1 ECE 201 – Fall 2009 Exam #2 October 26, 2009 Division 0101: Tan (11:30am) Division 0201: Clark (7:30 am) Division 0301: Elliott (1:30 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, ii, 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. Course Outcome Exam Questions Total Points Possible Minimum Points required to satisfy course outcome i 6-8 21 14 iii 1-5 35 21 iv 9-14 42 21 viii 5 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 = RC2 1. Which of the circuits, shown as (1) to (7), is equivalent to the following?3 2. For the circuit below, find Isc. (1) 0.6 A (2) 0.75 A (3) 1 A (4) 1.2 A (5) 2.4 A (6) 6 A (7) none of the above 3. A variable load RL is attached to a linear network “A” as shown. V and I are as labeled. The I-V plot of network “A” is shown on the right. What are Rth and Voc for network A? (1) th ocR 1 , V 1V=Ω= (2) th ocR 2 , V 4V=Ω= (3) th ocR 1 , V 5V=Ω= (4) th ocR 2 , V 10V=Ω= (5) th ocR 1 , V 5V=Ω=− (6) th ocR 2 , V 10V=Ω=− (7) None of the above4 4. Find the Thévenin equivalent resistance as seen at terminal A-B. (1) 1 Ω (2) 2 Ω (3) 3 Ω (4) 4 Ω (5) 5 Ω (6) 6 Ω (7) none of the above 5. In the circuit shown, RL is variable. Determine the maximum power delivered to RL in the circuit below. (1) −1.5 W (2) 0 W (3) 1.5 W (4) 5 W (5) 7.5 W (6) 11.25 W (7) 25 W5 6. The inductor voltage vL for t ≥ 0 is as shown below. It is also known that iL(∞) = 0A. The initial condition iL(0+) is: (1) −0.5 A (2) 0.5 A (3) −1.5 A (4) 1.5 A (5) −2.5 A (6) 2.5 A (7) none of the above 7. Determine the equivalent inductance eqL. (1) L (2) 2 L (3) 13L (4) 14L (5) 10 L (6) 56L (7) 710L (8) 18 L6 8. Determine the equivalent capacitance eqC. (1) 1F (2) 10F− (3) 20F (4) 4F (5) 18F (6) 6F (7) 34F (8) 163F 9. In the circuit below, the initial output voltage at t = 0 is ( )outV0 0−= volts. If ( ) ( ) ( )inV t 50u t 30 30u t 50= −− − volts, then what is the final output voltage ( )outV ∞? (1) 0V (2) 30V− (3) 4V (4) 50V (5) 10V (6) 30V (7) 10V− (8) 50V−7 10. Determine the correct expression for the Kirchhoff Current Law (KCL) applied at node A. (1) ( )132() ()0++ =AAVt dVtit CR dt (2) ( )( )1 332()() 0−+ + − =AAVtdit C Vt VtR dt (3) ( ) ( )1 332()0−− =AVtdit C VtR dt (4) ( )132() ()0−+ − =AAVt dVtit CR dt (5) 31211() () () 0++ =AA AdVt Vt C VtR R dt (6) ( )( )0133100tti V t dtC+=∫ (7) 212()+ARVtRR (8) ( ) ( )1 332()() 0−+ + − =AAVtdit C Vt VtR dt 11. For the circuit shown, the source, which has been off for a long time, turns on at t = 0. Determine V(0+). (1) −30V (2) −20V (3) −10V (4) 0 (5) 10V (6) 20V (7) 30V8 12. In the circuit shown, the current source is initially off, turns on to 5 mA at t = 0, and turns off again at t = 1 ms. Determine the current iC(t) at t = 1.5 ms. (1) 0.55mA e 3.03mA−−=− (2) ( )1 0.55mA 1 e e 1.92mA−−−− =− (3) 1.55mA e 1.12mA−−=− (4) 0 (5) 1.55mA e 1.12mA−= (6) ()1 0.55mA 1 e e 1.92mA−−+− = (7) 0.55mA e 3.03mA−+= 13. For the circuit shown, determine the time constant τ. (1) 0.267 µs (2) 1.00 µs (3) 1.27 µs (4) 1.40 µs (5) 1.80 µs (6) 2.00 µs (7) 2.20 µs9 14. In the circuit shown, the current and voltage at t = 0 are i(0) = 200mA and v(0) = 2V. Which of the following plots best represents the voltage
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