Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits

Size: px
Start display at page:

Download "Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits"

Transcription

1 Lecure 13 RC/RL Circuis, Time Dependen Op Amp Circuis

2 RL Circuis The seps involved in solving simple circuis conaining dc sources, resisances, and one energy-sorage elemen (inducance or capaciance) are:

3 1. Apply Kirchhoff s curren and volage laws o wrie he circui equaion.. If he equaion conains inegrals, differeniae each erm in he equaion o produce a pure differenial equaion. 3. Assume a soluion of he form K 1 K e s.

4 4. Subsiue he soluion ino he differenial equaion o deermine he values of K 1 and s. (Alernaively, we can deermine K 1 by solving he circui in seady sae) 5. Use he iniial condiions o deermine he value of K. 6. Wrie he final soluion.

5 RL Transien Analysis Find i( and he volage v( i( 0 for < 0 since he swich is open prior o 0 Apply KVL around he loop: V S i( R v( 0

6 RL Transien Analysis V S i( R v( 0 di( i( R L d V S

7 RL Transien Analysis di( i ( R L d V S Try i( K1 Ke s di( i ( R L d V S Try i( K1 Ke s RK 1 ( RK slk ) e V S s RK V K VS R 100V 50Ω 1 S 1 A RK slk 0 s L R

8 RL Transien Analysis L R e K i / ) ( 0 ) 0 ( 0 K K e K i L R e i / ) (

9 RL Transien Analysis Define τ L R i( e /τ

10 RL Transien Analysis v( V S i( R i( ( e ) 100e /τ /τ

11 RL Transien Analysis Transien sars by opening swich Prior o 0 inducor acs as a shor circui so ha: v( 0 for < 0 i( V S /R 1 for < 0

12 RL Transien Analysis Afer 0 curren circulaes hrough L and R, dissipaing energy in he resisance R / / / / 0 ) ( 0 ) ( 0 ) ( ) ( R L Ke L R RKe Ke L for Ke i R i d di L > τ τ τ τ τ τ τ

13 1 0 1 ) (0 R V K Ke R V i S S 0 ) ( / 1 > for e R V i S τ RL Transien Analysis

14 RL Transien Analysis 0 0 ) ( 0 ) ( ) ( / 1 / 1 < > for v for e R LV e R V d d L d di L v s S τ τ τ

15 RL Transien Analysis wih a Curren Source Afer he swich is opened, i R (0 ) A, I L (0 ) 0 Find v(, i R (, i L (

16 RL Transien Analysis wih a Curren Source Afer he swich is closed, i R (0 ) A, I L (0 ) 0 di ( v( L L 10iR ( and i R( il ( A d di ( L 10 i L d ( ( ))

17 RL Transien Analysis wih a Curren Source di ( L 10 i L d ( ( )) di L d ( 5i ( L 10 Try i L ( K K e 1 s

18 RL Transien Analysis wih a Curren Source dil ( d Try i L 5i ( 10 ( i L (0 ) 0 K 1 -K L K1 Ke s ske ( s s 5) Ke 5( K Ke s s ) 10 5K 10 s 5

19 RL Transien Analysis wih a Curren Source A i L ( K Ke 5 In seady sae he inducor becomes a shor circui i L ( ) K A i L ( e 5

20 RL Transien Analysis wih a Curren Source L e e d d d di L v ) ( ) ( ) ( L R e e i i 5 5 ) ( ) ( ) (

21 RL Transien Analysis Prior o 0 i(0) 100V/100Ω 1A Find i(, v(

22 RL Transien Analysis Prior o 0 i(0) 100V/100Ω 1A di( Afer 0: L i( R 100V d di( 00i( 100 d

23 RL Transien Analysis di( d 00i( 100 ry i( K 1 K e s sk e s 00( K 1 K e s ) 100 (00 s) K e s 00K s 00, τ 1/ 00 5ms

24 RL Transien Analysis i(0) 1 K K 100V i( ) K1 00Ω i( e A K 0.5A

25 RL Transien Analysis v( L di( d d d ( e 00 ) 100e 00

26 RC and RL Circuis wih General Sources Firs order differenial equaion wih consan coefficiens di( L Ri( v ( d L di( v ( i( R d R dx( τ x( f ( d Forcing funcion

27 RC and RL Circuis wih General Sources The general soluion consiss of wo pars.

28 The paricular soluion (also called he forced response) is any expression ha saisfies he equaion. dx( τ d x( f ( In order o have a soluion ha saisfies he iniial condiions, we mus add he complemenary soluion o he paricular soluion.

29 The homogeneous equaion is obained by seing he forcing funcion o zero. dx( τ x( 0 d The complemenary soluion (also called he naural response) is obained by solving he homogeneous equaion.

30 Sep-by-Sep Soluion Circuis conaining a resisance, a source, and an inducance (or a capaciance) 1. Wrie he circui equaion and reduce i o a firs-order differenial equaion.

31 . Find a paricular soluion. The deails of his sep depend on he form of he forcing funcion. 3. Obain he complee soluion by adding he paricular soluion o he complemenary soluion x c Ke -/τ which conains he arbirary consan K. 4. Use iniial condiions o find he value of K.

32 Transien Analysis of an RC Circui wih a Sinusoidal Source Ri( Ri( q( sin(00 C 1 C i( vc (0) sin(00 0

33 Transien Analysis of an RC Circui wih a Sinusoidal Source Take he derivaive: di( 1 R i( 400 cos(00 d C di( RC i( 400C cos(00 d Try a paricular soluion : i ( Acos(00 p Bsin(00

34 Transien Analysis of an RC Circui wih a Sinusoidal Source 5x10 3 di( d i( 400x10 6 cos(00 Asin(00 B cos(00 Acos(00 Bsin(00 400x10 6 cos(00 equaing he coefficiens for sin erms : A B 0 A B equaing he coefficiens for cos erms : B A Solving 400x10 : 6 A 00x μA B 00x μA i p ( 00 cos(00 00sin(00

35 Transien Analysis of an RC Circui wih a Sinusoidal Source The complemenary soluion is given by: i c ( Ke / τ τ RC (5kΩ)(1μ F) 5ms The complee soluion is given by he sum of he paricular soluion and he complemenary soluion: i( 00 cos(00 00sin(00 Ke / τ μa

36 Transien Analysis of an RC Circui wih a Sinusoidal Source Iniial condiions: sin(0 ) 0 v C (0 ) 1V v R (0 ) v C (0) 0 v R (0 ) -1V i(0 ) v R /R -1V/5000Ω -00μA 00cos(0)00sin(0)Ke 0 00 K K -400μA

37 Transien Analysis of an RC Circui wih a Sinusoidal Source i( 00 cos(00 00sin(00 400e /τ μa

38 Transien Analysis of an RC Circui wih a Sinusoidal Source Wha happens if we replace he source wih cos(00 and he capacior iniially uncharged v c (0)0?

39 Transien Analysis of an RC Circui wih a Sinusoidal Source Wha happens if we replace he source wih cos(00 and he capacior iniially uncharged v c (0)0? q( Ri( cos(00 C 1 Ri( C i( vc (0) cos(00 0

40 Transien Analysis of an RC Circui wih a Sinusoidal Source Take he derivaive: di( 1 R i( 400sin(00 d C di( RC i( 400C sin(00 d Try a paricular soluion : i ( Acos(00 p Bsin(00

41 Transien Analysis of an RC Circui wih a Sinusoidal Source 5x10 3 di( d i( 400x10 6 sin(00 Asin(00 B cos(00 Acos(00 Bsin(00 400x10 6 sin(00 equaing he coefficiens for sin erms : A B 400x10 6 equaing he coefficiens for cos erms : B A 0 Solving : A 00x μA B 00x10 i p 6 00μA ( 00 cos(00 00sin(00

42 Transien Analysis of an RC Circui wih a Sinusoidal Source The complemenary soluion is given by: i c ( Ke / τ τ RC (5kΩ)(1μ F) 5ms The complee soluion is given by he sum of he paricular soluion and he complemenary soluion: i( 00 cos(00 00sin(00 Ke / τ μa

43 Transien Analysis of an RC Circui wih a Sinusoidal Source Iniial condiions: cos(0 ) v C (0 ) 0V v R (0 ) v C (0) v R (0 ) i(0 ) v R /R V/5000Ω 400μA 00cos(0)-00sin(0)Ke 0 00 K K 00μA

44 Transien Analysis of an RC Circui wih a Sinusoidal Source i( 00 cos(00 00sin(00 00e /τ μa

45 Transien Analysis of an RC Circui wih an Exponenial Source Wha happens if we replace he source wih 10e - and he capacior is iniially charged o v c (0)5? q( Ri( 10e C Ri( 1 C i( vc (0) 0 10e

46 Transien Analysis of an RC Circui wih an Exponenial Source Take he derivaive: di( 1 R i( 10e d C di( RC i( 10Ce d Try a paricular soluion : i p ( Ae

47 Transien Analysis of an RC Circui wih an Exponenial Source RCAe Ae A RCA 10C A(1 RC) 10C 10Ce A 10C RC 1 (10)(x ) 0μA

48 Transien Analysis of an RC Circui wih an Exponenial Source The complemenary soluion is given by: i c ( Ke / τ τ RC (1M Ω)(μF) s The complee soluion is given by he sum of he paricular soluion and he complemenary soluion: i( 0e Ke / μa

49 Transien Analysis of an RC Circui wih an Exponenial Source Iniial condiions: 10e 0 10 v C (0 ) 5V v R (0 ) v C (0) 10 v R (0 ) 5 i(0 ) v R /R 5V/1MΩ 5μA 0e 0 Ke 0 0 K K -15μA

50 Transien Analysis of an RC Circui wih an Exponenial Source i( 0e 15e / μ A

51 Inegraors and Differeniaors Inegraors produce oupu volages ha are proporional o he running ime inegral of he inpu volages. In a running ime inegral, he upper limi of inegraion is.

52 i in v o v in R v c v c 0 q C 1 C v v 1 RC 1 RC o 0 i in d c 0 0 v v in in d d

53 v o () 1 v ()d in RC 0

54 If R 10 kω, C 0.1μF RC 0.1 ms v o () 1000 v () d 0 in

55 v o () 1000 v in() 0 d d 5000 for 0 < < 1ms ms 5d 0 1ms 5d for 1ms < < 3ms

56 Differeniaor Circui i in dq d C dv d in 0 R v o v o RC dv d in

57 Differeniaor Circui v o () RC dv d in

8. Basic RL and RC Circuits

8. Basic RL and RC Circuits 8. Basic L and C Circuis This chaper deals wih he soluions of he responses of L and C circuis The analysis of C and L circuis leads o a linear differenial equaion This chaper covers he following opics

More information

Voltage/current relationship Stored Energy. RL / RC circuits Steady State / Transient response Natural / Step response

Voltage/current relationship Stored Energy. RL / RC circuits Steady State / Transient response Natural / Step response Review Capaciors/Inducors Volage/curren relaionship Sored Energy s Order Circuis RL / RC circuis Seady Sae / Transien response Naural / Sep response EE4 Summer 5: Lecure 5 Insrucor: Ocavian Florescu Lecure

More information

Chapter 7 Response of First-order RL and RC Circuits

Chapter 7 Response of First-order RL and RC Circuits Chaper 7 Response of Firs-order RL and RC Circuis 7.- The Naural Response of RL and RC Circuis 7.3 The Sep Response of RL and RC Circuis 7.4 A General Soluion for Sep and Naural Responses 7.5 Sequenial

More information

Chapter 8 The Complete Response of RL and RC Circuits

Chapter 8 The Complete Response of RL and RC Circuits Chaper 8 The Complee Response of RL and RC Circuis Seoul Naional Universiy Deparmen of Elecrical and Compuer Engineering Wha is Firs Order Circuis? Circuis ha conain only one inducor or only one capacior

More information

First Order RC and RL Transient Circuits

First Order RC and RL Transient Circuits Firs Order R and RL Transien ircuis Objecives To inroduce he ransiens phenomena. To analyze sep and naural responses of firs order R circuis. To analyze sep and naural responses of firs order RL circuis.

More information

ECE 2100 Circuit Analysis

ECE 2100 Circuit Analysis ECE 1 Circui Analysis Lesson 35 Chaper 8: Second Order Circuis Daniel M. Liynski, Ph.D. ECE 1 Circui Analysis Lesson 3-34 Chaper 7: Firs Order Circuis (Naural response RC & RL circuis, Singulariy funcions,

More information

(b) (a) (d) (c) (e) Figure 10-N1. (f) Solution:

(b) (a) (d) (c) (e) Figure 10-N1. (f) Solution: Example: The inpu o each of he circuis shown in Figure 10-N1 is he volage source volage. The oupu of each circui is he curren i( ). Deermine he oupu of each of he circuis. (a) (b) (c) (d) (e) Figure 10-N1

More information

ES 250 Practice Final Exam

ES 250 Practice Final Exam ES 50 Pracice Final Exam. Given ha v 8 V, a Deermine he values of v o : 0 Ω, v o. V 0 Firs, v o 8. V 0 + 0 Nex, 8 40 40 0 40 0 400 400 ib i 0 40 + 40 + 40 40 40 + + ( ) 480 + 5 + 40 + 8 400 400( 0) 000

More information

Direct Current Circuits. February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 1

Direct Current Circuits. February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 1 Direc Curren Circuis February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 1 Ammeers and Volmeers! A device used o measure curren is called an ammeer! A device used o measure poenial difference

More information

Homework-8(1) P8.3-1, 3, 8, 10, 17, 21, 24, 28,29 P8.4-1, 2, 5

Homework-8(1) P8.3-1, 3, 8, 10, 17, 21, 24, 28,29 P8.4-1, 2, 5 Homework-8() P8.3-, 3, 8, 0, 7, 2, 24, 28,29 P8.4-, 2, 5 Secion 8.3: The Response of a Firs Order Circui o a Consan Inpu P 8.3- The circui shown in Figure P 8.3- is a seady sae before he swich closes a

More information

ECE 2100 Circuit Analysis

ECE 2100 Circuit Analysis ECE 1 Circui Analysis Lesson 37 Chaper 8: Second Order Circuis Discuss Exam Daniel M. Liynski, Ph.D. Exam CH 1-4: On Exam 1; Basis for work CH 5: Operaional Amplifiers CH 6: Capaciors and Inducor CH 7-8:

More information

INDEX. Transient analysis 1 Initial Conditions 1

INDEX. Transient analysis 1 Initial Conditions 1 INDEX Secion Page Transien analysis 1 Iniial Condiions 1 Please inform me of your opinion of he relaive emphasis of he review maerial by simply making commens on his page and sending i o me a: Frank Mera

More information

( ) = Q 0. ( ) R = R dq. ( t) = I t

( ) = Q 0. ( ) R = R dq. ( t) = I t ircuis onceps The addiion of a simple capacior o a circui of resisors allows wo relaed phenomena o occur The observaion ha he ime-dependence of a complex waveform is alered by he circui is referred o as

More information

7. Capacitors and Inductors

7. Capacitors and Inductors 7. Capaciors and Inducors 7. The Capacior The ideal capacior is a passive elemen wih circui symbol The curren-volage relaion is i=c dv where v and i saisfy he convenions for a passive elemen The capacior

More information

EEEB113 CIRCUIT ANALYSIS I

EEEB113 CIRCUIT ANALYSIS I 9/14/29 1 EEEB113 CICUIT ANALYSIS I Chaper 7 Firs-Order Circuis Maerials from Fundamenals of Elecric Circuis 4e, Alexander Sadiku, McGraw-Hill Companies, Inc. 2 Firs-Order Circuis -Chaper 7 7.2 The Source-Free

More information

CHAPTER 6: FIRST-ORDER CIRCUITS

CHAPTER 6: FIRST-ORDER CIRCUITS EEE5: CI CUI T THEOY CHAPTE 6: FIST-ODE CICUITS 6. Inroducion This chaper considers L and C circuis. Applying he Kirshoff s law o C and L circuis produces differenial equaions. The differenial equaions

More information

Section 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients

Section 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients Secion 3.5 Nonhomogeneous Equaions; Mehod of Undeermined Coefficiens Key Terms/Ideas: Linear Differenial operaor Nonlinear operaor Second order homogeneous DE Second order nonhomogeneous DE Soluion o homogeneous

More information

EE202 Circuit Theory II , Spring. Dr. Yılmaz KALKAN & Dr. Atilla DÖNÜK

EE202 Circuit Theory II , Spring. Dr. Yılmaz KALKAN & Dr. Atilla DÖNÜK EE202 Circui Theory II 2018 2019, Spring Dr. Yılmaz KALKAN & Dr. Ailla DÖNÜK 1. Basic Conceps (Chaper 1 of Nilsson - 3 Hrs.) Inroducion, Curren and Volage, Power and Energy 2. Basic Laws (Chaper 2&3 of

More information

R.#W.#Erickson# Department#of#Electrical,#Computer,#and#Energy#Engineering# University#of#Colorado,#Boulder#

R.#W.#Erickson# Department#of#Electrical,#Computer,#and#Energy#Engineering# University#of#Colorado,#Boulder# .#W.#Erickson# Deparmen#of#Elecrical,#Compuer,#and#Energy#Engineering# Universiy#of#Colorado,#Boulder# Chaper 2 Principles of Seady-Sae Converer Analysis 2.1. Inroducion 2.2. Inducor vol-second balance,

More information

( ) ( ) ( ) () () Signals And Systems Exam#1. 1. Given x(t) and y(t) below: x(t) y(t) (A) Give the expression of x(t) in terms of step functions.

( ) ( ) ( ) () () Signals And Systems Exam#1. 1. Given x(t) and y(t) below: x(t) y(t) (A) Give the expression of x(t) in terms of step functions. Signals And Sysems Exam#. Given x() and y() below: x() y() 4 4 (A) Give he expression of x() in erms of sep funcions. (%) x () = q() q( ) + q( 4) (B) Plo x(.5). (%) x() g() = x( ) h() = g(. 5) = x(. 5)

More information

CHAPTER 12 DIRECT CURRENT CIRCUITS

CHAPTER 12 DIRECT CURRENT CIRCUITS CHAPTER 12 DIRECT CURRENT CIUITS DIRECT CURRENT CIUITS 257 12.1 RESISTORS IN SERIES AND IN PARALLEL When wo resisors are conneced ogeher as shown in Figure 12.1 we said ha hey are conneced in series. As

More information

Basic Circuit Elements Professor J R Lucas November 2001

Basic Circuit Elements Professor J R Lucas November 2001 Basic Circui Elemens - J ucas An elecrical circui is an inerconnecion of circui elemens. These circui elemens can be caegorised ino wo ypes, namely acive and passive elemens. Some Definiions/explanaions

More information

EECE251. Circuit Analysis I. Set 4: Capacitors, Inductors, and First-Order Linear Circuits

EECE251. Circuit Analysis I. Set 4: Capacitors, Inductors, and First-Order Linear Circuits EEE25 ircui Analysis I Se 4: apaciors, Inducors, and Firs-Order inear ircuis Shahriar Mirabbasi Deparmen of Elecrical and ompuer Engineering Universiy of Briish olumbia shahriar@ece.ubc.ca Overview Passive

More information

Inductor Energy Storage

Inductor Energy Storage School of Compuer Science and Elecrical Engineering 5/5/ nducor Energy Sorage Boh capaciors and inducors are energy sorage devices They do no dissipae energy like a resisor, bu sore and reurn i o he circui

More information

Electrical Circuits. 1. Circuit Laws. Tools Used in Lab 13 Series Circuits Damped Vibrations: Energy Van der Pol Circuit

Electrical Circuits. 1. Circuit Laws. Tools Used in Lab 13 Series Circuits Damped Vibrations: Energy Van der Pol Circuit V() R L C 513 Elecrical Circuis Tools Used in Lab 13 Series Circuis Damped Vibraions: Energy Van der Pol Circui A series circui wih an inducor, resisor, and capacior can be represened by Lq + Rq + 1, a

More information

University of Cyprus Biomedical Imaging and Applied Optics. Appendix. DC Circuits Capacitors and Inductors AC Circuits Operational Amplifiers

University of Cyprus Biomedical Imaging and Applied Optics. Appendix. DC Circuits Capacitors and Inductors AC Circuits Operational Amplifiers Universiy of Cyprus Biomedical Imaging and Applied Opics Appendix DC Circuis Capaciors and Inducors AC Circuis Operaional Amplifiers Circui Elemens An elecrical circui consiss of circui elemens such as

More information

2.9 Modeling: Electric Circuits

2.9 Modeling: Electric Circuits SE. 2.9 Modeling: Elecric ircuis 93 2.9 Modeling: Elecric ircuis Designing good models is a ask he compuer canno do. Hence seing up models has become an imporan ask in modern applied mahemaics. The bes

More information

2.4 Cuk converter example

2.4 Cuk converter example 2.4 Cuk converer example C 1 Cuk converer, wih ideal swich i 1 i v 1 2 1 2 C 2 v 2 Cuk converer: pracical realizaion using MOSFET and diode C 1 i 1 i v 1 2 Q 1 D 1 C 2 v 2 28 Analysis sraegy This converer

More information

Lab 10: RC, RL, and RLC Circuits

Lab 10: RC, RL, and RLC Circuits Lab 10: RC, RL, and RLC Circuis In his experimen, we will invesigae he behavior of circuis conaining combinaions of resisors, capaciors, and inducors. We will sudy he way volages and currens change in

More information

( ) ( ) if t = t. It must satisfy the identity. So, bulkiness of the unit impulse (hyper)function is equal to 1. The defining characteristic is

( ) ( ) if t = t. It must satisfy the identity. So, bulkiness of the unit impulse (hyper)function is equal to 1. The defining characteristic is UNIT IMPULSE RESPONSE, UNIT STEP RESPONSE, STABILITY. Uni impulse funcion (Dirac dela funcion, dela funcion) rigorously defined is no sricly a funcion, bu disribuion (or measure), precise reamen requires

More information

EE100 Lab 3 Experiment Guide: RC Circuits

EE100 Lab 3 Experiment Guide: RC Circuits I. Inroducion EE100 Lab 3 Experimen Guide: A. apaciors A capacior is a passive elecronic componen ha sores energy in he form of an elecrosaic field. The uni of capaciance is he farad (coulomb/vol). Pracical

More information

Signal and System (Chapter 3. Continuous-Time Systems)

Signal and System (Chapter 3. Continuous-Time Systems) Signal and Sysem (Chaper 3. Coninuous-Time Sysems) Prof. Kwang-Chun Ho kwangho@hansung.ac.kr Tel: 0-760-453 Fax:0-760-4435 1 Dep. Elecronics and Informaion Eng. 1 Nodes, Branches, Loops A nework wih b

More information

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4.

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4. PHY1 Elecriciy Topic 7 (Lecures 1 & 11) Elecric Circuis n his opic, we will cover: 1) Elecromoive Force (EMF) ) Series and parallel resisor combinaions 3) Kirchhoff s rules for circuis 4) Time dependence

More information

Chapter 2: Principles of steady-state converter analysis

Chapter 2: Principles of steady-state converter analysis Chaper 2 Principles of Seady-Sae Converer Analysis 2.1. Inroducion 2.2. Inducor vol-second balance, capacior charge balance, and he small ripple approximaion 2.3. Boos converer example 2.4. Cuk converer

More information

ENGI 9420 Engineering Analysis Assignment 2 Solutions

ENGI 9420 Engineering Analysis Assignment 2 Solutions ENGI 940 Engineering Analysis Assignmen Soluions 0 Fall [Second order ODEs, Laplace ransforms; Secions.0-.09]. Use Laplace ransforms o solve he iniial value problem [0] dy y, y( 0) 4 d + [This was Quesion

More information

Physics 1502: Lecture 20 Today s Agenda

Physics 1502: Lecture 20 Today s Agenda Physics 152: Lecure 2 Today s Agenda Announcemens: Chap.27 & 28 Homework 6: Friday nducion Faraday's Law ds N S v S N v 1 A Loop Moving Through a Magneic Field ε() =? F() =? Φ() =? Schemaic Diagram of

More information

8.022 (E&M) Lecture 16

8.022 (E&M) Lecture 16 8. (E&M) ecure 16 Topics: Inducors in circuis circuis circuis circuis as ime Our second lecure on elecromagneic inducance 3 ways of creaing emf using Faraday s law: hange area of circui S() hange angle

More information

EE202 Circuit Theory II

EE202 Circuit Theory II EE202 Circui Theory II 2017-2018, Spring Dr. Yılmaz KALKAN I. Inroducion & eview of Fir Order Circui (Chaper 7 of Nilon - 3 Hr. Inroducion, C and L Circui, Naural and Sep epone of Serie and Parallel L/C

More information

dv 7. Voltage-current relationship can be obtained by integrating both sides of i = C :

dv 7. Voltage-current relationship can be obtained by integrating both sides of i = C : EECE202 NETWORK ANALYSIS I Dr. Charles J. Kim Class Noe 22: Capaciors, Inducors, and Op Amp Circuis A. Capaciors. A capacior is a passive elemen designed o sored energy in is elecric field. 2. A capacior

More information

ln 2 1 ln y x c y C x

ln 2 1 ln y x c y C x Lecure 14 Appendi B: Some sample problems from Boas Here are some soluions o he sample problems assigned for Chaper 8 8: 6 Soluion: We wan o find he soluion o he following firs order equaion using separaion

More information

Math 333 Problem Set #2 Solution 14 February 2003

Math 333 Problem Set #2 Solution 14 February 2003 Mah 333 Problem Se #2 Soluion 14 February 2003 A1. Solve he iniial value problem dy dx = x2 + e 3x ; 2y 4 y(0) = 1. Soluion: This is separable; we wrie 2y 4 dy = x 2 + e x dx and inegrae o ge The iniial

More information

8.022 (E&M) Lecture 9

8.022 (E&M) Lecture 9 8.0 (E&M) Lecure 9 Topics: circuis Thevenin s heorem Las ime Elecromoive force: How does a baery work and is inernal resisance How o solve simple circuis: Kirchhoff s firs rule: a any node, sum of he currens

More information

Some Basic Information about M-S-D Systems

Some Basic Information about M-S-D Systems Some Basic Informaion abou M-S-D Sysems 1 Inroducion We wan o give some summary of he facs concerning unforced (homogeneous) and forced (non-homogeneous) models for linear oscillaors governed by second-order,

More information

Complete solutions to Exercise 14(b) 1. Very similar to EXAMPLE 4. We have same characteristic equation:

Complete solutions to Exercise 14(b) 1. Very similar to EXAMPLE 4. We have same characteristic equation: Soluions 4(b) Complee soluions o Exercise 4(b). Very similar o EXAMPE 4. We have same characerisic equaion: 5 i Ae = + Be By using he given iniial condiions we obain he simulaneous equaions A+ B= 0 5A

More information

Chapter 10 INDUCTANCE Recommended Problems:

Chapter 10 INDUCTANCE Recommended Problems: Chaper 0 NDUCTANCE Recommended Problems: 3,5,7,9,5,6,7,8,9,,,3,6,7,9,3,35,47,48,5,5,69, 7,7. Self nducance Consider he circui shown in he Figure. When he swich is closed, he curren, and so he magneic field,

More information

RC, RL and RLC circuits

RC, RL and RLC circuits Name Dae Time o Complee h m Parner Course/ Secion / Grade RC, RL and RLC circuis Inroducion In his experimen we will invesigae he behavior of circuis conaining combinaions of resisors, capaciors, and inducors.

More information

ME 391 Mechanical Engineering Analysis

ME 391 Mechanical Engineering Analysis Fall 04 ME 39 Mechanical Engineering Analsis Eam # Soluions Direcions: Open noes (including course web posings). No books, compuers, or phones. An calculaor is fair game. Problem Deermine he posiion of

More information

ECE-205 Dynamical Systems

ECE-205 Dynamical Systems ECE-5 Dynamical Sysems Course Noes Spring Bob Throne Copyrigh Rober D. Throne Copyrigh Rober D. Throne . Elecrical Sysems The ypes of dynamical sysems we will be sudying can be modeled in erms of algebraic

More information

MEI STRUCTURED MATHEMATICS 4758

MEI STRUCTURED MATHEMATICS 4758 OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary General Cerificae of Educaion Advanced General Cerificae of Educaion MEI STRUCTURED MATHEMATICS 4758 Differenial Equaions Thursday 5 JUNE 006 Afernoon

More information

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc,

More information

Phys1112: DC and RC circuits

Phys1112: DC and RC circuits Name: Group Members: Dae: TA s Name: Phys1112: DC and RC circuis Objecives: 1. To undersand curren and volage characerisics of a DC RC discharging circui. 2. To undersand he effec of he RC ime consan.

More information

LabQuest 24. Capacitors

LabQuest 24. Capacitors Capaciors LabQues 24 The charge q on a capacior s plae is proporional o he poenial difference V across he capacior. We express his wih q V = C where C is a proporionaliy consan known as he capaciance.

More information

i L = VT L (16.34) 918a i D v OUT i L v C V - S 1 FIGURE A switched power supply circuit with diode and a switch.

i L = VT L (16.34) 918a i D v OUT i L v C V - S 1 FIGURE A switched power supply circuit with diode and a switch. 16.4.3 A SWITHED POWER SUPPY USINGA DIODE In his example, we will analyze he behavior of he diodebased swiched power supply circui shown in Figure 16.15. Noice ha his circui is similar o ha in Figure 12.41,

More information

Morning Time: 1 hour 30 minutes Additional materials (enclosed):

Morning Time: 1 hour 30 minutes Additional materials (enclosed): ADVANCED GCE 78/0 MATHEMATICS (MEI) Differenial Equaions THURSDAY JANUARY 008 Morning Time: hour 30 minues Addiional maerials (enclosed): None Addiional maerials (required): Answer Bookle (8 pages) Graph

More information

Chapter 8 The Complete Response of RL and RC Circuits

Chapter 8 The Complete Response of RL and RC Circuits Chaper 8 he Complee Response of R and RC Ciruis Exerises Ex 8.3-1 Before he swih loses: Afer he swih loses: 2 = = 8 Ω so = 8 0.05 = 0.4 s. 0.25 herefore R ( ) Finally, 2.5 ( ) = o + ( (0) o ) = 2 + V for

More information

Lectures 29 and 30 BIQUADRATICS AND STATE SPACE OP AMP REALIZATIONS. I. Introduction

Lectures 29 and 30 BIQUADRATICS AND STATE SPACE OP AMP REALIZATIONS. I. Introduction EE-202/445, 3/18/18 9-1 R. A. DeCarlo Lecures 29 and 30 BIQUADRATICS AND STATE SPACE OP AMP REALIZATIONS I. Inroducion 1. The biquadraic ransfer funcion has boh a 2nd order numeraor and a 2nd order denominaor:

More information

dv i= C. dt 1. Assuming the passive sign convention, (a) i = 0 (dc) (b) (220)( 9)(16.2) t t Engineering Circuit Analysis 8 th Edition

dv i= C. dt 1. Assuming the passive sign convention, (a) i = 0 (dc) (b) (220)( 9)(16.2) t t Engineering Circuit Analysis 8 th Edition . Assuming he passive sign convenion, dv i= C. d (a) i = (dc) 9 9 (b) (22)( 9)(6.2) i= e = 32.8e A 9 3 (c) i (22 = )(8 )(.) sin. = 7.6sin. pa 9 (d) i= (22 )(9)(.8) cos.8 = 58.4 cos.8 na 2. (a) C = 3 pf,

More information

Experimental Buck Converter

Experimental Buck Converter Experimenal Buck Converer Inpu Filer Cap MOSFET Schoky Diode Inducor Conroller Block Proecion Conroller ASIC Experimenal Synchronous Buck Converer SoC Buck Converer Basic Sysem S 1 u D 1 r r C C R R X

More information

Chapter 9 Sinusoidal Steady State Analysis

Chapter 9 Sinusoidal Steady State Analysis Chaper 9 Sinusoidal Seady Sae Analysis 9.-9. The Sinusoidal Source and Response 9.3 The Phasor 9.4 pedances of Passive Eleens 9.5-9.9 Circui Analysis Techniques in he Frequency Doain 9.0-9. The Transforer

More information

AC Circuits AC Circuit with only R AC circuit with only L AC circuit with only C AC circuit with LRC phasors Resonance Transformers

AC Circuits AC Circuit with only R AC circuit with only L AC circuit with only C AC circuit with LRC phasors Resonance Transformers A ircuis A ircui wih only A circui wih only A circui wih only A circui wih phasors esonance Transformers Phys 435: hap 31, Pg 1 A ircuis New Topic Phys : hap. 6, Pg Physics Moivaion as ime we discovered

More information

Differential Equations

Differential Equations Mah 21 (Fall 29) Differenial Equaions Soluion #3 1. Find he paricular soluion of he following differenial equaion by variaion of parameer (a) y + y = csc (b) 2 y + y y = ln, > Soluion: (a) The corresponding

More information

ELEG 205 Fall Lecture #13. Mark Mirotznik, Ph.D. Professor The University of Delaware Tel: (302)

ELEG 205 Fall Lecture #13. Mark Mirotznik, Ph.D. Professor The University of Delaware Tel: (302) ELEG 205 Fall 2017 Leure #13 Mark Miroznik, Ph.D. Professor The Universiy of Delaware Tel: (302831-4221 Email: mirozni@ee.udel.edu Chaper 8: RL and RC Ciruis 1. Soure-free RL iruis (naural response 2.

More information

EECE 301 Signals & Systems Prof. Mark Fowler

EECE 301 Signals & Systems Prof. Mark Fowler EECE 31 Signals & Sysems Prof. Mar Fowler Noe Se #1 C-T Signals: Circuis wih Periodic Sources 1/1 Solving Circuis wih Periodic Sources FS maes i easy o find he response of an RLC circui o a periodic source!

More information

Designing Information Devices and Systems I Spring 2019 Lecture Notes Note 17

Designing Information Devices and Systems I Spring 2019 Lecture Notes Note 17 EES 16A Designing Informaion Devices and Sysems I Spring 019 Lecure Noes Noe 17 17.1 apaciive ouchscreen In he las noe, we saw ha a capacior consiss of wo pieces on conducive maerial separaed by a nonconducive

More information

System of Linear Differential Equations

System of Linear Differential Equations Sysem of Linear Differenial Equaions In "Ordinary Differenial Equaions" we've learned how o solve a differenial equaion for a variable, such as: y'k5$e K2$x =0 solve DE yx = K 5 2 ek2 x C_C1 2$y''C7$y

More information

MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions

MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions MA 14 Calculus IV (Spring 016) Secion Homework Assignmen 1 Soluions 1 Boyce and DiPrima, p 40, Problem 10 (c) Soluion: In sandard form he given firs-order linear ODE is: An inegraing facor is given by

More information

Solutions to Assignment 1

Solutions to Assignment 1 MA 2326 Differenial Equaions Insrucor: Peronela Radu Friday, February 8, 203 Soluions o Assignmen. Find he general soluions of he following ODEs: (a) 2 x = an x Soluion: I is a separable equaion as we

More information

Laplace Transforms. Examples. Is this equation differential? y 2 2y + 1 = 0, y 2 2y + 1 = 0, (y ) 2 2y + 1 = cos x,

Laplace Transforms. Examples. Is this equation differential? y 2 2y + 1 = 0, y 2 2y + 1 = 0, (y ) 2 2y + 1 = cos x, Laplace Transforms Definiion. An ordinary differenial equaion is an equaion ha conains one or several derivaives of an unknown funcion which we call y and which we wan o deermine from he equaion. The equaion

More information

EE 301 Lab 2 Convolution

EE 301 Lab 2 Convolution EE 301 Lab 2 Convoluion 1 Inroducion In his lab we will gain some more experience wih he convoluion inegral and creae a scrip ha shows he graphical mehod of convoluion. 2 Wha you will learn This lab will

More information

LAPLACE TRANSFORM AND TRANSFER FUNCTION

LAPLACE TRANSFORM AND TRANSFER FUNCTION CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION Professor Dae Ryook Yang Spring 2018 Dep. of Chemical and Biological Engineering 5-1 Road Map of he Lecure V Laplace Transform and Transfer funcions

More information

UNIVERSITY OF CALIFORNIA AT BERKELEY

UNIVERSITY OF CALIFORNIA AT BERKELEY Homework #10 Soluions EECS 40, Fall 2006 Prof. Chang-Hasnain Due a 6 pm in 240 Cory on Wednesday, 04/18/07 oal Poins: 100 Pu (1) your name and (2) discussion secion number on your homework. You need o

More information

Chapter 5-4 Operational amplifier Department of Mechanical Engineering

Chapter 5-4 Operational amplifier Department of Mechanical Engineering MEMS08 Chaper 5-4 Operaional amplifier Deparmen of Mechanical Engineering Insrumenaion amplifier Very high inpu impedance Large common mode rejecion raio (CMRR) Capabiliy o amplify low leel signals Consisen

More information

Section 2.2 Charge and Current 2.6 b) The current direction is designated as the direction of the movement of positive charges.

Section 2.2 Charge and Current 2.6 b) The current direction is designated as the direction of the movement of positive charges. Chaper Soluions Secion. Inroducion. Curren source. Volage source. esisor.4 Capacior.5 Inducor Secion. Charge and Curren.6 b) The curren direcion is designaed as he direcion of he movemen of posiive charges..7

More information

Wall. x(t) f(t) x(t = 0) = x 0, t=0. which describes the motion of the mass in absence of any external forcing.

Wall. x(t) f(t) x(t = 0) = x 0, t=0. which describes the motion of the mass in absence of any external forcing. MECHANICS APPLICATIONS OF SECOND-ORDER ODES 7 Mechanics applicaions of second-order ODEs Second-order linear ODEs wih consan coefficiens arise in many physical applicaions. One physical sysems whose behaviour

More information

Section 3.8, Mechanical and Electrical Vibrations

Section 3.8, Mechanical and Electrical Vibrations Secion 3.8, Mechanical and Elecrical Vibraions Mechanical Unis in he U.S. Cusomary and Meric Sysems Disance Mass Time Force g (Earh) Uni U.S. Cusomary MKS Sysem CGS Sysem fee f slugs seconds sec pounds

More information

Continuous Time. Time-Domain System Analysis. Impulse Response. Impulse Response. Impulse Response. Impulse Response. ( t) + b 0.

Continuous Time. Time-Domain System Analysis. Impulse Response. Impulse Response. Impulse Response. Impulse Response. ( t) + b 0. Time-Domain Sysem Analysis Coninuous Time. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 1. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 2 Le a sysem be described by a 2 y ( ) + a 1

More information

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3. Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies

More information

THE BERNOULLI NUMBERS. t k. = lim. = lim = 1, d t B 1 = lim. 1+e t te t = lim t 0 (e t 1) 2. = lim = 1 2.

THE BERNOULLI NUMBERS. t k. = lim. = lim = 1, d t B 1 = lim. 1+e t te t = lim t 0 (e t 1) 2. = lim = 1 2. THE BERNOULLI NUMBERS The Bernoulli numbers are defined here by he exponenial generaing funcion ( e The firs one is easy o compue: (2 and (3 B 0 lim 0 e lim, 0 e ( d B lim 0 d e +e e lim 0 (e 2 lim 0 2(e

More information

6.2 Transforms of Derivatives and Integrals.

6.2 Transforms of Derivatives and Integrals. SEC. 6.2 Transforms of Derivaives and Inegrals. ODEs 2 3 33 39 23. Change of scale. If l( f ()) F(s) and c is any 33 45 APPLICATION OF s-shifting posiive consan, show ha l( f (c)) F(s>c)>c (Hin: In Probs.

More information

3. Alternating Current

3. Alternating Current 3. Alernaing Curren TOPCS Definiion and nroducion AC Generaor Componens of AC Circuis Series LRC Circuis Power in AC Circuis Transformers & AC Transmission nroducion o AC The elecric power ou of a home

More information

Università degli Studi di Roma Tor Vergata Dipartimento di Ingegneria Elettronica. Analogue Electronics. Paolo Colantonio A.A.

Università degli Studi di Roma Tor Vergata Dipartimento di Ingegneria Elettronica. Analogue Electronics. Paolo Colantonio A.A. Universià degli Sudi di Roma Tor Vergaa Diparimeno di Ingegneria Eleronica Analogue Elecronics Paolo Colanonio A.A. 2015-16 Diode circui analysis The non linearbehaviorofdiodesmakesanalysisdifficul consider

More information

Pulse Generators. Any of the following calculations may be asked in the midterms/exam.

Pulse Generators. Any of the following calculations may be asked in the midterms/exam. ulse Generaors ny of he following calculaions may be asked in he miderms/exam.. a) capacior of wha capaciance forms an RC circui of s ime consan wih a 0 MΩ resisor? b) Wha percenage of he iniial volage

More information

The problem with linear regulators

The problem with linear regulators he problem wih linear regulaors i in P in = i in V REF R a i ref i q i C v CE P o = i o i B ie P = v i o o in R 1 R 2 i o i f η = P o P in iref is small ( 0). iq (quiescen curren) is small (probably).

More information

6.01: Introduction to EECS I Lecture 8 March 29, 2011

6.01: Introduction to EECS I Lecture 8 March 29, 2011 6.01: Inroducion o EES I Lecure 8 March 29, 2011 6.01: Inroducion o EES I Op-Amps Las Time: The ircui Absracion ircuis represen sysems as connecions of elemens hrough which currens (hrough variables) flow

More information

d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3

d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3 and d = c b - b c c d = c b - b c c This process is coninued unil he nh row has been compleed. The complee array of coefficiens is riangular. Noe ha in developing he array an enire row may be divided or

More information

Analytic Model and Bilateral Approximation for Clocked Comparator

Analytic Model and Bilateral Approximation for Clocked Comparator Analyic Model and Bilaeral Approximaion for Clocked Comparaor M. Greians, E. Hermanis, G.Supols Insiue of, Riga, Lavia, e-mail: gais.supols@edi.lv Research is suppored by: 1) ESF projec Nr.1DP/1.1.1.2.0/09/APIA/VIAA/020,

More information

non-linear oscillators

non-linear oscillators non-linear oscillaors The invering comparaor operaion can be summarized as When he inpu is low, he oupu is high. When he inpu is high, he oupu is low. R b V REF R a and are given by he expressions derived

More information

10. State Space Methods

10. State Space Methods . Sae Space Mehods. Inroducion Sae space modelling was briefly inroduced in chaper. Here more coverage is provided of sae space mehods before some of heir uses in conrol sysem design are covered in he

More information

Chapter 1 Fundamental Concepts

Chapter 1 Fundamental Concepts Chaper 1 Fundamenal Conceps 1 Signals A signal is a paern of variaion of a physical quaniy, ofen as a funcion of ime (bu also space, disance, posiion, ec). These quaniies are usually he independen variables

More information

MEMS 0031 Electric Circuits

MEMS 0031 Electric Circuits MEMS 0031 Elecric Circuis Chaper 1 Circui variables Chaper/Lecure Learning Objecives A he end of his lecure and chaper, you should able o: Represen he curren and volage of an elecric circui elemen, paying

More information

V(z, t) t < L v. z = 0 z = vt z = L. I(z, t) z = L

V(z, t) t < L v. z = 0 z = vt z = L. I(z, t) z = L W.C.Chew ECE 35 Lecure Noes 12. Transiens on a Transmission Line. When we do no have a ime harmonic signal on a ransmission line, we have o use ransien analysis o undersand he waves on a ransmission line.

More information

Homework: See website. Table of Contents

Homework: See website. Table of Contents Dr. Friz Wilhelm page of 4 C:\physics\3 lecure\ch3 Inducance C circuis.docx; P /5/8 S: 5/4/9 9:39: AM Homework: See websie. Table of Conens: 3. Self-inducance in a circui, 3. -Circuis, 4 3.a Charging he

More information

Problem Set #1. i z. the complex propagation constant. For the characteristic impedance:

Problem Set #1. i z. the complex propagation constant. For the characteristic impedance: Problem Se # Problem : a) Using phasor noaion, calculae he volage and curren waves on a ransmission line by solving he wave equaion Assume ha R, L,, G are all non-zero and independen of frequency From

More information

DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 2008

DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 2008 [E5] IMPERIAL COLLEGE LONDON DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 008 EEE/ISE PART II MEng BEng and ACGI SIGNALS AND LINEAR SYSTEMS Time allowed: :00 hours There are FOUR quesions

More information

Silicon Controlled Rectifiers UNIT-1

Silicon Controlled Rectifiers UNIT-1 Silicon Conrolled Recifiers UNIT-1 Silicon Conrolled Recifier A Silicon Conrolled Recifier (or Semiconducor Conrolled Recifier) is a four layer solid sae device ha conrols curren flow The name silicon

More information

Math Final Exam Solutions

Math Final Exam Solutions Mah 246 - Final Exam Soluions Friday, July h, 204 () Find explici soluions and give he inerval of definiion o he following iniial value problems (a) ( + 2 )y + 2y = e, y(0) = 0 Soluion: In normal form,

More information

555 Timer. Digital Electronics

555 Timer. Digital Electronics 555 Timer Digial Elecronics This presenaion will Inroduce he 555 Timer. 555 Timer Derive he characerisic equaions for he charging and discharging of a capacior. Presen he equaions for period, frequency,

More information

first-order circuit Complete response can be regarded as the superposition of zero-input response and zero-state response.

first-order circuit Complete response can be regarded as the superposition of zero-input response and zero-state response. Experimen 4:he Sdies of ransiional processes of 1. Prpose firs-order circi a) Use he oscilloscope o observe he ransiional processes of firs-order circi. b) Use he oscilloscope o measre he ime consan of

More information

L1, L2, N1 N2. + Vout. C out. Figure 2.1.1: Flyback converter

L1, L2, N1 N2. + Vout. C out. Figure 2.1.1: Flyback converter page 11 Flyback converer The Flyback converer belongs o he primary swiched converer family, which means here is isolaion beween in and oupu. Flyback converers are used in nearly all mains supplied elecronic

More information

Structural Dynamics and Earthquake Engineering

Structural Dynamics and Earthquake Engineering Srucural Dynamics and Earhquae Engineering Course 1 Inroducion. Single degree of freedom sysems: Equaions of moion, problem saemen, soluion mehods. Course noes are available for download a hp://www.c.up.ro/users/aurelsraan/

More information