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UC BERKELEY EECS 40, Spring 2007
EECS 40, Spring 2007 Prof. Chang-Hasnain Midterm #
September 17, 2007 Total Time Allotted: 50 minutes Total Points: 100
- This is a closed book exam. However, you are allowed to bring one page (8.5” x 11”), single-sided notes.
- No electronic devices, i.e. calculators, cell phones, computers, etc.
- SHOW all the steps on the exam. Answers without steps will be given only a small percentage of credits. Partial credits will be given if you have proper steps but no final answers.
- Draw BOXES around your final answers.
- Remember to put down units. Points will be taken off for answers without units.
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Score:
Problem 1 (25 pts)
Problem 2 (43 pts):
Problem 3 (32 pts)
Total
For Evaluation Only.
Copyright (c) by Foxit Software Company, 2004
UC BERKELEY EECS 40, Spring 2007
- (25 pts) Node-Voltage Analysis
All voltages of the voltage sources, the currents of the current sources and the values of the resistors are given.
IS1 VS2 IS
R 4
R 2 VS
R 1
R 3
V 2
V 1
a) (7 pts) Does the current source I (^) S1 have impact on voltages V 1 and V 2? Justify your answer. (Hint: You do not need to solve the rest of the problem to do this part!!)
Is1 does not impact V1 and V2 since it is in parallel with Vs3 and thus has the same voltage as Vs3. Also if one sets Is1 to 0 then it is clearly seen that it has no effect.
b) (12 pts) Write KCL equation for node 1.
Is3-(V1—Vs2)/R1-V1/R2-(V1-V3)/R4=
c) (6 pts) Consider the two terminals surrounding VS4 a super node. Write a KCL equation for the branch connecting R 4 , R 3 and V (^) S4. -(V3-V1)/R4-V2/R3=
V 1
V 2
V 3
Node 1
For Evaluation Only.
Copyright (c) by Foxit Software Company, 2004
UC BERKELEY EECS 40, Spring 2007
d) (8 points) Solve for VB and i 2. Hint: both should be integers. If they are not, go back and check your work. Combining b and c 2(i2)-6=8(i2)+ 6(i2)=- I2=-2 A Plugging this back into c) you get: vb=-10 V
e) (5 points) Determine the value of i (^) 3. Using Current Divider equation: I3=(i2)(12/(12+24)) I3=-2/3 A
For Evaluation Only.
Copyright (c) by Foxit Software Company, 2004
UC BERKELEY EECS 40, Spring 2007
- (32 pts) Thévenin and Norton Equivalent Circuits (a) (18 pts) Looking into terminals across a and b. What are the open circuit voltage V (^) ab, and the Thevenin Resistance R (^) th?
Let Vb=0, Vab=Va (25-Va)/5+3=Va/ 8=Va (1/5+1/20)=Va (1/4)
Va=Vab=32 V
Thevenin Resistance Voltage source is short, current source is open.
5 Ω parallel with 20 Ω= 4 Ohm 4 Ohm in series with 4 Ohm = Rth=8ohms
For Evaluation Only.
Copyright (c) by Foxit Software Company, 2004
