# KING FAHD UNIVERSITY OF PETROLEUM &amp; MINERALS ELECTRICAL ENGINEERING DEPARTMENT

```KING FAHD UNIVERSITY OF PETROLEUM &amp; MINERALS
ELECTRICAL ENGINEERING DEPARTMENT
FIRST SEMESTER 2009-2010 (091)
Course Title:
Fundamentals of Electric Circuits
Course Number:
EE 204
Exam Type:
MAJOR EXAM II
Date:
December 23, 2009
Time:
07:00 pm – 8:30 pm (1 &amp; 1/2 hours)
Indicate your lab section number from the list below (lab section/instructor name/lab starting time/day)
51/Salim/2:10/S
52/Al-Saihati/8:00/U
53/Salhab/11:00/U
54/Al-Saihati/11:0/U
55/Alfaouri/2:10/U
56/Ali/2:10/U
57/Mahmoud/2:10/M
58/Akber/2:10/M
59/Al-Saihati/8:00/T
60/Salhab/11:00/T
61/Ismail/11:00/T
62/Alfaouri/2:10/T
63/Akber/2:10/T
64/Ghani/2:10/W
Student Name:
________________________________
Student ID:
________________________________
Section:
________________________________
Serial Number:
________________________________
Question 1:
16
Question 2:
20
Question 3:
24
Total:
60
Be neat, organized, and show all your work and results.
Question 1:
For the given ‘is’,
a) Find the equations for 'VC' for the following intervals.
(i) 0 &lt; t ≤ 1 (ii) 1 &lt; t &lt; ∞
b) Plot the result in part a.
c) Plot the voltage 'VR'.
Solution:
(a) Find the equivalent capacitance:
Ceq = [4 in series with 4] in parallel with 2 = [(4*4)/(4+4)]+2 = 2+2 = 4 F.
(i) 0 &lt; t ≤ 1
1
VC 
C eq
t
t
1
1 2 t 1 2
i
(

)
d



d


( )0  t
s

4 0
8
8

(ii) 1 &lt; t &lt; ∞
1
VC 
C eq
t
1
1
1
i
(

)
d


i
(

)
d


V
(1)

s
C
 s
4 0
8

(b) and (c)
Question 2:
Using only the superposition principle, determine the
following:
a) Thevenin's equivalent circuit that is connected to
the resistor RL.
b) RL that absorbs maximum power.
c) The maximum power absorbed by RL.
(a) We need to find RTh and VOC
Rth: Deactivate the two sources
RTh  [20 // 60]  5 
1200
 5  15  5  20 
80
VOC: use superposition
The voltage source circuit
48 3
  0.6 A
80 5
V OC 1  60I  60(0.6)  36 V
I 
Or using voltage division rule
V OC 1 
60
3
48  48  36 V
80
4
The current source circuit
V OC 2  6(20)  120 V
V OC V OC 1 V OC 2  36  120  84 V
Draw the equivalent circuit as shown
(b) RL  RTh  20 
(c) (2 marks for the power calculation)
Pmax 

V OC2
4RTh
(84) 2

4(20)
7056 441

 88.2 W
80
5
Question 3:
For the circuit shown below:
a) Write the three node equations for node A, B, and C and
put them in the standard system of equations.
b) Find the node voltages VA, VB, and VC.
c) What is the power dissipated in (absorbed by) R1.
d) Which method you prefer to use node voltage method or
mesh current method? Why?
Solution:
V A V B
 I1  I 2  0
a)
R1
V A V B  18  1 V A  1 V B  0 V C  18
(1)
V B V A V B V C V B  0


0
R1
R2
R3
V A  6V B  3V C  0
(2)
KVL for V1
0 VA  0 V B  1 V C  4
(3)
 1 1 0  VA  18
 1 6 3 V    0 

  B    (1 mark)
 0 0 1  VC   4 
b) Substitute from 3 into 2
V A  6V B  12
(4)
Solve (1) and (4)
V B  6V ,V A  24V
VA
24V
VB
6V
VC
4V
c) (2 marks)
V R 1 V A V B  18V
I R 1  I 1  I 2  3A
P  I R1  9*6  54W
2
R1
OR
P
V R21
 54W
R1
d) Mesh is better. As the current in two meshes are known by using KCL directly (I1, I2). We will have
only one equation in one unknown for the third mesh contains R2, R3, and V1. (2 marks)
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