ECE 201 Spring 2008 Exam 2 March 6 2008 Division 0101 Qi 7 30am Division 0201 Elliott 12 30 pm Division 0301 Capano 2 30 pm Division 0401 Jung 3 30 pm Instructions 1 DO NOT START UNTIL TOLD TO DO SO 2 Write your Name division professor and student ID PUID on your scantron sheet 3 This is a CLOSED BOOKS and CLOSED NOTES exam 4 There is only one correct answer to each question 5 Calculators are allowed but not necessary Please clear any formulas text or other information from your calculator memory prior to the exam 6 If extra paper is needed use back of test pages 7 Cheating will not be tolerated Cheating in this exam will result in an F in the course 8 If you cannot solve a question be sure to look at the other ones and come back to it if time permits 9 As described in the course syllabus we must certify that every student who receives a passing grade in this course has satisfied each of the course outcomes On this exam you have the opportunity to satisfy outcomes i iii iv and viii See the course syllabus for a complete description of each outcome On the chart below we list the criteria we use for determining whether you have satisfied these course outcomes Outcomes i and iii were covered on Exam 1 You only need to satisfy the outcomes once during the course Course Outcome Exam Questions Total Points Possible i iii iv viii 4 5 7 9 14 1 3 6 7 10 12 6 14 3 42 49 63 7 Minimum Points required to satisfy course outcome 21 28 35 7 If you fail to satisfy any of the course outcomes don t panic There will be more opportunities for you to do so Potentially useful formulas are x t x x t o x e t t o L R 1 RC o 1 LC 1 The Thevenin equivalent resistance RTh of the circuit below is in ohms 1 1 3 3 5 5 7 7 2 2 4 4 6 6 1 vz 2 Req vz 2 2 The Norton Equivalent circuit of the following is 3 3 In the circuit below the load resistor RL is selected to maximize the power it absorbs Assume that RL is chosen correctly The maximum power extracted from the source by RL is in watts 1 20 3 60 5 120 7 360 2 30 4 72 6 240 5 6A RL 10 Load 4 For the network shown below find the equivalent capacitance seen by R1 20 mF 20 mF 1 5 mF 5 25 mF 2 10 mF 6 30 mF R1 5 mF 15 mF 5 mF 15 mF 3 15 mF 7 35 mF 4 4 20 mF 5 The circuit below is equivalent to 1 2 3 4 5 6 7 a 1 33 H inductor in series with a 3 F capacitor a 1 33 H inductor in parallel with a 3 F capacitor a 3 H inductor in series with a 1 33 F capacitor a 3 H inductor in parallel with a 1 33 F capacitor a 3 H inductor in series with a 3 F capacitor a 1 33 H inductor in series with a 1 33 F capacitor a 3 H inductor in parallel with a 1 33 F capacitor 2H 2H 2F 2F 2H 2F 6 For the network shown below find C t for t 0 given that C 0 10V 10 k 30 k 1 2 10e t 0 2 V 5 10e t 0 2 V 30 k 2 2 10e t 0 4 V 6 10e t 0 4 V C t 3 10 5e t 0 2 V 7 5e t 0 2 V 5 30 k 4 10 5e t 0 4 V 8 5e t 0 4 V 7 For the circuit shown below find iL t for t 0 given that iL 0 1A Note This initial current is due to some other source not shown in the circuit 10 I 6A 1 3 2e t 0 001 A 5 2 3e t 0 005 A 20 10 2 3 2e t 0 001 A 6 2 3e t 0 005 A 10 mH 3 2e t 0 1A 7 3 3e t 0 005 A 4 3 2e t 0 1 A 8 In the following circuit determine the current ix 0 at t 0 1 200 mA 2 120 mA 3 80 mA 4 40 mA 5 0 6 40 mA 7 80 mA 8 120 mA 6 9 The switch which has been open for a long time closes at t 0 Determine VC the voltage across the capacitor a long time after the switch closes 1 0 V 2 1 V 3 1 5 V 4 2 0 V 5 2 5 V 6 3 0 V 7 4 0 V 8 5 0 V 10 Determine the time constant for the natural response of the circuit shown for t 0 1 100 sec 2 150 sec 3 200 sec 4 300 sec 5 500 sec 6 1 msec 7 2 msec 7 11 The current flow through an inductor is t 4 i t 1 12e How long would it take for i t to drop from 7A to 4A Hint Recall that An a Anb An a b 7 4 1 An2 2 An 5 An3 6 4An4 3 4An 7 4 7 4An2 8 4 4An 8 An 12 7 12 7 12 vc t for the above circuit is 0 t 1 t 1 1 3 2e t e t 1 V 2 2e t V If VS2 changes from 3u t V to 1 5u t V what would vc t be Hint use linearity 0 t 1 1 2 3 4 5 6 7 8 9 1 e V 1 5 1 5e t V 2 2e t V 1 e t V 1 e t V 2 2e t V 2 2e t V 1 e t V 1 e t V t t 1 1 5 2e t 0 5e t 1 V 1 5 0 5e t 0 5e t 1 V 3 2e t 0 5e t 1 V 2 e t e t 1 V 1 5 e t e t 1 V 2 e t e t 1 V 1 5 e t 0 5e t 1 V 3 2e t e t 1 V 1 5 0 5e t 0 5e t 1 V 9 13 At t 0 i 0 5mA and v 0 0 5V Determine the amplitude B where i t A cos t Bsin t and 1 667 105 rad sec 1 15mA 2 10mA 3 5mA 4 0 5 5mA 6 10mA 7 15mA 14 Determine the total energy stored in the circuit elements at t 10 s for the circuit shown in the preceding problem 1 2 5 nJ 2 22 5 nJ 3 25 0 nJ 4 50 nJ 5 100 nJ 6 225 nJ …
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