ECE 201 – Spring 2007 Exam #3 April 12, 2007 (7:00 – 8:00 pm) Division 0001: Prof. Elliott (11:30am) Division 0002: Prof. Tan (7:30 am) Division 0003: 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 Minimum Points required to satisfy course outcome iv 1-5 35 18 v 8-14 49 25 ix 6, 7 14 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.21. 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) 0.1 mH 0.4 mF + v c(t)_ (1) 12 A (2) 3 A (3) 0.25 A (4) 6 A (5) 4 A (6) 1.2 A (7) 0.3 A 32. 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 43. The RLC circuit below has a critically-damped response in the form of 4t 4tC1 2V(t) Ae Ate V−−=+ for t > 0. The value of R (in Ω) leading to this response is: (1) 0.125 (2) 0.25 (3) 0.5 (4) 4 (5) 2.5 (6) 8 (7) ∞ 54. Consider the circuit below with the initial conditions nd . The value of Li(0) 10A+= aCV(0) 10V+=Ct0dV (t)dt+= is (in V/s): (1) 10 (2) -10 (3) 0 (4) 20 (5) -20 (6) 5 (7) -5 65. 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: 15 Ω+ _ vc(t) 4 mF 25 Ω _ + 75u(t) V iL(t) 20 mH (1) 75 V (2) 0 V (3) 3 A (4) 4 A (5) 5 V (6) 6 A (7) 0 A 76. 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 (2) 20 (3) 30 (4) 40 (5) 50 (6) 60 (7) 0 87. 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) 2.5 mH inductor (2) 2.5 mH inductor in parallel with a 1 kΩ resistor (3) 1 kΩ resistor (4) 2.5 μF capacitor (5) 1 kΩ resistor in parallel with a 3V source (6) 2.5 μH inductor (7) none of the above 98. If vs(t) = 10 sin(2t) V, then is(t) equals (in A): (1) 5/√2 cos(2t + 36.9o) + _ 0.6 Ω 0.5 F 0.1 H vS(t) iS(t) (2) 10 cos(2t) (3) 5 sin(2t) (4) 5/√2 sin(2t) (5) 10 cos(2t – 36.9o) (6) 5 cos(2t + 45o) (7) none of the above 109. Given that ()()()( ) 5cos , ( ) 3sin , and ( ) 4sin 50== =−vt t vt t vt tωω ω12 3−°, the order in which these voltages lead one another, from one leading foremost to one lagging furthest behind, is: (1) (2) 123(), (), () vt vt vt132(), (), () vt vt vt(3) (4) 231(), (), () vt vt vt213(), (), () vt vt vt(5) (6) 312(), (), () vt vt vt321(), (), () vt vt vt(7) all in phase 1110. The voltage ()()( ) 5cos 10 5cos 10 120ovt t t=++ has a phasor representation in polar form given by: (1) 10 (2) 10 (3) 0o∠ 120o∠5260o∠ (4) 52 120o∠− (5) 56 (6) 5 (7) none of the above 0o∠ 120o∠− 1211. 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 1312. The circuit below is in steady state. The phasor voltage and current at ω = 10 rad/sec are as shown graphically. The values of R and L are, respectively, (1) 2.5Ω, 4.33 H (2) 5Ω, 0.433 H (3) 5Ω, 4.33 H (4) 2.5Ω, 0.433 H (5) 2.5Ω, 0.0433 H (6) 5Ω, 2.5 H (7) 2.5Ω, 2.5 H 1413. For the circuit below, determine the equivalent admittance, Yin(jω), (in S) assuming ω = 4 rad/sec. 0.05 F 0.25 Ω Yin (jω) (1) 0.25 – j20 (2) 0.25 + j5 (3) 0.25 + j20 (4) 4 + j0.2 (5) 4 + j5 (6) 4 - j0.2 (7) none of the above 1514. Determine the value of C for which the input current in the circuit below would become zero: (1) 1 mF (2) 2 mF (3) 3 mF (4) 4 mF (5) 5 mF (6) 0 F (7) infinite 1617SOLUTIONS: 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.
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