ECE 231 HW-2 Solution Set

# ECE 231 HW-2 Solution Set - ECE-231 Circuits and Systems I...

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Unformatted text preview: ECE-231 Circuits and Systems I Spring ’11 HW Set #2 Solutions Professor Stewart Personick ECEC-321 [email protected] Instructions • Send your solution set, via E-mail, to: [email protected] • The “subject ” of your E-mail must be : ECE 231 HW#2 (your last name) • Your solution set must be in the form of an attachment. I.e. a Microsoft Office file (.doc, .docx, .xls, .xlsx, .ppt, .pptx) or a legible scan of a handwritten file. • Your solution set must be sent to me by Problem #1 i1 i1 a b vab i1 Magnitude (Watts) Direction -5 Volts 2 Amperes 10W West 3 Volts -3 Amperes 9W West-4 Volts -1.5 Amperes 6W East 2 Volts 5.0 Amperes 10W East Box A Box B East For each of the four (4) cases, fill in the magnitude and the direction of power flow Problem #2 Box A i1(t) i1(t) a b Vab(t) is plotted below 10 Ohm Resistor 2 Volts 1 Volt 0 Volts Time (seconds) 10 Vab(t) 20 Plot* the current, i1(t), in Amperes, v. time, for 0<t<20 seconds Plot* the power, p(t), in Watts, delivered by Box A to the 10 Ohm resistor, v. time, for 0<t<20 seconds. Recall that: p(t) = i1(t) x Vab(t) Calculate the energy, in Joules, delivered to the resistor during the interval 0<t< 20 seconds. Recall that, by definition: this energy = the Integral of p(t) dt from t= 0 seconds to t=20 seconds *Plot neatly, on the next slide, with points for t=0 seconds to t=20 seconds, at 1 second intervals vab(t)= 2.8V -0.2V/s x t; for 4s<t<14s Solutions for Problem 2 Time (seconds) 1 p(t) 2 Time (seconds) 1 i1(t) 2 0.1A 0.2A 0.1W 0.4W The area under the curve of power v. time = 0.1W x 2s + 0.4W x 2s + 0.004W/s2 x (10s)3/3= 0.2J + 0.8J + 4J/3 = 2.33J 2.33J/14 s Problem #3...
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ECE 231 HW-2 Solution Set - ECE-231 Circuits and Systems I...

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