ABDCABCBECE201 -- Exam 1 February 9, 2010 Division 0101: Prof. Capano (9:30am) Division 0201: Prof. Tan (10:30 am) Division 0301: Prof. Jung (7:30 am) Division 0401: Prof. Capano (11:30am) INSTRUCTIONS: ♦ DO NOT OPEN THIS EXAM UNTIL TOLD TO DO SO. ♦ There are fourteen (14) multiple choice problems worth 7 points each. ♦ Students who properly identify themselves with name and id number on scantron sheet will receive two extra points. ♦ To maximize our assessment of your knowledge and understanding, do NOT dwell on a single problem. If you get stuck, move on to the next problem and return later, time permitting. ♦ This is a closed book, closed notes exam. ♦ Calculators are allowed but not necessary. All students are expected to abide by the usual ethical standards of the university, i.e., your answers must reflect only your own knowledge and reasoning ability. Students caught cheating will receive a grade of ‘F’ for the course. Course Outcomes: 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, and iii. 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 1,2,4,5,7,9,11,12 56 28 ii 6,8,10,13 28 14 iii 3,14 14 7 If you fail to satisfy any of the course outcomes, don’t panic. There will be more opportunities for you to do so later on.1 1. A quantity of positive charge passing through the resistor from “B” to “A” is q(t) = 4t (in Coulombs) Find VAB (in V): (1) 10 (2) −10 (3) 20 (4) −20 (5) 30 (6) −30 (7) 40 (8) −40 2. Determine the unknown current, i (in A): (1) 1 (2) −2 (3) 3 (4) −4 (5) 5 (6) −6 (7) 7 6 A -5 A -4 A i A B 10 Ω2 3. There is a circuit that has only resistors in it. It has two input ports: one for a voltage input (VIN) and another one for a current input (IIN). The output VOUT is measured at two different sets of input conditions, and the results are shown below in the table. Using the two measurement results, find VOUT when VIN = 15V and IIN = −3A. VIN IIN VOUT 20 V 2 A 2 V 10 V 4 A 6 V 15 V −3 A ? (1) 2 V (2) 4 V (3) 6 V (4) 8 V (5) −2 V (6) −4 V (7) −6 V3 4. Find the voltage VBC (in V): (1) 12 (2) 26 (3) 39 (4) 48 (5) 0 (6) 6 (7) 24 5. In the circuit below, it is known that the voltage source supplies a current of 4A to the resistors. ALL resistors have the same resistance, and the power dissipated by the first resistor is 192 W. Find the power delivered by the voltage source (in Watts): (1) 1152 (2) 288 (3) 312 (4) 440 (5) 572 (6) 625 (7) 750 8 Ω 16Ω 72 V + 5 Ω 10 Ω _ Ix 3Ix A B C D _ + 192 W _ + 4 Α VS4 6. Find the current ix in the circuit below (in A): (1) 0.33 (2) −0.33 (3) 2 (4) −4.3 (5) −0.9 (6) 3.33 (7) 0.75 7. For the circuit shown below, find Vout (in V). (1) Vout = 50 ⋅ Vin (2) Vout = 100 ⋅ Vin (3) Vout = 150 ⋅ Vin (4) Vout = 200 ⋅ Vin (5) Vout = 250 ⋅ Vin (6) Vout = 300 ⋅ Vin (7) Vout = 350 ⋅ Vin 20 Ω 100 Ω 2 A 50 V ix 300 ix A B B C _ + + _ 20 Ω 100/9 Ω5 8. For the circuit shown below, find VAB. (1) 2.5 V (2) 5.0 V (3) 7.5 V (4) 2.0 V (5) 4.0 V (6) 6.0 V (7) 8.0 V 9. For the circuit shown below, find Va. (1) 20 V (2) 40 V (3) 60 V (4) 80 V (5) 100 V (6) 120 V (7) 140 V6 10. For the circuit shown below, find IX. (1) 0.5 A (2) 1.0 A (3) 1.5 A (4) 2.0 A (5) 2.5 A (6) 3.0 A (7) 3.5 A 11. Find the value of R3 in the circuit below, so that I2 = 6 A. (1) 1 Ω (2) 2 Ω (3) 3 Ω (4) 4 Ω (5) 5 Ω (6) 6 Ω (7) 7 Ω (8) 8 Ω7 12. Find the equivalent resistance Req as seen by the voltage source VS. (1) 1 Ω (2) 2 Ω (3) 3 Ω (4) 4 Ω (5) 5 Ω (6) 6 Ω (7) 7 Ω (8) 8 Ω 13. The loop equations for the circuit shown below are as follows: =SS21IVII102025 For R2 = 20 Ω, find the value of R1. (1) 1 Ω (2) 2 Ω (3) 3 Ω (4) 4 Ω (5) 5 Ω (6) 6 Ω (7) 7 Ω (8) 8 Ω8 14. We are trying to calculate 1i using superposition. First, we calculate (1)1i with only the voltage source active as shown in Figure (1). Second, we calculate (2)1i with only the current source active as shown in Figure (2). Find (1)1i, (2)1i, and 1i. (1) (1)1i=1A, (2)1i=0.6A, and 1i=1.6A (2) (1)1i=2A, (2)1i=1.2A, and 1i=3.2A (3) (1)1i=1A, (2)1i=0.6A, and 1i=0.4A (4) (1)1i=2A, (2)1i=1.2A, and 1i=0.8A (5) (1)1i=1A, (2)1i=0.6A, and 1i=-0.4A (6) (1)1i=2A, (2)1i=1.2A, and 1i=-0.8A (7) (1)1i=1A, (2)1i=1.2A, and 1i=2.2A Figure (1) Figure
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