DIRECT CURRENT CIRCUITS


 Monica Douglas
 10 months ago
 Views:
Transcription
1 DRECT CURRENT CUTS ELECTRC POWER Consider the circuit shown in the Figure where bttery is connected to resistor R. A positive chrge dq will gin potentil energy s it moves from point to point b through the bttery. The chrge loses the sme mount of potentil energy s it moves from point c to point d through the resistor. This mount is given by d V R c b du = dqε = dqv Now du dq P = = V = V dt dt Using the formul V=R P = V = V R = R Exmple 7.7 You re given n electric heter mde of nichrome wire of resistnce 8 Ω. Find the current crried by the wire nd the power of the heter if it is connected to 0 V source. Solution = R V 0 = = 5A 8.0 The power cn be found using P ( 5) (8.0).8 KW = = R =
2 ELECTROMOTVE FOE Consider the circuit shown in the figure. Current will be estblished through the resistor R if potentil r b difference is mintined cross its ends. f the end b is t higher potentil thn the end, then chrge will move through the resistor from b to. For the current R to circulte round closed circuit, the chrge must jump from to b. This mens tht we need device tht cpble of pumping chrge from lower potentil to higher potentil. The function of such device is clled electromotive force, bbrevited emf, nd denoted by the symbol ε. The bttery nd the genertor re common emf devices. A source of emf cn be considered s chrge pump tht pumps chrges in direction opposite to the electrosttic force inside the source. t is exctly like wter pump tht pushes wter from lower to higher level opposite to the grvittionl force. The resistnce r is clled the internl resistnce of the bttery, nd R is clled the lod resistor. We shll ssume tht the connecting wire hve no resistnce. Any positive chrge moving from to b will gin potentil ε s it psses from the negtive terminl to the positive terminl of the bttery. However, it will lose potentil r s it psses through the internl resistor, where is the current in the circuit. Thus, the terminl voltge of the bttery, V = V b V, is given by V =ε r 8. From this Eqution it is cler tht the emf is equl to the terminl voltge of bttery in n open circuit, tht is when the current is zero. As the connecting wires hve no resistnce we conclude tht the voltge V must lso equl the potentil cross the lod resistor R, tht is V = R ε
3 From the bove two Equtions we get = ε R + r 8. Exmple 8. A bttery hs n emf 0f.0 V nd n internl resistnce of 0.05 Ω. ts terminls re connected to lod resistnce of 3.0 Ω. ) Find nd the terminl voltge of the bttery. b) Clculte the power delivered to R, r, nd by the bttery. Solution: ) ε.0 = = = 3.93A R + r V = ε r =.0 ( ) =.8V b) = R= ( 3.93) ( 3.0) = 46.3W P R P r = r = ( 3.93) ( 0.05) = 0.77W P ε = ε = = 47.W Note tht P = P R + P ε r
4 RESSTORS N SERES AND N PARALLEL When two resistors re connected together s shown we sid tht they re connected in series. As it is cler from the figure, ny chrge tht flows through R must equl the chrge tht flows through R, tht is the current is the sme in ech resistor. Since the potentil difference between nd b equls the sum of the potentil drop cross ech resistor we hve R ε R b = R + R = ( R ) 8.3 V eq + R where V eq is the potentil drop cross the equivlent resistor. Therefore we conclude tht R eq = R + R 8.4 The equivlent resistor of more thn two resistors connected in series is then R eq R + R + +L 8.5 = R3 R Now consider the two resistors connected s shown in the Figure. The potentil drops cross R nd R re equl nd must equl to the potentil drop cross ny equivlent resistor connected between nd b, tht is R b V eq = V = V 8.6 ε f nd re the currents pssing through R nd R, respectively, then the net current of the circuit is
5 = Using Ohm s lw nd Eqution 8.6, we get R = + eq R R 8.8 n generl if more tht two resistors re connected in prllel, then we hve = + + +L 8.9 R eq R R R 3 Exmple 8.4 Four resistors re connected s shown in. () Wht is the equivlent resistnce between points & c.? (b) Wht is the current in ech resistor if potentil difference of 4 V is mintined between & c.? 8.0 Ω 4.0 Ω 3.0 A 3.0 A b.0 A 6.0 V 6.0 Ω 3.0 Ω 6.0 V.0 A c Ω.0 Ω 3.0 A b 3.0 A c 36 V 6.0 V 4 Ω 3.0 A 4 V c
6 Solution: () The circuit cn be reduced, step by step, to single equivlent resistnce s shown in the Figure. The 8.0Ω nd the 4.0Ω re connected in series, nd so they cn be replced by n equivlent resistor of Ω. The 6.0Ω nd the 3.0Ω re connected in prllel, nd so they cn be replced by n equivlent resistor of.0 Ω. The equivlents re connected in series. The equivlent resistnce of the circuit is then R eq = +.0 = 4Ω (b) Since the Ω nd the.0ω re connected in series, they hve the sme current eq, which must equl to the current of the 4Ω resistor. Using Ohm s lw we get R eq eq = Req 48 = = 3.0A 4 Now the potentil difference cross the.0ω is V bc = R = 3.0(.0) = 6.0 V eq This potentil difference is the sme cross the 6.0Ω nd the 3.0Ω resistors due the prllel connection between them. So, we cn find the current pssing through the 3.0Ω resistor s 6.0 = = 3.0.0A nd the current through the 6.0Ω resistor s 6.0 = =.0 A 6.0 The current pssing through the 4.0Ω nd the 8.0Ω is the sme s tht pssing through the Ω (3.0 A) due the series connection between them.
7 Exmple 8.5 shown? Wht is equivlent resistnce between nd b in the Figure c Ω 5 Ω Ω d Ω Ω b 5 Ω Ω Ω Ω c,d Ω b 0.5 Ω 0.5 Ω c,d b 0.5 Ω b Solution There re no series or prllel connections in the system given. Now consider current entering the junction. Becuse of the symmetry in the circuit, the current in brnches c & d must be equl, nd hence the points c & d hve the sme potentil (V cd =0), tht is the circuit cn be reduced s in the figure.
8 Exmple 8.6 Three resistors re connected in prllel s shown. A potentil difference of 8 V is pplied cross points nd b. ) Find the current in ech resistor. b) Clculte the power delivered to ech resistor nd the totl power delivered to the combintion. 8 V 3Ω 6 Ω 3 9 Ω Solution: ) Since the resistors re connected in prllel, the potentil difference cross ech one is the sme nd equl to 8 V. Now V 8 = = = 6A R 3 V 8 = = = 3A R 6 V3 8 3 = = = A R 3 9 b) For the power delivered to ech resistor we pply P = R = ( 6) ( 3) = 08 W P = R = ( 3) ( 6) = 54 W P 3 = 3 R3 = ( ) ( 9) = 36 W P R = ( ) (.64) = 98 W eq = eq eq Note tht P eq = P + P + P3 b
9 8.3 KHHOFF S RULES  The sum of the currents entering ny junction must equl the sum of the currents leving tht junction. (A junction is ny point in circuit where current cn split)  The lgebric sum of the potentil differences cross ll the elements round ny loop must be zero. The first rule is n ppliction of the conservtion of chrge principle, while the second rule is n ppliction of the conservtion of energy principle. To pply the second rule we should know the following two remrks:  The chnge in potentil through ny resistor is negtive for move in the direction of the current nd positive for move opposite to the direction of the current. This is becuse the current through resistor moves from the end of higher potentil to tht of lower potentil.  The chnge in potentil through n idel bttery is positive for move from the negtive to the positive terminl of the bttery nd negtive for move in the opposite direction. Strtegy for solving problems using Kirchhoff s rules:  Drw circuit digrm nd lbel ll quntities, known nd unknown.  Assign direction for the current in ech prt of the circuit. Do not bother if your guess of current direction is incorrect; the result will hve negtive vlue. 3 Apply the first Kirchhoff s rule to ny junction in the circuit. n generl this rule is used one time fewer thn the number of junctions in the circuit. 4 Choose ny closed loop in the network, nd designte direction (clockwise or counterclockwise) to trverse the loop. 5 Strting from one point in the loop, go round the loop in the designted direction. Sum the potentil differences cross ll the elements of the chosen loop to zero. n doing so you should note the two remrks discussed bove, tht is, the potentil difference cross n emf is +ε if it is trversed from the negtive to the positive terminl nd ε if trversed in the opposite direction. The potentil
10 difference cross ny resistor is R if this resistor is trversed in the direction of the ssumed current nd +R if trversed in the opposite direction. 6 Choose nother loop nd repet the fifth step to get different eqution relting the unknown quntities. Continue until you hve s mny equtions s unknowns. 7 Solve these equtions simultneously for the unknowns. Exmple 8.8 n the circuit shown, find the current in the circuit nd the power delivered to ech resistor nd the power delivered by the V bttery. ε =6V R = 0 Ω R = 8 Ω Solution The directions of the currents re ssigned rbitrry s shown in the Figure. As it cler from the circuit there is one loop with no junctions. Now, we pply Kirchhoff s second rule to the loop nd trverse the loop in the clockwise direction, obtining ε ε R R = = = = = 0.33A The minus sign indictes tht the direction of is opposite the ssumed direction. To find the power delivered to ech resistor, we use = R = = R = P P 0.87 W. W ε =V
11 And for the power delivered by the bttery we hve P = ε = 0.33 = 4W Note tht P + P = = W Tht is hlf of the power supplied by the Vbttery is delivered to the resistors nd the other hlf is delivered to the 6Vbttery Exmple.4 ) Find the current in ech resistor in the figure shown. b) Clculte the potentil difference Vb V. 4V Solution ) f we pply Kirchhoff s first rule to the junction b we get 4Ω 6 Ω b = + 3 () Now pplying Kirchhoff s second rule to the upper loop trversing it clockwise we get 0V Ω = 0 () For the bottom loop trversing it in the clockwise direction gives = 0 (3) Substituting for 3 from Eq. () into Eq.(3) 0 6 ( + ) = = 0 (4) Dividing Eq.() by + 3 = 0 (5) Subtrcting Eqs. (4) & (5) = 0 or =.0 A
12 From Eq. (4) we get or = 3.0A nd from Eq. () we get or 3 =.0A The minus sign indictes tht & 3 should be reversed b) Strting t point, we follow pth towrd point b, dding potentil differences cross ll the elements we encounter. f we follow the pth through the middle bttery we obtin V V b = 0 6 = 0 6 = V The minus sign here mens tht V > V b. Try to follow nother pths from to b to verify tht they lso give the sme result. The CUTS Chrging Process The figure shows cpcitor, initilly unchrged, connected in series with ε C resistor. f the switch S is thrown t point t t = 0, the cpcitor will begin to chrge, creting current in the circuit. Let be the current in the circuit t some instnt during the chrging process, nd q be the chrge on the cpcitor t the sme instnt. Applying Kirchhoff s second rule to the circuit, we obtin q ε R = C S R Substituting for with dq = εc q dt = dq dt, in Eqution 8.0, nd rerrnging we obtin 8.
13 Noting tht the chrge on the cpcitor is initilly zero, i.e., q = 0 t t = 0, we cn integrte both sides of Eqution 8. s q dq = C q 0 ε t 0 dt εc q ln = εc t t q = ε C e 8. where e is the bse of the nturl logrithm. To find the current s function of time, we differentite Eqution 8. with respect to time to get t = ε e 8.3 R The quntity is clled the time constnt, τ, which defined s the time required for the current to decrese to e of its initil vlue. Equtions 8. nd 8.3, which re plotted in the following Figure, tell the following:  At t = 0, the chrge q is zero, s required, nd the initil current o is
14 q Q m o t t () (b) Figure.8 () The chrge versus time in chrging process for circuit. (b) The current versus time in chrging process for the sme circuit. o ε = 8.4 R tht is, the cpcitor cts s if it were wire with negligible resistnce (short circuit).  As t (fter long time), the chrge hs its mximum equilibrium vlue, Q m Q m = ε C 8.5 nd the current is zero, tht is the cpcitor cts s it were n open switch (open circuit). Dischrging Process Suppose tht the cpcitor is now fully chrged such tht its potentil difference is equl to the emf ε. f the switch is thrown to point t new time t = 0, the cpcitor begin to dischrge through the resistor. Let be the current in the circuit t some instnt during this process, nd q be the chrge on the cpcitor t the sme instnt. Applying Kirchhoff s rule to the loop, we get
15 q R = 0 C Substituting for with obtin 8.6 = dq dt (explin the negtive sign), nd rerrnge we dq = dt 8.7 q Using the initil condition, q = Q m t t = 0 we cn integrte the lst eqution to obtin q Q ln mx q Qm dq t = dt q 0 = t q t = Q m e 8.8 The current is the rte of decrese of the chrge on the cpcitor, tht is dq dt t o e = = 8.9 where Q o = 8.0 Exmple 8. A kω resistor nd 5µF cpcitor re connected, in series, with V bttery s shown. The cpcitor is initilly unchrged, nd the switch S is closed t t=0.
16 ) Find the time constnt of the circuit, nd the mximum chrge on the cpcitor. b) Wht is the time required for the current to drop to hlf its initil vlue? c) After being closed for long time, the switch is now opened t t=0, wht is the time required for the chrge nd for the energy to decrese to onefourth their mximum vlue. V S R 5 µf Solution ) The time constnt is τ = = 5 6 ( )( ) = 4.0s The mximum chrge is, Q m 6 ( )( ) 60µ C = ε C = = And the mximum current is ε o = = 5µ A R = b) Now we hve = o e t To find the time required for the current to drop to hlf its vlue, we substitute = into this eqution: o o = o e t Tking the logrithm of both sides, we hve
17 or t ln = t = ( ln ) =.8s c) n the dischrging process, the chrge vries with time ccording to q = Q m e t Substituting for q = Q, nd tking the logrithm of both sides we get 4 m or ln 4 = t = t ln 4 = 5.5s For the energy we hve q Q t t U = = e = U m e C C Substituting for U = 4 U m, nd tking the logrithm of both sides we get or ln 4 = t = t ln = 4.8s
18 Exmple (Extr) n the circuit shown, the cpcitor is initilly empty nd the switch S is closed t t=0. ) Find the current in ech brnch of the circuit t t=0. b) Clculte the mximum chrge on the cpcitor. 3V 6 Ω S µf 3 4Ω Solution ) At t=0, the cpcitor is treted s if it were wire with negligible resistnce. This mens tht the cpcitor mkes short circuit cross the 4Ω resistor. Therefore, we hve = 0 nd 3 = 3 = = A b) The mximum chrge is ttined fter long time ( t ). At this time the cpcitor is treted s if it were n open switch. So, 3 = 0 nd 3 = = = ( ) 3.A. To clculte the chrge on the cpcitor, we first wnt to find the potentil difference cross it. Applying Kirchhoff s second rule to right loop we find tht the potentil difference V cross the cpcitor is V = R ( 3.)( 4.0) V = = Now the mximum chrge is ( 6 0 )(.8) = C Q = CV = µ
200 points 5 Problems on 4 Pages and 20 Multiple Choice/Short Answer Questions on 5 pages 1 hour, 48 minutes
PHYSICS 132 Smple Finl 200 points 5 Problems on 4 Pges nd 20 Multiple Choice/Short Answer Questions on 5 pges 1 hour, 48 minutes Student Nme: Recittion Instructor (circle one): nme1 nme2 nme3 nme4 Write
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationPhysics Honors. Final Exam Review Free Response Problems
Physics Honors inl Exm Review ree Response Problems m t m h 1. A 40 kg mss is pulled cross frictionless tble by string which goes over the pulley nd is connected to 20 kg mss.. Drw free body digrm, indicting
More informationPOLYPHASE CIRCUITS. Introduction:
POLYPHASE CIRCUITS Introduction: Threephse systems re commonly used in genertion, trnsmission nd distribution of electric power. Power in threephse system is constnt rther thn pulsting nd threephse
More informationVersion 001 HW#6  Electromagnetism arts (00224) 1
Version 001 HW#6  Electromgnetism rts (00224) 1 This printout should hve 11 questions. Multiplechoice questions my continue on the next column or pge find ll choices efore nswering. rightest Light ul
More informationEMF Notes 9; Electromagnetic Induction ELECTROMAGNETIC INDUCTION
EMF Notes 9; Electromgnetic nduction EECTOMAGNETC NDUCTON (Y&F Chpters 3, 3; Ohnin Chpter 3) These notes cover: Motionl emf nd the electric genertor Electromgnetic nduction nd Frdy s w enz s w nduced electric
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationDesigning Information Devices and Systems I Spring 2018 Homework 8
EECS 16A Designing Informtion Devices nd Systems I Spring 2018 Homework 8 This homework is due Mrch 19, 2018, t 23:59. Selfgrdes re due Mrch 22, 2018, t 23:59. Sumission Formt Your homework sumission
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More informationdy ky, dt where proportionality constant k may be positive or negative
Section 1.2 Autonomous DEs of the form 0 The DE y is mthemticl model for wide vriety of pplictions. Some of the pplictions re descried y sying the rte of chnge of y(t) is proportionl to the mount present.
More informationPhysics 9 Fall 2011 Homework 2  Solutions Friday September 2, 2011
Physics 9 Fll 0 Homework  s Fridy September, 0 Mke sure your nme is on your homework, nd plese box your finl nswer. Becuse we will be giving prtil credit, be sure to ttempt ll the problems, even if you
More informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
More informationPhysics 2135 Exam 1 February 14, 2017
Exm Totl / 200 Physics 215 Exm 1 Ferury 14, 2017 Printed Nme: Rec. Sec. Letter: Five multiple choice questions, 8 points ech. Choose the est or most nerly correct nswer. 1. Two chrges 1 nd 2 re seprted
More informationLecture 7 notes Nodal Analysis
Lecture 7 notes Nodl Anlysis Generl Network Anlysis In mny cses you hve multiple unknowns in circuit, sy the voltges cross multiple resistors. Network nlysis is systemtic wy to generte multiple equtions
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More information10/25/2005 Section 5_2 Conductors empty.doc 1/ Conductors. We have been studying the electrostatics of freespace (i.e., a vacuum).
10/25/2005 Section 5_2 Conductors empty.doc 1/3 52 Conductors Reding Assignment: pp. 122132 We hve been studying the electrosttics of freespce (i.e., vcuum). But, the universe is full of stuff! Q: Does
More informationFundamentals of Electrical Circuits  Chapter 3
Fundmentls of Electricl Circuits Chpter 3 1S. For the circuits shown elow, ) identify the resistors connected in prllel ) Simplify the circuit y replcing prllel connect resistors with equivlent resistor.
More informationChapter 4 Contravariance, Covariance, and Spacetime Diagrams
Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz
More informationOverview. Before beginning this module, you should be able to: After completing this module, you should be able to:
Module.: Differentil Equtions for First Order Electricl Circuits evision: My 26, 2007 Produced in coopertion with www.digilentinc.com Overview This module provides brief review of time domin nlysis of
More informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationVersion 001 Exam 1 shih (57480) 1
Version 001 Exm 1 shih 57480) 1 This printout should hve 6 questions. Multiplechoice questions my continue on the next column or pge find ll choices before nswering. Holt SF 17Rev 1 001 prt 1 of ) 10.0
More informationJackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell
Jckson 2.26 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: The twodimensionl region, ρ, φ β, is bounded by conducting surfces t φ =, ρ =, nd φ = β held t zero
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationReference. Vector Analysis Chapter 2
Reference Vector nlsis Chpter Sttic Electric Fields (3 Weeks) Chpter 3.3 Coulomb s Lw Chpter 3.4 Guss s Lw nd pplictions Chpter 3.5 Electric Potentil Chpter 3.6 Mteril Medi in Sttic Electric Field Chpter
More informationElectric Potential. Concepts and Principles. An Alternative Approach. A Gravitational Analogy
. Electric Potentil Concepts nd Principles An Alterntive Approch The electric field surrounding electric chrges nd the mgnetic field surrounding moving electric chrges cn both be conceptulized s informtion
More information2. VECTORS AND MATRICES IN 3 DIMENSIONS
2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2dimensionl Vectors x A point in 3dimensionl spce cn e represented y column vector of the form y z zxis yxis z x y xxis Most of the
More information#6A&B Magnetic Field Mapping
#6A& Mgnetic Field Mpping Gol y performing this lb experiment, you will: 1. use mgnetic field mesurement technique bsed on Frdy s Lw (see the previous experiment),. study the mgnetic fields generted by
More informationa a a a a a a a a a a a a a a a a a a a a a a a In this section, we introduce a general formula for computing determinants.
Section 9 The Lplce Expnsion In the lst section, we defined the determinnt of (3 3) mtrix A 12 to be 22 12 21 22 2231 22 12 21. In this section, we introduce generl formul for computing determinnts. Rewriting
More informationContinuous Random Variables
STAT/MATH 395 A  PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is relvlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht
More informationPhysics 121 Sample Common Exam 1 NOTE: ANSWERS ARE ON PAGE 8. Instructions:
Physics 121 Smple Common Exm 1 NOTE: ANSWERS ARE ON PAGE 8 Nme (Print): 4 Digit ID: Section: Instructions: Answer ll questions. uestions 1 through 16 re multiple choice questions worth 5 points ech. You
More informationDesigning Information Devices and Systems I Fall 2016 Babak Ayazifar, Vladimir Stojanovic Homework 6. This homework is due October 11, 2016, at Noon.
EECS 16A Designing Informtion Devices nd Systems I Fll 2016 Bk Ayzifr, Vldimir Stojnovic Homework 6 This homework is due Octoer 11, 2016, t Noon. 1. Homework process nd study group Who else did you work
More informationPreSession Review. Part 1: Basic Algebra; Linear Functions and Graphs
PreSession Review Prt 1: Bsic Algebr; Liner Functions nd Grphs A. Generl Review nd Introduction to Algebr Hierrchy of Arithmetic Opertions Opertions in ny expression re performed in the following order:
More information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationDEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS
3 DEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS This chpter summrizes few properties of Cli ord Algebr nd describe its usefulness in e ecting vector rottions. 3.1 De nition of Associtive
More information13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS
33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in
More informationSOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014
SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.
More informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the righthnd side limit equls to the lefthnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationResistors. Consider a uniform cylinder of material with mediocre to poor to pathetic conductivity ( )
10/25/2005 Resistors.doc 1/7 Resistors Consider uniform cylinder of mteril with mediocre to poor to r. pthetic conductivity ( ) ˆ This cylinder is centered on the xis, nd hs length. The surfce re of the
More informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More informationDESCRIBING MOTION: KINEMATICS IN ONE DIMENSION
DESCRIBING MOTION: KINEMATICS IN ONE DIMENSION Responses to Questions. A cr speedometer mesures only speed. It does not give ny informtion bout the direction, so it does not mesure velocity.. If the velocity
More informationA. Limits  L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.
A. Limits  L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( x) lim where lim f x x! c g x ( ) = or lim f ( x) = limg( x) = ". ( ) x! c limg( x) = 0 x! c x! c
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationCalculus  Activity 1 Rate of change of a function at a point.
Nme: Clss: p 77 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit Rte of chnge of function t point. ) Strt Mths Helper Plus, then lod the file: Clculus
More informationModule 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur
Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 8 The Force Method of Anlysis: Bems Version CE IIT, Khrgpur Instructionl Objectives After reding
More informationMATHS NOTES. SUBJECT: Maths LEVEL: Higher TEACHER: Aidan Roantree. The Institute of Education Topics Covered: Powers and Logs
MATHS NOTES The Institute of Eduction 06 SUBJECT: Mths LEVEL: Higher TEACHER: Aidn Rontree Topics Covered: Powers nd Logs About Aidn: Aidn is our senior Mths techer t the Institute, where he hs been teching
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationMath 0230 Calculus 2 Lectures
Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More informationFundamentals of Analytical Chemistry
Homework Fundmentls of nlyticl hemistry hpter 9 0, 1, 5, 7, 9 cids, Bses, nd hpter 9(b) Definitions cid Releses H ions in wter (rrhenius) Proton donor (Bronsted( Lowry) Electronpir cceptor (Lewis) hrcteristic
More informationProblem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are:
(x + y ) = y + (x + y ) = x + Problem Set 9 Discussion: Nov., Nov. 8, Nov. (on probbility nd binomil coefficients) The nme fter the problem is the designted writer of the solution of tht problem. (No one
More informationWeek 10: Riemann integral and its properties
Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the
More informationStrategy: Use the Gibbs phase rule (Equation 5.3). How many components are present?
University Chemistry Quiz 4 2014/12/11 1. (5%) Wht is the dimensionlity of the threephse coexistence region in mixture of Al, Ni, nd Cu? Wht type of geometricl region dose this define? Strtegy: Use the
More informationConducting Ellipsoid and Circular Disk
1 Problem Conducting Ellipsoid nd Circulr Disk Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 (September 1, 00) Show tht the surfce chrge density σ on conducting ellipsoid,
More informationalong the vector 5 a) Find the plane s coordinate after 1 hour. b) Find the plane s coordinate after 2 hours. c) Find the plane s coordinate
L8 VECTOR EQUATIONS OF LINES HL Mth  Sntowski Vector eqution of line 1 A plne strts journey t the point (4,1) moves ech hour long the vector. ) Find the plne s coordinte fter 1 hour. b) Find the plne
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationRiemann Integrals and the Fundamental Theorem of Calculus
Riemnn Integrls nd the Fundmentl Theorem of Clculus Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University September 16, 2013 Outline Grphing Riemnn Sums
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationLinear Systems with Constant Coefficients
Liner Systems with Constnt Coefficients 4305 Here is system of n differentil equtions in n unknowns: x x + + n x n, x x + + n x n, x n n x + + nn x n This is constnt coefficient liner homogeneous system
More informationCS12N: The Coming Revolution in Computer Architecture Laboratory 2 Preparation
CS2N: The Coming Revolution in Computer Architecture Lortory 2 Preprtion Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes
More informationPART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point.
PART MULTIPLE CHOICE Circle the pproprite response to ech of the questions below. Ech question hs vlue of point.. If in sequence the second level difference is constnt, thn the sequence is:. rithmetic
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More information(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35
7 Integrtion º½ ÌÛÓ Ü ÑÔÐ Up to now we hve been concerned with extrcting informtion bout how function chnges from the function itself. Given knowledge bout n object s position, for exmple, we wnt to know
More informationExperiment 2: Equipotentials and Electric Fields
Chpter 4 Experiment 2: Equipotentils nd Electric Fields 4.1 Introduction One wy to look t the force between chrges is to sy tht the chrge lters the spce round it by generting n electric field E. Any other
More informationSome Methods in the Calculus of Variations
CHAPTER 6 Some Methods in the Clculus of Vritions 6. If we use the vried function ( α, ) α sin( ) + () Then d α cos ( ) () d Thus, the totl length of the pth is d S + d d α cos ( ) + α cos ( ) d Setting
More informationExam 2 Solutions ECE 221 Electric Circuits
Nme: PSU Student ID Numer: Exm 2 Solutions ECE 221 Electric Circuits Novemer 12, 2008 Dr. Jmes McNmes Keep your exm flt during the entire exm If you hve to leve the exm temporrily, close the exm nd leve
More information7.6 The Use of Definite Integrals in Physics and Engineering
Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 7.6 The Use of Definite Integrls in Physics nd Engineering It hs been shown how clculus cn be pplied to find solutions to geometric problems
More informationHT Module 2 Paper solution. Module 2. Q6.Discuss Electrical analogy of combined heat conduction and convection in a composite wall.
HT Module 2 Pper solution Qulity Solutions wwwqulitytutorilin Module 2 Q6Discuss Electricl nlogy of combined het conduction nd convection in composite wll M16Q1(c)5m Ans: It is frequently convient to
More informationUnique Solutions R. All about Electromagnetism. C h a p t e r. G l a n c e
5. C h p t e r t G l n c e When electric current is pssed through conductor, it produces mgnetic field round it. The first discovery of the connection between electricity nd mgnetism ws mde by H. C. Oersted.
More informationLecture 1: Electrostatic Fields
Lecture 1: Electrosttic Fields Instructor: Dr. Vhid Nyyeri Contct: nyyeri@iust.c.ir Clss web site: http://webpges.iust.c. ir/nyyeri/courses/bee 1.1. Coulomb s Lw Something known from the ncient time (here
More informationGRADE 4. Division WORKSHEETS
GRADE Division WORKSHEETS Division division is shring nd grouping Division cn men shring or grouping. There re cndies shred mong kids. How mny re in ech shre? = 3 There re 6 pples nd go into ech bsket.
More informationThe Thermodynamics of Aqueous Electrolyte Solutions
18 The Thermodynmics of Aqueous Electrolyte Solutions As discussed in Chpter 10, when slt is dissolved in wter or in other pproprite solvent, the molecules dissocite into ions. In queous solutions, strong
More informationInstructor(s): Acosta/Woodard PHYSICS DEPARTMENT PHY 2049, Fall 2015 Midterm 1 September 29, 2015
Instructor(s): Acost/Woodrd PHYSICS DEPATMENT PHY 049, Fll 015 Midterm 1 September 9, 015 Nme (print): Signture: On m honor, I hve neither given nor received unuthorized id on this emintion. YOU TEST NUMBE
More informationNew Expansion and Infinite Series
Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06073 HIKARI Ltd, www.mhikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationTest , 8.2, 8.4 (density only), 8.5 (work only), 9.1, 9.2 and 9.3 related test 1 material and material from prior classes
Test 2 8., 8.2, 8.4 (density only), 8.5 (work only), 9., 9.2 nd 9.3 relted test mteril nd mteril from prior clsses Locl to Globl Perspectives Anlyze smll pieces to understnd the big picture. Exmples: numericl
More informationHomework Assignment 3 Solution Set
Homework Assignment 3 Solution Set PHYCS 44 6 Ferury, 4 Prolem 1 (Griffiths.5(c The potentil due to ny continuous chrge distriution is the sum of the contriutions from ech infinitesiml chrge in the distriution.
More informationCalculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties
Clculus nd liner lgebr for biomedicl engineering Week 11: The Riemnn integrl nd its properties Hrtmut Führ fuehr@mth.rwthchen.de Lehrstuhl A für Mthemtik, RWTH Achen Jnury 9, 2009 Overview 1 Motivtion:
More informationdifferent methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).
Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different
More informationPHYSICS 211 MIDTERM I 21 April 2004
PHYSICS MIDERM I April 004 Exm is closed book, closed notes. Use only your formul sheet. Write ll work nd nswers in exm booklets. he bcks of pges will not be grded unless you so request on the front of
More informationChapter 4. Additional Variational Concepts
Chpter 4 Additionl Vritionl Concepts 137 In the previous chpter we considered clculus o vrition problems which hd ixed boundry conditions. Tht is, in one dimension the end point conditions were speciied.
More informationIntegrals along Curves.
Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the
More informationSummer MTH142 College Calculus 2. Section J. Lecture Notes. Yin Su University at Buffalo
Summer 6 MTH4 College Clculus Section J Lecture Notes Yin Su University t Bufflo yinsu@bufflo.edu Contents Bsic techniques of integrtion 3. Antiderivtive nd indefinite integrls..............................................
More informationFORM FIVE ADDITIONAL MATHEMATIC NOTE. ar 3 = (1) ar 5 = = (2) (2) (1) a = T 8 = 81
FORM FIVE ADDITIONAL MATHEMATIC NOTE CHAPTER : PROGRESSION Arithmetic Progression T n = + (n ) d S n = n [ + (n )d] = n [ + Tn ] S = T = T = S S Emple : The th term of n A.P. is 86 nd the sum of the first
More informationQuantum Physics II (8.05) Fall 2013 Assignment 2
Quntum Physics II (8.05) Fll 2013 Assignment 2 Msschusetts Institute of Technology Physics Deprtment Due Fridy September 20, 2013 September 13, 2013 3:00 pm Suggested Reding Continued from lst week: 1.
More informationUSA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year
1/1/21. Fill in the circles in the picture t right with the digits 18, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationNumerical integration
2 Numericl integrtion This is pge i Printer: Opque this 2. Introduction Numericl integrtion is problem tht is prt of mny problems in the economics nd econometrics literture. The orgniztion of this chpter
More informationChapter 2 Finite Automata
Chpter 2 Finite Automt 28 2.1 Introduction Finite utomt: first model of the notion of effective procedure. (They lso hve mny other pplictions). The concept of finite utomton cn e derived y exmining wht
More informationSTEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.
STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t
More information3.4 Numerical integration
3.4. Numericl integrtion 63 3.4 Numericl integrtion In mny economic pplictions it is necessry to compute the definite integrl of relvlued function f with respect to "weight" function w over n intervl [,
More informationPrecalculus Spring 2017
Preclculus Spring 2017 Exm 3 Summry (Section 4.1 through 5.2, nd 9.4) Section P.5 Find domins of lgebric expressions Simplify rtionl expressions Add, subtrct, multiply, & divide rtionl expressions Simplify
More informationStrong acids and bases. Strong acids and bases. Systematic Treatment of Equilibrium & Monoprotic Acidbase Equilibrium.
Strong cids nd bses Systemtic Tretment of Equilibrium & Monoprotic cidbse Equilibrium onc. (M) 0.0.00 .005.008 p Strong cids nd bses onc. (M) p 0.0.0.00 .0.005 5.0.008 8.0? We hve to consider utoprotolysis
More informationAlg. Sheet (1) Department : Math Form : 3 rd prep. Sheet
Ciro Governorte Nozh Directorte of Eduction Nozh Lnguge Schools Ismili Rod Deprtment : Mth Form : rd prep. Sheet Alg. Sheet () [] Find the vlues of nd in ech of the following if : ) (, ) ( 5, 9 ) ) (,
More informationMACsolutions of the nonexistent solutions of mathematical physics
Proceedings of the 4th WSEAS Interntionl Conference on Finite Differences  Finite Elements  Finite Volumes  Boundry Elements MACsolutions of the nonexistent solutions of mthemticl physics IGO NEYGEBAUE
More informationV. DEMENKO MECHANICS OF MATERIALS LECTURE 6 Plane Bending Deformation. Diagrams of Internal Forces (Continued)
V. DEMENKO MECHNCS OF MTERLS 015 1 LECTURE 6 Plne ending Deformtion. Digrms of nternl Forces (Continued) 1 Construction of ending Moment nd Shering Force Digrms for Two Supported ems n this mode of loding,
More informationUNIVERSITY OF MALTA DEPARTMENT OF CHEMISTRY. CH237  Chemical Thermodynamics and Kinetics. Tutorial Sheet VIII
UNIVERSITY OF MALTA DEPARTMENT OF CHEMISTRY CH237  Chemicl Thermodynmics nd Kinetics Tutoril Sheet VIII 1 () (i) The rte of the rection A + 2B 3C + D ws reported s 1.0 mol L 1 s 1. Stte the rtes of
More informationTopic 1 Notes Jeremy Orloff
Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble
More informationMotion. Acceleration. Part 2: Constant Acceleration. October Lab Phyiscs. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.
Motion ccelertion Prt : Constnt ccelertion ccelertion ccelertion ccelertion is the rte of chnge of elocity. =  o t = Δ Δt ccelertion = =  o t chnge of elocity elpsed time ccelertion is ector, lthough
More informationFBR Neutronics: Breeding potential, Breeding Ratio, Breeding Gain and Doubling time
FBR eutronics: Breeding potentil, Breeding Rtio, Breeding Gin nd Doubling time K.S. Rjn Proessor, School o Chemicl & Biotechnology SASTRA University Joint Inititive o IITs nd IISc Funded by MHRD Pge 1
More information