ECE 201 Spring 2007 Exam 3 April 12 2007 7 00 8 00 pm Division 0001 Division 0002 Division 0003 Prof Elliott 11 30am Prof Tan 7 30 am Prof Capano 4 30 pm Instructions 1 DO NOT START UNTIL TOLD TO DO SO 2 Write your NAME DIVISION PROFESSOR S NAME and PUID on your scantron sheet 3 This is a CLOSED BOOKS and CLOSED NOTES exam 4 Calculators are allowed 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 ABET requires that every student who receives a passing grade in this course has satisfied each of the course outcomes as listed in the course syllabus On this exam you have the opportunity to satisfy outcomes iv v and ix See the course syllabus for a complete description of each outcome The criteria used for determining whether you have satisfied these course outcomes are listed below Course Outcome Exam Questions Total Points Possible iv v ix 1 5 8 14 6 7 35 49 14 Minimum Points required to satisfy course outcome 18 25 7 If you fail to satisfy any of the course outcomes on this exam you will likely have an additional opportunity on the final exam 2 1 At t t0 the current in the circuit below is zero and the voltage across the capacitor is 3V At a later time instant t t1 the capacitor voltage is measured to be zero The circuit current at t t1 is iL t 0 4 mF vc t 0 1 mH 1 12 A 5 4 A 2 3 A 6 1 2 A 3 0 25 A 7 0 3 A 3 4 6 A 2 In the circuit shown below VC and iL are respectively 1 15V and 0A 2 0V and 2A 3 0V and 3A 4 15V and 2A 5 20V and 2A 6 20V and 3A 7 15V and 3A 4 3 The RLC circuit below has a critically damped response in the form of VC t A1e 4t A 2 te 4t V for t 0 The value of R in leading to this response is 1 0 125 2 0 25 3 0 5 4 4 5 2 5 6 8 7 5 4 Consider the circuit below with the initial conditions i L 0 10A and VC 0 10V The value of 1 10 5 20 dVC t is in V s dt t 0 2 10 6 5 3 0 7 5 6 4 20 5 For the circuit below the response for the inductor current iL t is desired If the general solution is of the form x t xn t XF select the final value XF in appropriate units iL t 25 75u t V 4 mF 1 75 V 5 5 V vc t 2 0 V 6 6 A 15 3 3 A 7 0 A 4 4 A 7 20 mH 6 The circuit below has an independent voltage source V1 t 20u t V Assuming an ideal OpAmp the output voltage Vout t for t 0 is in V 1 10 5 50 2 20 6 60 3 30 7 0 8 4 40 7 Consider the circuit to the right The op amp is ideal and the element marked X is unknown When a voltage Vin t is applied the output is What is the X element 1 2 3 4 5 6 7 2 5 mH inductor 2 5 mH inductor in parallel with a 1 k resistor 1 k resistor 2 5 F capacitor 1 k resistor in parallel with a 3V source 2 5 H inductor none of the above 9 8 If vs t 10 sin 2t V then is t equals in A 1 5 2 cos 2t 36 9o 0 6 iS t 2 10 cos 2t 3 5 sin 2t 4 5 2 sin 2t 0 5 F vS t 0 1 H 5 10 cos 2t 36 9o 6 5 cos 2t 45o 7 none of the above 10 9 Given that v1 t 5cos t v2 t 3sin t and v3 t 4sin t 50 the order in which these voltages lead one another from one leading foremost to one lagging furthest behind is v1 t v2 t v3 t v1 t v3 t v2 t 1 2 v2 t v3 t v1 t v2 t v1 t v3 t 3 4 v3 t v1 t v2 t v3 t v2 t v1 t 5 6 7 all in phase 11 10 The voltage v t 5cos 10t 5cos 10t 120o has a phasor representation in polar form given by 1 10 0o 5 5 60o 2 6 10 120o 5 120o 3 5 2 60o 4 7 none of the above 12 5 2 120o 11 For the circuit below determine V3 in V 1 3 4j V 2 4 3j V 3 1 j V 4 3 2j V 5 jV 6 1 5j V 7 cannot be determined 13 12 The circuit below is in steady state The phasor voltage and current at 10 rad sec are as shown graphically The values of R and L are respectively 1 2 5 4 33 H 2 5 0 433 H 3 5 4 33 H 4 2 5 0 433 H 5 2 5 0 0433 H 6 5 2 5 H 7 2 5 2 5 H 14 13 For the circuit below determine the equivalent admittance Yin j in S assuming 4 rad sec Yin j 0 05 F 0 25 1 0 25 j20 4 4 j0 2 7 none of the above 2 0 25 j5 5 4 j5 15 3 0 25 j20 6 4 j0 2 14 zero Determine the value of C for which the input current in the circuit below would become 1 1 mF 5 5 mF 2 2 mF 6 0 F 3 3 mF 7 infinite 16 4 4 mF SOLUTIONS 1 4 2 7 3 3 4 2 5 3 6 6 7 4 8 5 9 5 10 5 11 3 12 4 13 4 14 1 17
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