ECE 201 – Fall 2008Exam #2October 28, 2008Division 0101: Clark (7:30am)Division 0201: Elliott (10:30 pm)Division 0301: Capano (3:30 pm)Division 0401: Qi (4:30 pm)Instructions1. 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 inthis course has satisfied each of the course outcomes. On this exam, you have the opportunity tosatisfy outcomes i, iii, iv, and viii. (See the course syllabus for a complete description of eachoutcome.) On the chart below, we list the criteria we use for determining whether you have satisfiedthese course outcomes. Outcome i is a repeat. We use this outcome result only if you did not satisfy itpreviously.CourseOutcomeExamQuestionsTotal Points PossibleMinimum Pointsrequired to satisfycourse outcomei 6-9 28 14iii 1-4 28 14iv 9-14 42 21viii 5 7 7If you fail to satisfy any of the course outcomes, don’t panic. There will be more opportunities for youto do so.Potentially useful formulas are:( )ot t /ox(t) x( ) x(t ) x( ) e- - t+� �= � + - �� � = L/R = RCo1LCw =121. Suppose that two experiments are performed on the circuit to the right and the results are tabulated below. VsIsIoutExperiment 1 2 V 3 A 14 AExperiment 2 1 V -1 A 2 AThen in the third experiment, if we choose Vs=3 V and Is=1 A, the output Iout (in A) will be (1) 2 (2) 4 (3) (4) 8(5) 10 (6) 1(7) 2. The current I in the circuit to the right is (1) 1 A(2) A (3) A(4) 4 A(5) 5 A(6) 6 A (7) -5 A 33. Find the Thevenin equivalent resistance RTH for the network below. (1) 1 (2) 1.6 (3) 2 (4) 2.5 (5) 3 (7) 7 (8) 8 (9) 9 4. For the following network, determine the short-circuit current (i.e. the current iA under short circuit condition) at the terminals A-B.(1) 90 mA (2) 60 mA (3) 40 mA(4) 20 mA (5) 0 mA (6) -20 mA(7) -40 mA (8) 60 mA (9) -90 mA45. Find the maximum possible power transferred to the load resistor, RL. (1) 6 W (2) 9 W (3) 18 W (4) 36 W (5) 72 W144 W512 VI1RL4 2I16. The voltage vL(t) across an inductor ( L = 4mH) is shown in the plot below. The current at t = 0 is 0. Which plot best represents the current iL(t) through the inductor?67. In the circuit shown, the voltage source is vs(t) = 4 cos (t) V. Then vL(t) is(1) 3 cos (t) V (2) cos (t) V(3) 6 sin (t) V (4) 6 cos (t) V(5) sin (t) V (6) sin (t) V(7) 12 sin (t) V8. Find the equivalent inductance of the network shown.(1) 1H(2) 2 H(3) 3 H (4) 4 H(5) 8 H(6) 12 H(7) 16 H (8) 24 H7vL( t )+2 F0 . 2 5 H+_vs( t )ix3 ixvC( t )+_9. Find the time constant , in seconds for Vout(t).(1) 1sec(2) 2sec(3) 3sec(4) 4sec(5) 5sec(6) 6sec(7) 7sec(8) 8sec10. The following circuit consists of a switch and a step current source given by 2u(-t-1) A.Determine the current iL(t) (in A), for t > -1 second. (1) ( )( )2 12 1te u t- ++(2) ( )( )1122 1te u t-- -- -(3) ( )( )1144 1te u t-+- -(4) ( )( )4 1112te u t- --(5) ( )( )3148 1te u t+- -(6) ( )( )4 14 1te u t-+(7) ( )( )2 12 1te u t- - -- +(8) ( )( )112314te u t-- -- -8( )2 1u t- -4H5H1H11W8W( )Li t1sect =-1W11. For the circuit shown, the switch, which has been closed for a long time, opens at t = 0. Which expression yields the voltage v(t) valid for t > 0?(1) 5000t / s5V e-(2) 5000t / s5V e--(3) ( )5000t / s5V 1 e--(4) ( )5000t / s5V 1 e-- -(5) 5000t / s10V e-(6) 5000t / s10V e--(7) ( )5000t / s10V 1 e--(8) ( )5000t / s10V 1 e-- -12. For the following network, determine ( )50i 0+.(1) 138.6 mA (2) 100 mA(3) 120 mA (4) 37.5 mA(5) 54.5 mA (6) 200 mA913. In a first-order RC circuit excited by a voltage source, the capacitor voltage is represented as / 2( ) 4 6tcv t e-= + VFind the time (in s) required for the capacitor voltage to change from 8 V to 5 V.(Hint: Recall that an a nb nb- =l l l.)(1) 2 ln(3/6)(2) 2 ln(4)(3) 4 ln(4)(4) 6 ln(13/4)(5) 2 ln(8/5)14 The switch changes from position “1” to “2” at time t = 0. The angular frequency, , for vc(t) for t > 0 is: (1) 116 rad/s (2) 18 rad/s (3) 14 rad/s(4) 12 rad/s (5) 1 rad/s (6) 2 rad/s(7) 4 rad/s (8) 8 rad/s (9) 16 rad/s 10122 V2 F2 F4
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