# King Fahd University of Petroleum &amp; Minerals Electrical Engineering Department

```King Fahd University
of Petroleum &amp; Minerals
Electrical Engineering Department
EE 204 Fundamentals of Electric Circuits
Second Semester (102)
Exam 2
Wednesday, 27 April 2011
6:30 pm – 8:00 pm
Name:
ID:
Section:
Instructors
Sections
Dr. Qureshi
Dr. Hammi
Dr. Wajih Abul Al-Saud
Dr. Abdulmalik Zidouri
1, 4, 6
2
3
5
7
8, 10
9, 11
Problem
Out of
Score
1
30
2
30
3
40
Total
100
Good Luck
Problem 1
Write ALL node-voltage equations in terms of THE NODE VOLTAGES ONLY that are
necessary to solve for the node voltages in the following circuit (DO NOT SOLVE THEM).
Write your equations in the box below.
v2
7
8
12 V
v3
v4
2
v1
9A
5
6
4
v5
10 V
3
v 1  10 V
v 2 v 3  12 V
v 2 v 1 v 2 v 4 v 3 v 4 v 3 v 5 v 3 v 1




9  0
8
7
5
4
2
v 4 v 2 v 4 v 5 v 4 v 3


9  0
7
6
5
v 5 v 4 v 5 v 3 v 5

 0
6
4
3
Problem 2
Using the MESH-CURRENT METHOD on the following circuit, find:
a) The mesh currents
b) The value of the current ix .
2
i1
5A
1
4V
i2
2
i3
ix
(a)
Mesh (2)
4+(1)(i 2  i1 )  (2)(i 2  i 3 )  0
  i1  3i 2  2i3  4 (1)
Current Restriction
i1  i3  5 (2)
Super Mesh (1) and (3)
2i1  4i3  (2)(i3  i 2 )  (1)(i1  i 2 )  0
 3i1  3i 2  6i3  0 (3)
 1 3 2   i1   4 
  1 0 1 i 2    5 
 3 3 6   i3   0 
 i1   4 
Solving i 2    2 
i3   1
Super Mesh (1), (2) and (3) can also be used
2i1  4i3  4  0
(b)
i x  i3  i 2  1  2  3 A
4
Problem 3
Given the following circuit:
a) Find the value of the resistor RL that will absorb maximum power from the circuit.
b) Using the NODE-VOLTAGE METHOD, find the maximum power that will be
absorbed by RL .
6
4
3
120 V
6A
2
RL
a) Let us find the resistance seen by RL , which is equal to the Thevenin resistance seen
by RL
6
4
3
2
RTh = [ (6 || 3) + 4 ] || 2 = 1.5 
For maximum power to be transferred to RL, Set
RL = RTh = 1.5 
-
b) Method 1: (Find vTh = vOC )
6
v1
v2
3
120 V
v 1  120 V
v 2 v 1 v 2 v 3 v 2

 0
6
4
3
v 3 v 2 v 3
 6  0
4
2
Solving the above equations gives
v 3  19 V
v Th  v 3  19 V
Pmax
v 
 Th
2
4R Th
19   60.167 W

4 1.5 
2
4
v3
6A
2
vTh
Method 2: (Find vL )
6
v1
v2
3
120 V
v 1  120 V
v 2 v 1 v 2 v 3 v 2

 0
6
4
3
v 3 v 2 v 3
v
 6 3  0
4
2
1.5
Solving the above equations gives
v 3  9.5 V
v L  v 3  9.5 V
Pmax
v 
 L
RL
2
 9.5

1.5
2
 60.167 W
4
v3
6A
2
vL
RL
```