{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw11sols - COLORADO STATE UNIVERSITY DEPARTMENT OF PHYSICS...

Info icon This preview shows pages 1–24. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
Image of page 11

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 12
Image of page 13

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 14
Image of page 15

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 16
Image of page 17

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 18
Image of page 19

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 20
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: COLORADO STATE UNIVERSITY DEPARTMENT OF PHYSICS Spring Semester 2008 Ph142 - Physics for Scientists and Engineers PROBLEM SET 11 - CHAPTER 29 - AC CIRCUITS - PART I DUE AT THE START OF RECITATION CLASS ON 17 APRIL 2008 Some guidelines for problem solving (whether handed in and graded or not): Focus is or understanding and implementation, not “plug and chug” problems. Almost every problem will involve some level of conceptual understanding and some challenge in the mathematical set-up and parameterization. Required format: (1) Do you work on grid style engineering paper. (2) Use one side of the page only. (3) Start each problem on a new page and state the problem number clearly. (4) Put your name, section number, problem set number, and due date on every page. (5) Include proper and reasonably professional looking diagrams or sketches for each problem. (6) Assemble in problem order (#1, #2, #3, etc.) and staple your pages in the upper left corner. Note #1: For assigned problems, your work (here and on future problem sets) should be accompanied by: (a) A clear problem statement in words, diagrams, and equations (not a regurgitation of the problem text in Tipler and Mosoa, necessarily, but adequate for some one to know what the problem is without going back to the book). (b) Good clear well labeled diagrams. (c) Well developed equations with all parameters defined. (d) Graphs with clearly defmed and labeled axes, etc. Note #2: You Should parameterize all problems. That is, assign algebraic parameters to all relevant variables and constants, set the problem up algebraically, and solve algebraically. Only then (if numerical evaluations are required), should you begin to plug in numbers. When you do plug in numbers, if it is a simple exercise, do as much of the evaluation as you can by hand (not by calculator). Multiply and cancel simple factors, add and subtract powers often, etc, to obtain a simplified numerical expression for evaluation. Often, you will fmd that you do not even need to do a calculator evaluation! If it is a complicated evaluation, you may wish to use some appropriate software, such as MathCad, etc. Note #4: A Problem Set will generally have from five to ten problems. If your recitation instructor decides to grade problems in some fashion, he or she may give partial credit, when deserved. However, there will be no credit for nonsense. Everything you say must make sense. Take care to do some intelligent work on ALL the problems. One completely missing problem can have a serious effect on your understanding and (if graded) on your Problem Set grade. Note #5. D Jn’t skimp on pages. Start each problem on a new page. Note #6. Start on a given problem set AS SOON AS IT IS MADE AVAILABLE. Work AT your solutions every day. Develop your solutions. Only when you are done, should you “write up” your solution set. When you do write it up, do so with understanding. Do not simply copy from your scratch notes. If you simply copy, without understanding, your graphs will not make sense, your equations will not make sense, and probably nothing will make sense. The problems (a) Tipler and Mosca, Problem 29-01. Why is this question somewhat ambiguous? (b) Tipler and Mosca, Problem 29-02. (c) Tipler and Mosca, Problem 29-04. How can you explain this intuitively. What happens in the zero and infinite frequency limits? (d) Tipler and Mosca, Problem 29-05. How can you explain this intuitively. What happens in the zero and infinite frequency limits? (e) Show that inductive and capacitive reactance has units of ohms. Variation on Tipler and Mosca, Problem 29-33. This is similar to the LC circuit analysis developed in class, this time with the capacitor initially charged to some iQo and an open switch in the circuit. Take the convention for positive current indicated by the down arrow on the lefi leg of the circuit in Fig. 29-30. Close the switch at time t: 0 and analyze the response. Write the Kirchoff loop equation. Discem the proper connection between the time dependent charge on the capacitor [taken as iq(r) such that q(0) = Q0] and the current i(t) . Use your loop equation to obtain a differential equation for q(t), apply initial conditions, and solve for q(t) and the frequency of the response. Sketch the corresponding q(t) and 1'0) responses. Sketch the capacitive and inductive energies as a function of time as well. Do tie analysis for a ac voltage source with a frequency to (angular, of course) driving an inductance L . Show that the voltage leads the current by 90°. Show that the ratio of the peak voltage to the peak current is equal to the inductive reactance X L = (:31. . Justify this result in the low and high frequency limits. Do the analysis for a ac voltage source with a frequency to (angular, of course) driving a capacitance C . Show that the voltage lags the current by 90°. Show that the ratio of the peak voltage to the peak current is equal to the capacitive reactance X c = 1/ roC . Justify this result in the low and high fi'equency limits. Harmonic function equivalence. Show that the forms yC (t) = X cos(a:t+6x), ys(t) = X sin(wt+ ax), and yABU) = Acos(art)+ B sin(a)t) are mathematically equivalent. That is, one can take yC (t) = y .4 3 (t) and obtain X and 9x in terms of A and B , and likewise, one can take ys (t): yAB (t) and obtain X and ¢x in terms of A and B. Variation on Tipler and Mosca, Problem 29—40. Work (a) and (b) as stated. For (0), the equations are correct but the question is poorly stated. Work out expressions for the phase 6 and the impedance Z . Add part (d): Check that the response makes sense in the limits at —-) 0 and a) —> no. Please pay attention to this problem. In Part II, we will apply the method of phasors to solve this same problem in a much easier way. Hint: Note that the voltage, £(t) = 80 cos a): , is common to both elements. The current through the inductor lags this voltage by 90° and is given by [L (t: = (60 / X L ) sin cot . The current through the resistor is in phase with the drive voltage. Just write these down, add them up, and apply the trig identity cos(a — fl) = cos a cos [3 + siua sin ,6 . Variation on the high pass filter problem, Tipler and Mosca, Problems 29-44 and 29-45. This situation is ill posed as stated. From the known fact that the ac current through a capacitor leads the voltage by 90°, the conunon current for an R and C in series will lead the overall applied input voltage by some angle between zero and 90°. The output voltage, which is across the resistor in this problem, will be in phase with the current and will lead the input voltage by this same amount. The Tipler and Mosca notation implies that the output voltage will lag the input voltage. The rigorous solution, of course, would yield a negative 5 . In order to emphasize the physics, work this problem in a slightly different way. Consider the circuit in Fig. 29-35 and start with a common current of the form 1(1) = 10 cos cat for the series R and C elements. Then work backwards to obtain the input voltage in the form, VIN = Vm pEAK cos(rnt—6). Determine VH / VIN pEAK and c3 . Do this by adding the voltage across the resistor, VR = [OH cos(a)t) = VH (in Tipler, this "H" is for "high pass"), and the voltage across the capacitor, Vc = 10X c sin(r:ot) (noting that the sin function lags the cos function by 90 °), and applying once again the trig. identity cos(a — fl) = cosacos ,B+sinasinfl. This will give you the phase factor (5' , which will now be a positive value since we have already taken the signs into account physically, and the ratio VH / VIN FEAK = 11’1l1+l/(coRC)2 as in the text. Check limits. Note that in the high frequency limit, the capacitive reactance Xc goes to zero and the capacitor looks like a short. In the opposite, low fi’equency limit, the capacitor looks like an open circuit. What are the corresponding implications of these limits for this filter? GNQ chlel 6N9 Panda. ._ l H T — /CO Sec, .— 1/? poSV‘hVG awd wejod‘we Vgeqlff one Separq+4d L7 CL. \nalF Vac/VdeL ' / TquKx-er : Q 1/2.? : /I.ZO BOW POx§ HWY! N€a~ peakf are \l. 1/ WeaK \leL/Pj _ Wk? Wen/ll ”W Chef-e owe 0mm“ We 011191” 2 ON? IWILS’LW‘ "Ii—MN}: JPO Cowndvn (31¢le +146 VQJLTIVQ peqmr P05“ Treaocsw r. T :- I/P double CL. Vfrmf +(‘fl—cucf Vo . "Wm" Jhovlcl \M {Sawfly obV1 ouJ‘. {L [O TIM Z°I-Okl ><=WL L. (See \eefi—wp/ 01 Jvex'l‘, or deny-e LLOL/nrekp. T E: EOCQSC—UT‘ : L. i_ d+ gh £0 CUJQJT‘ M L. J; = $2 Srwco'l' CuL. w .10 XL: E'a/T : 031— Double "(‘10 Praqvwr Mcq/ OLOLJble ‘l’L-LQ {NdvUQTLUQ rpficmwce‘ |N+U1JHV9I9 E FD’I'Q lmsl/w‘r Fae-0 preqvvwcg/ hue indium in? mvrwfi‘ Lon-5‘ +0 Chawj'e $3M; BacK PM'F 50”“ ‘0‘35‘4 'E‘ 1713344" £0 [NCV‘P/J-f-Pf WI]?! ‘pr‘eQUV-IUCV’ Ia iLCd) TH”! Z7-o§ *@lw .—[- é 2“ 85 C05 0J1— N T_ V; :- 60 CGJQJT g, .Q :C, (_ V V: 9 chv C.~ QM) —'—= C Se (‘3sz IM) M (593pr =~CEacoswafl~ W Io 80 e. X g 80 ._. A ...__——- .— c._ _. _. Ia éawc. (MC; [F freak/LON? 15 ClOt/jalr—OA/ XC (1“)le 191/ Q )Qqc'i‘or ca? +wo_ L5 AN 65va cwwfll‘, Xc :- 90 . X; ~3‘40Uli deem do" 60 1MC'1’IOAJ‘9J‘, — + KL eqrv \L L Z 56 CCW .. +620 “Q0 CVOM 5 1—90 (211' + (M) c mm) —— L—rm— .99) :o d+ C— Crew New I (2%) : 389) N6T {Mm/Ur]: d1“ (SUV curred if Plenum/5 IN+0 'bqf “EC? Via+e F £QL4)_ QIJ) L w “5‘ (1162/3) __ __ __L GOP) d+ LC :0 Sakunw QM)?- Hchu'f +Bchw'f‘ JINCP er’) 11? Q0 od‘ i: O B 9 00 Ed): 433;!) d—r — Rod (oxafi— 509$:wa —.... Bur lfo) :70 Sc) H :O Q(j() 3 Q0 CGJ (0+ S©\U+16’U qLeLJ-r Q00 "2' Q0 (0560—1— §LQ : IQ) = "' COQOStn/Cd'l" M 51:61 3 _. (Alfie @sz : —_‘. (Q 6350.)“?— di“ LC K- 1 CO 2: J. ‘93 00 H La \ILC, \ 2.. We KNVW Dc 2 2 C VC VC. .. «3.9) : 32 (<3wa C. C LSQ T” J— C .492- Co‘SlC‘JT—z '1 00; (51sz Z CL Z E: V s -‘ LIL: .1 L wzQZSA/(wr L 2_ 2. 6 [OUT L001: L '1 2 J LC C :0 «Q s _1 gimzwk Z c: ”if" U: T” —' fiazECdsofl“ +314» 011:] Z c; " Q62. '- CO NIT! EN€T§7 box/N015 LWLCK can/~21 431%? b~9+vvepfl L. aux/d C, 3. If VL: £5 CdJ-C‘UT éaCaSOJ‘f' L- _._. dink) " d+ LH‘) = ESCOIw1—df L : :2 51M 00+ W L VoH—csp L I Vét‘k‘flsfi [PALS- +HQ cur‘raflf‘. YES? 2 60 2:. (4)" :5 XL ILo 50/”,— lNclucxr'Ne re ruff‘qrv 16, Cu -—7 o \S‘kncnri‘, XL- ‘3' O . ln/Lucwv- (ool’cflrke Q. (A) -—-'-> NJ (Ndch '9 017%: err—cur?“ (XL'V‘a‘) {HO LL (1: Q/V V : Q! C a: cv 5: 2663““. QM) : C Ea Co: c4221“ [15‘ clp‘rumad IszJQOI') _~EOCOCS1NOU+ (LC (aw-70 Xe “7 04 DC (Vim/NOW vision!” ”H’IVBVLS‘L? CL Q¢pOLCtTorr fli Co—7w Xc’? O Ll‘jsI/Pp( GIL/d LUSH?!” prfiqVflfl/Cy Curribn—‘(J Flaw more 904,1), '("hru CL. cmpam +o p. 5‘] \H3 than = >< (aJ(ou'{‘-t- 67‘) (gm) —. >< gwmh (25X) (153(9‘) ":7 H (oswi‘l‘BSon/CUj— = H (GJW++BSINCU+ XCOSQ)‘ '3 H --)(S::v 92‘: +13 X 1059* + fights, _... Hz+B& X2": 92* BL : HZ": 62- §1N OX — E C05 Gk .— fl” TAN GA " ”fi/Q YgH) : >< Smurf Cw Q + XCusz 3ng 50 R : XS‘NQSF B = X (osgx X31/fi7‘4c B7— +AN®K: @159 far/13> @x : +Ar~:t fi/B ,IF {yarn/Magnify: arc 01/2051”, PropJHyj +5105? +1?th fofrmf fife all 4'th Same, (o, \[49 Tm Zfl~“/O N R L E: E?) (05001— (q) The V9 L5 6: comm voé’tL/aj-Q (CC) (115 CUT acrUJJ 196+“? Comporvprvfir. (g7) VL = 85C¢5w+ We (rm/aw ‘Fm/ «Arvz/vdvrror/ ‘h’l Q COW?” [92f ’FPIG Udl'Lc‘)? 19:4 (for: 1:) -'T-" 60% XL: wL IL/Jr) ‘3’ SE2 (o5(6u+-— ‘20? \HL, (IA/Q (Li 4144? CIA/OEIMJJf ”J Prob. 3.) (C) W9 exporq‘ 177a ’{hp '{d’éql Colfrvafi ITH) : fa (03001” 4' :0 (MOUT' ‘70“) R x1. [5“ a}: h“? £017»; ITM) = :0 co; (Mr -~ 3) LAM-ff? Ia Gun/0] § are +0 be H (wart +B$uucuT ~== C Cw («fl—S) C Codaflr" S) 4- C[C0J60+Q55 + qufig‘ng C6358; H C Suvéz B C cos-5.: 56/ Cswg : a R /><L C149 ('K CL) #7 O L (Jon-1 MM § -—-7 910° Q) -—-7 00 X1. —§ 0‘4! R (Lg/MINHJ‘J‘ \A\%\\ “((2) '7. $1 mef Fxl+0f E: 53 Coéw'l— Tlm; ‘5 c: 59 (Haj CtrCUlTI 50 Ifj) {p COM/1410M. l/UP (LON? KNOW! "by? QMQII’f‘x/do of +149 ([0959 ,. IN FhQI‘Q VR :. IGXR (0560'!— { ujo\\><2// 50% ’F (“3. (v lag]: RM C) Xg: I/WC 0F Cot/Vf'e \Hq Le‘T VT = ongccsfw+ +32% mow- W VbHaéflS MVJT 01ch £9! \fer'tef Compowvm—FJ. Tflcd CowcnoN L/ fame Lacak, V '5 CCdJCQ/ihg’) T : C (6‘; 0J1~ Cof§~ CSM/OUTSINg ”V R : Io XK CQJ'CU'T + onc Sim/OUT C c055 = YKIQ ("gm/g: XCIQ W2— [P am!) (“55- LUJIW ; Ibf) (~0an 5/30 but g 10+): 5" Cod/(m + 8) W ”P hand. han if ox L'HJ"? parfpr‘FW’" 05 00 Leta/mu" (01’5":- HEW )Ia"> SMK F'" 05- W LECOMMJ‘ 5M4“ ,Lc/ “70 [1 —7l “70 RIM) 80R (lede +0? 1 mg R "l" /U7\CL de‘a’M ["1 R \ 5f wee—KC CJKC>PI w wac (<1 $0M“? M pr070/ 1901 MUC") QQJl-Q Y“. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern