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ECE 201 - Spring 2010 Final Exam, Exams of Electrical Circuit Analysis

Purdue UniversityElectrical Circuit Analysis

This is the final exam for the ece 201 course offered in the spring semester of 2010. The exam consists of 23 multiple choice questions and is closed book and closed notes. The exam covers topics such as circuit analysis, thevenin equivalents, capacitors, and power in ac circuits. Calculators are allowed, but not necessary. Cheating is not tolerated and will result in an f in the course.

Typology: Exams

2011/2012

Uploaded on 04/27/2012

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koofers-user-l5m 🇺🇸

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ECE 201 – Spring 2010
Final Exam
May 7, 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
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. Problem 1 is worth 5 points, Problem 2 is worth 6
points, and the rest of the problems are worth 9 points each.
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
Minimum correct answers required
to satisfy the course outcome
i
1, 2
1
ii
3
1
iii
4, 5
1
iv
6, 7, 8
1
v
11, 12, 13, 14, 16
2
vi
15, 17
1
vii
18, 19, 20, 21
2
viii
22, 23
1
ix
9, 10
1
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff

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ECE 201 – Spring 2010

Final Exam

May 7, 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

  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. Problem 1 is worth 5 points, Problem 2 is worth 6 points, and the rest of the problems are worth 9 points each.
  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.
  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 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

Minimum correct answers required to satisfy the course outcome i 1, 2 1 ii 3 1 iii 4, 5 1 iv 6, 7, 8 1 v 11, 12, 13, 14, 16 2 vi 15, 17 1 vii 18, 19, 20, 21 2 viii 22, 23 1 ix 9, 10 1

  1. (5pts) The current IR in the circuit below is:
(1) −10A (2) 10A (3) −100A (4) 100A
(5) −1A (6) 1A
  1. (6 pts) If the transconductance (gm) equals 0.002 S, find the current Is.

(1) 1A (2) 2mA (3) −2A (4) −4mA

(5) 8mA (6) 4A (7) 0

  1. (9 pts) The Thevenin equivalent of the circuit below is:

(1) a single 6Ω resistor

(2) a short circuit

(3) an open circuit

(4) a single 6 A independent source

(5) a single 6 V independent source

  1. (9 pts) In the circuit below, the switch has been at the top position for a long time. It is suddenly opened at t = − 4 sec. The voltage vc (t) across the capacitor for t ≥ − 4 sec is (in V):

(1) (^) 5e −100(t^ +4) (2)

(^1) (t 4) 10e^3

− + (3) (^) −5e −(t^ −4) (4)

(^1) (t 4) 10e^3

− −

(5) 5e −100(t^ −4) (6) −5e−3(t^ −4)

  1. (9 pts) Find R (in Ω) for the capacitor voltage v (^) c (t) for t > 0 to be critically damped.
  1. (9 pts) The circuit below has initial conditions, i (^) L(0−) = 8 A and v (^) C (0−) = 20 V. Find the

value of

dvc ( 0 )

dt

(in V/s).

  1. (9 pts) Determine the frequency ω, in rad/s, for which the input impedance Zin (jω) is purely resistive.
  1. (9 pts) Find V out.
(1) 50 ∠ 0 ° V (2) 50 ∠ 90 ° V
(3) 100 ∠ 0 ° V (4) 100 ∠ 90 ° V
(5) 150 ∠ 0 ° V (6) 150 ∠ 90 ° V
(7) 200 ∠ 0 ° V
  1. (9 pts) For the following circuit, determine the value V th and Z th of the Thevenin equivalent network.

V th Z th (1) (^60) ∠ 0 ° j (2) (^60) ∠ 0 ° j (3) 60 ∠ 90 ° j (4) (^90) ∠ 0 ° j (5) (^90) ∠ 0 ° j (6) (^90) ∠ 90 ° j (7) (^110) ∠ 0 ° j

  1. (9 pts) The circuit below is in sinusoidal steady state. The average power absorbed by the capacitor is (in mW):
  1. (9 pts) The current phasors in the circuit below are expressed in effective (i.e., root-mean- square) value. The value of the impedance Z is (in Ω):

(1) 6 – j3 (2) 6 + j3 (3) −10 + j20 (4) −10 – j

(5) 25 (6) j

  1. (9 pts) The load shown below consists of one resistor and one capacitor. Using the voltage and current waveforms shown below, compute the average power absorbed by the resistor in the load. Assume ω=2π rad/sec.

− (^40) 0.125 0.25 0.375 0.5 0.625 0.75 0.875 1 1.125 1.25 1.375 1.5 1.625 1.75 1.875 2

− 2

0

2

4

υ ( )t i t( )

t

(1) 2 W (2) 4 W
W
W

(5) 2 + j2 VA (6) 2 – j2 VA

(7) 4 + j2 VA (8) 4 – j2 VA

  1. (9 pts) A voltage source rated at 240V, 60Hz and 100kVA is used to operate two equally- rated, 40kW motors. Find the minimum power factor (pf) rating for the motors that can be operated by the available voltage source. Neglect line losses.
  1. (9 pts) Assuming the answer to problem 20 is pf = 0.5 lagging (i.e., inductive loading), find the reactive power (in kVAR) for each of the 40 kW motors.

(1) 80.0 (2) 25.0 (3) 36.71 (4) 43.

(5) 56.57 (6) 69.28 (7) 75.

  1. (9 pts) In the circuit below, v (t)s = 50 2 cos(2000t)V. Find C (in mF) such that the

maximum average power is absorbed by R (^) L, and find the maximum average power absorbed by R (^) L (in W).

(1) 1.25 mF, 125 W (2) 2.5 mF, 500 W (3) 3.75 mF, 250 W (4) 1.25 mF, 500 W

(5) 2.5 mF, 125 W (6) 3.75 mF, 500 W (7) 2.5 mF, 250 W

  1. (9 pts) Find the load impedance (ZL) that will absorb maximum power from the source network in the dotted line. Assume ω=2× 106 rad/sec.

(1) (^) 1.68 + j1.44 Ω (2) (^) 1.68 – j1.44 Ω (3) (^) 1.92 + j1.44 Ω (4) (^) 1.92 – j1.44 Ω (5) (^) 1.92 + j1.68 Ω (6) (^) 1.92 – j1.68 Ω (7) (^) 1.92 + j1.92 Ω