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Unformatted text preview: NAME: 8 . ma g23 ECEBox# Sowno”;
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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‘ Maowt}: Brooms 3 f 0 At t = 2.5 seconds, how much power is being dissipated by the load? [3]
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g) Based on these graphs, explain why this component cannot be a resistor. [3] _Two £56045 '
C')_ DEVICE MES [Var 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 andGum Laws ”0
‘5’ For the circuit shown, solve for Va, vb, and vc. Place your answers in the boxes provided. [10] KW 1006’s:
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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 Mosr 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“ «1er = 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
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OHMJ U144): K“ 9"“: [6'74 f ’4.’ Resistors in Series and Parallel [15
a) Determine REQ in the circuit shown. [6] /‘ \
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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
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