ECE 201 Fall 2009 Exam 3 November 17 2009 Division 0101 Tan 11 30am Division 0201 Clark 7 30 am Division 0301 Elliott 1 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 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 v and ix 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 Course Outcome Exam Questions Total Points Possible i ii iii iv v ix 4 8 2 13 9 11 1 5 10 11 14 6 10 14 14 14 42 28 35 Minimum Points required to satisfy course outcome 7 7 7 21 14 14 If you fail to satisfy any of the course outcomes don t panic There will be more opportunities for you to do so 10 You will find formulas on the final page of this exam 1 1 Determine ic 0 1 0 3A 2 0 2A 3 0 1A 4 0 5 0 1A 6 0 2A 1 7 5V 2 5V 3 2 5V 4 0 5 2 5V 6 5V 7 0 3A 2 Determine vL 0 7 7 5V 2 Problems 3 5 are based on the following circuit Each question is independent of the others and can be solved in any order The current iL t in the circuit shown is known to be i L t 5 1 5 e 500t 0 5 e 1500t A for t 0 where t is in seconds The current is t is I1 for t 0 and I2 for t 0 as shown in the plot on the right 3 Determine I1 and I2 1 I1 1A I2 1A 2 I1 1A I2 4A 3 I1 1A I2 5A 4 I1 4A I2 1A 5 I1 4A I2 4A 6 I1 4A I2 5A 7 I1 8A I2 1A 8 I1 8A I2 4A 9 I1 8A I2 5A 4 Determine the voltage v t for t 0 1 0 375 1 125e 500t 1 5e 1500t V 2 5 1 5e 500t 0 5e 1500t V 4 3 1 5e 500t 1 5e 1500t V 0 6e 500t 0 27e 1500t mV 5 5 1 5e 500t 0 5e 1500t mV 6 1 5e 500t 0 5e 1500t mV 7 10e 500t 10e 1500t V 8 3 2 5 1 5e 500t 1 0e 1500t V Problems 3 5 are based on the following circuit Each question is independent of the others and can be solved in any order The current iL t in the circuit shown is known to be i L t 5 1 5 e 500t 0 5 e 1500t A for t 0 where t is in seconds The current is t is I1 for t 0 and I2 for t 0 The current is t is I1 for t 0 and I2 for t 0 as shown in the plot on the right 5 Determine C 1 10mF 2 1000F 3 500F 4 1mF 5 4mF 6 0 1mF 7 3 75mF 6 Assuming an ideal Op Amp the output voltage Vout t for t 0 is 1 0V 2 10V 3 10V 4 20V 5 20V 6 30V 7 30V 4 7 In the circuit below find the value of V2 given that Vo 5V 1 1V 2 2V 3 3V 4 4V 5 5V 6 6V 7 7V 8 Assume an ideal operational amplifier and that the capacitor is uncharged at t 0 Find the correct expression for the output voltage Vo t in volts for t 0 where t is in seconds 1 t 2 t 3 3 t2 4 t2 3 5 2t2 6 5t2 7 5t2 3 5 9 In the Op Amp circuit below the Th venin equivalent resistance Rth at the inverting input is 1 3k 2 2k 3 1k 4 0k 5 1k 6 2k 7 3k 10 The Op Amp circuit shown below has the response vout t 50 50e t 0 015 u t V Find the value of R1 1 0 75k 2 1k 3 1 5k 4 3 0k 5 3 333k 6 10k 7 15k 6 11 Determine the value of the inductor L such that the left and right networks share the same impedance Z eq for 2 rad sec 1 1 25 H 2 2 H 3 10 H 4 3 5 H 5 8 25 H 6 7 H 7 4 H 8 5 5 H 12 Given VS t 8cos 2t determine the phasor voltage VR 1 4 2 tan 1 1 V 2 4 tan 1 4 V 3 2 2 tan 1 4 V 4 8 tan 1 1 2 V 5 16 tan 1 1 4 V 6 7 6 tan 1 1 8 V 8 2 6 tan 1 1 V 7 2 tan 1 3 4 V 13 Determine the phase angle of iR where i1 t 10sin t i2 t 6 cos t and i3 t 2 cos t 45 11 1 tan 1 7 3 2 tan 1 5 5 3 tan 1 9 4 4 tan 1 13 5 60 6 45 7 0 8 30 14 Given the phasor A 3 6 j determine its magnitude and phase angle 1 45 tan 1 1 2 2 3 5 tan 1 1 2 3 3 5 tan 1 2 4 3 5 tan 1 2 5 9 tan 1 2 6 3 tan 1 1 2 2 7 3 5 tan 1 1 2 8 9 tan 1 2 2 8 Useful formulas x t x x t o x e t t o L R RC x t x A cos d t Bsin d t e t x t x A Bt e t x t x Aes1t Bes 2 t s1 s 2 b b 2 4c 1 for s 2 bs c 0 where c LC 2 R 2L series b 1 2 parallel 2RC s1 2 2 o2 4c b 2 d o2 2 2 o 1 LC 9
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