ECE 201 – Spring 2008 Exam #2 March 6, 2008 Division 0101: Qi (7:30am) Division 0201: Elliott (12:30 pm) Division 0301: Capano (2:30 pm) Division 0401: Jung (3:30 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). Please clear any formulas, text, or other information from your calculator memory prior to the exam. 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, 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. Outcomes i and iii were covered on Exam #1. You only need to satisfy the outcomes once during the course Course Outcome Exam Questions Total Points Possible Minimum Points required to satisfy course outcome i 4,5, 7-9, 14 42 21 iii 1-3, 6, 7, 10, 12 49 28 iv 6-14 63 35 viii 3 7 7 If you fail to satisfy any of the course outcomes, don’t panic. There will be more opportunities for you to do so. Potentially useful formulas are: ()ott /ox(t) x( ) x(t ) x( ) e−− τ+⎡⎤=∞+ −∞⎣⎦ τ = L/R τ = RC o1LCω= 11. The Thevenin equivalent resistance, RTh, of the circuit below is (in ohms): (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 7 Reqvz−+2Ω1Ωvz 22. The Norton Equivalent circuit of the following is 33. In the circuit below, the load resistor RL is selected to maximize the power it absorbs. Assume that RL is chosen correctly. The maximum power extracted from the source by RL is (in watts): (1) 20 (2) 30 (3) 60 (4) 72 (5) 120 (6) 240 (7) 360 6 A10Ω5ΩRLLoad 4. For the network shown below, find the equivalent capacitance seen by R1. (1) 5 mF (2) 10 mF (3) 15 mF (4) 20 mF (5) 25 mF (6) 30 mF (7) 35 mF R1 15 mF15 mF5 mF5 mF20 mF 20 mF 45. The circuit below is equivalent to (1) a 1.33-H inductor in series with a 3-F capacitor (2) a 1.33-H inductor in parallel with a 3-F capacitor (3) a 3-H inductor in series with a 1.33-F capacitor (4) a 3-H inductor in parallel with a 1.33-F capacitor (5) a 3-H inductor in series with a 3-F capacitor (6) a 1.33-H inductor in series with a 1.33-F capacitor (7) a 3-H inductor in parallel with a 1.33-F capacitor 2 H2F2 H2 H2 F2 F 6. For the network shown below, find υC(t) for t > 0 given that υC(0) = 10V. 10 kΩ 30 kΩ 30 kΩ 30 kΩ υC(t) (1) (2-10e-t/0.2)V (2) (2-10e-t/0.4)V (3) (10-5e-t/0.2)V (4) (10-5e-t/0.4)V (5) 10e-t/0.2 V (6) 10e-t/0.4 V (7) 5e-t/0.2 V (8) 5e-t/0.4 V 57. For the circuit shown below, find iL(t) for t > 0 given that iL(0) = 1A. (Note: This initial current is due to some other source not shown in the circuit.) 10Ω 20Ω I = 6A 10Ω 10 mH (1) (3-2e-t/0.001)A (2) (3+2e-t/0.001)A (3) -2e-t/0.1A (4) (3-2e-t/0.1)A (5) (2-3e-t/0.005)A (6) (2+3e-t/0.005)A (7) (3-3e-t/0.005)A 8. In the following circuit, determine the current ix(0+) at t = 0+. (1) -200 mA (2) -120 mA (3) -80 mA (4) -40 mA (5) 0 (6) 40 mA (7) 80 mA (8) 120 mA 69. The switch, which has been open for a long time, closes at t = 0. Determine VC(∞), the voltage across the capacitor a long time after the switch closes. (1) 0 V (2) 1 V (3) 1.5 V (4) 2.0 V (5) 2.5 V (6) 3.0 V (7) 4.0 V (8) 5.0 V 10. Determine the time constant τ for the natural response of the circuit shown, for t > 0. (1) 100 μsec (2) 150 μsec (3) 200 μsec (4) 300 μsec (5) 500 μsec (6) 1 msec (7) 2 msec 711. The current flow through an inductor is t4i(t) 1 12e−=+. How long would it take for i(t) to drop from 7A to 4A? (Hint: Recall that an a nb nb−=AAA.) (1) An2 (2) 7n4A (3) 74n4A (4) 124n7A (5) An3 (6) 4An4 (7) 4An2 (8) 12n7A 812. vc(t) for the above circuit is: 0 ≤ t < 1 t ≥ 1 ()t22e V−−1) ()t1t32e e V−−−⎡⎤−−⎢⎥⎣⎦ If VS2 changes from 3u(t) V to 1.5u(t) V, what would vc(t) be? (Hint: use linearity!) 0 ≤ t < 1 t ≥ 1 ()t1e V−−1) ()t1t1.5 2e 0.5e V−−−⎡⎤−−⎢⎥⎣⎦ 2) ()t1.5 1.5e V−− ()t1t1.5 0.5e 0.5e V−−−⎡⎤−−⎢⎥⎣⎦3) ()t22e V−− ()t1t32e 0.5e V−−−⎡⎤−−⎢⎥⎣⎦ 4) ()t1e V−− ()t1t2e e V−−−⎡⎤−−⎢⎥⎣⎦ 5) ()t1e V−− ()t1t1.5 e e V−−−⎡⎤−−⎢⎥⎣⎦ 6) ()t22e V−− ()t1t2e e V−−−⎡⎤−−⎢⎥⎣⎦ 7) ()t22e V−− ()t1t1.5 e 0.5e V−−−⎡⎤−−⎢⎥⎣⎦ 8) ()t1e V−− ()t1t32e e V−−−⎡⎤−−⎢⎥⎣⎦ 9) ()t1e V−− ()t1t1.5 0.5e 0.5e V−−−⎡⎤−−⎢⎥⎣⎦ 913. At t = 0, i(0) = 5mA and v(0) = 0.5V. Determine the amplitude B, where , and rad/sec. () ()i(t) A cos t Bsin t=ω+ω51.667 10ω= × (1) -15mA (2) -10mA (3) -5mA (4) 0 (5) 5mA (6) 10mA (7) 15mA 14. Determine the total energy stored in the circuit elements at t = 10 μs for the circuit shown in the preceding problem. (1) 2.5 nJ (2) 22.5 nJ (3) 25.0 nJ (4) 50 nJ (5) 100 nJ (6) 225 nJ (7) 250 nJ
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