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c08_sol_exam1

c08_sol_exam1 - NAME 8 ma g23 ECEBox Sow-no” $5 0 Problem...

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Unformatted text preview: NAME: 8 . ma g23 ECEBox# Sow-no”; $5, 0 Problem Score Points 1 Hr 25 /€. 3 2 7:? 10 3 fi 10 4 E 25 5 _//i 15 6 .6 15 AVE, : 8’3. 3 ECE 2011 — Term C - 2008 Introduction to Electrical and Computer Engineering Exam 1 This is a closed book test. No notes of any kind. You may use calculators. Show [email protected] Partial credit may be given. \_—_—— When in doubt, let common sense prevail. Don’t get entangled in unnecessary algebra. As in real life, problems may include more information than required. You will have 50 minutes to complete this exam. NOTE: BE SURE TO CLEARLY MARK ALL ANSWERS! 1. Basic Electrical uantities £1 a) For each of the electrical quantities given in the chart below, identify the Standard Units associated with it, as well as the Fundamental Units that it can be broken down into. [4] |—Quantity Standard Units Fundamental Units Charge C 00L OMQ, C OULOM 3 Current 9 M ‘0ng Coulombs / Second Voltage L— Volt 3'0” LES /COI)LOM8__| Power w ,4 1—..- J‘oo an”: ken“: ‘1 Energy 3"“) L C J?) U LE The graphs below indicate the current i (t) and voltage v(t) for the circuit element shown, and pertain to questions ([9) through (g). i(t) Wl W) tSEC b) On the axes provided, sketch the power p(t) as a function of time. NOTE: BE SURE TO INDICATE KEY FEA TURES OF YOUR GRAPH! [6] p(t) 6w ~--_71 tSEC c) Based on these graphs, how much charge has passed through the element after 3 seconds? NOTE: Please place your answer in the box provided. [3] A264 0mm ((4) 15 m manur- lc+2g+lc39c 1/C d) How much power is dissipated in the load at t = 0.5 seconds? [3] WM» came on {war- (B) P (urea) : 3(4) 3(4) e) How much energy has been supplied by the source at the end of 1.5 seconds? [3] 40210 Umm poms»: curwic Is 77+}? Eomcy‘ Mao-wt}: Brooms 3 f 0 At t = 2.5 seconds, how much power is being dissipated by the load? [3] u AT rms Hut) 7715 £049 4 IS Aer)”; (,[g‘ A sowing ’. ..'. 2-r - .— P( M) ' 3&0 Gamma ‘3 ‘4) (on r 3» Supra: 03} g) Based on these graphs, explain why this component cannot be a resistor. [3] _Two £56045 '- C')_ DEVICE MES [Va-r MLA Lam Var cummum. 1:04: “A €04.er V chflvépg LJ/_ T7"'_€. 13: Is. ”of-:ifflf wvl ...=—.—.- f. C23 Dimer: scarring: mtg mom [AL/Lpfi7"{> Egg/57m W 258W WMM 2. Kirchhoff" 5 Voltage and-Gum Laws ”0| ‘5’ For the circuit shown, solve for Va, vb, and vc. Place your answers in the boxes provided. [10] KW 1006’s: 0) ~lo -—”Ua +2030 "UK: )ov CZ? ~26 +25 +VL:° vb?“ m WV Wm .3 Kirchhost w Current Lawigcontinued. ..1 :5 For the circuit shown, solve for id, ib, and 1'6. [10] 6A @ KCL ”"551 (AMY/M But 14/ 77975 “My, ” 50w or 6mm 69771214; «Door (9 Mos-r awn 044mm: Emma ma”) 0 : (q +964 -'- (a = *5“ ® (/A +5.5 == 6A 5» = OM (3 4A +52"? 16¢ '.- (it: 8/4 Li. Circuits l25| a) Using KCL, KVL, and Ohm’s Law, determine ix in the circuit shown. [2’] / 0 0) AT la n. REX/11’! ) <2) .'. Va} 3 0.§A|(§:"01L) 3 9V (9).: (ix : 0.57“ «1-er = 0.794 6 T” 1‘ )9: Determine in the circuit shown. [8] ELlM/A/W 9” v69 ('1- + I ..., H. Circuits (continued... 1 / >/ 5) Determine vx and ix in the circuit shown. LL01 KVL ~20V +'Ux+$"l/x:0 {Ukr'alov WK: £21: 3-33V OHMJ U144): K“ 9"“: [6'74 f ’4.’ Resistors in Series and Parallel [15| a) Determine REQ in the circuit shown. [6] /‘ \ /' I ~ \ 109 r \ \ \ \ \‘ REQ= \ REQ_(7 I 129’ I Jo} QIL b) Determine REQ between the two points a and b in the circuit shown. Assume the value of each resistor is 100 Q. [9]; REQZ /00_rz_ - Afiw J: (Cwmw leECWO~> . BY SVWWV) co‘fifiw IA) MIDDLE WW", MOST 3E 2W]. V: (3/23/00 + "IE/2. (I on) L ‘ ‘ «£20 I OD : D J: -01) {4y /0 42 lLTms Is A u/mmm: zeros: C/ecwr- Lwlw m1: @212“: IS EAL/Mm“ / T31: CLfiKE‘A/T LS ?f/Zo Tflfiwm 7,??? ”MILE EYE/I507? (6 [1 IEEE Code of Ethics US] List at least 5 of the 10 IEEE Codes of Ethics: E EC :0“: Co 0 If 0): Ja‘rrh a] (1) (2) (3) (4) (5) ...
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