ECE 201 Spring 2011 Exam 2 Thursday March 10 2011 Division 0101 Prof Tan 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 Calculators are allowed but not necessary 5 If extra paper is needed use back of test pages 6 Cheating will not be tolerated Cheating in this exam will result in an F in the course 7 If you cannot solve a question be sure to look at the other ones and come back to it if time permits 8 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 5 6 1 3 11 7 10 12 13 4 15 30 45 7 5 x t x x t o x e t t o L R RC Minimum Points required to satisfy course outcome 7 5 15 22 5 7 5 2 1 LC 1 For the circuit shown below find the Thevenin equivalent resistance Req in k as seen by the variable load RL 1 2 3 2 2 5 3 3 5 4 2 7 5 5 7 6 10 7 7 2 8 5 2 For the circuit shown below find ISC in mA the short circuit current for the Norton equivalent circuit 1 30 2 30 3 50 4 50 5 100 6 100 7 150 8 150 3 Find the Norton equivalent circuit seen by the load resistor 1 Isc 4A R th 4 2 Isc 4A R th 5 3 Isc 2A R th 4 4 Isc 2A R th 5 5 Isc 4A R th 4 6 Isc 4A R th 5 7 Isc 2A R th 4 8 Isc 2A R th 5 4 For the circuit shown below find the load resistance RL for maximum power transfer and the corresponding PLmax 1 1k 2 2k 3 3k 4 4k 5 1k 6 2k 7 3k 8 4k 10W 10W 10W 10W 20W 20W 20W 20W 5 The current through a 1F capacitor for t 0 is as shown below If vc 0V find vc 0 in V the initial condition for the capacitor at t 0 1 1 2 2 3 3 4 4 5 1 6 2 7 3 8 4 6 For the circuit shown below find i Leq in H the equivalent inductance and ii v3 t in V the voltage drop across inductor L3 1 2 1 0 4sin 2 t 2 1 2 0 4cos 2 t 3 1 2 1 6sin 2 t 4 1 2 1 6cos 2 t 5 3 32 0 4sin 2 t 6 3 32 0 4cos 2 t 7 3 32 1 6sin 2 t 8 3 32 1 6cos 2 t 7 For the circuit shown below find vo t for t 0 vO t vin t 10u t 4u t 1 vo t 3 vo t 5 vo t 7 vo t 6 7 3t e 3 3 6 7 3t e 5 5 6 7 6t e 3 3 6 7 6t e 5 5 2 4 6 8 6 7 3t vo t e 3 3 6 7 3t vo t e 5 5 6 7 6t vo t e 3 3 6 7 6t vo t e 5 5 8 In the circuit below the switch opens at t 0 sec Find the voltage across the resistor vr t for t 0 in V 1 6e 10t 2 12e 10t 3 12e 8 33t 4 6e 10t 5 12e 10t 6 6e 8 33t 7 12e 8 33t 8 6e 8 33t 9 In the circuit below vc 0 0V Find vc t for t 0 in V 1 2e 3t 2 2e t 4 3 2e 1 5t 2 4 2e 3t 1 5 2e 1 5t 6 2e t 7 2e t 2 8 2e 1 1 10 The switch in the circuit below has been open for a long time before it is closed at time t 0 Find ix 0 in A ix in A and the time constant in s 1 2 3 4 5 6 7 8 ix 0 ix 0 ix 0 ix 0 ix 0 ix 0 ix 0 ix 0 0 A 1 A 0 A 1 A 0 A 1 A 0 A 1 A ix 1 5 A ix 1 5 A ix 1 5 A ix 1 5 A ix 1 0 A ix 1 0 A ix 1 0 A ix 1 0 A 4 5 s 4 5 s 2 5 s 2 5 s 4 5 s 4 5 s 2 5 s 2 5 s 11 Consider the 1st order circuit below The response vc t for t 0 under two conditions are shown in the table Find vc t for t 0 under condition 3 in V Condition vc 0 vs t vc t 1 5V 0 5e 2t V 2 0 10u t V 4 1 e 2t V 3 10 V 15u t V 1 4 e 2t 2 4 e 2t 3 5 e 2t 4 5 e 2t 5 6 4e 2t 6 6 4e 2t 7 9 4e 2t 8 9 4e 2t 12 For the circuit shown below what is the time it takes for vC t to rise from 0V to 5V in second Assume vC 0 0V R 2k 10V C 1mF 1 1 ln 1 5 2 1 ln 2 0 3 1 ln 2 5 4 1 ln 3 0 5 2 ln 1 5 6 2 ln 2 0 7 2 ln 2 5 8 2 ln 3 0 13 For the circuit shown below the switch is closed for a long time before it opens at time t 0 Find vC 0 iL 0 and for t 0 L 2H C 2F 1 2 3 4 5 6 7 8 vC 0 0V vC 0 0V vC 0 0V vC 0 0V vC 0 5V vC 0 5V vC 0 5V vC 0 5V iL 0 0 5A iL 0 0 5A iL 0 0 5A iL 0 0 5A iL 0 0 0A iL 0 0 0A iL 0 0 0A iL 0 0 0A 0 25 rad s 0 50 rad s 0 75 rad s 1 00 rad s 0 25 rad s 0 …
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