Figure Circuit for Question 1. Figure Circuit for Question 2


 Shannon Bates
 3 years ago
 Views:
Transcription
1 Exercises 10.7 Exercises Multiple Choice 1. For the circuit of Figure the time constant is A. 0.5 ms µs 2, 000 s D. 0.2 ms 4 Ω 2 Ω 12 Ω 1 mh 12u 0 () t V Figure Circuit for Question 1 2. For the circuit of Figure the time constant is A. 50 ms 100 ms 190 ms D ms 4 KΩ 5 KΩ 5u 0 () t A 6 KΩ 10 KΩ 10 µf Figure Circuit for Question 2 3. The forced response component of the inductor current for the circuit of Figure is i Lf A. 16 A 10 A Circuit Analysis I with MATLAB Applications 1033
2 Chapter 10 Forced and Total Response in RL and RC Circuits 6 A D. 2 A 16u 0 () t A 4 Ω 12 Ω 5 Ω 1 mh Figure Circuit for Question 3 4. The forced response component of the capacitor voltage for the circuit of Figure is v Cf A. 10 V 2 V 32 3 V D. 8 V 16u 0 () t V 4 KΩ 12 KΩ 2 KΩ 1 µf Figure Circuit for Question 4 5. For the circuit of Figure i L ( 0 ) 2 A. For t > 0 the total response of i L () t is A. 6 A 6e 5000t A 6 6e 5000t A D. 6 4e 5000t A Circuit Analysis I with MATLAB Applications
3 Exercises 16u 0 () t A 4 Ω 12 Ω 5 Ω 1 mh i L () t Figure Circuit for Question 5 6. For the circuit of Figure v C ( 0 ) 5 V. For t > 0 the total response of v C () t is A. 12 V 10 5e 500t V 12 7e 200t V D. 12 7e 200t V 16u 0 () t V 4 KΩ 12 KΩ 2 KΩ 1 µf v C () t Figure Circuit for Question 6 7. For the circuit of Figure i L ( 0 ) 2 A. For t > 0 the total response of v L () t is A. 20e 5000t V 20e 5000t V 32e 8000t V D. 32e 8000t V Circuit Analysis I with MATLAB Applications 1035
4 Chapter 10 Forced and Total Response in RL and RC Circuits 16u 0 () t A 4 Ω 12 Ω 5 Ω 1 mh v L () t Figure Circuit for Question 7 8. For the circuit of Figure v C ( 0 ) 5 V. For t > 0 the total response i C () t is A. 1400e 200t A 1.4e 200t A 3500e 500t A D. 3.5e 500t A 16u 0 () t V 4 KΩ 12 KΩ 2 KΩ 1 µf i C () t Figure Circuit for Question 8 9. The waveform of Figure can be expressed as A. D. 3tu 0 () t A 3u [ 0 () t ] 3u [ 0 ( t 3) ] A 3t[ u 0 () t u 0 ( t 1) ] ( 1.5t 4.5) [ u 0 ( t 1) u 0 ( t 3) ] A 3t[ u 0 () t u 0 ( t 3) ] ( 1.5t 4.5) [ u 0 () t u 0 ( t 3) ] A Circuit Analysis I with MATLAB Applications
5 Exercises 3 i L () t A t ( s) Figure Waveform for Question The waveform of Figure can be expressed as A. D. 21 ( e αt e βt )u 0 () t V ( 2 2e αt )[ u 0 () t u 0 ( t 2) ] ( 2e βt )[ u 0 ( t 2) u 0 ( t 3) ] V ( 2 2e αt )[ u 0 () t u 0 ( t 2) ] ( 2e βt )[ u 0 ( t 2) u 0 ( t 3) ] V ( 2 2e αt )[ u 0 () t ] ( 2e βt )[ u 0 ( t 3) ] V 2 1 v C () t V 2 2e αt 1 2 2e βt 3 t ( s) Figure Waveform for Question 10 Problems 1. In the circuit of Figure 10.54, the voltage source () t varies with time as shown by the waveform of Figure Compute, sketch, and express v LOAD () t as a sum of unit step functions for 0 t 5 s. Circuit Analysis I with MATLAB Applications 1037
6 Chapter 10 Forced and Total Response in RL and RC Circuits 6 Ω 6 Ω 12 Ω () t 10 Ω v LOAD () t Figure Circuit for Problem 1 () t (V) Figure Waveform for Problem 1 t(s) 2. In the circuit of Figure 10.56, () t 15u 0 () t 30u 0 ( t 2) V. Compute i L () t for t > 0. R () t 3 KΩ L i L () t 1 mh Figure Circuit for Problem 2 3. In the circuit of Figure (a), the excitation () t is a pulse as shown in Figure (b). a. Compute i L () t for 0< t< 0.3 ms b. Compute and sketch i L () t for all t > Circuit Analysis I with MATLAB Applications
7 Exercises i L () t () t 3 Ω 1 mh 24 V 6 Ω 6 Ω () t (a) 0.3 (b) t (ms) Figure Circuit and waveform for Problem 3 4. In the circuit of Figure 10.58, switch S has been open for a very long time and closes at t 0. Compute and sketch i L () t and i SW () t for t > 0. 4 Ω t 0 i SW () t 6 Ω S 20 V 8 Ω 1 H i L () t Figure Circuit for Problem 4 5. For the circuit of Figure 10.59, compute v C () t for t > 0. 10u 0 () t A 38 Ω 24 V 2 Ω 50 µf vc () t Figure Circuit for Problem 5 6. For the circuit of Figure 10.60, compute v out () t for t > 0 in terms of R, C, and v in u 0 () t given that v C ( 0 ) 0 Circuit Analysis I with MATLAB Applications 1039
8 Chapter 10 Forced and Total Response in RL and RC Circuits v in u 0 () t R C v out () t Figure Circuit for Problem 6 7. In the circuit of Figure 10.61, switch S has been open for a very long time and closes at t 0. Compute and sketch v C () t and v R3 () t for t> 0. R 1 2 R KΩ t 0 100u 0 ( t) V S 50 V 25 KΩ C 1 µf R KΩ v C () t v R3 () t Figure Circuit for Problem 7 8. For the circuit of Figure 10.62, it is given that v C ( 0 ) 5 V. Compute i C () t for t > 0. Hint: Be careful in deriving the time constant for this circuit. i C () t C 1 F v C () t R 1 18 Ω R 2 12 Ω 10i C () t Figure Circuit for Problem Circuit Analysis I with MATLAB Applications
9 Answers to Exercises 10.8 Answers to Exercises Multiple Choice 1. D 2. B 3. C 4. E 5. E 6. C 7. D 8. A 9. C 10. B 12 V 6 4e 8000t A Problems 1. We replace the given circuit shown below with its Thevenin equivalent. () t x 6 Ω 6 Ω 12 Ω 10 Ω y v LOAD () t v TH 10 Ω 10 Ω v LOAD () t () t () t (V) t(s) For the Thevenin equivalent voltage at different time intervals is as shown below. Circuit Analysis I with MATLAB Applications 1041
10 Chapter 10 Forced and Total Response in RL and RC Circuits and v LOAD () t v TH () t v 18 s () t v 18 s () t v 18 s () t v 18 s () t v 20 TH () t 0.5v TH () t V 0< t< 1 s V 1< t< 3 s ( 60) 40 V 3< t< 4 s V t > 4 s V 0 < t < 1 s V 1 < t < 3 s 0.5 ( 40) 20 V 3< t< 4 s V t > 4 s The waveform of the voltage across the load is as shown below. v LOAD () t ( V) t ( s) The waveform above can now be expressed as a sum of unit step functions as follows: v LOAD () t 20u 0 () t 20u 0 ( t 1) 40u 0 ( t 1) 40u 0 ( t 3) 20u 0 ( t 4) 20u 0 ( t 3) 20u 0 ( t 4) 20u 0 ( t 4) 20u 0 () t 20u 0 ( t 1) 60u 0 ( t 3) 40u 0 ( t 4) 2. The circuit at t is as shown below and since we are not told otherwise, we will assume that i L ( 0 ) 0 0 For t > 0 we let i L1 () t be the inductor current when the 15u 0 () t voltage source acts alone and i L2 () t when the 30u 0 ( t 2) voltage source acts alone. Then, i L TOTAL () t i L1 () t i L2 () t Circuit Analysis I with MATLAB Applications
11 Answers to Exercises 3 kω 1 mh il 0 ( 0 ) 0 ( ) 0 For 0 < t < 2 s the circuit is as shown below. 3 KΩ 15 V i L1() t 1 mh Then, i L1 () t i L1f i L1n where i L1f ma and i 3 KΩ L1n A 1 e ( R L)t Ae t Thus, i L1 () t 5 A 1 e t ma and using the initial condition i L ( 0 ) i L ( 0 ) 0, we get i L1 ( 0) 5 A 1 e 0 ma or A 1 5. Therefore, i L1 () t 5 5e t (1) Next, with the 30u 0 ( t 2) voltage source acts alone the circuit is as shown below. 30 V 3 KΩ i L () t 1 mh Then, i L2 () t i L2f i L2n, i L2f ma and i 3 KΩ L2n A 2 e ( R L) ( t 2) Be ( t 2) Thus, i L2 () t 10 A 2 e ( t 2) ma and the initial condition at t 2 is found from (1) above as i L1 t 6 t 5 5e ma. Therefore, 2 s 5 10 A or 2 e ( 2 2) ma A 2 15 i L2 i t 2 s L1 t 2 s and i L2 () t 10 15e t 2 ( ) ma (2) Circuit Analysis I with MATLAB Applications 1043
12 Chapter 10 Forced and Total Response in RL and RC Circuits Therefore, the total current when both voltage sources are present is the summation of (1) and (2), that is, i L TOTAL () t i L1 () t i L2 () t 5 5e t 10 15e t 2 ( ) ma e t 15e ( t 2) ma a. For this circuit () t 24[ u 0 () t u 0 ( t 0.3) ] and since we are not told otherwise, we will assume that i L ( 0 ) 0. For 0< t< 0.3 ms the circuit and its Thevenin equivalent are as shown below. 3 Ω 6 Ω 6 Ω 1 mh i L () t () t () t 24[ u 0 () t u 0 ( t 0.3) ] v TH 8 Ω 1 mh i L () t v TH () t () t 16[ u 0 () t u 0 ( t 0.3) ] Then, and at t 0 i L () t i Lf i Ln 16 8 A 1 e ( R L)t 2 Ae 8000t i L ( 0) i L ( 0 ) 0 2 A 1 e 0 or A 1 2 and thus for 0 < t < 0.3 ms i L () t 2 2e 8000t (1) b. For t > 0.3 ms the circuit is as shown below. For this circuit i L () t A 2 e ( R L) ( t 0.3) A 2 e 8000 ( t 0.3 ) (2) and A 2 is found from the initial condition at t 0.3 ms, that is, with (1) above we get 3 Ω () t 0 6 Ω 6 Ω 1 mh i L () t 8 Ω 1 mh i L () t Circuit Analysis I with MATLAB Applications
13 i L 2 2e t 0.3 ms and by substitution into (2) above or A Therefore for t > 0.3 ms 2 2e A i L 1.82 A t 0.3 ms 2 e ( ) i L () t 1.82e ( t 0.3 ms ) The waveform for the inductor current i L () t for all t > 0 is shown below i L ( A) Answers to Exercises 0.3 t ( ms) 4. At t 0 the circuit is as shown below where i L ( 0 ) 20 ( 4 6) 2 A and thus the initial condition has been established. 4 Ω 6 Ω 20 V i L 0 ( ) For and t> 0 the circuit and its Thevenin equivalent are as shown below where 8 v TH V 4 8 R 8 4 TH Ω 8 4 Circuit Analysis I with MATLAB Applications 1045
14 Chapter 10 Forced and Total Response in RL and RC Circuits 4 Ω 6 Ω 20 V 8 Ω Closed Switch 1 H i L () t R TH v TH 1 H 40 3 V 26 3 Ω i L () t Then, and A i L () t i Lf i 40 3 ( Ln Ae R L )t Ae is evaluated from the initial condition, i.e., ( )t i L ( 0 ) i L ( 0 ) Ae 0 from which A 6 13 and thus for t > 0 i L () t e ( 26 3)t e 8.67t (1) Next, to find () t we observe that this current flows also through the 8 Ω resistor and this can be found from i SW v 8 Ω shown on the circuit below. 4 Ω i SW () t 6 Ω 20 V 8 Ω v 8 Ω 1 H i L () t Now, and v 8 Ω v 6 Ω v L () t 6i L () t L di L dt 6( e 8.67t d ) ( e 8.67t ) dt e 8.67t e 8.67t e 8.67t Circuit Analysis I with MATLAB Applications
15 Answers to Exercises i SW v () t i 8 Ω e 8.67t 8 Ω e 8.67t 8 8 Therefore, from the initial condition, (1) and (2) above we have (2) i L ( 0 ) 2 i L ( ) 1.54 i SW ( 0 ) i SW ( ) 1.16 and with these values we sketch i L () t and i SW () t as shown below. i L () t A i SW () t A t t 5. At t 0 the circuit is as shown below where v C ( 0 ) 24 V and thus the initial condition has been established. 38 Ω 24 V 2 Ω 50 µf vc () t The circuit for source. t > 0 is shown below where the current source has been replaced with a voltage 38 Ω 24 V 2 Ω 20 V 50 µf vc () t 4 V 40 Ω 50 µf v C () t Now, v C () t v Cf v Cn 4 Ae 1 ( RC) t 4 Ae 500t Circuit Analysis I with MATLAB Applications 1047
16 Chapter 10 Forced and Total Response in RL and RC Circuits and with the initial condition v C ( 0 ) v C ( 0 ) 24 V 4 Ae 0 from which A 20 we get v C () t 4 20e 500t 6. For t > 0 the op amp circuit is as shown below. v in () t R v C v out () t Application of KCL at the minus () input yields and since or v 0 v v in C dv C R dt C dv C dt dv C dt v in R v in RC Integrating both sides and observing that v out () t v C () t we get v out () t v in t k RC where k is the constant of integration of both sides and it is evaluated from the given initial condition. Then, or k 0. Therefore, v C ( 0 ) v C ( 0 ) 0 0 k v out () t v in t u RC 0 () t and v in RC is the slope as shown below Circuit Analysis I with MATLAB Applications
17 Answers to Exercises slope v in RC 7. At t 0 the circuit is as shown below where v C ( 0 ) 150 V and thus the initial condition has been established. 175 KΩ 150 V 1 µf v C 0 ( ) The circuit for t> 0 is shown below where the voltage source 1 is absent for all positive time and the 50 KΩ is shorted out by the closed switch. 125 KΩ 50 V 1 µf v C () t For the circuit above v C () t v Cf v Cn 50 Ae t RC) 50 Ae 8t and with the initial condition v C ( 0 ) v C ( 0 ) Ae 0 from which A 100 and thus for t> 0 V v C () t e 8t To find v R3 () t we will first find i C () t from the circuit below where i C () t C dv C ( e 8t ) dt Circuit Analysis I with MATLAB Applications 1049
18 Chapter 10 Forced and Total Response in RL and RC Circuits 50 V 25 KΩ 1 µf 100 KΩ v C () t v R3 () t Then, v R3 () t ( 100 KΩ)i C 10 5 ( e 8t ) 80e 8t V The sketches below show v C () t and v R3 () t as they approach their final values. 150 v C () t V v R3 () t V t t 8. For this circuit we cannot short the dependent source and therefore we cannot find by combining the resistances R 1 and R 2 in parallel combination in order to find the time constant τ R Instead, we will derive the time constant from the differential equation of (9.9) of the previous chapter, that is, R eq From the given circuit shown below dv C dt v C RC v C i C () t C 1 F v C () t R 1 18 Ω R 2 12 Ω 10i C () t or v i C 10i C v C C R 1 R Circuit Analysis I with MATLAB Applications
19 Answers to Exercises or or C dv C dt vc C dv C dt R 1 R 2 R dv C C R 1 R vc 0 dt R 1 R 2 R 1 dv C dt R 1 R R 1 R v C 0 C R 1 and from this differential equation we see that the coefficient of v C is and thus R eq C τ v C () t Ae t and with the given initial condition v C ( 0 ) V 0 A 5 V we get Then, using v C () t 5e t we find that for t > 0 and the minus () sign indicates that the i C C dv C dt i C () t ( 1) ( e t ) e t i C () t direction is opposite to that shown. Circuit Analysis I with MATLAB Applications 1051
Basic 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 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 informationElectric 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 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 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 informationECE Spring 2015 Final Exam
ECE 20100 Spring 2015 Final Exam May 7, 2015 Section (circle below) Jung (1:30) 0001 Qi (12:30) 0002 Peleato (9:30) 0004 Allen (10:30) 0005 Zhu (4:30) 0006 Name PUID Instructions 1. DO NOT START UNTIL
More informationProblem Set 5 Solutions
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
More informationAlternating Current Circuits. Home Work Solutions
Chapter 21 Alternating Current Circuits. Home Work s 21.1 Problem 21.11 What is the time constant of the circuit in Figure (21.19). 10 Ω 10 Ω 5.0 Ω 2.0µF 2.0µF 2.0µF 3.0µF Figure 21.19: Given: The circuit
More informationFirstorder transient
EIE209 Basic Electronics Firstorder transient Contents Inductor and capacitor Simple RC and RL circuits Transient solutions Constitutive relation An electrical element is defined by its relationship between
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 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 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 informationSinusoidal Response of RLC Circuits
Sinusoidal Response of RLC Circuits Series RL circuit Series RC circuit Series RLC circuit Parallel RL circuit Parallel RC circuit RL Series Circuit RL Series Circuit RL Series Circuit Instantaneous
More 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 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 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 informationElectronics. Basics & Applications. group talk Daniel Biesinger
Electronics Basics & Applications group talk 23.7.2010 by Daniel Biesinger 1 2 Contents Contents Basics Simple applications Equivalent circuit Impedance & Reactance More advanced applications  RC circuits
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 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 informationAC Circuits Homework Set
Problem 1. In an oscillating LC circuit in which C=4.0 μf, the maximum potential difference across the capacitor during the oscillations is 1.50 V and the maximum current through the inductor is 50.0 ma.
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 informationET4119 Electronic Power Conversion 2011/2012 Solutions 27 January 2012
ET4119 Electronic Power Conversion 2011/2012 Solutions 27 January 2012 1. In the singlephase rectifier shown below in Fig 1a., s = 1mH and I d = 10A. The input voltage v s has the pulse waveform shown
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 informationDEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING RUTGERS UNIVERSITY
DEPARTMENT OF EECTRICA AND COMPUTER ENGINEERING RUTGERS UNIVERSITY 330:222 Principles of Electrical Engineering II Spring 2002 Exam 1 February 19, 2002 SOUTION NAME OF STUDENT: Student ID Number (last
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 informationENGR4300 Spring 2009 Test 2. Name: SOLUTION. Section: 1(MR 8:00) 2(TF 2:00) 3(MR 6:00) (circle one) Question I (20 points): Question II (20 points):
ENGR43 Test 2 Spring 29 ENGR43 Spring 29 Test 2 Name: SOLUTION Section: 1(MR 8:) 2(TF 2:) 3(MR 6:) (circle one) Question I (2 points): Question II (2 points): Question III (17 points): Question IV (2 points):
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 informationEXPERIMENT 07 TO STUDY DC RC CIRCUIT AND TRANSIENT PHENOMENA
EXPERIMENT 07 TO STUDY DC RC CIRCUIT AND TRANSIENT PHENOMENA DISCUSSION The capacitor is a element which stores electric energy by charging the charge on it. Bear in mind that the charge on a capacitor
More informationErrata to LINEAR CIRCUITS, VOL1 DECARLO/LIN, EDITION 3 CHAPTERS 1 11 (updated 8/23/10)
Errata to LINEAR CIRCUITS, VOL1 DECARLO/LIN, EDITION 3 CHAPTERS 1 11 (updated 8/23/10) page location correction 42 Ch1, P1, statement (e) Figure P.1.3b should be Figure P.1.1b 46 Ch1, P19, statement (c)
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 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 informationSinusoidal Steady State Analysis (AC Analysis) Part I
Sinusoidal Steady State Analysis (AC Analysis) Part I Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/
More 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 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 informationFirst Order RC and RL Transient Circuits
First Order R and RL Transient ircuits Objectives To introduce the transients phenomena. To analyze step and natural responses of first order R circuits. To analyze step and natural responses of first
More informationLECTURE 8 Fundamental Models of PulseWidth Modulated DCDC Converters: f(d)
1 ECTURE 8 Fundamental Models of PulseWidth Modulated DCDC Converters: f(d) I. QuasiStatic Approximation A. inear Models/ Small Signals/ Quasistatic I V C dt AmpSec/Farad V I dt VoltSec/Henry 1. Switched
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 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 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 informationEECE 2150 Circuits and Signals, Biomedical Applications Final Exam Section 3
EECE 2150 Circuits and Signals, Biomedical Applications Final Exam Section 3 Instructions: Closed book, closed notes; Computers and cell phones are not allowed You may use the equation sheet provided but
More informationEE101 IMPERIAL COLLEGE LONDON DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 2013 ANALYSIS OF CIRCUITS. Tuesday, 28 May 10:00 am
EE101 IMPERIAL COLLEGE LONDON DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 2013 ExamHeader: EEE/EIE PART I: MEng, Beng and ACGI ANALYSIS OF CIRCUITS Tuesday, 28 May 10:00 am Time allowed:
More informationCircuit AnalysisII. Circuit AnalysisII Lecture # 5 Monday 23 rd April, 18
Circuit AnalysisII Capacitors in AC Circuits Introduction ü The instantaneous capacitor current is equal to the capacitance times the instantaneous rate of change of the voltage across the capacitor.
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 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 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 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 informationEECE 2510 Circuits and Signals, Biomedical Applications Final Exam Section 3. Name:
EECE 2510 Circuits and Signals, Biomedical Applications Final Exam Section 3 Instructions: Closed book, closed notes; Computers and cell phones are not allowed Scientific calculators are allowed Complete
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 informationSingle Phase Parallel AC Circuits
Single Phase Parallel AC Circuits 1 Single Phase Parallel A.C. Circuits (Much of this material has come from Electrical & Electronic Principles & Technology by John Bird) n parallel a.c. circuits similar
More informationSchedule. ECEN 301 Discussion #20 Exam 2 Review 1. Lab Due date. Title Chapters HW Due date. Date Day Class No. 10 Nov Mon 20 Exam Review.
Schedule Date Day lass No. 0 Nov Mon 0 Exam Review Nov Tue Title hapters HW Due date Nov Wed Boolean Algebra 3. 3.3 ab Due date AB 7 Exam EXAM 3 Nov Thu 4 Nov Fri Recitation 5 Nov Sat 6 Nov Sun 7 Nov Mon
More 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 informationELEC 2501 AB. Come to the PASS workshop with your mock exam complete. During the workshop you can work with other students to review your work.
It is most beneficial to you to write this mock midterm UNDER EXAM CONDITIONS. This means: Complete the midterm in 3 hour(s). Work on your own. Keep your notes and textbook closed. Attempt every question.
More informationSinusoidal SteadyState Analysis
Chapter 4 Sinusoidal SteadyState Analysis In this unit, we consider circuits in which the sources are sinusoidal in nature. The review section of this unit covers most of section 9.1 9.9 of the text.
More informationSingleTimeConstant (STC) Circuits This lecture is given as a background that will be needed to determine the frequency response of the amplifiers.
SingleTimeConstant (STC) Circuits This lecture is given as a background that will be needed to determine the frequency response of the amplifiers. Objectives To analyze and understand STC circuits with
More informationRLC Series Circuit. We can define effective resistances for capacitors and inductors: 1 = Capacitive reactance:
RLC Series Circuit In this exercise you will investigate the effects of changing inductance, capacitance, resistance, and frequency on an RLC series AC circuit. We can define effective resistances for
More informationInitial conditions. Necessity and advantages: Initial conditions assist
Initial conditions Necessity and advantages: Initial conditions assist To evaluate the arbitrary constants of differential equations Knowledge of the behavior of the elements at the time of switching Knowledge
More informationLaboratory 7: Charging and Discharging a Capacitor Prelab
Phys 132L Fall 2018 Laboratory 7: Charging and Discharging a Capacitor Prelab Consider a capacitor with capacitance C connected in series to a resistor with resistance R as shown in Fig. 1. Theory predicts
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 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 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 informationPHYS 241 EXAM #2 November 9, 2006
1. ( 5 points) A resistance R and a 3.9 H inductance are in series across a 60 Hz AC voltage. The voltage across the resistor is 23 V and the voltage across the inductor is 35 V. Assume that all voltages
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 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 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 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 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 informationHomwork AC circuits. due Friday Oct 8, 2017
Homwork AC circuits due Friday Oct 8, 2017 Validate your answers (show your work) using a Matlab Live Script and Simulink models. Include these computations and models with your homework. 1 Problem 1 Worksheet
More information1. /25 2. /30 3. /25 4. /20 Total /100
Circuit Exam 2 Spring 206. /25 2. /30 3. /25 4. /20 Total /00 Name Pleae write your name at the top of every page! Note: ) If you are tuck on one part of the problem, chooe reaonable value on the following
More informationScanned by CamScanner
Scanned by CamScanner Scanned by CamScanner t W I w v 6.00fall 017 Midterm 1 Name Problem 3 (15 pts). F the circuit below, assume that all equivalent parameters are to be found to the left of port
More informationREACTANCE. By: Enzo Paterno Date: 03/2013
REACTANCE REACTANCE By: Enzo Paterno Date: 03/2013 5/2007 Enzo Paterno 1 RESISTANCE  R i R (t R A resistor for all practical purposes is unaffected by the frequency of the applied sinusoidal voltage or
More information(d) describe the action of a 555 monostable timer and then use the equation T = 1.1 RC, where T is the pulse duration
Chapter 1  Timing Circuits GCSE Electronics Component 2: Application of Electronics Timing Circuits Learners should be able to: (a) describe how a RC network can produce a time delay (b) describe how
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 informationI(t) R L. RL Circuit: Fundamentals. a b. Specifications: E (emf) R (resistance) L (inductance) Switch S: a: current buildup. b: current shutdown
RL Circuit: Fundamentals pecifications: E (emf) R (resistance) L (inductance) witch : a: current buildup a b I(t) R L b: current shutdown Timedependent quantities: I(t): instantaneous current through
More informationLab 4 RC Circuits. Name. Partner s Name. I. Introduction/Theory
Lab 4 RC Circuits Name Partner s Name I. Introduction/Theory Consider a circuit such as that in Figure 1, in which a potential difference is applied to the series combination of a resistor and a capacitor.
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 informationThe RLC circuits have a wide range of applications, including oscillators and frequency filters
9. The RL ircuit The RL circuits have a wide range of applications, including oscillators and frequency filters This chapter considers the responses of RL circuits The result is a secondorder differential
More informationInductors. Hydraulic analogy Duality with capacitor Charging and discharging. Lecture 12: Inductors
Lecture 12: nductors nductors Hydraulic analogy Duality with capacitor Charging and discharging Robert R. McLeod, University of Colorado http://hilaroad.com/camp/projects/magnet.html 99 Lecture 12: nductors
More informationLECTURE 8 RC AND RL FIRSTORDER CIRCUITS (PART 1)
CIRCUITS by Ulaby & Maharbiz LECTURE 8 RC AND RL FIRSTORDER CIRCUITS (PART 1) 07/18/2013 ECE225 CIRCUIT ANALYSIS All rights reserved. Do not copy or distribute. 2013 National Technology and Science Press
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 informationTransient response of RC and RL circuits ENGR 40M lecture notes July 26, 2017 ChuanZheng Lee, Stanford University
Transient response of C and L circuits ENG 40M lecture notes July 26, 2017 ChuanZheng Lee, Stanford University esistor capacitor (C) and resistor inductor (L) circuits are the two types of firstorder
More informationConverter System Modeling via MATLAB/Simulink
Converter System Modeling via MATLAB/Simulink A powerful environment for system modeling and simulation MATLAB: programming and scripting environment Simulink: block diagram modeling environment that runs
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 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 informationLecture 6: Impedance (frequency dependent. resistance in the s world), Admittance (frequency. dependent conductance in the s world), and
Lecture 6: Impedance (frequency dependent resistance in the s world), Admittance (frequency dependent conductance in the s world), and Consequences Thereof. Professor Ray, what s an impedance? Answers:
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 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 informationChapter 11 AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode
Chapter 11 AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode Introduction 11.1. DCM Averaged Switch Model 11.2. SmallSignal AC Modeling of the DCM Switch Network 11.3. HighFrequency
More informationCircuits. Fawwaz T. Ulaby, Michel M. Maharbiz, Cynthia M. Furse. Solutions to the Exercises
Circuits by Fawwaz T. Ulaby, Michel M. Maharbiz, Cynthia M. Furse Solutions to the Exercises Chapter 1: Circuit Terminology Chapter 2: Resisitive Circuits Chapter 3: Analysis Techniques Chapter 4: Operational
More informationConsider a simple RC circuit. We might like to know how much power is being supplied by the source. We probably need to find the current.
AC power Consider a simple RC circuit We might like to know how much power is being supplied by the source We probably need to find the current R 10! R 10! is VS Vmcosωt Vm 10 V f 60 Hz V m 10 V C 150
More informationDesigning Information Devices and Systems II Fall 2018 Elad Alon and Miki Lustig Discussion 5A
EECS 6B Designing Information Devices and Systems II Fall 208 Elad Alon and Miki Lustig Discussion 5A Transfer Function When we write the transfer function of an arbitrary circuit, it always takes the
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 informationa. Clockwise. b. Counterclockwise. c. Out of the board. d. Into the board. e. There will be no current induced in the wire
Physics 1B Winter 2012: Final Exam For Practice Version A 1 Closed book. No work needs to be shown for multiplechoice questions. The first 10 questions are the makeup Quiz. The remaining questions are
More informationPossible
Department of Electrical Engineering and Computer Science ENGR 21. Introduction to Circuits and Instruments (4) ENGR 21 SPRING 24 FINAL EXAMINATION given 5/4/3 Possible 1. 1 2. 1 3. 1 4. 1 5. 1 6. 1 7.
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 informationFACULTY OF BIOSCIENCES AND MEDICAL ENGINEERING
Fakulti: FACULTY OF BIOSCIENCES AND MEDICAL ENGINEERING Semakan Nama Matapelajaran: Laboratory 1 Tarikh Keluaran Kod Matapelajaran : SMBE 2712 Pindaan Terakhir No. Prosedur : : 2013 : 2017 : FACULTY OF
More informationMassachusetts Institute of Technology Department of Electrical Engineering and Computer Science Electronic Circuits Fall 2000.
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.002 Electronic Circuits Fall 2000 Final Exam Please write your name in the space provided below, and circle
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 informationCHAPTER FOUR MUTUAL INDUCTANCE
CHAPTER FOUR MUTUAL INDUCTANCE 4.1 Inductance 4.2 Capacitance 4.3 SerialParallel Combination 4.4 Mutual Inductance 4.1 Inductance Inductance (L in Henry is the circuit parameter used to describe an inductor.
More informationMODULE I. Transient Response:
Transient Response: MODULE I The Transient Response (also known as the Natural Response) is the way the circuit responds to energies stored in storage elements, such as capacitors and inductors. If a capacitor
More informationu (t t ) + e ζωn (t tw )
LINEAR CIRCUITS LABORATORY OSCILLATIONS AND DAMPING EFFECT PART I TRANSIENT RESPONSE TO A SQUARE PULSE Transfer Function F(S) = ω n 2 S 2 + 2ζω n S + ω n 2 F(S) = S 2 + 3 RC ( RC) 2 S + 1 RC ( ) 2 where
More information