ECEELECTRICAL and COMPUTER ENGINEERING DEPARTMENTUniversity of Utah

              ECE 1270 - Introduction to Electrical and Computer Engineering


GRADES

Course Info      Study Guides      Homework      Communications Assignments      Labs      Matlab      Practice Exams      


Instructor                            ECE
Dr. Angela Rasmussen
Email: ece1270@comcast.net
Office: MEB 3254

Class Time:  EMCB 102, T 9:35am-11:30am and W 9:35am-10:30am

Syllabus and Schedule

Course Procedure | (pdf) (grading, notebooks, reports, etc.)


Learning Objectives/Study Guides                            ECE


Homework                            ECE

Homework

Modifications

Answers:

Due

 

 

 

 

Problem Set #1    Solution

 

1.a.  i1=-1.5A, i2=-0.5A, Vo=-20V, b.  p_2A=-130W, p_10ohm=40W, p_30ohm=67.5W, p_40ohm=10W, p_50ohm=12.5W.  2.  i1=-2A, i2=-1A, Vo=-10V.  3.  vx=-500V, i1=30mA, power=-30W.

T: May 22,2007

Problem Set #2    Solution

Take v1 to be across resistor R1 for #3. 

1.  V1=60V. 2.  i1=150microA.

3.  V1=(iaR3-va)*(R1/(R1+R2+R3)).

4.  V1=-Va*R1/(R1+(1+alpha)R2) 5.  vo=vs+isR2

W: May 23, 2007

Problem Set #3    Solution

#4 Hint:  remember to replace Vx with an equation containing your mesh current variables.

1.  Vo=150/67V, i1=50/67A;  2.  Vx=(-6/44)V,    Vy=-30V, power=36/85,184W; 3.  V1=2V, V2=5V, iL=1mA; 4.  i1=(21/8k)A, power=-21mW

W: June 6, 2007

Problem Set #4    Solution

#4 Change the dependent current source into a dependent voltage source of value aix.

4.  Rth=R1+R2+alpha; Vth=isR1; 5.  power=-27mW;

W: June 6, 2007

Problem Set #5    Solution

 

1. R2=4k, R3=2k; 2.  i(t)=e^(-9tsec); 3.  vo(t)=24+36e^(-t/12); 4.  vo(t)=27-9e^(-t/(3/8))

W: 6/27/2007

Problem Set #6    Solution

 

1.a. wL=5.625nJ, b. i(t)=3.75e-3e^(-t/250n); 2.a. Vc(t)=-Vs+Vs(R2/(R1+R2))*[1-e^(-t/(R1||R2+R3)C]; 3.RL=1.8k, power=293.9mW; 4.i=is/(1-alphaR2) 

W: 6/27/2007

Problem Set #7    Solution

 

1.  Zab=(3.8+0.6j)ohms; 2.  omega=447 rad/sec.; 3.  I=4e^(-j39); 4.  i_x(t)=1.92cos(4t+14)A

W: 7/18/2007

Problem Set #8    Solution

Note that the magnitude of a complex number is taken without a sign. (e.g. take the absolute value of the vector length).  (i.e.  It is the vector length not the direction of the vector)

1.  a.  .352 –j0.936; b.  2e^(-j30); c.  3e^j150 or 3e^-j210; d.  10; e.  2;  2.  a.  R=30k or L=1.2H (this is a huge value – not real practical); 3.  R=30k => Io=3sqrt(2)mA; L=1.2H => Io=12sqrt(2)mA.  4.  Vth=5e^j0 V, Z_Th=1.25k-j3k 

W:  7/18/2007

Problem Set #9    Solution

 

1.  a.  vo=-Vs1(R2/R1)+Vs2(1+R2/R1); 2. a.R2=300kohm, b.Vo=vsum – Vdelta(1+2R2/R1), c.Rin=R3; 3.  a.C=0.1microF; 5.a.  Vth=12, Rth=2.4k, b.  RL=2.4k, c. Pmax=15mW; 6. a.  I1=1/2(1+j)milliamp, b. i(t)=sqrt(2)/2cos(20kt+45)milliamp

W:  8/1/2007


Communications Assignments                            ECE

·  CLEAR Center:

·  Writing Assignments:

 

·  Writing Center Help:

  • Location: 1st floor of Marriott Library.
  • Phone # to make reservations: 587-9122

·  Oral Presentation:

  • Oral presentation order (handed out in class only)

Labs                            ECE


Matlab and Labview                            ECE


Practice Exams/Solutions                            ECE