Engineering Circuit Analysis 8th Edition Chapter Nine Exercise Solutions
|
|
- Tracey Peters
- 5 years ago
- Views:
Transcription
1 Engnrng rcu naly 8h Eon hapr Nn Exrc Soluon. = KΩ, = µf, an uch ha h crcu rpon oramp. a For Sourc-fr paralll crcu: For oramp or b H 9V, V / hoo = H.7.8 ra / V ,
2 . = nh, = mf, h crcu ourc-fr paralll crcu. a o ha h crcu ju barly oramp. For Sourc-fr paralll crcu: For oramp or <.Ω hoo.. 9 E b P, / n /.8 ra / ra / p p E n n /.8,
3 . = mf, = mh, choo uch ha h crcu a Ju barly oramp; 5 ra /.5. 5 b Ju barly unramp;.5. 8 c rcally amp..5 Do your anwr for par a chang f h ror olranc %? %?. % unramp. % oramp. % unramp... 8 oramp So, y h anwr for par a wll chang f h ror olranc % or % Incra h xponnal ampng coffcn for par c by %. I h crcu now unramp, oramp, or ll crcally amp? Explan.
4 ol nw.. 5 Snc h crcu now oramp.
5 . alcula α, ω,, an for a ourc-fr paralll crcu f a = Ω, =. H, an =.5 mf; α = 8 b = nh, = pf, an % of h alu rqur o mak h crcu unramp.. ra / unramp 5.8 %.58. oramp c alcula h ampng rao for h crcu of par a an b a.7. b.
6 5. No k_ ror, = µf, >H, mr long pc of WG of ol coppr wr. plac h ror wh mr of WG coppr wr. From Tabl., WG of ol coppr wr ha a ranc of 5.7 Ω/f. Thu, h wr ha a ranc of cm n f 5.7 m m.5cm n f 977 ra / oramp.99,
7 8. ourc-fr paralll crcu hang, ra /, 5. a V H / b c c c Gn an ach nally or mj of nrgy w w c, J J
8 7. = 5 Ω, = μf, an uch ha crcally amp. a Drmn. 5 H Y, h alu larg for a prn-crcu boar moun componn. For xampl, H nucor maurng cm all 8 cm w 8 cm p. b a ror n paralll o h xng componn uch ha h ampng rao qual o. 5 5 nw // nw 5 5 nw 55. nw c Incrang h ampng rao furhr la o an oramp crcu nc Whn ncrang h ampng rao w ar ncrang α
9 8. =KΩ, = 7mH, = nf. nally charg an rong 7.mJ a ,.5 c / oramp ra b hrough h ror for > V V w ,
10 c -.78
11 9. 5 V a clar all; clo all; clc; =[:.:.5]; c=*xp-*; c=5*xp-*; c=c-c; plo,c,'-ro' hol on plo,c,'-.b' hol on plo,c,'.-g' hlg = lgn'*xp-*','5*xp-*',''; gr on xlabl'' ylabl''
12 b
13 .. a clar all; clo all; clc; =[:.:.5]; l=.*xp-*; l=.*xp-*; l=l-l; plo,l,'-ro' hol on plo,l,'-.b' hlg = lgn'.*xp-*','.*xp-*'; gr on xlabl'' ylabl''
14 b c w ] [ c /.5 5 w H H oramp F l ra
15 =[:.:.5]; l=.*xp-*; l=.*xp-*; l=l-l; plo,l,'-ro' hol on plo,l,'-.b' l=l-l; hol on plo,l,'*g' ylabl'' hlg = lgn'.*xp-*','.*xp-*',''; fgur wl=.5*l.^; plo,wl gr on xlabl'' ylabl'w'
16 . a..7 7 ln7 8, max m m m m m b.9.. %. 8 max max c Th m corrponng o h ror aborbng.5 W. c ln.5 ln P
17
18 . oramp ra / rplac h nucor wh hor crcu an h capacor wh opn crcu. Th olag powr upply wll b on. V k k V ,
19 . a V ra , , , 9..,, / 5., b V V , ,.99.99,.99,
20 c % max max max 9.. max % max max max 9.. max 9..
21
22 . a = boh h ourc an h 8 Ω ar rmo lang h mf, 5 mh, an Ω ror n paralll. a b c // V f g h V
23 5. for > a V oramp ra , /.7 5 5
24 b V V 5 max max max max %
25 . a a, 5 a V
26 V V oramp ra , / 5 5.
27 b V m m m m , c max ln % 5 max
28 7. V m V m oramp ra ].9 [ , 8. 5., c /
29 , c / w V m oramp ra b.7 c.9 c max max max max w J w J w w
30
31 ..75,.8 a w, u o h prnc of h nucor. Prformng mh analy: Mh wo:..5 Solng for,.5,.97 oramp.v Th ra / c 5
32 V , , w b
33 . = 8H, = µf a, rcal ampng k 8 b 5 c r 5 5 r Show ha r a oluon o r r r r r r r r
34 . H F V. F V V. H V. V. V. H F V
35 . = Ω, = 5.µH, = 8µF a ra / b W coul nr mak α xacly qual o bcau n pracc unuual o oban componn ha ar clor han prcn of hr pcf alu. c or mj, nally charg w.5.5, V / V m m V w V c mj V /
36 % max 8.5V max.9.85v
37
38 5. = Ω, = pf. a.8 8 H b In pracc unuual o oban componn ha ar clor han % of hr pcf alu. c, w pj w ,, V 5 5 [ P ] P V - n.757 W V /
39 . u m a o ha V, h currn ourc on, h nucor can b ra a a hor crcu, an a an opn crcu. Thu appar acro an gn by m b m V h currn ourc ha urn lf off an hor. W ar lf wh a paralll crcu compr of 5, F, mh
40 V m m V m crcally amp ra , 5 c / c for V V V m mn c max mn mn max
41
42 7. u o ha , h currn ourc on, h nucor can b ra a a hor crcu, an a an opn crcu. Thu appar acro an gn by 5 // 5 V m 5.m. 5 5.V V V, 5 5
43 8. a rcally ampng H.8 b 99. J m w w V V ,..,
44 J m w w V V , ,..,
45 9. a, Th 8 5.5H
46 . Unramp a,, b nf, mh ra c 58. / c If h ampng rao ncra h crcu bcom l unramp nc α wll b largr han ra / c 5.5. ra /
47 . pf H k 8, 7, a,, ra unramp ra / 7. /c b n co n 7. co c nj w V n V w.55. n co n co
48 ..5k, mh, mf a.5..ra /.ra / c b co n. co. c, 9V...85,.85[..9 m max..85 n... n. n. co... 9V co.. n.. n.]
49 ...85 n. for =.5kΩ
50 For = 5kΩ.85 n..85 9,,. n. 9, n. co. n co /. c / V ra unramp ra
51 For = 5kΩ.85 n..85 9,,. n. 9, n. co. n co /. c / V ra unramp ra
52 ., a = kω, = mh, = mf.5. ra / c unramp. ra / co n.5 co. n., 9V co. n. 9.5.,, co..7n.
53 b = Ω. n9.5 9 co9.5.,, n9.5 9 co , n9.5 co9.5 n co / 9.5 c / V ra unramp ra c
54
55 5. a = boh h ourc an h Ω ar rmo lang h.5µf, mh, an 5Ω ror n paralll. a b c // V f g h V
56 . for >.88, V..5 unramp ra / co co 5.7 co n n 5.7V n 5.7 co 5. n ra / c
57 7. w wll ha h followng crcu: 5 V 5.5 w wll ha h followng crcu:
58 5 58 ra / c unramp 9. ra / co n 5 co9. n 9..5, V 5 co 9. n ,.5, co 9..95n c
59 b.7v.c max..7 max.5 max
60 9. For <, w from h crcu blow ha h capacor an h ror ar hor by h prnc of h nucor. Hnc, - =.5 an - = V. Whn h.5- ourc urn off a =, w ar lf wh a paralll crcu. a 5 5 unramp 5 ra / b co n co n.5.5 5ra / c.5co n co 7.5n n co,,..5co.n c.5co.n w
61 . M 5 a ra unramp ra / 5 c / 5 5 b.5n 5.5co ,, 5.5 co 5 5 n 5 5.5n 5.5 co 5 n 5.5 co n 5 co 5 n co c.7 c.5 %.5.5n 5.5 co 5.5n 5.5 co 5 max % max max max 5 5 w w w w
62
63 . mf, 5mH a rcally amp rpon b ju barly unramp. 5 c paralll.5 5ra / c 5ra / c.5 rcally amp.5 5 ra / c Unramp
64 ., mf, mh V, ra / c unramp co n 5 ra / 5 co5 n 5 5 n n m n 5.9 m n 5.9 m n 5.
65 ., mf, mh nw / / 5 77 ra / c unramp co n ra / 5 co n 5 co n 5.78 V 5 co.78n, m -.59V
66 ., mf, mh V, m ra / c unramp co n 5 ra / 5 co5 n 5 m 5 co5 n co 5.999n 5,
67 5. mh mf k,,, V a,...., / , c / oramp ra
68 . H F,.5, a.5 b c /..5 ra c , oramp.97, , 79.99
69 5., mf, H a 5. ra / c b u u m m m mv mv m m mv m m mv mv
70
71 5. only h rgh-han currn ourc ac.,, m, m m V, m V Durng h nral from = o = +, h lf-han currn ourc bcom ac. V, h lf han no: 5 m, m5m 5m, 5m, 5m 5V, 5 V 5 9 5, h rgh han no:,. 5 / V / 5 / 5 / 5 V / 7 V /
72 5. 8 u V a V 8V 8V b
73 5. u V a 5 b V V c V ra / oramp V
74 ., m h a r wh =, or a paralll wh = ra / c Th rpon form a follow: co n n co n 7.7 In gnng h op amp ag, w fr wr h ffrnal quaon: 9 ', Tak h ra of boh : On pobl oluon : 5 9 5
75
76 ., m h a r wh =, or a paralll wh = ra / c 9 Th rpon form a follow: co n m co n n co 9 co 7.7 In gnng h op amp ag, w fr wr h ffrnal quaon: 9 ', Tak h ra of boh : 9 9 On pobl oluon :
77
78 ., m, H / / PH, 5F h a r wh =, or a paralll wh = / / H q 9. ra / c 8 Th rpon form a follow: co n m co n n co 8 co. In gnng h op amp ag, w fr wr h ffrnal quaon: 9 ' ', Tak h ra of boh : 9 5 On pobl oluon :
79
80 . crcu k,.mf.v a. b On pobl oluon
81 . crcu, 5H a 5 b /, /.5 8
82
ELEC 201 Electric Circuit Analysis I Lecture 9(a) RLC Circuits: Introduction
//6 All le courey of Dr. Gregory J. Mazzaro EE Elecrc rcu Analy I ecure 9(a) rcu: Inroucon THE ITADE, THE MIITAY OEGE OF SOUTH AOINA 7 Moulre Sree, harleon, S 949 V Sere rcu: Analog Dcoery _ 5 Ω pf eq
More informationConventional Hot-Wire Anemometer
Convnonal Ho-Wr Anmomr cro Ho Wr Avanag much mallr prob z mm o µm br paal roluon array o h nor hghr rquncy rpon lowr co prormanc/co abrcaon roc I µm lghly op p layr 8µm havly boron op ch op layr abrcaon
More informationt=0 t>0: + vr - i dvc Continuation
hapr Ga Dlay and rcus onnuaon s rcu Equaon >: S S Ths dffrnal quaon, oghr wh h nal condon, fully spcfs bhaor of crcu afr swch closs Our n challng: larn how o sol such quaons TUE/EE 57 nwrk analys 4/5 NdM
More informationConsider a system of 2 simultaneous first order linear equations
Soluon of sysms of frs ordr lnar quaons onsdr a sysm of smulanous frs ordr lnar quaons a b c d I has h alrna mar-vcor rprsnaon a b c d Or, n shorhand A, f A s alrady known from con W know ha h abov sysm
More informationSummary: Solving a Homogeneous System of Two Linear First Order Equations in Two Unknowns
Summary: Solvng a Homognous Sysm of Two Lnar Frs Ordr Equaons n Two Unknowns Gvn: A Frs fnd h wo gnvalus, r, and hr rspcv corrspondng gnvcors, k, of h coffcn mar A Dpndng on h gnvalus and gnvcors, h gnral
More informationFinal Exam : Solutions
Comp : Algorihm and Daa Srucur Final Exam : Soluion. Rcuriv Algorihm. (a) To bgin ind h mdian o {x, x,... x n }. Sinc vry numbr xcp on in h inrval [0, n] appar xacly onc in h li, w hav ha h mdian mu b
More information2/20/2013. EE 101 Midterm 2 Review
//3 EE Mderm eew //3 Volage-mplfer Model The npu ressance s he equalen ressance see when lookng no he npu ermnals of he amplfer. o s he oupu ressance. I causes he oupu olage o decrease as he load ressance
More informationChapter 9 Transient Response
har 9 Transn sons har 9: Ouln N F n F Frs-Ordr Transns Frs-Ordr rcus Frs ordr crcus: rcus conan onl on nducor or on caacor gornd b frs-ordr dffrnal quaons. Zro-nu rsons: h crcu has no ald sourc afr a cran
More informationChapter 13 Laplace Transform Analysis
Chapr aplac Tranorm naly Chapr : Ouln aplac ranorm aplac Tranorm -doman phaor analy: x X σ m co ω φ x X X m φ x aplac ranorm: [ o ] d o d < aplac Tranorm Thr condon Unlaral on-dd aplac ranorm: aplac ranorm
More informationMa/CS 6a Class 15: Flows and Bipartite Graphs
//206 Ma/CS 6a Cla : Flow and Bipari Graph By Adam Shffr Rmindr: Flow Nwork A flow nwork i a digraph G = V, E, oghr wih a ourc vrx V, a ink vrx V, and a capaciy funcion c: E N. Capaciy Sourc 7 a b c d
More informationPHYS ,Fall 05, Term Exam #1, Oct., 12, 2005
PHYS1444-,Fall 5, Trm Exam #1, Oct., 1, 5 Nam: Kys 1. circular ring of charg of raius an a total charg Q lis in th x-y plan with its cntr at th origin. small positiv tst charg q is plac at th origin. What
More informationChapter 7 Stead St y- ate Errors
Char 7 Say-Sa rror Inroucon Conrol ym analy an gn cfcaon a. rann ron b. Sably c. Say-a rror fnon of ay-a rror : u c a whr u : nu, c: ouu Val only for abl ym chck ym ably fr! nu for ay-a a nu analy U o
More informationV A. V-A ansatz for fundamental fermions
Avan Parl Phy: I. ak nraon. A Thory Carfl analy of xprnal aa (pary volaon, nrno hly pn hang n nlar β-ay, on ay propr oghr w/ nvraly fnally l o h -A hory of (nlar wak ay: M A A ( ( ( ( v p A n nlon lpon
More informationFrequency Response. Response of an LTI System to Eigenfunction
Frquncy Rsons Las m w Rvsd formal dfnons of lnary and m-nvaranc Found an gnfuncon for lnar m-nvaran sysms Found h frquncy rsons of a lnar sysm o gnfuncon nu Found h frquncy rsons for cascad, fdbac, dffrnc
More informationCmd> data<-matread("hwprobs.dat","exmpl8.2") This is the data shown in Table 8.5, the copper and diet factors.
Sa 5303 (Ohlr): Facoral Hanou 3 Cm> aa
More informationR th is the Thevenin equivalent at the capacitor terminals.
Chaper 7, Slun. Applyng KV Fg. 7.. d 0 C - Takng he derae f each erm, d 0 C d d d r C Inegrang, () ln I 0 - () I 0 e - C C () () r - I 0 e - () V 0 e C C Chaper 7, Slun. h C where h s he Theenn equalen
More informationFL/VAL ~RA1::1. Professor INTERVI of. Professor It Fr recru. sor Social,, first of all, was. Sys SDC? Yes, as a. was a. assumee.
B Pror NTERV FL/VAL ~RA1::1 1 21,, 1989 i n or Socil,, fir ll, Pror Fr rcru Sy Ar you lir SDC? Y, om um SM: corr n 'd m vry ummr yr. Now, y n y, f pr my ry for ummr my 1 yr Un So vr ummr cour d rr o l
More informationAdvanced Queueing Theory. M/G/1 Queueing Systems
Advand Quung Thory Ths slds ar rad by Dr. Yh Huang of Gorg Mason Unvrsy. Sudns rgsrd n Dr. Huang's ourss a GMU an ma a sngl mahn-radabl opy and prn a sngl opy of ah sld for hr own rfrn, so long as ah sld
More informationA capacitor consists of two conducting plates, separated by an insulator. Conduction plates: e.g., Aluminum foil Insulator: air, mica, ceramic, etc
3//7 haper 6 apacors and Inducors Makng preparaon for dynamc crcus, whch hae far more applcaons han he sac crcus we hae learned so far. 6. apacors Sore energy n elecrc feld nsulaor onducng plaes A capacor
More informationEE"232"Lightwave"Devices Lecture"16:"p7i7n"Photodiodes"and" Photoconductors"
EE"232"Lgwav"Dvcs Lcur"16:"p77n"Pooos"an" Pooconucors" Rang:"Cuang,"Cap."15"(2 n E) Insrucor:"Mng"C."Wu Unvrsy"of"Calforna,"Brkly Elcrcal"Engnrng"an"Compur"Scncs"Dp. EE232$Lcur$16-1 Rvrs"bas%p""n%juncon
More informationBethe-Salpeter Equation Green s Function and the Bethe-Salpeter Equation for Effective Interaction in the Ladder Approximation
Bh-Salp Equaon n s Funcon and h Bh-Salp Equaon fo Effcv Inacon n h Ladd Appoxmaon Csa A. Z. Vasconcllos Insuo d Físca-UFRS - upo: Físca d Hadons Sngl-Pacl Popagao. Dagam xpanson of popagao. W consd as
More informationChapter 7 AC Power and Three-Phase Circuits
Chaper 7 AC ower and Three-hae Crcu Chaper 7: Oulne eance eacance eal power eacve power ower n AC Crcu ower and Energy Gven nananeou power p, he oal energy w ranferred o a load beween and : w p d The average
More informationPhysics 160 Lecture 3. R. Johnson April 6, 2015
Physics 6 Lcur 3 R. Johnson April 6, 5 RC Circui (Low-Pass Filr This is h sam RC circui w lookd a arlir h im doma, bu hr w ar rsd h frquncy rspons. So w pu a s wav sad of a sp funcion. whr R C RC Complx
More informationThe Exile Began. Family Journal Page. God Called Jeremiah Jeremiah 1. Preschool. below. Tell. them too. Kids. Ke Passage: Ezekiel 37:27
Faily Jo Pag Th Exil Bg io hy u c prof b jo ou Shar ab ou job ab ar h o ay u Yo ra u ar u r a i A h ) ar par ( grp hav h y y b jo i crib blo Tll ri ir r a r gro up Allo big u r a i Rvi h b of ha u ha a
More informationy cos x = cos xdx = sin x + c y = tan x + c sec x But, y = 1 when x = 0 giving c = 1. y = tan x + sec x (A1) (C4) OR y cos x = sin x + 1 [8]
DIFF EQ - OPTION. Sol th iffrntial quation tan +, 0
More informationPeriod vs. Length of a Pendulum
Gaphcal Mtho n Phc Gaph Intptaton an Lnazaton Pat 1: Gaphng Tchnqu In Phc w u a vat of tool nclung wo, quaton, an gaph to mak mol of th moton of objct an th ntacton btwn objct n a tm. Gaph a on of th bt
More informationMath 34A. Final Review
Math A Final Rviw 1) Us th graph of y10 to find approimat valus: a) 50 0. b) y (0.65) solution for part a) first writ an quation: 50 0. now tak th logarithm of both sids: log() log(50 0. ) pand th right
More informationEE243 Advanced Electromagnetic Theory Lec # 10: Poynting s Theorem, Time- Harmonic EM Fields
Appl M Fall 6 Nuruhr Lcur # r 9/6/6 4 Avanc lcromagnc Thory Lc # : Poynng s Thorm Tm- armonc M Fls Poynng s Thorm Consrvaon o nrgy an momnum Poynng s Thorm or Lnar sprsv Ma Poynng s Thorm or Tm-armonc
More informationEXERCISE - 01 CHECK YOUR GRASP
DIFFERENTIAL EQUATION EXERCISE - CHECK YOUR GRASP 7. m hn D() m m, D () m m. hn givn D () m m D D D + m m m m m m + m m m m + ( m ) (m ) (m ) (m + ) m,, Hnc numbr of valus of mn will b. n ( ) + c sinc
More informationLast time. Resistors. Circuits. Question. Quick Quiz. Quick Quiz. ( V c. Which bulb is brighter? A. A B. B. C. Both the same
Last tim Bgin circuits Rsistors Circuits Today Rsistor circuits Start rsistor-capacitor circuits Physical layout Schmatic layout Tu. Oct. 13, 2009 Physics 208 Lctur 12 1 Tu. Oct. 13, 2009 Physics 208 Lctur
More informationWave Phenomena Physics 15c
Wv hnon hyscs 5c cur 4 Coupl Oscllors! H& con 4. Wh W D s T " u forc oscllon " olv h quon of oon wh frcon n foun h sy-s soluon " Oscllon bcos lr nr h rsonnc frquncy " hs chns fro 0 π/ π s h frquncy ncrss
More informationCharging of capacitor through inductor and resistor
cur 4&: R circui harging of capacior hrough inducor and rsisor us considr a capacior of capacianc is conncd o a D sourc of.m.f. E hrough a rsisr of rsisanc R, an inducor of inducanc and a y K in sris.
More informationLecture 1: Growth and decay of current in RL circuit. Growth of current in LR Circuit. D.K.Pandey
cur : Growh and dcay of currn in circui Growh of currn in Circui us considr an inducor of slf inducanc is conncd o a DC sourc of.m.f. E hrough a rsisr of rsisanc and a ky K in sris. Whn h ky K is swichd
More informationElectrochemistry L E O
Rmmbr from CHM151 A rdox raction in on in which lctrons ar transfrrd lctrochmistry L O Rduction os lctrons xidation G R ain lctrons duction W can dtrmin which lmnt is oxidizd or rducd by assigning oxidation
More informationNAME: ANSWER KEY DATE: PERIOD. DIRECTIONS: MULTIPLE CHOICE. Choose the letter of the correct answer.
R A T T L E R S S L U G S NAME: ANSWER KEY DATE: PERIOD PREAP PHYSICS REIEW TWO KINEMATICS / GRAPHING FORM A DIRECTIONS: MULTIPLE CHOICE. Chs h r f h rr answr. Us h fgur bw answr qusns 1 and 2. 0 10 20
More informationLoad Equations. So let s look at a single machine connected to an infinite bus, as illustrated in Fig. 1 below.
oa Euatons Thoughout all of chapt 4, ou focus s on th machn tslf, thfo w wll only pfom a y smpl tatmnt of th ntwok n o to s a complt mol. W o that h, but alz that w wll tun to ths ssu n Chapt 9. So lt
More informationEE 119 Homework 6 Solution
EE 9 Hmwrk 6 Slutin Prr: J Bkr TA: Xi Lu Slutin: (a) Th angular magniicatin a tlcp i m / th cal lngth th bjctiv ln i m 4 45 80cm (b) Th clar aprtur th xit pupil i 35 mm Th ditanc btwn th bjctiv ln and
More informationChapter 5 Transient Analysis
hpr 5 rs Alyss Jsug Jg ompl rspos rs rspos y-s rspos m os rs orr co orr Dffrl Equo. rs Alyss h ffrc of lyss of crcus wh rgy sorg lms (ucors or cpcors) & m-ryg sgls wh rss crcus s h h quos rsulg from r
More informationFigure Circuit for Question 1. Figure Circuit for Question 2
Exercises 10.7 Exercises Multiple Choice 1. For the circuit of Figure 10.44 the time constant is A. 0.5 ms 71.43 µs 2, 000 s D. 0.2 ms 4 Ω 2 Ω 12 Ω 1 mh 12u 0 () t V Figure 10.44. Circuit for Question
More informationH is equal to the surface current J S
Chapr 6 Rflcion and Transmission of Wavs 6.1 Boundary Condiions A h boundary of wo diffrn mdium, lcromagnic fild hav o saisfy physical condiion, which is drmind by Maxwll s quaion. This is h boundary condiion
More informationFirst-order piecewise-linear dynamic circuits
Frs-order pecewse-lnear dynamc crcus. Fndng he soluon We wll sudy rs-order dynamc crcus composed o a nonlnear resse one-por, ermnaed eher by a lnear capacor or a lnear nducor (see Fg.. Nonlnear resse one-por
More informationCh. 9 Common Emitter Amplifier
Ch. 9 Common mttr mplfr Common mttr mplfr nput and put oltags ar 180 o -of-phas, whl th nput and put currnts ar n-phas wth th nput oltag. Output oltag ( V ) V V V C CC C C C C and V C ar -of-phas 10 μ
More informationA Simple Representation of the Weighted Non-Central Chi-Square Distribution
SSN: 9-875 raoa Joura o ovav Rarch Scc grg a Tchoogy (A S 97: 7 Cr rgaao) Vo u 9 Sbr A S Rrao o h Wgh No-Cra Ch-Squar Drbuo Dr ay A hry Dr Sahar A brah Dr Ya Y Aba Proor D o Mahaca Sac u o Saca Su a Rarch
More informationREPETITION before the exam PART 2, Transform Methods. Laplace transforms: τ dτ. L1. Derive the formulas : L2. Find the Laplace transform F(s) if.
Tranform Mhod and Calculu of Svral Variabl H7, p Lcurr: Armin Halilovic KTH, Campu Haning E-mail: armin@dkh, wwwdkh/armin REPETITION bfor h am PART, Tranform Mhod Laplac ranform: L Driv h formula : a L[
More informationAdditional Math (4047) Paper 2 (100 marks) y x. 2 d. d d
Aitional Math (07) Prpar b Mr Ang, Nov 07 Fin th valu of th constant k for which is a solution of th quation k. [7] Givn that, Givn that k, Thrfor, k Topic : Papr (00 marks) Tim : hours 0 mins Nam : Aitional
More informationΕρωτήσεις και ασκησεις Κεφ. 10 (για μόρια) ΠΑΡΑΔΟΣΗ 29/11/2016. (d)
Ερωτήσεις και ασκησεις Κεφ 0 (για μόρια ΠΑΡΑΔΟΣΗ 9//06 Th coffcnt A of th van r Waals ntracton s: (a A r r / ( r r ( (c a a a a A r r / ( r r ( a a a a A r r / ( r r a a a a A r r / ( r r 4 a a a a 0 Th
More informationTransfer function and the Laplace transformation
Lab No PH-35 Porland Sa Univriy A. La Roa Tranfr funcion and h Laplac ranformaion. INTRODUTION. THE LAPLAE TRANSFORMATION L 3. TRANSFER FUNTIONS 4. ELETRIAL SYSTEMS Analyi of h hr baic paiv lmn R, and
More informationMathcad Lecture #4 In-class Worksheet Vectors and Matrices 1 (Basics)
Mh Lr # In-l Workh Vor n Mri (Bi) h n o hi lr, o hol l o: r mri n or in Mh i mri prorm i mri mh oprion ol m o linr qion ing mri mh. Cring Mri Thr r rl o r mri. Th "Inr Mri" Wino (M) B K Poin Rr o
More information10. If p and q are the lengths of the perpendiculars from the origin on the tangent and the normal to the curve
0. If p and q ar h lnghs of h prpndiculars from h origin on h angn and h normal o h curv + Mahmaics y = a, hn 4p + q = a a (C) a (D) 5a 6. Wha is h diffrnial quaion of h family of circls having hir cnrs
More informationECE Circuit Theory. Final Examination. December 5, 2008
ECE 212 H1F Pg 1 of 12 ECE 212 - Circuit Theory Final Examination December 5, 2008 1. Policy: closed book, calculators allowed. Show all work. 2. Work in the provided space. 3. The exam has 3 problems
More informationHomework: Introduction to Motion
Homwork: Inroducon o Moon Dsanc vs. Tm Graphs Nam Prod Drcons: Answr h foowng qusons n h spacs provdd. 1. Wha do you do o cra a horzona n on a dsancm graph? 2. How do you wak o cra a sragh n ha sops up?
More informationSearch sequence databases 3 10/25/2016
Sarch squnc databass 3 10/25/2016 Etrm valu distribution Ø Suppos X is a random variabl with probability dnsity function p(, w sampl a larg numbr S of indpndnt valus of X from this distribution for an
More informationChapter 12 Introduction To The Laplace Transform
Chapr Inroducion To Th aplac Tranorm Diniion o h aplac Tranorm - Th Sp & Impul uncion aplac Tranorm o pciic uncion 5 Opraional Tranorm Applying h aplac Tranorm 7 Invr Tranorm o Raional uncion 8 Pol and
More informationSPH4U Electric Charges and Electric Fields Mr. LoRusso
SPH4U lctric Chargs an lctric Fils Mr. LoRusso lctricity is th flow of lctric charg. Th Grks first obsrv lctrical forcs whn arly scintists rubb ambr with fur. Th notic thy coul attract small bits of straw
More informationMidterm exam 2, April 7, 2009 (solutions)
Univrsiy of Pnnsylvania Dparmn of Mahmaics Mah 26 Honors Calculus II Spring Smsr 29 Prof Grassi, TA Ashr Aul Midrm xam 2, April 7, 29 (soluions) 1 Wri a basis for h spac of pairs (u, v) of smooh funcions
More information3 ) = 10(1-3t)e -3t A
haper 6, Sluin. d i ( e 6 e ) 0( - )e - A p i 0(-)e - e - 0( - )e -6 W haper 6, Sluin. w w (40)(80 (40)(0) ) ( ) w w w 0 0 80 60 kw haper 6, Sluin. i d 80 60 40x0 480 ma haper 6, Sluin 4. i (0) 6sin 4-0.7
More informationLecture 2: Current in RC circuit D.K.Pandey
Lcur 2: urrn in circui harging of apacior hrough Rsisr L us considr a capacior of capacianc is conncd o a D sourc of.m.f. E hrough a rsisr of rsisanc R and a ky K in sris. Whn h ky K is swichd on, h charging
More informationHigh Energy Physics. Lecture 5 The Passage of Particles through Matter
High Enrgy Physics Lctur 5 Th Passag of Particls through Mattr 1 Introduction In prvious lcturs w hav sn xampls of tracks lft by chargd particls in passing through mattr. Such tracks provid som of th most
More information(b) i(t) for t 0. (c) υ 1 (t) and υ 2 (t) for t 0. Solution: υ 2 (0 ) = I 0 R 1 = = 10 V. υ 1 (0 ) = 0. (Given).
Problem 5.37 Pror to t =, capactor C 1 n the crcut of Fg. P5.37 was uncharged. For I = 5 ma, R 1 = 2 kω, = 5 kω, C 1 = 3 µf, and C 2 = 6 µf, determne: (a) The equvalent crcut nvolvng the capactors for
More informationLaPlace Transform in Circuit Analysis
LaPlac Tranform in Circui Analyi Obciv: Calcula h Laplac ranform of common funcion uing h dfiniion and h Laplac ranform abl Laplac-ranform a circui, including componn wih non-zro iniial condiion. Analyz
More informationThe Mathematics of Harmonic Oscillators
Th Mhcs of Hronc Oscllors Spl Hronc Moon In h cs of on-nsonl spl hronc oon (SHM nvolvng sprng wh sprng consn n wh no frcon, you rv h quon of oon usng Nwon's scon lw: con wh gvs: 0 Ths s sos wrn usng h
More informationCS 491 G Combinatorial Optimization
CS 49 G Cobinatorial Optiization Lctur Nots Junhui Jia. Maiu Flow Probls Now lt us iscuss or tails on aiu low probls. Thor. A asibl low is aiu i an only i thr is no -augnting path. Proo: Lt P = A asibl
More informationn r t d n :4 T P bl D n, l d t z d th tr t. r pd l
n r t d n 20 20 :4 T P bl D n, l d t z d http:.h th tr t. r pd l 2 0 x pt n f t v t, f f d, b th n nd th P r n h h, th r h v n t b n p d f r nt r. Th t v v d pr n, h v r, p n th pl v t r, d b p t r b R
More informationC From Faraday's Law, the induced voltage is, C The effect of electromagnetic induction in the coil itself is called selfinduction.
Inducors and Inducanc C For inducors, v() is proporional o h ra of chang of i(). Inducanc (con d) C Th proporionaliy consan is h inducanc, L, wih unis of Hnris. 1 Hnry = 1 Wb / A or 1 V sc / A. C L dpnds
More informationChapter 5 The Laplace Transform. x(t) input y(t) output Dynamic System
EE 422G No: Chapr 5 Inrucor: Chung Chapr 5 Th Laplac Tranform 5- Inroducion () Sym analyi inpu oupu Dynamic Sym Linar Dynamic ym: A procor which proc h inpu ignal o produc h oupu dy ( n) ( n dy ( n) +
More information3 2x. 3x 2. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Math B Intgration Rviw (Solutions) Do ths intgrals. Solutions ar postd at th wbsit blow. If you hav troubl with thm, sk hlp immdiatly! () 8 d () 5 d () d () sin d (5) d (6) cos d (7) d www.clas.ucsb.du/staff/vinc
More informationSource code. where each α ij is a terminal or nonterminal symbol. We say that. α 1 α m 1 Bα m+1 α n α 1 α m 1 β 1 β p α m+1 α n
Adminitrivia Lctur : Paring If you do not hav a group, pla pot a rqut on Piazzza ( th Form projct tam... itm. B ur to updat your pot if you find on. W will aign orphan to group randomly in a fw day. Programming
More informationLecture 4: Parsing. Administrivia
Adminitrivia Lctur 4: Paring If you do not hav a group, pla pot a rqut on Piazzza ( th Form projct tam... itm. B ur to updat your pot if you find on. W will aign orphan to group randomly in a fw day. Programming
More informationChapter 2 Homework Solution P2.2-1, 2, 5 P2.4-1, 3, 5, 6, 7 P2.5-1, 3, 5 P2.6-2, 5 P2.7-1, 4 P2.8-1 P2.9-1
Chapter Homework Solution P.-1,, 5 P.4-1, 3, 5, 6, 7 P.5-1, 3, 5 P.6-, 5 P.7-1, 4 P.8-1 P.9-1 P.-1 An element ha oltage and current i a hown in Figure P.-1a. Value of the current i and correponding oltage
More informationInstructors Solution for Assignment 3 Chapter 3: Time Domain Analysis of LTIC Systems
Inrucor Soluion for Aignmn Chapr : Tim Domain Anali of LTIC Sm Problm i a 8 x x wih x u,, an Zro-inpu rpon of h m: Th characriic quaion of h LTIC m i i 8, which ha roo a ± j Th zro-inpu rpon i givn b zi
More informationSensors and Actuators Introduction to sensors
Snsrs and Actuatrs Intrductin t snsrs Sandr Stuijk (s.stuijk@tu.nl) Dpartmnt f Elctrical Enginring Elctrnic Systms APAITIVE IUITS (haptr., 7., 9., 0.6,.,.) apaciti snsr capacitanc dpnds n physical prprtis
More informationP a g e 5 1 of R e p o r t P B 4 / 0 9
P a g e 5 1 of R e p o r t P B 4 / 0 9 J A R T a l s o c o n c l u d e d t h a t a l t h o u g h t h e i n t e n t o f N e l s o n s r e h a b i l i t a t i o n p l a n i s t o e n h a n c e c o n n e
More informationBoyce/DiPrima 9 th ed, Ch 2.1: Linear Equations; Method of Integrating Factors
Boc/DiPrima 9 h d, Ch.: Linar Equaions; Mhod of Ingraing Facors Elmnar Diffrnial Equaions and Boundar Valu Problms, 9 h diion, b William E. Boc and Richard C. DiPrima, 009 b John Wil & Sons, Inc. A linar
More informationUNIT #5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Answr Ky Nam: Da: UNIT # EXPONENTIAL AND LOGARITHMIC FUNCTIONS Par I Qusions. Th prssion is quivaln o () () 6 6 6. Th ponnial funcion y 6 could rwrin as y () y y 6 () y y (). Th prssion a is quivaln o
More information,. *â â > V>V. â ND * 828.
BL D,. *â â > V>V Z V L. XX. J N R â J N, 828. LL BL D, D NB R H â ND T. D LL, TR ND, L ND N. * 828. n r t d n 20 2 2 0 : 0 T http: hdl.h ndl.n t 202 dp. 0 02802 68 Th N : l nd r.. N > R, L X. Fn r f,
More informationProblem 1: Consider the following stationary data generation process for a random variable y t. e t ~ N(0,1) i.i.d.
A/CN C m Sr Anal Profor Òcar Jordà Wnr conomc.c. Dav POBLM S SOLIONS Par I Analcal Quon Problm : Condr h followng aonar daa gnraon proc for a random varabl - N..d. wh < and N -. a Oban h populaon man varanc
More informationD t r l f r th n t d t t pr p r d b th t ff f th l t tt n N tr t n nd H n N d, n t d t t n t. n t d t t. h n t n :.. vt. Pr nt. ff.,. http://hdl.handle.net/2027/uiug.30112023368936 P bl D n, l d t z d
More informationEAcos θ, where θ is the angle between the electric field and
8.4. Modl: Th lctric flux flows out of a closd surfac around a rgion of spac containing a nt positiv charg and into a closd surfac surrounding a nt ngativ charg. Visualiz: Plas rfr to Figur EX8.4. Lt A
More informationCHAPTER-2. S.No Name of the Sub-Title No. 2.5 Use of Modified Heffron Phillip's model in Multi- Machine Systems 32
9 HAPT- hapr : MODIFID HFFON PHILLIP MODL.No Nam of h ub-tl Pag No.. Inroucon..3 Mollng of Powr ym Hffron Phllp Mol.4 Mof Hffron Phllp Mol 7.5 U of Mof Hffron Phllp mol n Mul- Machn ym 3 HAPT-.. Inroucon
More informationMath 656 March 10, 2011 Midterm Examination Solutions
Math 656 March 0, 0 Mdtrm Eamnaton Soltons (4pts Dr th prsson for snh (arcsnh sng th dfnton of snh w n trms of ponntals, and s t to fnd all als of snh (. Plot ths als as ponts n th compl plan. Mak sr or
More information46 D b r 4, 20 : p t n f r n b P l h tr p, pl t z r f r n. nd n th t n t d f t n th tr ht r t b f l n t, nd th ff r n b ttl t th r p rf l pp n nt n th
n r t d n 20 0 : T P bl D n, l d t z d http:.h th tr t. r pd l 46 D b r 4, 20 : p t n f r n b P l h tr p, pl t z r f r n. nd n th t n t d f t n th tr ht r t b f l n t, nd th ff r n b ttl t th r p rf l
More informationELEN E4830 Digital Image Processing
ELEN E48 Dgal Imag Procssng Mrm Eamnaon Sprng Soluon Problm Quanzaon and Human Encodng r k u P u P u r r 6 6 6 6 5 6 4 8 8 4 P r 6 6 P r 4 8 8 6 8 4 r 8 4 8 4 7 8 r 6 6 6 6 P r 8 4 8 P r 6 6 8 5 P r /
More informationMATHEMATICS PAPER IB COORDINATE GEOMETRY(2D &3D) AND CALCULUS. Note: This question paper consists of three sections A,B and C.
MATHEMATICS PAPER IB COORDINATE GEOMETRY(D &D) AND CALCULUS. TIME : hrs Ma. Marks.75 Not: This qustion papr consists of thr sctions A,B and C. SECTION A VERY SHORT ANSWER TYPE QUESTIONS. 0X =0.If th portion
More informationExam 1. It is important that you clearly show your work and mark the final answer clearly, closed book, closed notes, no calculator.
Exam N a m : _ S O L U T I O N P U I D : I n s t r u c t i o n s : It is important that you clarly show your work and mark th final answr clarly, closd book, closd nots, no calculator. T i m : h o u r
More informationCHAPTER CHAPTER14. Expectations: The Basic Tools. Prepared by: Fernando Quijano and Yvonn Quijano
Expcaions: Th Basic Prpard by: Frnando Quijano and Yvonn Quijano CHAPTER CHAPTER14 2006 Prnic Hall Businss Publishing Macroconomics, 4/ Olivir Blanchard 14-1 Today s Lcur Chapr 14:Expcaions: Th Basic Th
More informationWave Superposition Principle
Physcs 36: Was Lcur 5 /7/8 Wa Suroson Prncl I s qu a common suaon for wo or mor was o arr a h sam on n sac or o xs oghr along h sam drcon. W wll consdr oday sral moran cass of h combnd ffcs of wo or mor
More information9. Simple Rules for Monetary Policy
9. Smpl Ruls for Monar Polc John B. Talor, Ma 0, 03 Woodford, AR 00 ovrvw papr Purpos s o consdr o wha xn hs prscrpon rsmbls h sor of polc ha conomc hor would rcommnd Bu frs, l s rvw how hs sor of polc
More informationWhy is a E&M nature of light not sufficient to explain experiments?
1 Th wird world of photons Why is a E&M natur of light not sufficint to xplain xprimnts? Do photons xist? Som quantum proprtis of photons 2 Black body radiation Stfan s law: Enrgy/ ara/ tim = Win s displacmnt
More informationJ = 1 J = 1 0 J J =1 J = Bout. Bin (1) Ey = 4E0 cos(kz (2) (2) (3) (4) (5) (3) cos(kz (1) ωt +pπ/2) (2) (6) (4) (3) iωt (3) (5) ωt = π E(1) E = [E e
) ) Cov&o for rg h of olr&o for gog o&v r&o: - Look wv rog&g owr ou (look r&o). - F r wh o&o of fil vor. - I h CCWLHCP CWRHCP - u &l & hv oo g, h lr- fil vor r ou rgh- h orkrw for RHCP! 3) For h followg
More informations-domain Circuit Analysis
Domain ircui Analyi Operae direcly in he domain wih capacior, inducor and reior Key feaure lineariy i preerved c decribed by ODE and heir I Order equal number of plu number of Elemenbyelemen and ource
More informationME 200 Thermodynamics I Spring 2014 Examination 3 Thu 4/10/14 6:30 7:30 PM WTHR 200, CL50 224, PHY 112 LAST NAME FIRST NAME
M 00 hrodynac Sprng 014 xanaton 3 hu 4/10/14 6:30 7:30 PM WHR 00, CL50 4, PHY 11 Crcl your dvon: PHY 11 WHR 00 WHR 00 CL50 4 CL50 4 PHY 11 7:30 Joglkar 9:30 Wagrn 10:30 Gor 1:30 Chn :30 Woodland 4:30 Srcar
More informationELECTROMAGNETIC INDUCTION CHAPTER - 38
. (a) CTOMAGNTIC INDUCTION CHAPT - 38 3 3.dl MT I M I T 3 (b) BI T MI T M I T (c) d / MI T M I T. at + bt + c s / t Volt (a) a t t Sc b t Volt c [] Wbr (b) d [a., b.4, c.6, t s] at + b. +.4. volt 3. (a)
More informationThe Hyperelastic material is examined in this section.
4. Hyprlastcty h Hyprlastc matral s xad n ths scton. 4..1 Consttutv Equatons h rat of chang of ntrnal nrgy W pr unt rfrnc volum s gvn by th strss powr, whch can b xprssd n a numbr of dffrnt ways (s 3.7.6):
More informationEconomics 302 (Sec. 001) Intermediate Macroeconomic Theory and Policy (Spring 2011) 3/28/2012. UW Madison
Economics 302 (Sc. 001) Inrmdia Macroconomic Thory and Policy (Spring 2011) 3/28/2012 Insrucor: Prof. Mnzi Chinn Insrucor: Prof. Mnzi Chinn UW Madison 16 1 Consumpion Th Vry Forsighd dconsumr A vry forsighd
More informationAR(1) Process. The first-order autoregressive process, AR(1) is. where e t is WN(0, σ 2 )
AR() Procss Th firs-ordr auorgrssiv procss, AR() is whr is WN(0, σ ) Condiional Man and Varianc of AR() Condiional man: Condiional varianc: ) ( ) ( Ω Ω E E ) var( ) ) ( var( ) var( σ Ω Ω Ω Ω E Auocovarianc
More informationLM A F LABL Y H FRMA H P UBLCA B LV B ACCURA ALL R PC H WVR W C A AU M RP BLY FR AY C QUC RUL G F RM H U HR F H FRMA C A HR UBJC CHA G WHU C R V R W H
H R & C C M RX700-2 Bx C LM A F LABL Y H FRMA H P UBLCA B LV B ACCURA ALL R PC H WVR W C A AU M RP BLY FR AY C QUC RUL G F RM H U HR F H FRMA C A HR UBJC CHA G WHU C R V R W H PUBLCA M AY B U CRP RA UCH
More informationSECTION where P (cos θ, sin θ) and Q(cos θ, sin θ) are polynomials in cos θ and sin θ, provided Q is never equal to zero.
SETION 6. 57 6. Evaluation of Dfinit Intgrals Exampl 6.6 W hav usd dfinit intgrals to valuat contour intgrals. It may com as a surpris to larn that contour intgrals and rsidus can b usd to valuat crtain
More information10.5 Linear Viscoelasticity and the Laplace Transform
Scn.5.5 Lnar Vclacy and h Lalac ranfrm h Lalac ranfrm vry uful n cnrucng and analyng lnar vclac mdl..5. h Lalac ranfrm h frmula fr h Lalac ranfrm f h drvav f a funcn : L f f L f f f f f c..5. whr h ranfrm
More information4 4 N v b r t, 20 xpr n f th ll f th p p l t n p pr d. H ndr d nd th nd f t v L th n n f th pr v n f V ln, r dn nd l r thr n nt pr n, h r th ff r d nd
n r t d n 20 20 0 : 0 T P bl D n, l d t z d http:.h th tr t. r pd l 4 4 N v b r t, 20 xpr n f th ll f th p p l t n p pr d. H ndr d nd th nd f t v L th n n f th pr v n f V ln, r dn nd l r thr n nt pr n,
More informationCircuit Transients time
Circui Tranin A Solp 3/29/0, 9/29/04. Inroducion Tranin: A ranin i a raniion from on a o anohr. If h volag and currn in a circui do no chang wih im, w call ha a "ady a". In fac, a long a h volag and currn
More information