KING FAHD UNIVERSITY OF PETROLEUM & MINERALS
ELECTRICAL ENGINEERING DEPARTMENT
EE 202
EXAM I
DATE: Wednesday 3/10/2012
TIME: 6:00 PM-7:30 PM
ID#
Name
KEY
Section#
Maximum
Score
Q1
Q2
Q3
TOTAL
30
35
35
30
Score
Problem #1 (30)
(a)
The equivalent resistance
R eq
between the two terminals A and B is
2
2
A
R eq 
2
2
B
(I) 2 
(II) 2.8 
(III) 1 
(IV) 2.5 
(V) 1.2 
(VI) 3 
(VII) 1.5 
(VIII) 3.3 
(b)
In the circuit shown below , the value of

v1
Vx
is
5

Vx
-
+
-
10 V
0.4 v 1
(I) 10 V
(II) - 10 V
(III) 4 V
(V) - 6 V
(VI) 0 V
(VII) 6 V
(IV) - 4 V
(VIII) 5 V
is
(c) In the circuit shown below , the value of
and the power absorb by
R 1 are
-3 A
3A
R2
is
R3
+
-
10 V
R1
(I) 6 A , - 30 W
(II) - 6 A , - 30 W
(III) 6 A , 30 W
(IV) - 6 A , 30 W
(V) 3 A , - 30 W
(VI) - 6 A , - 60 W
(VII) - 3 A , 60 W
(VIII) 3 A , 60 W
(d) In the circuit shown below , the value of
v
2
is
1
1A
10 V
+
-
v
1
2
1
+
-
5V
(I) 2 V
(II) - 2 V
(III) - 5 V
(IV) 5 V
(V) - 3 V
(VI) 3 V
(VII) 0.5 V
(VIII) - 0.5 V
(e) In the circuit shown below , the value of
2
i
is
4
7
6A
i
3
24
A
5
24
(V) A
5
(I)
(II) -
1
6
A
5
(VI) - 8 A
6
A
5
(IV) 2 A
(VII) 8 A
(VIII) - 2 A
(III)
Problem #2 (35)
10 i x A
I3
R
R
 Vx -
I1
R
I2
3 Vx V
+
-
110 V
ix
+
-
R
For the circuit shown above :
(a) If R = 1  write the mesh currents equations necessary to solve for the mesh
currents
I,I ,I
1
2
and put the equations in a matrix form (DO NOT SOLVE THE
3
EQUATIONS)
(a) Solution
R= 1 
KVL on Mesh 1
1I 1 + 1(I 1 - I 3 ) + 110 + 1I 1 = 0
 3I 1 - I 3  - 110
KVL on Mesh 2
1(I 2 - I 3 ) + 3v x - 110 = 0
Since v x  1(I 2 - I 3 )
 4I 2 - 4I 3  110
Mesh 3
10i x  I 3
Since i x  -I 1
 10I 1  I 3  0
 3 0 -1  I 1   -110 
  0 4 -4   I 2    110 

  

10 0 1   I 3   0 
If R = 5 in the previous circuit the mesh currents are found to be
(b)
I1  -1.7 A I 2 = 22.4 A
I 3 16.9 A
Find the power absorbed or delivered ( indicate it) by the voltage sources ?
(b) Solution
R= 5 
P
110

110  I 1 - I 2  (3)
OR
P
110
(deliver)  110  I 2 - I 1 
 110  -1.7 - 22.4   110  -1.7 - 22.4 
= - 2651 W ( Absorbing ) (
OR 2651 W ( Delivering )
P
3v x

3v
x
I 2 
OR
P
3v x
(deliver) 
 3(5)  I 2 - I 3  I 2   15  22.4 - 16.9  22.4 
 1848 W (Absorbing)
OR - 1848 W ( Delivering )
- 3v x  I 
2
Problem #3 (35)
3
 Vx
Supper node 1
20 V
v1
2
I
 -
6
v2
Supper node 2
-
3 Vx V
v3
 -
4
10 A
v4
R
For the circuit shown above
(I)
If R = 1  write the node voltage equations necessary to solve for the node
voltages v , v , v , v and put the equations in a matrix form (DO
1
2
3
4
NOT SOLVE THE EQUATIONS)
(I) Solution
R= 1 
KCL on Supper Node 1
v 1 v 1 -v 4
v 2 -v 3
+
- 10 +
=0
2
3
6
 5v 1 +v 2 - v 3 - 2v 4  60
KCL on Supper Node 2
v 3 -v 2 v 3 v 4 v 4 -v 1
+ + +
=0
6
4 1
3
 -4v 1 - 2v 2  5v 3  16v 4  0
Since v 3 - v 4  3V x  3 v 1 - v 4 
 3v 1 - v 3 - 2v 4  0
Since v 1 - v 2  20
 5 1 -1 -2  v 1   60 
 -4 -2 5 16  v 2   0 
    
 
 3 0 -1 -2  v 3   0 
 1 -1 0 0  v   20 

 4  
(II) If If R = 12  in the previous circuit and the nodes voltages now are :
v 1  26.67 V , v 2 = 6.67 V , v 3=
- 32 V
, v 4  56 V
Find the power absorbed or delivered (indicate it) by the independent voltage
source P
?
20 V
(II) Solution
R= 12 
P  20I
20
(3)
KCL on node v1 we have

P (deliver) 
OR
20
- 20I
v 1 v 1 -v 4
+
+I =0
2
3
26.67
26.67 - 56
+
+I =0
2
3
 I = - 3.56 A
P
20

20I
 20(-3.56) = - 71.167 W (Absorbing)
OR 71.167 W (Delivering)