EENG223 Circuit Theory I / INFE221 Electric Circuits – Midterm Exam
Faculty of Engineering
ELECTRICAL AND ELECTRONIC ENGINEERING DEPARTMENT
EENG223 Circuit Theory I
INFE221 – Electrical Circuits
Midterm Exam
Spring 2009-10
20 April 2010
Duration: 90 minutes
Instructor: O. Kukrer
Solve all 5 Problems
STUDENT’S
NUMBER
NAME
SURNAME
GROUP NO.
Problem
Points
1
2
3
4
5
TOTAL
20
20
20
20
20
100
EENG223 Circuit Theory I / INFE221 Electric Circuits – Midterm Exam
1. In the circuit of Figure 1, find the current io using nodal analysis.
2Ω
4Ω
v1
v2
Figure 1
30 V
KCL eqn. for node 1:
KCL eqn. for node 2:
Solving (1) and (2)
+
_
io
3io
3Ω
2A
v1  30
v v
 3i0  1 2  0  3v1  v2  12i0  60
2
4
v2
i0 
 3v1  3v2  60 (1)
3
v2  v1 v2
  2  0   3v1  7v2  24 (2)
4
3
v
 v1  11.6 V v2  8.4 V
i0  2  2.8 A
3
EENG223 Circuit Theory I / INFE221 Electric Circuits – Midterm Exam
2. In the circuit of Figure 2,
(a) Find the voltage Vx using mesh analysis. (15 pts)
(b) Find the power supplied by the 4-A current source. (10 pts)
4Ω
Figure 2
10 Ω
C
supermesh
+ Vx 70 V
+
_
i2
i1
+
_
4Vx
4A
(a)
KVL eqn. for the supermesh:
70  4i1  10i2  4Vx  0
Vx  4i1
KCL at node C :
 20i1  10i2  70
(1)
i2  i1  4
(1)  30i1  40  70  i1  1 A
Vx  4i1  4 V
,
i2  5 A
(b) Voltage across the 4-A current source VC  70  4i1  66 V  P  4  VC  264 W
EENG223 Circuit Theory I / INFE221 Electric Circuits – Midterm Exam
3.
Use superposition to find the current i0 in the circuit in Figure 3.
i0
6Ω
Figure 3
2A
8Ω
2Ω
+
3i0
_
24 V
1. Apply the 2A-source, turn off the 24V-source:
i01
6Ω
Nodal analysis:
2A
v1
8Ω
v2
KCL eqn. for node 1:
2Ω
3i01
KCL eqn. for node 2:
i01 
(1) & (2)  v1  9 V
v1  v2
6
v1
v v
 2  1 2  0  4v1  v2  12 (1)
2
6
v2  v1
v
 2  3i01  2  0
6
8
 8v1  5v2  48 (2)
v2  24 V
i01 
9  24
 2.5 A
6
2. Apply the 24V -source, turn off the 2A -source:
i02
6Ω
KCL eqn. for node 2:
8Ω
v2
2Ω
i02  
3i02
+
_
v2
8
v2
v  24
 3i02  2
0
8
8
 v2  24 V  i02  3 A
24 V
 i0  i01  i02  0.5 A
EENG223 Circuit Theory I / INFE221 Electric Circuits – Midterm Exam
4. Use source transformation to find the voltage v0 in the circuit in Figure 4.
6v0
-
Figure 4
6Ω
+
3Ω
6Ω
+
v0
-
+
_
20V
v0
Transform the dependent voltage source (6v0) in
series with the 6Ω resistance to a current source:
6Ω
2Ω
3Ω
6Ω
+
v0
-
2v0
Transform the dependent current source (v0) in
parallel with the 2Ω resistance (equivalent of
6 3 ) to a voltage source:
-
6Ω
KVL for the loop :
8i  2v0  20  0
+
_
+
v0
-
20 V
2Ω
+
+
_
i
v0  6i  i  1 A  v0  6 V
20 V
EENG223 Circuit Theory I / INFE221 Electric Circuits – Midterm Exam
5. Obtain the Thevenin equivalent of the circuit in Figure 5 with respect to the terminals a-b.
5Ω
a
Figure 5
5ix
+ -
+
voc
_
iin
ix
4Ω
10 A
b
Since the current iin is zero (terminals a-b are open)
ix  10 A  voc  5ix  4ix  90 V
 VTh  90 V
To find RTh, turn off the 10A-source, then apply
a test source between a-b:
vt  9it  5ix  0  ix  it
a
5ix
+ -
it
 vt  14it
vt
 RTh  14 
4Ω
+
-
b
Thevenin equivalent circuit:
14 Ω
a
+
-
b
5Ω
90 V
ix