ECE 201 Spring 2010 Exam 2 Tuesday March 9 2010 Division 0101 Prof Capano 9 30am Division 0201 Prof Tan 10 30 am Division 0301 Prof Jung 7 30 am Division 0401 Prof Capano 11 30am 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 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 If you fail to satisfy any of the course outcomes don t panic There will be more opportunities for you to do so Course Outcome Exam Questions Total Points Possible i iii iv viii 7 8 1 5 14 9 14 6 14 42 42 7 Potentially useful formula x t x x t o x e t t o L R 1 Minimum Points required to satisfy course outcome 7 21 21 7 RC 1 Using source transformations find the voltage v in V 1 12 2 24 3 36 5 56 6 60 7 72 4 48 2 The Norton equivalent network for a more complicated circuit has a load resistor RL attached to it as shown A plot if iL v relationship is also given Find VOC and RTH for the corresponding Thevenin equivalent network 1 12V 6 2 24V 4 3 36V 6 5 24V 8 6 48V 2 7 60V 5 2 4 48V 4 3 Find the correct Thevenin equivalent network for the circuit shown below 1 2 3 4 5 6 7 3 4 For the circuit below the short circuit current ISC in the Norton equivalent circuit is in A 1 0 2 0 5 3 1 5 0 5 6 1 7 1 5 4 1 5 5 The Thevenin equivalent resistance RTH for the circuit shown below is in 1 1 2 2 3 3 5 5 6 6 7 7 4 4 4 6 Find the maximum power transferred to the load resistor for the circuit below in W 1 150 2 300 3 450 5 750 6 900 7 1050 4 600 7 In the circuit below the switch has been closed at the left position as shown for a very long time Find the instantaneous stored energy in the capacitor at t 0 sec in J 1 1 2 2 3 3 5 5 6 6 7 7 5 4 4 8 Compute the equivalent capacitance Ceq of the capacitor network shown below 1 2 F 2 4 F 3 6 F 5 10 F 6 12 F 7 14 F 4 8 F 9 The switch in the circuit shown below was opened at t 0 sec Find iL t for t 0 in A 1 10e 10t 2 5e 0 2t 3 20e 0 2t 5 20e 5t 6 10e 20t 7 5e 2 5t 6 4 5e 2t 10 The switch in the circuit shown below was closed for a long time It opened at t 0 sec The initial condition of the inductor current was i L 0 1A Find the inductor current at t 3 sec iL t 3 sec in A 1 0 2 e 1 3 e 2 5 e 4 6 e 5 7 1 4 e 3 11 In the circuit shown below the switch was moved from the left position to the right position at t 0 sec Find vc t for t 0 in V 1 40 e 2t 2 20 20e 0 5t 3 30 20e 0 5t 5 30 20e 2t 6 30 5e t 7 20 10e t 7 4 40 10e 2t 12 For the circuit shown below the input voltage is vin t 18u t 9u t V Compute the input current iin t for t 0 in A 1 1 10 3 1 0 10 3e t 2 2 10 3 1 0 10 3e t 3 1 10 3 1 5 10 3e t 4 2 10 3 1 5 10 3e t 5 1 10 3 2 0 10 3e t 6 2 10 3 2 0 10 3e t 7 1 10 3 2 5 10 3e t 13 Assuming C 0 0 V what is the time it takes for C t to rise to 12 1 e 2 V 1 1 2 2 3 3 5 5 6 6 7 7 8 4 4 14 For a linear network consisting of resistors and dependent sources the output voltage C t is computed using the given inputs and initial condition as shown below Initial condition C 0 10 C t 10e t 0 06 t t 0 06 0 06 50 1 e 30 1 e u t 0 5 Now the input current is increased from 0 25u t to 0 5u t as shown below Compute C t Initial condition C 0 10 C t t t 50 1 e 0 06 30 1 e 0 06 u t 0 5 t t t 20e 0 06 50 1 e 0 06 30 1 e 0 06 u t 0 5 t t t 10e 0 06 100 1 e 0 06 30 1 e 0 06 u t 0 5 t t t 10e 0 06 50 1 e 0 06 60 1 e 0 06 u t 0 5 t t t 20e 0 06 100 1 e 0 06 60 1 e 0 06 u t 0 5 t t t 20e 0 06 100 1 e 0 06 30 1 e 0 06 u t 0 5 t t t 10e 0 06 100 1 e 0 06 60 1 e 0 06 u t 0 5 1 C t 10e 2 C t 3 C t 4 C t 5 C t 6 C t 7 C t t 0 06 9
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