Exam 1 Solutions - Carnegie Mellon University Department of...

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Unformatted text preview: Carnegie Mellon University Department of Electrical and Computer Engineering 18-100 Fall 2007 Introduction to Electrical and Computer Engineering Exam 1: September 27, 2007 Closed Book Write your answers in the spaces provided. Do your work as neatly as possible. Cross out any work that is not pertinent to your final solution. The grader reserves the right to mark solutions as incorrect if s/he cannot follow your train of thought in the problem, or read your handwriting. BE CLEAR AND CONCISE IN YOUR ANSWERS. Show all work! Answers that have incorrect units (or no units at all) will have points taken off. Problem 1 (25 points) Problem 2 (25 points) Problem 3 (25 points) Problem 4 (25 points) Total = SDLUTtlTDNS Name (Lab) Section (A: Monday, B: Tuesday, C: Wednesday, D: Thursday, E: Friday) Problem 1: (25 points) Nodal Analysis Consider the circuit shown below: V 3 a) (4 points) Label the voltage and current on all of the devices in the circuit consistent with devices that dissipate power. Be thorough in your labeling (as we were in lecture) so that we know exactly to which device you refer when writing your equations in the next part. Use the given current directions for the resistors. b) (15 points) Solve for the node voltages at the nodes A, B, and C. Use the technique of Nodal Analysis. Please place a box around the equations that you have set up to solve, and place your final answers in the spaces provided on the next page. ’VM' Va ‘- 1V6 'WL '— 32mm 1,) 1 WVWS‘NQ VA= VB= W" = SVA Jr We 45%"ng qw Z Wu“ “)(er6 ”Wk 4 c) (6 points) Calculate the power dissipated in all of the elements in the circuit. ‘9 1 u} V9" UN)1L ‘ 7% ‘ ”I?” jg=0-w 71 1. PL: EL : Y1]... '— (leL :flDN‘) “4’ {LL lb» 1 P} _, E 7.. kvm'VL) (l 0U” (9)) 4» \r) {.3 —— 2 _.___..» I w ‘13 “um 9% 3V1}: V2- ; [Egg- : +qu 7c; 1:; .VA P; Vx 17L ZVX'T']: \b‘ ET “5.. OmV‘) Problem 2: (25 points) Superposition In this problem we will calculate the value of the cunent IL using the technique of superposition. Iy=15mA 6 b) (8 points) Find by, the contribution to IL from the current source 1y. ILy = 7 c) (9 points) Find 11.2, the contribution to IL from the voltage source Vz. Sum your results from parts a) and b) with the results of this part to get the total current, IL. ECHO ’1qu J 0%"). I), Viv Wk “Afl'a‘f‘: T - ‘21:]: a 3L 5'. 0 amp“ L3: {LL [CW 1,! ani om: {votuA L" : Org—YA 1L2: +DqumA 1L: an MA 8 Problem 3: (25 points) Thevenin and Norton Equivalents For the circuit below, you will find the Thevenin and Norton equivalent circuits as seen from terminals A and B. All resistors have a value of 1 k9. R1 R2 .a) (8 points) Find the value of Voc, the open-circuit voltage, as seen from terminals A and B. You must calculate Voc directly using one of the circuit solving techniques discussed in class (i.e., you may not use the “resistor-trick” as shown in class to calculate this quantity). "VDL‘NM Way VW-y 09ch 1L, V'uy J 53f: V2 1 ,— Voc=VTH= 33V W—lgv "' VDLYS-ljv 9 b) (8 points) Find the value of 150, the short-circuit current, as would flow in a wire connected across terminals A and B. You must calculate Isc directly using one of the circuit solving techniques discussed in class (i.e., you may not use the “resistor-trick” as shown in class to calculate this quantity). in: gym“), use SFWE mwi‘u P5 FMbt® “JV {Ni P \r'iflis' figTVIEng®Or® ft)!” .IHSL‘X fitx‘yi. IN ll!» iSLYf‘ “W“ ‘y—*1/§}’V —. 45...!“ 3 ELY 3m “ ”i "‘13 , f5!— = lr—A "2“}; v‘firvfl w c) (3 points) Find the value of the Thevenianorton resistance. You may use the resistor trick here if you like. V av 0L ______ # l I, 9" (twflfi ”if: ‘t/th’ 1W” 3'5”“ 'Dtr’ Jew W} P ‘7 M:- ‘Hllnfi fit ”14%“ = om + U“ “MW" ‘2'. 3.375” CPtUHi RTH= RN: 303 “J‘- d) (6 points total, 3 points each) Draw the Thevenin and Norton equivalent circuits for this circuit as seen from terminals A and B. ’35“ {1 1 1 Problem 4: (25 points) Below we have a circuit to which we wish to attach a load resistor, RL. Rooting through our pile of spare resistors, we select three possible values of R1. and we’d ultimately like to use the one that gives us the greatest power dissipation in RL. IySImA R1=10kQ The three resistors we chose are: RLl is red—black-red R12 is gray-black-red RL3 is brown-red—orange List the values of the three resistors here (the resistor color code is given on the last page of the exam): {Van - 1 W1 x 13w“ 1 DleL‘ 0 1. (mu-t D MD L ,5 (1450 h _ M0 ‘1 oM’f’VE 1m 31"” 1’1 ILA: m2: Yll’v | M: I} ”\w 12 Now, calculate the power that would be dissipated in each of the three resistors. Which one dissipates the most power? (Hint: You do not need to solve the entire circuit three separate times.) EWIU~ MM?” F3345 nxmw TMCLC , 3 \ \J in: 1'. WW 1“” filhl'fl‘p: Urn-i- }D] /iflL~. = Awr 61L» =Xl‘m K. Mg) ‘ Eryd he 3}» 8\/ IN» ”V‘- 1% ,_ 3%; p“: @L'ypimu VL 1mm: 1” :3le L“) 1 two- (LL yT “HQ, V13 - rm) 19:... s p —. 0193020. EH 1119» L3 My * - f3)“ PL1=7O?A\J m: H mm EaLP/WFJT 15m P13 = RM? MVJ ...
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