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Unformatted text preview: EE-202 Exam III April 15, 2010 Name: SOLUTION (Period)(Please print clearly) Student ID: _________________ CIRCLE YOUR DIVISION Morning 8:30 MWF Afternoon 3:30 MWFINSTRUCTIONS There are 9 multiple choice worth 5 points each and there are two workout problems worth 55 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. Do not open, begin, or peek inside this exam until you are instructed to do so.EE-202, Ex 3 Sp 09 page 2 MULTIPLE CHOICE 1. The resonant frequency of the circuit (R=4 Ω, C=132F, L1=0.25 H, L2=1 H, M=38H) below is (in rad/s): (1) 32 (2) 256 (3) 225 (4) 15 (5) 16 (6) 64 (7) 8 (8) none of above Solution 1. Leq=L1+L2-2M=1.25-0.75=0.5 H. ϖr=1LeqC=64=8 r/s. Ans. (7)2. Assume k=0.5 , L1=1 H, L2=1 H, and M=kL1L2. The zero-state response iout(t) to the input vin(t)=6u(t) V at t=4 is (in A): (1) 10 (2) 12 (3) 20 (4) 16 (5) 32 (6) 6 (7) 8 (8) none of above Solution 2. M=kL1L2=0.5=0.5 H. =V2= -0.5sIout+sI2⇒I2=0.5Iout. Vin(s)=sIout(s)-0.5sI2(s)=[s-0.25s]Iout(s)=0.75sIout. Thus Iout(s)=Vin(s)0.75s=8s2⇒iout(t)=8tu(t) ....
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