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Unformatted text preview: EE-202 Exam II March 9, 2010 Name: ___________DASH___________________ (Please print clearly) Student ID: _________________ CIRCLE YOUR DIVISION MORNING 8:30 MWFAFTERNOON 12:30 MWFINSTRUCTIONS There are 12 multiple choice worth 5 points each and there is 1 workout problem worth 40 points. This is a closed book, closed notes exam. No scrap paper or calculators are permitted. A transform table will be handed out separately. Carefully mark your multiple choice answers on the scantron form. Work on multiple choice problems and marked answers in the test booklet will not be graded. Nothing is to be on the seat beside you. When the exam ends, all writing is to stop. This is not negotiable. No writing while turning in the exam/scantron or risk an F in the exam. All students are expected to abide by the customary ethical standards of the university, i.e., your answers must reflect only your own knowledge and reasoning ability. As a reminder, at the very minimum, cheating will result in a zero on the exam and possibly an F in the course.Communicating with any of your classmates, in any language, by any means, for any reason, at any time between the official start of the exam and the official end of the exam is grounds for immediate ejection from the exam site and loss of all credit for this exercise. Exam 2, ECE-202, Sp 10 1.In the circuit below L1 H, C0.25 F, and the initial inductor current is iL(0)4 A while vC(0)0 . Then vC(t)(in V): (1) 8sin(4t)u(t) (2) 8sin(2t)u(t) (3) 4sin(2t)u(t) (4) 8sin(4t)u(t) (5) 8sin(2t)u(t) (6) 4sin(4t)u(t) (7) 16sin(2t)u(t) (8) 16sin(2t)u(t) (9) None of above Solution 1. VC(s) 4s4s4s16s24 82s24. ANSWER (5) 2. Rs0.5 , C0.25 F, gm 1 S, and R2 , then the transfer function of the circuit below is H(s): (1)1.5s2s2(2)2s21.5s2(3)21.5s2s2(4)s21.5s2(5)1.5s7s2(6)s2 (7)1s2(8) None of these Solution 2.IinGsgm1R1CsVoutGsgmsRs1CVout1s2s410.50....
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This document was uploaded on 04/02/2010.
- Spring '06