Problem Set 2 Solutions


 Adam Kelly
 1 years ago
 Views:
Transcription
1 University of Cliforni, Berkeley Spring 2012 EE 42/100 Prof. A. Niknej Prolem Set 2 Solutions Plese note tht these re merely suggeste solutions. Mny of these prolems n e pprohe in ifferent wys. 1. In prolems like this, you my fin it helpful to rerw the iruit to ignore ll the irrelevnt terminls n suh tht everything is retngulr. Note tht sine we re fining eq from terminls n in eh of these, we must simplify the iruits towr these terminls. () Terminls n re onnete together, while e n f re irrelevnt. If you rerw the iruit s elow, you n see tht the two resistors in the first olumn re in prllel s, they re onnete together t oth pirs of terminls (ue to the short). This is lso true for the seon olumn of resistors. So eh of the pirs n e rerwn s one resistor with 1 2. e f /2 /2 It is esy to see tht ll the resistors re in series, so eq = 3. () Terminls e n f re shorte. Intuitively, eq shoul only inlue the two inner resistors, sine the other four re shorte out. Tht is, ny urrent flowing from to woul only go through 2 of resistne n proee etween e n f vi the short. e f e f 2 2 () The short etween n e gets ri of the one resistor etween the two noes. After tht, everything else simplifies s efore.
2 e f e f 2 8/3 2. The key to this prolem is tht the ler is infinite. Using this ft, we n onlue tht the equivlent resistne in shoul not hnge if we remove one rung of the ler. So we n reple the ler, sve for the first rung, with one resistor in. in in Now we n write n eqution in terms of n in n solve. in in = ( in ) = 2 in 2 in 2 in 2 2 = 0 Sine this is qurti eqution, there re two solutions. We know tht resistne nnot e negtive, so we isr the negtive solution. The positive solution is in = (1 3) 3. We hve two unknown noes, so we ll hve two nol equtions, one t noe 1 n the other t noe 2. V 1 V 2 V = 0 V 2 V 1 V = 0 Solving this system gives us V 1 = V n V 2 = 74.4 V. We n lso o this using superposition. Here we split the prolem into two suiruits, eh one with ifferent urrent soure zeroe out. ell tht the zeroe out urrent soure eomes n open iruit. Ω 1 2 8A 30Ω 20Ω 2
3 V 1 V 2 V 2 V 1 V = 0 V 2 20 = 0 This first iruit gives us V 1 = V n V 2 = 19.2 V. Ω Ω 20Ω 3A V 1 V 2 V 2 V 1 V 1 30 = 0 V = 0 The solutions re V 1 = 7.2 V n V 2 = 55.2 V. Clerly, the sum of these with the solutions for the first suiruit yiels the sme nswers s the originl iruit. 4. There re five noes in this iruit, lele elow. Pling the referene noe t v E reues the numer of unknowns to three, sine now v E = 0 V n v D = 5 V. Ω A B C Ω D 5V 18Ω 75Ω 1.2A E 30Ω The three nol equtions re the following: V A V C V A V B 18 V A 5 = 0 43V A 25V B 9V C = 45 V B V A 18 V C V A V B V C 75 V C V B = 0 25V A 31V B 6V C = 540 V C 30 = 0 3V A 2V B 10V C = 0 The solutions re V A = 32.8 V, V B = 47.6 V, n V C = 19.3 V. The rnh urrents n e foun using Ohm s lw n KCL: I AC = 0.27 A, I AB = 0.82 A, I BC = 0.38 A, I AD = 0.56 A, I CE = 0.64 A, I DE = 0.56 A. 3
4 5. The iruit hs four unknown noes, lele elow. 100Ω 0Ω 1V V A V B V C 5V 10 V x 80Ω V x 5Ω V x 90Ω In ition, we must lso write supernoe eqution t the 1 oure. The epenent urrent soure oes not present prolem, sine we tret V x s one of the unknowns. The nol equtions t V A n V x re V A V A 80 V A V B 0 10V x = 0 V x V C 5 The two equtions t the supernoe re V x 90 = 0 V B V A 0 V C V x 5 V B V C = 1 The solution to these equtions is V A = V, V B = V, V C = V, V x = V. 6. Note tht the left prt of the iruit is inepenent of the right prt. Sine there is no omplete loop onneting the two prts, no urrent n flow etween them; the wire onneting them only serves s ommon groun. The equivlent resistne on the left is = 9 Ω. So the totl urrent from is just Vs. Also, we hve tht I 9 x is just hlf of tht, sine the two prllel rnhes hve equl resistne. So I x = Vs. 18 On the right hlf, the two resistors re in prllel, so the equivlent resistne is just 2 Ω. Sine the urrent going through them is given y AI x = AVs, the voltge ross 18 them is V out = I eq = AVs Vout. Solving for A n plugging in 9 V in = 9, we hve tht A = The steps to fining the equivlent iruits re to fin V o, I s, n/or eq. Sine there is epenent soure in this iruit, it is esier to fin the first two. V o is simply the voltge ifferene etween n. By inspetion, this is just = 0 V o = (100 kω)βi x = (100 kω)β 10 kω = 10β I s is foun y shorting the terminls n solving for I. Note tht shorting out the 100 kω resistor will leve us with I = βi x = β Vs (note the sign!). 10 kω 4
5 Thus, our esire vlues re V T h = 10β n I N = βvs. Also, the Thévenin n 10 kω Norton resistne is just given y T h = N = V T h I N = 100 kω. 8. () The mximum power trnsfer theorem tells us tht the lo resistne shoul e equl to the Thévenin resistne of the iruit. In this se, it is trivil to see tht tht L = T h = s. If you wnt to review the erivtion of this result, plese onsult the leture notes or textook. () We wnt the power ross the lo, P L, to e equl to 80% of the power elivere y the voltge soure, P s. The former n e foun using P L = I 2 L, wheres the ltter is P s = I. In this iruit, I = So putting everything together, we hve Solving for L gives us L = 4 s. s L Vs 2 L ( s L ) = 0.8 Vs 2 2 s L () Power is epenent on oth voltge n urrent. While the nswer to prt () gives us higher voltge ross L n greter frtion of the totl power, the totl power is tully muh smller in this se. This is euse the urrent is 5 times smller ue to the greter resistne, while the voltge ross L is only 1.6 times lrger. 9. Intuitively we know tht losing the swith puts 2 n 3 in prllel, lowering the effetive resistne. So it must e in the open se tht v o tkes the greter voltge vlue, 26 V. In this senrio, the iruit looks like the following: Ω 39V 2 v o = 26V Solving for this iruit, using voltge ivier, nol nlysis, or otherwise, gives us 2 = 100 Ω. Now if we lose the swith, v o = 24 V. The vlue of 2 remins unhnge, so ll tht s remining is to solve for 3. Ω 39V 100 Ω v o 3 5
6 We n write nol eqution t v o = 24 V: The solution is 3 = 400 Ω. v o 39 v o 100 v o 3 = () When the rige is lne, no urrent flows ross the glvnometer. By KCL, the urrent ross 1 n 2 re the sme, s well s the urrent ross 3 n 4. In ition, the voltges v n v must e equl ue to the 0 urrent. 1 G 3 I 1 I 2 2 x By KVL, the voltge rops ross 1 n 3 must then e the sme. The sme resoning goes for 2 n x. I 1 1 = I 2 3 I 1 2 = I 2 x Sustituting for the urrents n solving for x, we hve x = () Given our vlues for 1 n 3, we n mesure x = 2 2. Sine the highest vlue tht 2 n tke is 1000 Ω, the mximum resistne tht n e mesure for x is just 0 Ω. As 2 omes in inrements of 10 Ω, the mesurement n e me with n ury to within 20 Ω. () The Thévenin equivlent is foun y fining the voltge n equivlent resistne ross the open terminls n. Note tht here we nnot ssume the rige is lne, s we re looking t the generl se. v o is simply the ifferene in potentil etween noes n, oth of whih n e foun vi voltge ivier: 2 v = 1 2 v = 3 x ( 2 V T h = v v = ) x x The equivlent resistne n e foun y zeroing out the voltge soure, mking it short, n fining eq. 6 x
7 1 3 2 x We hve 1 n 2 in prllel with eh other, n 3 n x in prllel with eh other. These two pirs re in series, so the equivlent resistne is T h = ( 1 2 ) ( 3 x ). 7
1. Area under a curve region bounded by the given function, vertical lines and the x axis.
Ares y Integrtion. Are uner urve region oune y the given funtion, vertil lines n the is.. Are uner urve region oune y the given funtion, horizontl lines n the y is.. Are etween urves efine y two given
More informationMaximum area of polygon
Mimum re of polygon Suppose I give you n stiks. They might e of ifferent lengths, or the sme length, or some the sme s others, et. Now there re lots of polygons you n form with those stiks. Your jo is
More informationCHAPTER 31 CAPACITOR
. Given tht Numer of eletron HPTER PITOR Net hrge Q.6 9.6 7 The net potentil ifferene L..6 pitne v 7.6 8 F.. r 5 m. m 8.854 5.4 6.95 5 F... Let the rius of the is R re R D mm m 8.85 r r 8.85 4. 5 m.5 m
More informationVersion 001 CIRCUITS holland (1290) 1
Version CRCUTS hollnd (9) This printout should hve questions Multiplechoice questions my continue on the next column or pge find ll choices efore nswering AP M 99 MC points The power dissipted in wire
More informationGRADE 4. Fractions WORKSHEETS
GRADE Frtions WORKSHEETS Types of frtions equivlent frtions This frtion wll shows frtions tht re equivlent. Equivlent frtions re frtions tht re the sme mount. How mny equivlent frtions n you fin? Lel eh
More informationRatio and Proportion
Rtio nd Proportion Rtio: The onept of rtio ours frequently nd in wide vriety of wys For exmple: A newspper reports tht the rtio of Repulins to Demorts on ertin Congressionl ommittee is 3 to The student/fulty
More informationLesson 2.1 Inductive Reasoning
Lesson.1 Inutive Resoning Nme Perio Dte For Eerises 1 7, use inutive resoning to fin the net two terms in eh sequene. 1. 4, 8, 1, 16,,. 400, 00, 100, 0,,,. 1 8, 7, 1, 4,, 4.,,, 1, 1, 0,,. 60, 180, 10,
More informationAnswer, Key Homework 10 David McIntyre 1
Answer, Key Homework 10 Dvid McIntyre 1 This printout should hve 22 questions, check tht it is complete. Multiplechoice questions my continue on the next column or pge: find ll choices efore mking your
More informationAngles 2.1. Exercise 2.1... Find the size of the lettered angles. Give reasons for your answers. a) b) c) Example
2.1 Angles Reognise lternte n orresponing ngles Key wors prllel lternte orresponing vertilly opposite Rememer, prllel lines re stright lines whih never meet or ross. The rrows show tht the lines re prllel
More informationModule 5. Threephase AC Circuits. Version 2 EE IIT, Kharagpur
Module 5 Threehse A iruits Version EE IIT, Khrgur esson 8 Threehse Blned Suly Version EE IIT, Khrgur In the module, ontining six lessons (7), the study of iruits, onsisting of the liner elements resistne,
More information50 MATHCOUNTS LECTURES (10) RATIOS, RATES, AND PROPORTIONS
0 MATHCOUNTS LECTURES (0) RATIOS, RATES, AND PROPORTIONS BASIC KNOWLEDGE () RATIOS: Rtios re use to ompre two or more numers For n two numers n ( 0), the rtio is written s : = / Emple : If 4 stuents in
More informationThe remaining two sides of the right triangle are called the legs of the right triangle.
10 MODULE 6. RADICAL EXPRESSIONS 6 Pythgoren Theorem The Pythgoren Theorem An ngle tht mesures 90 degrees is lled right ngle. If one of the ngles of tringle is right ngle, then the tringle is lled right
More informationChapter. Contents: A Constructing decimal numbers
Chpter 9 Deimls Contents: A Construting deiml numers B Representing deiml numers C Deiml urreny D Using numer line E Ordering deimls F Rounding deiml numers G Converting deimls to frtions H Converting
More informationAssign, Ten Homework 9 Due: Dec 11 2003, 2:00 pm Inst: Richard Saenz 1
Assign, Ten Homework 9 Due: De 11 2003, 2:00 pm Inst: ihr Senz 1 This printout shoul he 34 questions. Multiplehoie questions my ontinue on the next olumn or pge fin ll hoies efore mking your seletion.
More informationMATH PLACEMENT REVIEW GUIDE
MATH PLACEMENT REVIEW GUIDE This guie is intene s fous for your review efore tking the plement test. The questions presente here my not e on the plement test. Although si skills lultor is provie for your
More informationPROJECTILE MOTION PRACTICE QUESTIONS (WITH ANSWERS) * challenge questions
PROJECTILE MOTION PRACTICE QUESTIONS (WITH ANSWERS) * hllenge questions e The ll will strike the ground 1.0 s fter it is struk. Then v x = 20 m s 1 nd v y = 0 + (9.8 m s 2 )(1.0 s) = 9.8 m s 1 The speed
More informationc b 5.00 10 5 N/m 2 (0.120 m 3 0.200 m 3 ), = 4.00 10 4 J. W total = W a b + W b c 2.00
Chter 19, exmle rolems: (19.06) A gs undergoes two roesses. First: onstnt volume @ 0.200 m 3, isohori. Pressure inreses from 2.00 10 5 P to 5.00 10 5 P. Seond: Constnt ressure @ 5.00 10 5 P, isori. olume
More information1 Fractions from an advanced point of view
1 Frtions from n vne point of view We re going to stuy frtions from the viewpoint of moern lger, or strt lger. Our gol is to evelop eeper unerstning of wht n men. One onsequene of our eeper unerstning
More informationCS99S Laboratory 2 Preparation Copyright W. J. Dally 2001 October 1, 2001
CS99S Lortory 2 Preprtion Copyright W. J. Dlly 2 Octoer, 2 Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes to oserve logic
More informationVectors Summary. Projection vector AC = ( Shortest distance from B to line A C D [OR = where m1. and m
. Slr prout (ot prout): = osθ Vetors Summry Lws of ot prout: (i) = (ii) ( ) = = (iii) = (ngle etween two ientil vetors is egrees) (iv) = n re perpeniulr Applitions: (i) Projetion vetor: B Length of projetion
More informationRotating DC Motors Part II
Rotting Motors rt II II.1 Motor Equivlent Circuit The next step in our consiertion of motors is to evelop n equivlent circuit which cn be use to better unerstn motor opertion. The rmtures in rel motors
More information1. Definition, Basic concepts, Types 2. Addition and Subtraction of Matrices 3. Scalar Multiplication 4. Assignment and answer key 5.
. Definition, Bsi onepts, Types. Addition nd Sutrtion of Mtries. Slr Multiplition. Assignment nd nswer key. Mtrix Multiplition. Assignment nd nswer key. Determinnt x x (digonl, minors, properties) summry
More information, and the number of electrons is 19. e e 1.60 10 C. The negatively charged electrons move in the direction opposite to the conventional current flow.
Prolem 1. f current of 80.0 ma exists in metl wire, how mny electrons flow pst given cross section of the wire in 10.0 min? Sketch the directions of the current nd the electrons motion. Solution: The chrge
More information2 DIODE CLIPPING and CLAMPING CIRCUITS
2 DIODE CLIPPING nd CLAMPING CIRCUITS 2.1 Ojectives Understnding the operting principle of diode clipping circuit Understnding the operting principle of clmping circuit Understnding the wveform chnge of
More informationFAULT TREES AND RELIABILITY BLOCK DIAGRAMS. Harry G. Kwatny. Department of Mechanical Engineering & Mechanics Drexel University
SYSTEM FAULT AND Hrry G. Kwtny Deprtment of Mechnicl Engineering & Mechnics Drexel University OUTLINE SYSTEM RBD Definition RBDs nd Fult Trees System Structure Structure Functions Pths nd Cutsets Reliility
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationState the size of angle x. Sometimes the fact that the angle sum of a triangle is 180 and other angle facts are needed. b y 127
ngles 2 CHTER 2.1 Tringles Drw tringle on pper nd lel its ngles, nd. Ter off its orners. Fit ngles, nd together. They mke stright line. This shows tht the ngles in this tringle dd up to 180 ut it is not
More informationReasoning to Solve Equations and Inequalities
Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing
More informationAppendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:
Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you
More informationGENERAL OPERATING PRINCIPLES
KEYSECUREPC USER MANUAL N.B.: PRIOR TO READING THIS MANUAL, YOU ARE ADVISED TO READ THE FOLLOWING MANUAL: GENERAL OPERATING PRINCIPLES Der Customer, KeySeurePC is n innovtive prout tht uses ptente tehnology:
More informationPROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive
More informationLecture 15  Curve Fitting Techniques
Lecture 15  Curve Fitting Techniques Topics curve fitting motivtion liner regression Curve fitting  motivtion For root finding, we used given function to identify where it crossed zero where does fx
More informationSECTION 72 Law of Cosines
516 7 Additionl Topis in Trigonometry h d sin s () tn h h d 50. Surveying. The lyout in the figure t right is used to determine n inessile height h when seline d in plne perpendiulr to h n e estlished
More informationVolumes by Cylindrical Shells: the Shell Method
olumes Clinril Shells: the Shell Metho Another metho of fin the volumes of solis of revolution is the shell metho. It n usull fin volumes tht re otherwise iffiult to evlute using the Dis / Wsher metho.
More informationHomework 3 Solutions
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.
More informationUnit 5 Section 1. Mortgage Payment Methods & Products (20%)
Unit 5 Setion 1 Mortgge Pyment Methos & Prouts (20%) There re tully only 2 mortgge repyment methos ville CAPITAL REPAYMENT n INTEREST ONLY. Cpitl Repyment Mortgge Also lle Cpitl & Interest mortgge or repyment
More informationExample 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
More informationSquare Roots Teacher Notes
Henri Picciotto Squre Roots Techer Notes This unit is intended to help students develop n understnding of squre roots from visul / geometric point of view, nd lso to develop their numer sense round this
More informationOn Equivalence Between Network Topologies
On Equivlene Between Network Topologies Tre Ho Deprtment of Eletril Engineering Cliforni Institute of Tehnolog tho@lteh.eu; Mihelle Effros Deprtments of Eletril Engineering Cliforni Institute of Tehnolog
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments  they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationBinary Representation of Numbers Autar Kaw
Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse rel number to its binry representtion,. convert binry number to n equivlent bse number. In everydy
More informationPractice Test 2. a. 12 kn b. 17 kn c. 13 kn d. 5.0 kn e. 49 kn
Prtie Test 2 1. A highwy urve hs rdius of 0.14 km nd is unnked. A r weighing 12 kn goes round the urve t speed of 24 m/s without slipping. Wht is the mgnitude of the horizontl fore of the rod on the r?
More informationLectures 8 and 9 1 Rectangular waveguides
1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves
More information1 Numerical Solution to Quadratic Equations
cs42: introduction to numericl nlysis 09/4/0 Lecture 2: Introduction Prt II nd Solving Equtions Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mrk Cowlishw Numericl Solution to Qudrtic Equtions Recll
More informationSolution to Problem Set 1
CSE 5: Introduction to the Theory o Computtion, Winter A. Hevi nd J. Mo Solution to Prolem Set Jnury, Solution to Prolem Set.4 ). L = {w w egin with nd end with }. q q q q, d). L = {w w h length t let
More informationReal Analysis HW 10 Solutions
Rel Anlysis HW 10 Solutions Problem 47: Show tht funtion f is bsolutely ontinuous on [, b if nd only if for eh ɛ > 0, there is δ > 0 suh tht for every finite disjoint olletion {( k, b k )} n of open intervls
More informationSection 54 Trigonometric Functions
5 Trigonometric Functions Section 5 Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form
More informationWords Symbols Diagram. abcde. a + b + c + d + e
Logi Gtes nd Properties We will e using logil opertions to uild mhines tht n do rithmeti lultions. It s useful to think of these opertions s si omponents tht n e hooked together into omplex networks. To
More information1.2 The Integers and Rational Numbers
.2. THE INTEGERS AND RATIONAL NUMBERS.2 The Integers n Rtionl Numers The elements of the set of integers: consist of three types of numers: Z {..., 5, 4, 3, 2,, 0,, 2, 3, 4, 5,...} I. The (positive) nturl
More informationIf two triangles are perspective from a point, then they are also perspective from a line.
Mth 487 hter 4 Prtie Prolem Solutions 1. Give the definition of eh of the following terms: () omlete qudrngle omlete qudrngle is set of four oints, no three of whih re olliner, nd the six lines inident
More informationSOLVING EQUATIONS BY FACTORING
316 (560) Chpter 5 Exponents nd Polynomils 5.9 SOLVING EQUATIONS BY FACTORING In this setion The Zero Ftor Property Applitions helpful hint Note tht the zero ftor property is our seond exmple of getting
More informationRegular Sets and Expressions
Regulr Sets nd Expressions Finite utomt re importnt in science, mthemtics, nd engineering. Engineers like them ecuse they re super models for circuits (And, since the dvent of VLSI systems sometimes finite
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More informationTallahassee Community College. Simplifying Radicals
Tllhssee Communit College Simplifing Rdils The squre root of n positive numer is the numer tht n e squred to get the numer whose squre root we re seeking. For emple, 1 euse if we squre we get 1, whih is
More informationFactoring Polynomials
Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles
More informationMath Review 1. , where α (alpha) is a constant between 0 and 1, is one specific functional form for the general production function.
Mth Review Vribles, Constnts nd Functions A vrible is mthemticl bbrevition for concept For emple in economics, the vrible Y usully represents the level of output of firm or the GDP of n economy, while
More informationMath 314, Homework Assignment 1. 1. Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.
Mth 4, Homework Assignment. Prove tht two nonverticl lines re perpendiculr if nd only if the product of their slopes is. Proof. Let l nd l e nonverticl lines in R of slopes m nd m, respectively. Suppose
More informationLINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES
LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of
More informationEQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in pointdirection nd twopoint
More informationWHAT HAPPENS WHEN YOU MIX COMPLEX NUMBERS WITH PRIME NUMBERS?
WHAT HAPPES WHE YOU MIX COMPLEX UMBERS WITH PRIME UMBERS? There s n ol syng, you n t pples n ornges. Mthemtns hte n t; they love to throw pples n ornges nto foo proessor n see wht hppens. Sometmes they
More informationCS 316: Gates and Logic
CS 36: Gtes nd Logi Kvit Bl Fll 27 Computer Siene Cornell University Announements Clss newsgroup reted Posted on wepge Use it for prtner finding First ssignment is to find prtners P nd N Trnsistors PNP
More informationNotes on Excess Burden (EB) most efficient lowest deadweight loss excess burden nondistorting tax system benchmark
Notes on Exess Buren (EB) Our gol is to lulte the exess uren of tx system. This will llow us to juge one tx system ginst nother. All txes use inome effets simply euse they tke money wy tht oul hve een
More informationBoğaziçi University Department of Economics Spring 2016 EC 102 PRINCIPLES of MACROECONOMICS Problem Set 5 Answer Key
Boğziçi University Deprtment of Eonomis Spring 2016 EC 102 PRINCIPLES of MACROECONOMICS Prolem Set 5 Answer Key 1. One yer ountry hs negtive net exports. The next yer it still hs negtive net exports n
More informationGraphs on Logarithmic and Semilogarithmic Paper
0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl
More informationForensic Engineering Techniques for VLSI CAD Tools
Forensi Engineering Tehniques for VLSI CAD Tools Jennifer L. Wong, Drko Kirovski, Dvi Liu, Miorg Potkonjk UCLA Computer Siene Deprtment University of Cliforni, Los Angeles June 8, 2000 Computtionl Forensi
More informationStrong acids and bases
Monoprotic AcidBse Equiliri (CH ) ϒ Chpter monoprotic cids A monoprotic cid cn donte one proton. This chpter includes uffers; wy to fi the ph. ϒ Chpter 11 polyprotic cids A polyprotic cid cn donte multiple
More informationWeek 11  Inductance
Week  Inductnce November 6, 202 Exercise.: Discussion Questions ) A trnsformer consists bsiclly of two coils in close proximity but not in electricl contct. A current in one coil mgneticlly induces n
More information1 GSW IPv4 Addressing
1 For s long s I ve een working with the Internet protools, people hve een sying tht IPv6 will e repling IPv4 in ouple of yers time. While this remins true, it s worth knowing out IPv4 ddresses. Even when
More information. At first sight a! b seems an unwieldy formula but use of the following mnemonic will possibly help. a 1 a 2 a 3 a 1 a 2
7 CHAPTER THREE. Cross Product Given two vectors = (,, nd = (,, in R, the cross product of nd written! is defined to e: " = (!,!,! Note! clled cross is VECTOR (unlike which is sclr. Exmple (,, " (4,5,6
More informationThe art of Paperarchitecture (PA). MANUAL
The rt of Pperrhiteture (PA). MANUAL Introution Pperrhiteture (PA) is the rt of reting threeimensionl (3D) ojets out of plin piee of pper or ror. At first, esign is rwn (mnully or printe (using grphil
More informationMath 135 Circles and Completing the Square Examples
Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for
More informationAssuming all values are initially zero, what are the values of A and B after executing this Verilog code inside an always block? C=1; A <= C; B = C;
B26 Appendix B The Bsics of Logic Design Check Yourself ALU n [Arthritic Logic Unit or (rre) Arithmetic Logic Unit] A rndomnumer genertor supplied s stndrd with ll computer systems Stn KellyBootle,
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationHeron s Formula for Triangular Area
Heron s Formul for Tringulr Are y Christy Willims, Crystl Holom, nd Kyl Gifford Heron of Alexndri Physiist, mthemtiin, nd engineer Tught t the museum in Alexndri Interests were more prtil (mehnis, engineering,
More informationExperiment 6: Friction
Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht
More informationClause Trees: a Tool for Understanding and Implementing Resolution in Automated Reasoning
Cluse Trees: Tool for Understnding nd Implementing Resolution in Automted Resoning J. D. Horton nd Brue Spener University of New Brunswik, Frederiton, New Brunswik, Cnd E3B 5A3 emil : jdh@un. nd spener@un.
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More informationPhysics 43 Homework Set 9 Chapter 40 Key
Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nmwide region t x
More informationSOLVING QUADRATIC EQUATIONS BY FACTORING
6.6 Solving Qudrti Equtions y Ftoring (6 31) 307 In this setion The Zero Ftor Property Applitions 6.6 SOLVING QUADRATIC EQUATIONS BY FACTORING The tehniques of ftoring n e used to solve equtions involving
More informationCurve Sketching. 96 Chapter 5 Curve Sketching
96 Chpter 5 Curve Sketching 5 Curve Sketching A B A B A Figure 51 Some locl mximum points (A) nd minimum points (B) If (x, f(x)) is point where f(x) reches locl mximum or minimum, nd if the derivtive of
More informationOrthopoles and the Pappus Theorem
Forum Geometriorum Volume 4 (2004) 53 59. FORUM GEOM ISSN 15341178 Orthopoles n the Pppus Theorem tul Dixit n Drij Grinerg strt. If the verties of tringle re projete onto given line, the perpeniulrs from
More informationSParameters for Three and Four Two Port Networks
the ehnology Interfe/pring 2007 Mus, diku, nd Akujuoi Prmeters for hree nd Four wo Port Networks rhn M. Mus, Mtthew N.O. diku, nd Cjetn M. Akujuoi Center of Exellene for Communition ystems ehnology Reserh
More informationMA 15800 Lesson 16 Notes Summer 2016 Properties of Logarithms. Remember: A logarithm is an exponent! It behaves like an exponent!
MA 5800 Lesson 6 otes Summer 06 Rememer: A logrithm is n eponent! It ehves like n eponent! In the lst lesson, we discussed four properties of logrithms. ) log 0 ) log ) log log 4) This lesson covers more
More informationHomework #6 Chapter 7 Homework Acids and Bases
Homework #6 Chpter 7 Homework Acids nd Bses 18. ) H O(l) H 3O (q) OH (q) H 3 O OH Or H O(l) H (q) OH (q) H OH ) HCN(q) H O(l) H 3O (q) CN (q) H 3 O HCN CN Or HCN(q) H (q) CN (q) H CN HCN c) NH 3(q) H O(l)
More informationH SERIES. Area and Perimeter. Curriculum Ready. www.mathletics.com
Are n Perimeter Curriulum Rey www.mthletis.om Copyright 00 3P Lerning. All rights reserve. First eition printe 00 in Austrli. A tlogue reor for this ook is ville from 3P Lerning Lt. ISBN 7886307 Ownership
More informationand thus, they are similar. If k = 3 then the Jordan form of both matrices is
Homework ssignment 11 Section 7. pp. 24925 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If
More informationThe Stirling Engine: The Heat Engine
Memoril University of Newfounln Deprtment of Physis n Physil Oenogrphy Physis 2053 Lortory he Stirling Engine: he Het Engine Uner no irumstnes shoul you ttempt to operte the engine without supervision:
More informationUnderstanding Basic Analog Ideal Op Amps
Appliction Report SLAA068A  April 2000 Understnding Bsic Anlog Idel Op Amps Ron Mncini Mixed Signl Products ABSTRACT This ppliction report develops the equtions for the idel opertionl mplifier (op mp).
More informationExample A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding
1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde
More informationEnd of term: TEST A. Year 4. Name Class Date. Complete the missing numbers in the sequences below.
End of term: TEST A You will need penil nd ruler. Yer Nme Clss Dte Complete the missing numers in the sequenes elow. 8 30 3 28 2 9 25 00 75 25 2 Put irle round ll of the following shpes whih hve 3 shded.
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More information4.11 Inner Product Spaces
314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces
More informationAREA OF A SURFACE OF REVOLUTION
AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.
More informationInterior and exterior angles add up to 180. Level 5 exterior angle
22 ngles n proof Ientify interior n exterior ngles in tringles n qurilterls lulte interior n exterior ngles of tringles n qurilterls Unerstn the ie of proof Reognise the ifferene etween onventions, eﬁnitions
More informationCalculus of variations. I = F(y, y,x) dx (1)
MT58  Clculus of vritions Introuction. Suppose y(x) is efine on the intervl, Now suppose n so efines curve on the ( x,y) plne. I = F(y, y,x) x (1) with the erivtive of y(x). The vlue of this will epen
More informationP.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn
33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of
More informationFurther applications of area and volume
2 Further pplitions of re n volume 2A Are of prts of the irle 2B Are of omposite shpes 2C Simpson s rule 2D Surfe re of yliners n spheres 2E Volume of omposite solis 2F Error in mesurement Syllus referene
More informationHow to Graphically Interpret the Complex Roots of a Quadratic Equation
Universit of Nersk  Linoln DigitlCommons@Universit of Nersk  Linoln MAT Em Epositor Ppers Mth in the Middle Institute Prtnership 7007 How to Grphill Interpret the Comple Roots of Qudrti Eqution Crmen
More informationIntersection Problems
Intersetion Prolems Determine pirs of interseting ojets? C A B E D Complex shpes forme y oolen opertions: interset, union, iff. Collision etetion in rootis n motion plnning. Visiility, olusion, renering
More information