Chapter E  Problems


 Loreen Bailey
 9 months ago
 Views:
Transcription
1 Chpter E  Prolems Blinn College  Physics Terry Honn Prolem E.1 A wire with dimeter d feeds current to cpcitor. The chrge on the cpcitor vries with time s QHtL = Q 0 sin w t. Wht re the current nd current density in the wire s functions of time? Solution to E.1 The current is the the time derivtive of the chrge. The current density is just the current per re. I = Q t = Q 0 w cos w t J = I A = Q 0 w cos w t p d 2 ë4 Prolem E.2 The current s function of time vries s IHtL = 15 sinh120 p tl in SI units. Wht is the totl chrge tht flows etween 0 nd 1 s. (This is hlf cycle of stndrd 60 Hz AC frequency.) 120 Solution to E.2 Q = Ÿ IHtL t = Ÿ 0 1ê sinh120 p tl t = p cosh120 p tl 0 1ê120 = 1 4 p C = C Prolem E.3 The dimeter of stndrd 12 guge wire is 2.05 mm. Suppose in stndrd houshold wiring 15 A current flows through 20 m length of copper wire. () Wht is the resistnce in the wire nd wht is the voltge drop cross the wire? () For ech copper tom there is one free electron ville for conduction. The density (mss/volume) of copper is 8920 kgëm 3 nd its tomic mss is u. Use the conversion 6.02µ10 26 u = 1 kg to find the mss of ech tom. From tht find the numer of toms per volume; this for copper is the sme s the numer of chrge crriers per volume. Clculte the drift velocity of the free electrons nd find the totl time it tkes one electron to migrte the full length of the wire. Solution to E.3 () The resistivity of copper is 1.7µ108 Wÿm. The crosssectionl re of the wire is The voltge drop cross wire is given y Ohm's lw. A = p H ê2L 2 = µ106 m 2 R = r { A = 1.7µ108 µ20 = = W A
2 2 Chpter E  Prolems V = I R = 15µ = 1.55 V () I = n q v d A q is the chrge of the chrge crriers, for metl q = e. A is the crosssectionl re, found ove. v d is the drift velocity nd n is the numer of chrge crriers per volume. The mss of ech tom is: 1 kg m tom = uµ 6.02µ10 26 = u µ1025 kg Since there is one electron per tom of copper tht contriutes to conduction, n is the sme s the numer of toms per volume. Since the density is the mss per volume we cn write: n = rêm tom. n = free electrons volume = I = n q v d A ï v d = free electrons µ toms tom volume = 1µ The time to trvel the length of the wire is given y { = v d t r = m tom 8920 kgëm µ1028 = µ1025 m 3 I n e A = 15 n 1.60µ1019 = A µ104 = 3.36µ104 m s t = { v d = 20 v d = s = 16.5 hr Note tht the electricl signl trvels much fster thn tht; tht speed is on the order of the speed of light. Prolem E.4 Suppose wire with restnce R is stretched y 30% keeping the volume of the wire fixed. Wht is the new resistnce of the wire? Solution to E.4 If wire is stretched its volume stys constnt so tht A { = A {. Thus { = 1.30 { nd Since R = r { ê A we get the rtio: A A = { A = 1.30 ï { A = R R = { ê{ A êa = ê1.30 = = 1.69 ï R = 1.69 R Prolem E.5 When the temperture of wire is decresed y 10 C its resistnce decreses y 6 %. Wht is its temperture coefficient? Solution to E.5 D R = R 0 ï D R R 0 = D T ï = H10L ï = 0.006êC Prolem E.6 The precise definition of the temperture coefficient is = 1 r Show tht if is constnt we get: r = r 0 e HTT 0 L. Use the pproximtion tht for smll x (x ` 1), e 1 + x, to show tht for smll D T = T  T 0 : r T. r = r 0 e H r + HT  T 0 LD.
3 Chpter E  Prolems 3 Solution to E.6 = 1 r Integrting this nd using r = r 0 t T = T 0 gives r T = ln r T ï ln r = T Ÿ r0 r ln r = Ÿ T0 T T ï ln r  ln r0 = HT  T 0 L. Exponentiting this gives: r = r 0 e HTT 0 L. If D T = T  T 0 is smll then x = HT  T 0 L is smll nd we cn sue the expnsion. r = r 0 e HTT 0 r + HT  T 0 LD Prolem E.7 A hir dryer desiigned for stndrd US outlet hs power rting of 1500 W. () How much current does it drw? () Wht is its resistnce? Solution to E.7 For stndrd outlet we hve V rms = 120 V. This is clled the rootmensqured voltge. We will discuss this in detil in the AC circuit chpter. For now we just need to know tht for purely resistive AC circuit we cn tret it like DC if we use the rms voltge nd current nd the verge power. () P = V I ï I = P V = = 12.5 A V = 120 V nd P = 1500 W () P = V2 R ï R = V2 P = = 9.6 W Prolem E.8 Suppose purely resistive Ohmic device designed for stndrd outlet (V = 120 V) is operted t too low voltge, 105 V. By wht percent would the power of the device e decresed? Solution to E.8 The power decreses y 23.4 %. P = V2 R ï P P = J V V N2 = J N2 = Prolem E.9 kwÿhr is the unit of energy used in electric ills. Wht is 1 kwÿhr in J? If the electric utility chrges $0.09 per kwÿhr, then wht is the cost of running 100 W ul constntly for 31 dy month? Solution to E.9
4 4 Chpter E  Prolems kwÿh = 1000 W µ3600 s = 3.6µ10 6 J Do the clcultion of cost using just kwÿh insted of J. For 100 W ul we hve Power = 0.10 kw. 24 hr Energy = Powerµtime = 0.10 kwµ31 dysµ = 74.4 kwÿhr dy $0.09 Cost = EnergyµRte = 74.4 kwÿhrµ = $6.70 kwÿhr Prolem E.10 A student tries to determine the internl resistnce of ttery. A voltmeter connected cross the ttery without lod reds the EMF E. When the ttery is connected cross known lod resistnce R 0, the voltmeter reds cross the ttery reds V 0. Wht is the internl resistnce r? Solution to E.10 V t = E  I r The voltmeter cross the ttery reds it's terminl voltge V t = V 0. This is cross the resistnce R 0 giving the current. V 0 = I R 0 ï I = V 0 R 0 V t = E  I r ï V 0 = E  V 0 R 0 r ï r = R 0 E V 01 Prolem E.11 An unknown resistnce R is connected cross fixed voltge source. When 200 W resistor is plced in prllel with it the current delivered y the source qudruples. Wht is the resistnce? Solution to E.11 Qudrupling the current t fixed voltge implies 1/4 the resistnce. R eq = R 4 = J 1 R N1 ï 4 R = 1 R ï R = 600 W Prolem E.12 12W 10W 15W 90V 20W Complete the tle elow with the voltge cross, the current through nd the totl power dissipted in ech resistor. V I 12 W 10 W 15 W 20 W
5 Chpter E  Prolems 5 P Also, give the totl current nd power delivered y the ttery. How re these vlues relted to those in the tle? Solution to E.12 Once one element in column is known the other two cn esily e found using V = I R nd P = V I. The 20 W resistor is connected cross the 90 V source. V 20 = 90 ï I 20 = V 20 = 90 R = 4.5 ï P 20 = V 20 I 20 = 90µ4.5 = 405 The equivlent resistnce of the other three resistors is: 12 + J N1 = 18. All of the current through this equivlent flows through the 12 W resistor. I 12 = I 18 = = 5 ï V 12 = I 12 R 12 = 5µ12 = 60 ï P 12 = V 12 I 12 = 60µ5 = 300 The voltges cross the other two prllel resistors re equl V 10 = V 15 = 90  V 12 = = 30 ï I 10 = V 10 R 10 = = 3 nd I 15 = V 15 R 15 = = 2 ï P 10 = V 10 I 10 = 30µ3 = 90 nd P 15 = V 15 I 15 = 30µ2 = W 10 W 15 W 20 W V 60 V 30 V 30 V 90 V I 5 A 3 A 2 A 4.5 A P 300 W 90 W 60 W 405 W The overll equivlent is: R eq = J N1 = 180 = The totl current delivered y the ttery is 19 I tot = V tot = 90 = 9.5 A nd the totl power delivered y the ttery is R eq 180ê19 P tot = V tot I tot = 90µ9.5 = 855 W. Note tht the totl current is the sum of the currents through the two rnches 9.5 = nd the totl power is the sum of ll powers 855 = Prolem E.13 Suppose resistnce R is put on ech edge of cue. Wht is the equivlent resistnce etween opposite corners of the cue? Solution to E.13 c d c d c Using symmetry we cn identify corners s hving the sme voltge. Nodes lelled y c re one resistor wy from node nd must hve the sme voltge. Similrly nodes lelled y d should hve the sme voltge. We my then, since they re d
6 6 Chpter E  Prolems nd must hve the sme voltge. Similrly nodes lelled y d should hve the sme voltge. We my then, since they re t the sme voltge, short out the c nodes nd short out the d nodes. There re three resistors etween nd c; these re in prllel nd thus hve n equivlent of Rê3. There re six resistors etween c nd d, which thus hve n equivlent of Rê6. The threee etween d nd give Rê3. The resulting three resistors re in series giving n overll equivlent of: R eq = R 3 + R 6 + R 3 = 5 6 R Prolem E.14 Wht is the equivlent resistnce etween nd? 12W 8W 16W 12W 24W 96W Solution to E.14 12W 3 16W 24W 48W 1 72W 96W There re no resistors in this network tht re either in series or prllel, so we must use the node reduction formul. Applying this to the three internl resistors we get: R = J N1 = 4 R 12 = R 1 R 2 = 12µ24 = 72 R 4 R 13 = R 1 R 3 = 12µ8 = 24 R 4 2 R 23 = R 2 R 3 = 24µ8 = 48 R 4 Now identify tht the 24 W nd 12 W resistors re in prllel s re the 48 W nd 16 W resistors re in series giving: The 20 W, 72 W nd 96 W resistors re then in prllel: J N1 + J N1 = = 20. resistors. The resulting two
7 Chpter E  Prolems 7 R eq = J N1 = W. Prolem E.15 Wht is the equivlent resistnce etween nodes nd? 1W 1W 3W 1W 5W Solution to E.15 This is n exmple where no two resistors re in series or prllel so we must use the nodereduction method. The three 1 W resistors re the esiest to eliminte (insted of the 1 W, 3 W nd 5 W ) ecuse the new resistors re equl. R = = R 1 R 2 R 3 3 ï R ij = R i R j = 1µ1 R 1ê3 = 3 1 3W 2 3W 3W 3W 3 5W Comine the pir of 3 W resistors in the lower left in prllel nd lso comine the 3 W nd 5 W resistors in the lower right in prllel. The resulting two resistors re then in series nd this is then in prllel with the 3 W t the top. J N1 + J N1 = 27 8 ï R eq = K ê8 O1 = W. Prolem E.16 4mF 9V 6mF 12mF 18mF () Wht is the equivlent cpcitnce cross the 9 V source? () Complete the tle elow with the voltge cross ech cpcitor nd the chrge on ech. V Q 18 mf 6 mf 4 mf 12 mf Solution to E.16
8 8 Chpter E  Prolems () The 4 mf nd 12 mf cpcitors re in series, so their equivlent is: J N1 = 3 mf. This is in prllel with the 6 mf cpcitor, giving = 9 mf. This, in turn, is in series with the 18 mf giving C eq = J N1 = 6 mf. () The chrge on the 18 mf is the sme s on the equivlent. Q 18 = Q eq = C eq V tot = 6µ9 = 54 mc. This is the chrge delivered y the ttery. The voltge cross it is V 18 = Q 18 C 18 = = 3 V. The voltge cross the 6 mf cpcitor nd the equivlent 3 mf cpcitor is: V 3 = V 6 = V tot  V 18 = 93 = 6 V. The chrge on the 6 mf cpcitor is: Q 6 = C 6 V 6 = 6µ6 = 36 mc The chrges on the 4 mf nd 12 mf cpcitors re equl to tht on their equivlent: Q 4 = Q 12 = Q 3 = C eq V 3 = 3µ6 = 18 mc. The voltges cross the 4 mf nd 12 mf cpcitors cn now e found: V 4 = Q 4 C 4 = 18 4 = 4.5 V nd V 12 = Q 12 C 12 = = 1.5 V 18 mf 6 mf 4 mf 12 mf V 3 V 6 V 4.5 V 1.5 V Q 54 mc 36 mc 18 mc 18 mc Prolem E.17 Wht is the equivlent cpcitnce etween nd? 2mF 8mF 3mF 3mF 9mF 6mF Solution to E.17 A node is point of constnt voltge in circuit. The two wires without components re nodes. To simplify this circuit shrink these wires to point. It follows, then, tht the 2 mf nd 8 mf cpcitors re in prllel (giving = 10), s re the 3 mf, 3 mf nd 9 mf (giving = 15). The two equivlents re then in series J N1 = 6 nd this is in prllel with the 6 mf cpcitor. The overll equivlent is then: C eq = = 12 mf. Prolem E.18 Give set of 5 liner equtions tht cn e solved for the currents. You need not solve the equtions.
9 Chpter E  Prolems 9 2W I 4 2V 3W 4W 9W I 3 6V I 5 I 2 I 1 8V 7V 7W 5W 5V Solution to E.18 The junction rule gives: I 1 = I 2 +I 3 + I 5 I 3 + I 5 = I 4 I 2 + I 4 = I 1 Recll tht ny one of the junction rule equtions is the sum of the others, so to get n independent set eliminte one of these three equtions. The loop rule gives: 79 I 24 I 1 +8 = I 4 +9 I I 3 +6 = I 35 I I 5 = 0
Hints for Exercise 1 on: Current and Resistance
Hints for Exercise 1 on: Current nd Resistnce Review the concepts of: electric current, conventionl current flow direction, current density, crrier drift velocity, crrier numer density, Ohm s lw, electric
More informationPotential Changes Around a Circuit. You must be able to calculate potential changes around a closed loop.
Tody s gend: Potentil Chnges Around Circuit. You must e le to clculte potentil chnges round closed loop. Electromotive force (EMF), Terminl Voltge, nd Internl Resistnce. You must e le to incorporte ll
More informationDesigning Information Devices and Systems I Discussion 8B
Lst Updted: 20181017 19:40 1 EECS 16A Fll 2018 Designing Informtion Devices nd Systems I Discussion 8B 1. Why Bother With Thévenin Anywy? () Find Thévenin eqiuvlent for the circuit shown elow. 2kΩ 5V
More informationI1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
2 The Prllel Circuit Electric Circuits: Figure 2 elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is
More informationDIRECT CURRENT CIRCUITS
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
More informationPhysics 1402: Lecture 7 Today s Agenda
1 Physics 1402: Lecture 7 Tody s gend nnouncements: Lectures posted on: www.phys.uconn.edu/~rcote/ HW ssignments, solutions etc. Homework #2: On Msterphysics tody: due Fridy Go to msteringphysics.com Ls:
More informationPhysics 1502: Lecture 7 Today s Agenda
1 Physics 1502: Lecture 7 Tody s gend nnouncements: Lectures posted on: www.phys.uconn.edu/~rcote/ HW ssignments, solutions etc. Homework #2: On Msterphysics tody: due Fridy Go to msteringphysics.com Ls:
More informationUnit 3: Direct current and electric resistance Electric current and movement of charges. Intensity of current and drift speed. Density of current in
Unit 3: Direct current nd electric resistnce Electric current nd movement of chrges. ntensity of current nd drift speed. Density of current in homogeneous currents. Ohm s lw. esistnce of homogeneous conductor
More information332:221 Principles of Electrical Engineering I Fall Hourly Exam 2 November 6, 2006
2:221 Principles of Electricl Engineering I Fll 2006 Nme of the student nd ID numer: Hourly Exm 2 Novemer 6, 2006 This is closedook closednotes exm. Do ll your work on these sheets. If more spce is required,
More informationPhysics 202, Lecture 10. Basic Circuit Components
Physics 202, Lecture 10 Tody s Topics DC Circuits (Chpter 26) Circuit components Kirchhoff s Rules RC Circuits Bsic Circuit Components Component del ttery, emf Resistor Relistic Bttery (del) wire Cpcitor
More information196 Circuit Analysis with PSpice: A Simplified Approach
196 Circuit Anlysis with PSpice: A Simplified Approch i, A v L t, min i SRC 5 μf v C FIGURE P7.3 () the energy stored in the inductor, nd (c) the instntneous power input to the inductor. (Dul of Prolem
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 information200 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 informationThings to Memorize: A Partial List. January 27, 2017
Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors  Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved
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 informationReading from Young & Freedman: For this topic, read the introduction to chapter 24 and sections 24.1 to 24.5.
PHY1 Electricity Topic 5 (Lectures 7 & 8) pcitors nd Dielectrics In this topic, we will cover: 1) pcitors nd pcitnce ) omintions of pcitors Series nd Prllel 3) The energy stored in cpcitor 4) Dielectrics
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 informationExam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B
PHY 249, Fll 216 Exm 1 Solutions nswer 1 is correct for ll problems. 1. Two uniformly chrged spheres, nd B, re plced t lrge distnce from ech other, with their centers on the x xis. The chrge on sphere
More informationDesigning Information Devices and Systems I Spring 2018 Homework 7
EECS 16A Designing Informtion Devices nd Systems I Spring 2018 omework 7 This homework is due Mrch 12, 2018, t 23:59. Selfgrdes re due Mrch 15, 2018, t 23:59. Sumission Formt Your homework sumission should
More informationName Class Date. Match each phrase with the correct term or terms. Terms may be used more than once.
Exercises 341 Flow of Chrge (pge 681) potentil difference 1 Chrge flows when there is between the ends of conductor 2 Explin wht would hppen if Vn de Grff genertor chrged to high potentil ws connected
More informationSample Exam 5  Skip Problems 13
Smple Exm 5  Skip Problems 13 Physics 121 Common Exm 2: Fll 2010 Nme (Print): 4 igit I: Section: Honors Code Pledge: As n NJIT student I, pledge to comply with the provisions of the NJIT Acdemic Honor
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 informationHomework Assignment 9 Solution Set
Homework Assignment 9 Solution Set PHYCS 44 3 Mrch, 4 Problem (Griffiths 77) The mgnitude of the current in the loop is loop = ε induced = Φ B = A B = π = π µ n (µ n) = π µ nk According to Lense s Lw this
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 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 informationA, Electromagnetic Fields Final Exam December 14, 2001 Solution
304351, Electrognetic Fiels Finl Ex Deceer 14, 2001 Solution 1. e9.8. In chpter9.proles.extr.two loops, e of thin wire crry equl n opposite currents s shown in the figure elow. The rius of ech loop is
More informationSection 20.1 Electric Charge and Static Electricity (pages )
Nme Clss Dte Section 20.1 Electric Chrge nd Sttic (pges 600 603) This section explins how electric chrge is creted nd how positive nd negtive chrges ffect ech other. It lso discusses the different wys
More informationLecture 5 Capacitance Ch. 25
Lecture 5 pcitnce h. 5 rtoon  pcitnce definition nd exmples. Opening Demo  Dischrge cpcitor Wrmup prolem Physlet Topics pcitnce Prllel Plte pcitor Dielectrics nd induced dipoles oxil cle, oncentric
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 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 informationSummary of equations chapters 7. To make current flow you have to push on the charges. For most materials:
Summry of equtions chpters 7. To mke current flow you hve to push on the chrges. For most mterils: J E E [] The resistivity is prmeter tht vries more thn 4 orders of mgnitude between silver (.6E8 Ohm.m)
More informationDesigning Information Devices and Systems I Anant Sahai, Ali Niknejad. This homework is due October 19, 2015, at Noon.
EECS 16A Designing Informtion Devices nd Systems I Fll 2015 Annt Shi, Ali Niknejd Homework 7 This homework is due Octoer 19, 2015, t Noon. 1. Circuits with cpcitors nd resistors () Find the voltges cross
More informationPhys102 General Physics II
Phys1 Generl Physics II pcitnce pcitnce pcitnce definition nd exmples. Dischrge cpcitor irculr prllel plte cpcitior ylindricl cpcitor oncentric sphericl cpcitor Dielectric Sls 1 pcitnce Definition of cpcitnce
More informationUniversity of Alabama Department of Physics and Astronomy. PH126: Exam 1
University of Albm Deprtment of Physics nd Astronomy PH 16 LeClir Fll 011 Instructions: PH16: Exm 1 1. Answer four of the five questions below. All problems hve equl weight.. You must show your work for
More informationChapter 4: Techniques of Circuit Analysis. Chapter 4: Techniques of Circuit Analysis
Chpter 4: Techniques of Circuit Anlysis Terminology NodeVoltge Method Introduction Dependent Sources Specil Cses MeshCurrent Method Introduction Dependent Sources Specil Cses Comprison of Methods Source
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationCAPACITORS AND DIELECTRICS
Importnt Definitions nd Units Cpcitnce: CAPACITORS AND DIELECTRICS The property of system of electricl conductors nd insultors which enbles it to store electric chrge when potentil difference exists between
More informationHomework Assignment 6 Solution Set
Homework Assignment 6 Solution Set PHYCS 440 Mrch, 004 Prolem (Griffiths 4.6 One wy to find the energy is to find the E nd D fields everywhere nd then integrte the energy density for those fields. We know
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 informationSUMMER KNOWHOW STUDY AND LEARNING CENTRE
SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18
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 informationElectricity and Magnetism
PHY472 Dt Provided: Formul sheet nd physicl constnts Dt Provided: A formul sheet nd tble of physicl constnts is ttched to this pper. DEPARTMENT OF PHYSICS & Autumn Semester 20092010 ASTRONOMY DEPARTMENT
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 informationHung problem # 3 April 10, 2011 () [4 pts.] The electric field points rdilly inwrd [1 pt.]. Since the chrge distribution is cylindriclly symmetric, we pick cylinder of rdius r for our Gussin surfce S.
More informationPhysics 1B: Review for Final Exam Solutions
Physics B: eview for Finl Exm s Andrew Forrester June 6, 2008 In this worksheet we review mteril from the following chpters of Young nd Freedmn plus some dditionl concepts): Chpter 3: Periodic Motion Chpter
More informationPhysics 3323, Fall 2016 Problem Set 7 due Oct 14, 2016
Physics 333, Fll 16 Problem Set 7 due Oct 14, 16 Reding: Griffiths 4.1 through 4.4.1 1. Electric dipole An electric dipole with p = p ẑ is locted t the origin nd is sitting in n otherwise uniform electric
More informationVersion 001 HW#6  Electromagnetic Induction arts (00224) 1 3 T
Version 001 HW#6  lectromgnetic Induction rts (00224) 1 This printout should hve 12 questions. Multiplechoice questions my continue on the next column or pge find ll choices efore nswering. AP 1998
More informationSummary: Method of Separation of Variables
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section
More informationThis chapter will show you. What you should already know. 1 Write down the value of each of the following. a 5 2
1 Direct vrition 2 Inverse vrition This chpter will show you how to solve prolems where two vriles re connected y reltionship tht vries in direct or inverse proportion Direct proportion Inverse proportion
More informationSection 6: Area, Volume, and Average Value
Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find
More informationSpecial Relativity solved examples using an Electrical Analog Circuit
1115 Specil Reltivity solved exmples using n Electricl Anlog Circuit Mourici Shchter mourici@gmil.com mourici@wll.co.il ISRAE, HOON 5454855 Introduction In this pper, I develop simple nlog electricl
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 informationFig. 1. OpenLoop and ClosedLoop Systems with Plant Variations
ME 3600 Control ystems Chrcteristics of OpenLoop nd ClosedLoop ystems Importnt Control ystem Chrcteristics o ensitivity of system response to prmetric vritions cn be reduced o rnsient nd stedystte responses
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 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 informationProblem Set 4: Mostly Magnetic
University of Albm Deprtment of Physics nd Astronomy PH 102 / LeClir Summer 2012 nstructions: Problem Set 4: Mostly Mgnetic 1. Answer ll questions below. Show your work for full credit. 2. All problems
More informationChapter 6 Continuous Random Variables and Distributions
Chpter 6 Continuous Rndom Vriles nd Distriutions Mny economic nd usiness mesures such s sles investment consumption nd cost cn hve the continuous numericl vlues so tht they cn not e represented y discrete
More informationCS683: calculating the effective resistances
CS683: clculting the effective resistnces Lecturer: John Hopcroft Note tkers: June Andrews nd JenBptiste Jennin Mrch 7th, 2008 On Ferury 29th we sw tht, given grph in which ech edge is lelled with resistnce
More informationCHAPTER 20: Second Law of Thermodynamics
CHAER 0: Second Lw of hermodynmics Responses to Questions 3. kg of liquid iron will hve greter entropy, since it is less ordered thn solid iron nd its molecules hve more therml motion. In ddition, het
More informationELE B7 Power Systems Engineering. Power System Components Modeling
Power Systems Engineering Power System Components Modeling Section III : Trnsformer Model Power Trnsformers CONSTRUCTION Primry windings, connected to the lternting voltge source; Secondry windings, connected
More informationContinuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom
Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive
More information7. Indefinite Integrals
7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find
More informationChapter 3 Single Random Variables and Probability Distributions (Part 2)
Chpter 3 Single Rndom Vriles nd Proilit Distriutions (Prt ) Contents Wht is Rndom Vrile? Proilit Distriution Functions Cumultive Distriution Function Proilit Densit Function Common Rndom Vriles nd their
More informationMATH 253 WORKSHEET 24 MORE INTEGRATION IN POLAR COORDINATES. r dr = = 4 = Here we used: (1) The halfangle formula cos 2 θ = 1 2
MATH 53 WORKSHEET MORE INTEGRATION IN POLAR COORDINATES ) Find the volume of the solid lying bove the xyplne, below the prboloid x + y nd inside the cylinder x ) + y. ) We found lst time the set of points
More informationMultiplying integers EXERCISE 2B INDIVIDUAL PATHWAYS. 6 ì 4 = 6 ì 0 = 4 ì 0 = 6 ì 3 = 5 ì 3 = 4 ì 3 = 4 ì 2 = 4 ì 1 = 5 ì 2 = 6 ì 2 = 6 ì 1 =
EXERCISE B INDIVIDUAL PATHWAYS Activity B Integer multipliction doc69 Activity B More integer multipliction doc698 Activity B Advnced integer multipliction doc699 Multiplying integers FLUENCY
More informationReview of Gaussian Quadrature method
Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge
More informationu( t) + K 2 ( ) = 1 t > 0 Analyzing Damped Oscillations Problem (Meador, example 218, pp 4448): Determine the equation of the following graph.
nlyzing Dmped Oscilltions Prolem (Medor, exmple 218, pp 4448): Determine the eqution of the following grph. The eqution is ssumed to e of the following form f ( t) = K 1 u( t) + K 2 e!"t sin (#t + $
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 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 informationMatrix Algebra. Matrix Addition, Scalar Multiplication and Transposition. Linear Algebra I 24
Mtrix lger Mtrix ddition, Sclr Multipliction nd rnsposition Mtrix lger Section.. Mtrix ddition, Sclr Multipliction nd rnsposition rectngulr rry of numers is clled mtrix ( the plurl is mtrices ) nd the
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More informationPhysics 24 Exam 1 February 18, 2014
Exm Totl / 200 Physics 24 Exm 1 Februry 18, 2014 Printed Nme: Rec. Sec. Letter: Five multiple choice questions, 8 points ech. Choose the best or most nerly correct nswer. 1. The totl electric flux pssing
More information1.2 What is a vector? (Section 2.2) Two properties (attributes) of a vector are and.
Homework 1. Chpters 2. Bsis independent vectors nd their properties Show work except for fillinlnksprolems (print.pdf from www.motiongenesis.com Textooks Resources). 1.1 Solving prolems wht engineers
More informationFundamental Theorem of Calculus
Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under
More informationSTRAND J: TRANSFORMATIONS, VECTORS and MATRICES
Mthemtics SKE: STRN J STRN J: TRNSFORMTIONS, VETORS nd MTRIES J3 Vectors Text ontents Section J3.1 Vectors nd Sclrs * J3. Vectors nd Geometry Mthemtics SKE: STRN J J3 Vectors J3.1 Vectors nd Sclrs Vectors
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More informationOn the diagram below the displacement is represented by the directed line segment OA.
Vectors Sclrs nd Vectors A vector is quntity tht hs mgnitude nd direction. One exmple of vector is velocity. The velocity of n oject is determined y the mgnitude(speed) nd direction of trvel. Other exmples
More informationragsdale (zdr82) HW2 ditmire (58335) 1
rgsdle (zdr82) HW2 ditmire (58335) This printout should hve 22 questions. Multiplechoice questions my continue on the next column or pge find ll choices before nswering. 00 0.0 points A chrge of 8. µc
More informationLesson 8. Thermomechanical Measurements for Energy Systems (MENR) Measurements for Mechanical Systems and Production (MMER)
Lesson 8 Thermomechnicl Mesurements for Energy Systems (MEN) Mesurements for Mechnicl Systems nd Production (MME) A.Y. 2056 Zccri (ino ) Del Prete Mesurement of Mechnicl STAIN Strin mesurements re perhps
More information2.4 Linear Inequalities and Interval Notation
.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or
More informationLecture 2: January 27
CS 684: Algorithmic Gme Theory Spring 217 Lecturer: Év Trdos Lecture 2: Jnury 27 Scrie: Alert Julius Liu 2.1 Logistics Scrie notes must e sumitted within 24 hours of the corresponding lecture for full
More informationState space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies
Stte spce systems nlysis (continued) Stbility A. Definitions A system is sid to be Asymptoticlly Stble (AS) when it stisfies ut () = 0, t > 0 lim xt () 0. t A system is AS if nd only if the impulse response
More information5.4 The QuarterWave Transformer
3/4/7 _4 The Qurter Wve Trnsformer /.4 The QurterWve Trnsformer Redg Assignment: pp. 7376, 443 By now you ve noticed tht qurterwve length of trnsmission le ( = λ 4, β = π ) ppers often microwve engeerg
More informationMathematics. Area under Curve.
Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationHOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016
HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M
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 informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More information7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus
7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e
More information4 VECTORS. 4.0 Introduction. Objectives. Activity 1
4 VECTRS Chpter 4 Vectors jectives fter studying this chpter you should understnd the difference etween vectors nd sclrs; e le to find the mgnitude nd direction of vector; e le to dd vectors, nd multiply
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 informationBob Brown Math 251 Calculus 1 Chapter 5, Section 4 1 CCBC Dundalk
Bo Brown Mth Clculus Chpter, Section CCBC Dundlk The Fundmentl Theorem of Clculus Informlly, the Fundmentl Theorem of Clculus (FTC) sttes tht differentition nd definite integrtion re inverse opertions
More informationQuadratic Forms. Quadratic Forms
Qudrtic Forms Recll the Simon & Blume excerpt from n erlier lecture which sid tht the min tsk of clculus is to pproximte nonliner functions with liner functions. It s ctully more ccurte to sy tht we pproximte
More information1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE
ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More information4.1. Probability Density Functions
STT 1 4.14. 4.1. Proility Density Functions Ojectives. Continuous rndom vrile  vers  discrete rndom vrile. Proility density function. Uniform distriution nd its properties. Expected vlue nd vrince of
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 informationIMPORTANT. Read these directions carefully:
Physics 208: Electricity nd Mgnetism Finl Exm, Secs. 506 510. 7 My. 2004 Instructor: Dr. George R. Welch, 415 EngineeringPhysics, 8457737 Print your nme netly: Lst nme: First nme: Sign your nme: Plese
More informationProblems for HW X. C. Gwinn. November 30, 2009
Problems for HW X C. Gwinn November 30, 2009 These problems will not be grded. 1 HWX Problem 1 Suppose thn n object is composed of liner dielectric mteril, with constnt reltive permittivity ɛ r. The object
More information