ECE 201 Fall 2009 Final Exam

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1 ECE 01 Fall 009 Final Exam December 16, 009 Division 0101: Tan (11:30am) Division 001: Clark (7:30 am) Division 0301: Elliott (1:30 pm) Instructions 1. DO NOT START UNTIL TOLD TO DO SO.. 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. Formulas are given on the final page of this exam. 8. Cheating will not be tolerated. Cheating in this exam will result in an F in the course. 9. If you cannot solve a question, be sure to look at the other ones and come back to it if time permits. 10. 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 all outcomes. (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. You only need to satisfy the outcomes once during the course, so any outcomes that you satisfied previously will remain satisfied, independent of your performance on this exam. Course Outcome Exam Questions i 1-3, 5 ii 16 1 iii 4, 9, 18, 7 iv v vi 19- vii 3-6 viii 7 1 ix Minimum correct answers required to satisfy the course outcome 1

2 1. Determine the voltage V z. (1) 15V () 10V (3) 7.5V (4) 5V (5) 0 (6) 5V (7) 7.5V (8) 10V (9) 15V. For the circuit shown below, the current I is: (1) 0A () A (3) 5A (4) 6A (5) 6A (6) 10A (7) 10A

3 3. The voltage V x in the circuit shown below is: (1) 33V () 1V (3) 9V (4) 3V (5) 3V (6) 9V (7) 1V (8) 33V 4. The circuit shown below has two independent sources V 1 and I. The output voltage V out has been measured for two combinations of input sources: V 1 I V out V 5 A 1 V 4 V.5 A 1 V What would the value of V out be for V 1 = 1 V and I = 1 A? (1) 1 V () V (3) 3 V (4) 4 V (5) 5 V (6) 6 V (7) 0 V 3

4 5. The capacitor is uncharged at t = 0 (i.e. V c (0 + ) = 0). Determine V c at t = sec. (1) 10 V () 8.65 V (3) 5 V (4) 1.35 V (5) 0 (6) 1.35 V (7) 5 V (8) 8.65 V (9) 10 V 6. The switch in the RL circuit shown below has been open for a long time and closes at t=0 s. Find the inductor current i L (t) for t > 0. (1) ( e 10t ) A () ( + e 10t ) A (3) ( 4 4e 10t ) A (4) ( 4 4e 10t ) (5) ( 4 e 10t ) A (6) ( 4 e 10t ) + A (7) none of the above + A 4

5 7. Determine the rate of change in the inductor current at t = 0 + +, il '0 ( ) circuit below. di ( t) L = in the dt + t= 0 (1) 0 A/s () A/s (3) 3 A/s (4) 5 A/s (5) -5 A/s (6) -3 A/s (7) - A/s (8) -1 A/s 8. Given an input voltage of 4V, what is the output voltage V out of this circuit containing an ideal operational amplifier? (1) 0 () 5 V (3) 10 V (4) 0 V (5) 40 V (6) -0 V (7) -10 V (8) -5 V 5

6 9. Assume an ideal operational amplifier and that the capacitor is uncharged at t = 0. Find the correct expression for the output voltage V o (t) in volts for t > 0 where t is in seconds. (1) t () t + 3 (3) t + 1/4 (4) 4t + (5) 4t + 3 (6) t + 4 (7) t + 1/4 10. In the Op Amp circuit below, the Thévenin equivalent resistance R th at the inverting input is: (1) 3 kω () kω (3) 1 kω (4) 0.5 kω (5) 0 kω (6) 0.5 kω (7) 1 kω (8) kω (9) 3 kω 6

7 11. The current phasor I in the circuit below is: (1) j100 A () j10 A (3) j1 A (4) 0 A (5) j1 A (6) j10 A (7) j100 A 1. At ω = π rad/s, the phasor current through the element shown below is I = ( 45 )Arms. Determine the voltage v(t) across the element. (1) 0 () sin ( π t) V (3) cos( π t) V (4) sin ( π t 45 ) V (5) cos( π t 45 ) V (6) ( π+ ) (7) cos( π+ t 45 ) V sin t 45 V 7

8 13. The phasor current I for the node sketched below has a magnitude of: (1) 0 A () 1 A (3) 3 A (4) A (5) 5 A (6) 7 A (7) 1 A 14. The mesh current phasor I 1 in the circuit below is: (1) 1 0 A () 0 A (3) j1 A (4) j A (5) (1+j) A (6) (1+j) A (7) (+j) A (8) (+j) A 8

9 15. At ω = rad/s, the impedance Z TH (jω) = ( + jω) shown in the circuit below can be realized with: (1) a single 1Ω resistor () a single Ω resistor (3) a single 1Η inductor (4) a single H inductor (5) a 1Ω resistor in series with a 1Η inductor (6) a 1Ω resistor in series with a Η inductor (7) a Ω resistor in series with a 1Η inductor (8) a Ω resistor in series with a Η inductor 9

10 16. The circuit below contains a floating voltage source. The voltage phasors at nodes A and B are V A and V B, respectively. The correct nodal equation involving the super node (containing both nodes A and B) is: (1) V A = 5 0 V () V B =10 0 V (3) (1+j)V A + jv B = 5 0 V (4) (1+j)V A + (1+j)V B = 15 0 V (5) (+j)v A + jv B = 5 0 V (6) (+j)v A + (+j)v B = 15 0 V (7) (1+j)V A + jv B = 5 0 V (8) (1+j)V A + (1+j)V B = 15 0 V 17. In the circuit shown below, let i s (t) =.5 cos(ωt) A, with the frequency ω variable. As ω is varied, we measure the amplitude of the voltage across the capacitor as shown in the plot to the right. Determine C. (1) 1 µf () µf (3) 5 µf (4) 10 µf (5) 0 µf (6) 50 µf (7) 100 µf (8) 00 µf 10

11 18. Find the Thévenin equivalent of the following network. The value of the source V th and impedance Z th are: (1) V th = 5V, Z th = 5Ω () V th = j5v, Z th = (5+j10)Ω (3) V th = 5V, Z th = 5Ω (4) V th = (5-j5)V, Z th = (5-j10)Ω (5) V th = j5v, Z th = 5jΩ (6) V th = 5V, Z th = (5+j5)Ω (7) V th = 5V, Z th = (5-j10)Ω (8) V th = 5V, Z th = 5jΩ 19. In the circuit below, the average power dissipated by the 1Ω resistor is 4W. If the magnitude of the source voltage phasor is increased to 3V, determine the average power dissipated by the 1Ω resistor. [Hint: Use linearity. You do not need to solve for X c!] (1) W () 4W (3) 5W (4) 6W (5) 9W (6) 1W (7) 15W 11

12 0. In the circuit shown below, the voltage across the resistor is v(t) = 4 + cos (πt) V. Find the effective (rms) value of the voltage. (1) V () V (3) 3V (4) 4V (5) 6V (6) V (7) 3V (8) 3 V 1. The current in the circuit shown below is i(t) = cos(ωt)+5cos(ωt) A. The average power absorbed by the inductor is: (1) 0 W () 0.5 W (3) 1W (4) 5W (5) 5 W (6) ( 1+ 5 ) W (7) cannot be determined 1

13 . The input current to the load Z load is i( t) = 5cos( 30 ) absorbed by the load? t A. What is the average power (1) 0 W () 5 W (3) 10 W (4) 0 W (5) 40 W (6) 80 W (7) 160 W (8) 30 W 3. A load (as shown) draws a complex power S = (40 + j30) VA. Determine the power factor of this load. (1) 1.0 leading () 0.8 leading (3) 0.6 leading (4) 0.4 leading (5) 1.0 lagging (6) 0.8 lagging (7) 0.6 lagging (8) 0.4 lagging 13

14 4. A load (as shown) draws a complex power S = (40 + j30) VA. Find the apparent power delivered to this load. (1) 50 VA () 30j VAR (3) 30 VAR (4) (40-30j) VA (5) 40 VA (6) ( j) -1 VAR 5. A load (as shown) draws a complex power S = (40 + j30) VA. Determine the value of C which, when placed in parallel with the original load, will minimize the current I. Use ω = 100 rad/s. (1) 30 µf () 100 µf (3) 300 µf (4) 1000 µf (5) 3000 µf (6) 10,000 µf (7) 30,000 µf 14

15 6. In the circuit shown below, Vs is a 60 Hz, 10 V rms generator. The load consumes 30 kw of power at a power factor of cos45 lagging. A capacitor is placed in parallel with the load to improve the power factor to cos30 lagging. It is known that without the capacitor, Q old Load = 30 kvar, and with the capacitor, Qnew Load, C = 17.3 kvar. What is the value of C? (1) 35 mf () 14.7 mf (3).3 mf (4) 0 mf (5).3 mf (6) 14.7 mf (7) 35 mf 7. Find the load impedance Z L that will absorb the maximum power from the source. (1) 5 (-1-j) Ω () 5 (1-j) Ω (3) 5 (1+j) Ω (4) 5 (-1+j) Ω (5) 50 (-1-j) Ω (6) 50 (1-j) Ω (7) 50 (1+j) Ω (8) 50 (-1+j) Ω 15

16 Useful formulas + + ( o ) x(t) = x( ) + x(t o ) x( ) e t t / τ τ = L/R τ = RC ( ) t x(t) = x( ) + Acosω dt + Bsinω dt e σ t x(t) = x( ) + ( A + Bt) e σ ( st 1 st) x(t) = x( ) + Ae + Be R / L σ= 1 RC (series) (parallel) s 1, = σ ± σ ω o b ± b 4c s 1,s = for s + bs + c = 0 4c b ω d = = ω σ o ω o = 1 LC 1 cos(x) cos(y) = cos(x + y) + cos(x y) [ ] 1 cos(x) sin(y) = sin(x + y) sin(x y) [ ] 1 sin(x) sin(y) = cos(x y) cos(x + y) sin(x) = cos( x 90 ) [ ] 16

ECE Spring 2017 Final Exam

ECE Spring 2017 Final Exam ECE 20100 Spring 2017 Final Exam May 2, 2017 Section (circle below) Qi (12:30) 0001 Tan (10:30) 0004 Hosseini (7:30) 0005 Cui (1:30) 0006 Jung (11:30) 0007 Lin (9:30) 0008 Peleato-Inarrea (2:30) 0009 Name