EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
Faculty of Engineering
ELECTRICAL AND ELECTRONIC ENGINEERING DEPARTMENT
EENG223 Circuit Theory I
INFE221 – Electrical Circuits
FINAL EXAM
Spring 2008-09
23 June 2009
Duration: 120 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
25
105
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
1. In the circuit of Figure 1,
(a) Find the current io using nodal analysis. (10 pts)
(b) Find the current io using mesh analysis. (10 pts)
io
12 V
12 Ω
8Ω
+
Figure 1
5Ω
+
vx
5vx
6A
(a)
supernode
KCL for the supernode :
io
12 V
v1
5Ω
+
v1
 6  5v x  0 v x  v1  v1  1.25 V
5
v2  v1  12  13.25 V
12 Ω
8Ω
v2
+
vx
v3
5vx
6A
KCL for node 3 :
v3  v2 v3  v1
122v1  3v2

 0  v3 
 38.45 V
8
12
5
v v
i0  1 3  3.1 A
12
5v x 
(b)
io
KVL eqn. can be written only for mesh 1:
12 Ω
i1
12 V
12  12i1  8(i1  i3 )  0
+
8Ω
5Ω
+
vx
5vx
i2
6A
i3
(1)
i3  5v x  5( 5i2 )  25i2
i3  i2  6 A
 i2  0.25 A
Substitute in (1)
 i3  6.25 A
 i1  i0  3.1 A
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
2. In the circuit of Figure 2,
(a) Find R so that maximum power is transferred to the resistance R. (15 pts)
(b) Find this maximum power. (5 pts)
2A
Figure 2
4Ω
6Ω
1Ω
10 V
+
6Ω
R
+
8V
(a) Thevenin equivalent at the terminals of R must be found.
Voc:
By source transformation:
2A
4Ω

1Ω
10 V
+
Voc
+
6Ω
KVL: 10  2  4  8I 0  0 
RTh:
4Ω
2V
4Ω
6Ω
+
8V
1Ω
10 V
+
+
Voc
I 0  1 A  Voc  10  4 I 0  6 V
6Ω
1Ω
RTh
RTh  4 (1  6 6)  4 4  2
6Ω
 For maximum power transfer
R  RTh  2 
(b) Maximum power transferred: Pmax 
VTh2
 4.5 W
4 RTh
+
3Ω
I0
4V
+
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
3. In the circuit in Figure 3, the op-amps are ideal
(a) Find the voltage v0 when R f   (open circuit) . (10 pts)
(b) Find the voltage v0 when R f  12 k . (10 pts)
40 kΩ
Figure 3
2 kΩ
0.4 V
Rf
10 kΩ
v0
+
10 kΩ
(a) When R f  
40 kΩ
v01  20  0.4  8 V
2 kΩ
+
0.4 V
10 kΩ
v+2
v01
v0
10
v01  4 V
20
 v0  v2  4 V

v 2 
10 kΩ
(b) When R f  12 k , the first amplifier may be considered as a summing amplifier, with inputs which
are the 0.4 V and v0.
40 
10
 40
v01     0.4  v0   (8  v0 ) V
12 
3
 2
10
5
 v2  v01  (4  v0 ) V
20
3
8
 v0  v2 
v0  4 V  v0  1.5 V
3
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
4. The switch in the circuit in Figure 4 has been closed for a long time. It is opened at t = 0. Find
the capacitor voltage v(t) for t > 0.
t=0
300 Ω
Figure 4
+
50 V
+
v(t)
2i0
_
100 Ω
i0
0.1 F
At t = 0- the circuit is under dc conditions:
300 Ω
KVL for the loop:
3i0
50 V
+
+
-
2i0
_
100 Ω
i0
v(0 )
50  300(3i0 )  100i0  0
 i0  0.05 A  v(0 )  100i0  5 V
When the switch is opened at t = 0:
v(0 )  v(0 )  5 V
A
+
2i0
KCL at node A: iC  3i0
iC
Also, iC  C
100 Ω
v(t)
i0
dv
v
, i0 
dt
100
dv
3
dv

v 
 0.3v
dt
100
dt
 v(t )  v(0).e 0.3t  5e 0.3t V
 0.1
0.1 F
OR the solution could be found by determining the time constant as   ReqC , where Req is found from
the circuit
A
+
2i0
vt
100 Ω
it
i0
vt  100i0
 Req 
vt 100


it
3
KCL:
it  3i0
 
100
10
 0.1 
s  v(t )  v (0).e  t /  5e 3t /10 V
3
3
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
5.The switch in the circuit in Figure 5 has been closed for a long time. It is opened at t = 0.
(a) Find the current i(t) for t > 0. (20 pts)
(b) Find the total energy dissipated in the 120 Ω resistor after the switch is opened. (5 pts)
t=0
10 μF
Figure 5
200 Ω
i(t)
10 A
100 mH
120 Ω
(a) At t = 0- the circuit is under dc conditions:
+ v(0 )
10 A
200 Ω
i(0 )
120 Ω
i(0 )  10 A
v(0 )  0 V
When the switch is opened at t = 0, the resulting circuit is a source-free RLC circuit:
10 μF
At t = 0+
i(0 )  i(0 )  10 A .
+ v(t)
i(t)
100 mH
120 Ω
1
1

 103 rad./s
6
LC
0.1  10  10
R
120 


 600 / s,   < 0
2 L 2  0.1 H
0 
Therefore, the response is under-damped. The solution for the current has the form
i(t )  et  B1 cos(d t )  B2 sin(d t )  ;
d  02   2  800 rad./s
To find the unknown coefficients, the initial conditions are applied. First the initial condition for the time
derivative of current must be determined.
di
di 
1
 v (t )  Ri (t )  0 
(0 )    v (0 )  Ri (0  )   10.(0  1200)  12000 A/s
dt
dt
L
i(0)  B1  10 A
KVL:
L
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
di
   e  t   B1 cos(d t )  B2 sin(d t )   e  t  d B1 sin(d t )  d B2 cos(d t ) 
dt
di
12000  600( 10)
t0 
(0)   B1  d B2  B2 
 7.5 A
dt
800
i(t )  e600t  10cos(800t )  7.5sin(800t )  A
(b) The energy dissipated in the resistor is the total energy stored in the inductor and the capacitor at
t = 0+ ,
Wtotal 
1 2  1 2 
1
Li (0 )  Cv (0 )   0.1  102  5 J
2
2
2