 2 R L C Properties Slide 3 Ideal Voltage Source *Current depends d on
More informationLab 10: DC RC circuits
Name: Lab 10: DC RC circuits Group Members: Date: TA s Name: Objectives: 1. To understand current and voltage characteristics of a DC RC circuit 2. To understand the effect of the RC time constant Apparatus:
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 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 informationPhys 2025, First Test. September 20, minutes Name:
Phys 05, First Test. September 0, 011 50 minutes Name: Show all work for maximum credit. Each problem is worth 10 points. Work 10 of the 11 problems. k = 9.0 x 10 9 N m / C ε 0 = 8.85 x 101 C / N m e
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 informationEnergy Storage Elements: Capacitors and Inductors
CHAPTER 6 Energy Storage Elements: Capacitors and Inductors To this point in our study of electronic circuits, time has not been important. The analysis and designs we have performed so far have been static,
More informationPhysics 248, Spring 2009 Lab 7: Capacitors and RCDecay
Name Section Physics 248, Spring 2009 Lab 7: Capacitors and RCDecay Your TA will use this sheet to score your lab. It is to be turned in at the end of lab. To receive full credit you must use complete
More informationPhysics 115. General Physics II. Session 24 Circuits Series and parallel R Meters Kirchoff s Rules
Physics 115 General Physics II Session 24 Circuits Series and parallel R Meters Kirchoff s Rules R. J. Wilkes Email: phy115a@u.washington.edu Home page: http://courses.washington.edu/phy115a/ 5/15/14 Phys
More informationUniversity of TN Chattanooga Physics 1040L 8/18/2012 PHYSICS 1040L LAB LAB 4: R.C. TIME CONSTANT LAB
PHYSICS 1040L LAB LAB 4: R.C. TIME CONSTANT LAB OBJECT: To study the discharging of a capacitor and determine the time constant for a simple circuit. APPARATUS: Capacitor (about 24 μf), two resistors (about
More informationAP Physics C. Inductance. Free Response Problems
AP Physics C Inductance Free Response Problems 1. Two toroidal solenoids are wounded around the same frame. Solenoid 1 has 800 turns and solenoid 2 has 500 turns. When the current 7.23 A flows through
More informationChapter 19 Lecture Notes
Chapter 19 Lecture Notes Physics 2424  Strauss Formulas: R S = R 1 + R 2 +... C P = C 1 + C 2 +... 1/R P = 1/R 1 + 1/R 2 +... 1/C S = 1/C 1 + 1/C 2 +... q = q 0 [1e t/(rc) ] q = q 0 e t/(rc τ = RC
More informationExperiment Guide for RC Circuits
GuideP1 Experiment Guide for RC Circuits I. Introduction 1. Capacitors A capacitor is a passive electronic component that stores energy in the form of an electrostatic field. The unit of capacitance is
More informationOperational amplifiers (Op amps)
Operational amplifiers (Op amps) v R o R i v i Av i v View it as an ideal amp. Take the properties to the extreme: R i, R o 0, A.?!?!?!?! v v i Av i v A Consequences: No voltage dividers at input or output.
More informationCircuit Theory Chapter 7 Response of FirstOrder RL and R Circuits
140310 Circuit Theory Chapter 7 Response of FirstOrer RL an R Circuits 140310 Circuit Theory Chapter 7 Response of FirstOrer RL an RC Circuits Chapter Objectives Be able to etermine the natural response
More informationChapter 8. Capacitors. Charging a capacitor
Chapter 8 Capacitors You can store energy as potential energy by pulling a bowstring, stretching a spring, compressing a gas, or lifting a book. You can also store energy as potential energy in an electric
More informationHomework 2. Due Friday (5pm), Feb. 8, 2013
University of California, Berkeley Spring 2013 EE 42/100 Prof. K. Pister Homework 2 Due Friday (5pm), Feb. 8, 2013 Please turn the homework in to the drop box located next to 125 Cory Hall (labeled EE
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 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 informationHomework Assignment 08
Homework Assignment 08 Question 1 (Short Takes) Two points each unless otherwise indicated. 1. Give one phrase/sentence that describes the primary advantage of an active load. Answer: Large effective resistance
More informationVer 3537 E1.1 Analysis of Circuits (2014) E1.1 Circuit Analysis. Problem Sheet 1 (Lectures 1 & 2)
Ver 3537 E. Analysis of Circuits () Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet (Lectures & ). [A] One of the following circuits is a series circuit and the other is a parallel circuit.
More informationECE 212H1F Circuit Analysis October 20, :1519: Reza Iravani 02 Reza Iravani 03 Ali NabaviNiaki. (Nonprogrammable Calculators Allowed)
Please Print Clearly Last Name: First Name: Student Number: Your Tutorial Section (CIRCLE ONE): 01 Thu 10:00 12:00 HA403 02 Thu 10:00 12:00 GB412 03 Thu 15:00 17:00 GB412 04 Thu 15:00 17:00 SF2202 05 Fri
More informationP114 University of Rochester NAME S. Manly Spring 2010
Exam 2 (March 23, 2010) Please read the problems carefully and answer them in the space provided. Write on the back of the page, if necessary. Show your work where indicated. Problem 1 ( 8 pts): In each
More informationUniversity Physics (PHY 2326)
Chapter 23 University Physics (PHY 2326) Lecture 5 Electrostatics Electrical energy potential difference and electric potential potential energy of charged conductors Capacitance and capacitors 3/26/2015
More informationCoulomb s constant k = 9x10 9 N m 2 /C 2
1 Part 2: Electric Potential 2.1: Potential (Voltage) & Potential Energy q 2 Potential Energy of Point Charges Symbol U mks units [Joules = J] q 1 r Two point charges share an electric potential energy
More information[1] (b) Fig. 1.1 shows a circuit consisting of a resistor and a capacitor of capacitance 4.5 μf. Fig. 1.1
1 (a) Define capacitance..... [1] (b) Fig. 1.1 shows a circuit consisting of a resistor and a capacitor of capacitance 4.5 μf. S 1 S 2 6.3 V 4.5 μf Fig. 1.1 Switch S 1 is closed and switch S 2 is left
More informationEE 321 Analog Electronics, Fall 2013 Homework #3 solution
EE 32 Analog Electronics, Fall 203 Homework #3 solution 2.47. (a) Use superposition to show that the output of the circuit in Fig. P2.47 is given by + [ Rf v N + R f v N2 +... + R ] f v Nn R N R N2 R [
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 informationDesigning Information Devices and Systems I Spring 2017 Babak Ayazifar, Vladimir Stojanovic Midterm 2. Exam location: 145 Dwinelle, last SID# 2
EECS 16A Designing Information Devices and Systems I Spring 2017 Babak Ayazifar, Vladimir Stojanovic Midterm 2 Exam location: 145 Dwinelle, last SID# 2 PRINT your student ID: PRINT AND SIGN your name:,
More information