Faculty of Engineering
ELECTRICAL AND ELECTRONIC ENGINEERING DEPARTMENT
EEE 223 Circuit Theory I
Instructor:
M. K. Uyguroğlu
Midterm EXAMINATION
November 25, 2004
Duration : 90 minutes
Number of Problems: 6
Good Luck
STUDENT’S
NUMBER
NAME
SURNAME
Problem Points
1 20
2 15
3 20
4 15
5 15
6 15
TOTAL 100
EEE 223 Circuit Theory I – Midterm Examination
1) The voltage and current at the terminals of the circuit element are shown in Fig. P1.
a) Sketch the power versus t plot for 0 ≤ t ≤ 10 s.
b) Calculate the energy delivered to the circuit element at t = 1, 6, and 10 s.
i
+ v -
i(A)
20
20t 0 < t <1
20 A 1< t < 3
i (t ) = −20t + 80 3<t <5
20t − 120 5<t <7
1 2 3 4 5 6 7 8 9 10 t(s)
20 A t >7
5V 0 < t <1
-20 0 1< t < 3
v(V) v(t ) = −5V 3 < t < 5
5 5V 5<t <7
0 t >7
100t 0 < t <1
1 2 3 4 5 6 7 8 9 10 t(s) 0 1< t < 3
p (t ) = 100t − 400 3<t <5
100t − 600 5<t <7
-5 0 t >7
p(W)
t
100
w(t ) = ∫ pdt
0
w(1s ) = 50 J
w(6s ) = 0 J
w(10 s ) = 50 J
1 2 3 4 5 6 7 8 9 10 t(s)
-100
M. K. Uyguroğlu November 25, 2004
EEE 223 Circuit Theory I – Midterm Examination
2. The current in the 9 Ω resistor in the circuit in Fig. P2 is 1 A, as shown.
a) Find vg
b) Find the power dissipated in the 20 Ω resistor.
1.2 A 20 Ω
+ 24 V-
8A 10 Ω 4.8 A 5Ω 4A 4Ω
+ 80 V- 2A + 24 V- 1A + 16 V- 3A
+ 1A + 1A
40 Ω
vg = 144 V 32 Ω 64 V + 25 Ω 25 V 9 Ω +
2Ω
40V - 9V
-
- -
3Ω 1Ω 9V
- 15 V +
5A
Figure P2
b) p20 Ω = 1.2 × 24 = 28.8W
M. K. Uyguroğlu November 25, 2004
EEE 223 Circuit Theory I – Midterm Examination
3. Use the nodal analysis to find v in the circuit in Fig. P3.
supernode
ix
12 V 1Ω v+ix = v+v1/7.5 v 2.5 Ω v1
+ ix
12 V 2.5 Ω 10 Ω v 7.5 Ω 4.8 A
Figure P3
KCL at v1 :
1 1 1
+ v1 − v = 4.8 multiply both sides of the equation by 7.5 yields:
2.5 7.5 2.5
4v1 − 3v = 36 (1.1)
KCL at the supernode:
1 v1 1 1 1
1 + v + − 12 + + v − v1 = 0 multiply both sides of the equation by
2.5 7.5 10 2.5 2.5
10 yields:
v
14 v + 1 − 120 + 5v − 4v1 = 0
7.5
(1.2)
16
19v − v1 = 120
7.5
By using Eqns. (1.1) and (1.2) v can be obtained as:
v =8V (1.3)
M. K. Uyguroğlu November 25, 2004
EEE 223 Circuit Theory I – Midterm Examination
4. Use mesh analysis to find the power dissipated in the 1 Ω resistor in the circuit in Fig. P4.
2Ω
Supermesh
2A i1
2Ω
10 V 2Ω 6V
i3
i2 = i1 + 2
1Ω
Figure P4
KVL around i3 :
( 2 + 2 + 1) i3 − 2(i1 + 2) − 2i1 = 6
(1.4)
5i3 − 4i1 = 10
KVL around the supermesh:
( 2 + 2 ) i1 − 2i3 + ( 2 )( i1 + 2 ) − 2i3 = 10
(1.5)
−4i3 + 6i1 = 6
By using Eqns. (1.4) and (1.5) i3 can be obtained as:
i3 = 6 A (1.6)
Then the power dissipated by 1 Ω resistor is:
p1Ω = ( i3 ) .1 = 36 W
2
(1.7)
M. K. Uyguroğlu November 25, 2004
EEE 223 Circuit Theory I – Midterm Examination
5. Find the Thevenin equivalent circuit with respect to the terminals a, b for the circuit in
Fig. P5.
3 ix
Figure P5
KCL at voc :
2Ω voc 1 1 1 voc
24 V + voc − .24 = −4 − 3ix = −4 − 3
a 2 8 2 8
ix voc = 8V = Vth
24 V 4A 8Ω
3 ix = 0
isc = 12 − 4 = 8 A
Therefore
12 A 2Ω voc 8
a Rth = = = 1Ω
isc 8
+24 V - ix = 0 isc
24 V 4A 8Ω
Rth = 1 Ω
a
Vth = 8 V
M. K. Uyguroğlu November 25, 2004
EEE 223 Circuit Theory I – Midterm Examination
6. Find v0 in the op amp circuit in Fig.P6.
12 kΩ 144 kΩ
0.5 V
18 kΩ v1 0A
0.3 V
0A
9 kΩ +
0.6 V 62 kΩ
v0
v1
24 kΩ _
0.8 V
72 kΩ
Figure P6
KCL at the non-inverting terminal:
1 1 1 0.6 0.8
+ + v1 − − =0
9 24 72 9 24
multiply both sides by 72 yields: (1.10)
12v1 = 7.2
v1 = 0.6 V
KCL at the inverting terminal:
1 1 1 0.5 0.3 v0
+ + v1 − − − =0
(1.11)
12 18 144 12 18 144
v0 = 21(0.6) − 6 − 2.4 = 4.2 V
M. K. Uyguroğlu November 25, 2004
