1 ECE 201 – Spring 2009 Exam #2 March 10, 2009 Division 0101: Elliott (9:30am) Division 0201: Capano (10:30 am) Division 0301: Jung (11:30 am) Division 0401: Capano (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). 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. Outcome i is a repeat. We use this outcome result only if you did not satisfy it previously. Course Outcome Exam Questions # of Questions Minimum correct answers required to satisfy course outcome i 2-6 5 3 iii 1, 2, 5, 7, 8 5 3 iv 7-13 7 3 viii 1 1 1 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 = RC2 1. The network shown delivers power to a load resistor RL. Determine the resistor RL that absorbs the maximum power from the source, and the power absorbed. RL(Ω) PL,opt(W) (1) 6 6 (2) 6 3 (3) 6 1.5 (4) 10 22.5 (5) 10 15 (6) 10 10 (7) 25 56.25 (8) 25 12.5 (9) 25 10 2. Determine the equivalent capacitance of the network shown. (1) 21.2 µF (2) 25.0 µF (3) 20.83 µF (4) 4.0 µF (5) 1.13 µF (6) 4.81 µF (7) 30.83 µF (8) 31.2 µF (9) 35.0 µF3 3. The 0.1A cos 1000t current source drives the single loop network shown. The initial capacitor voltage is zero (vc(0) = 0). Determine the voltage VL(t) across the inductor. (1) -4V sin 1000t (2) -3V sin 1000t (3) -2V sin 1000t (4) -1V sin 1000t (5) 0 (6) +1V sin 1000t (7) +2V sin 1000t (8) +3V sin 1000t (9) +4V sin 1000t 4. Find vout (in V): [Vout at t = 0 is zero.] (1) sin (ωt) (2) 2 sin (ωt) (3) 3 sin (ωt) (4) 4 sin (ωt) (5) 6 sin (ωt) (6) 36 sin (ωt) (7) 72 sin (ωt)4 5. Find the instantaneous energy stored in the 3H inductor at t = 0 s (in J): (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 7 6. A voltage source vs(t) = 10V cos 20t drives the capacitor as shown. Determine the voltage Vout(t). (1) +4V cos (20t) (2) +3V cos (20t) (3) +2V cos (20t) (4) +1V cos (20t) (5) 0 (6) -1V cos (20t) (7) -2V cos (20t) (8) -3V cos (20t) (9) -4V cos (20t)5 7. For the circuit shown below, find υR(t) for t ≥ 0. Assume iL(0) = 10mA. [Hint: Calculate iL(t) first.] (1) -0.1·e-t/0.01 [V] (2) -0.1·e-t/0.001 [V] (3) -0.2·e-t/0.01 [V] (4) -0.2·e-t/0.001 [V] (5) -0.3·e-t/0.01 [V] (6) -0.3·e-t/0.001 [V] (7) 0.3·e-t/0.01 [V] 8. Find the value for vL(0+) (in V): (1) 12 (2) 20 (3) 8 (4) 16 (5) 4 (6) 24 (7) 0 30Ω 60Ω 40 mH 20Ω υR(t) + -6 9. For the circuit shown below, assume the switch has been closed for a long time before it opens at t = 0. Find vc(0+). (1) 6V (2) -6V (3) 0 (4) 5.56V (5) -4V (6) 18V (7) 10.8V 10. In the circuit below, 0.4tcv (t) 20 10e V−= − for t ≥ 0. Also, 0.4tci (t) 0.4e−= for t ≥ 0. Find the value of R (in Ω): (1) 25 (2) 8 (3) 36 (4) 4 (5) 5 (6) 15 (7) 487 11. Find the inductor current for t ≥ 0 (in A): (1) 0.1t15 13e−− (2) 0.4t2 5e−− (3) 0.1t3 5e−+ (4) 2.5t2 5e−− (5) 10t5 3e−+ (6) 10t15 5e−− (7) 2.5t3 5e−− 12. For the circuit shown below, find zero input response (source free response) and zero state response for t ≥ 0. Zero input response (Source free response) Zero state response (1) υC(t) = 8·e-t/0.0016 + 16·(1-e-t/0.0016) (2) υC(t) = 8·e-t/0.0016 + -16·(1-e-t/0.0016) (3) υC(t) = -8·e-t/0.0016 + 16·(1-e-t/0.0016) (4) υC(t) = -8·e-t/0.0016 + -16·(1-e-t/0.0016) (5) υC(t) = 16·e-t/0.0016 + 8·(1-e-t/0.0016) (6) υC(t) = 16·e-t/0.0016 + -8·(1-e-t/0.0016) (7) υC(t) = -16·e-t/0.0016 + 8·(1-e-t/0.0016) (8) υC(t) = -16·e-t/0.0016 + -8·(1-e-t/0.0016) υin(t) = -10·u(-t)+20·u(t)8 13. The switch in the figure shown below is originally open at t=0. It closes when υC ≥ 8V and opens when υC ≤ 6V. Assume υC(0) = 0. The capacitor voltage is: ab0.5ta2.5(t t )abc0.5(t t )bc10 V (1 e ) 0 t t2V 6Ve t t tv (t)10V 4Ve t t t−−−−−− <<+ <<=− << as plotted to the right. Calculate batt− (1) ~0.062 s (2) ~0.162 s (3) ~0.262 s (4) ~0.362 s (5) ~0.462 s (6) ~0.562 s (7) ~0.662 s υC(t) 1 kΩ 10 V 0.5 mF 4 kΩ 8 V 6
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