Download EENG 223 Final Exam Spr09-10 soln

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 2009-10
10 June 2010
Duration: 120 minutes
Instructor: O. Kukrer
Solve all 6 Problems
STUDENT’S
NUMBER
NAME
SURNAME
GROUP NO.
Problem
Points
1
2
3
4
5
6
TOTAL
15
15
15
20
15
20
100
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
Q1. In the circuit of Figure 1, find the current io using mesh analysis (Other methods will not be
accepted). (15 pts)
io
8Ω
2Ω
Figure 1
6Ω
ix
4Ω
+
-
3ix
14 V
SOLUTION:
(a)
io
8Ω
i1
ix
4Ω
2Ω
6Ω
i2
14 V
+
-
3ix
i3
KVL eqn. for mesh 1:
16i1  2i2  6i3  0
(1)
6i2  2i1  14  0
(2)
KVL eqn. for mesh 2:
In meshes 2 and 3, i3  3ix
ix  i2
 i3  3i2
5
(1)  16i1  2i2  18i2  0  i1  i2
4
5
(2)  6i2  i2  14  i2  4 A
2
i1  5 A  i0  i1  5 A
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
Q2. In the circuit of Figure 2, use source transformation to find the voltage vx. (Other methods will
not be accepted) (15 pts)
4Ω
+ vx Figure 2
15 V
+
-
3vx
2Ω
SOLUTION:
Transform the dependent current source to a voltage source
4Ω
2Ω
+ vx 15 V
i
+
-
+
-
6vx
KVL for the loop:
15  6i  6v x  0
vx  4i  30i  15  i  0.5 A
 vx  2 V
OR a more difficult way:
Transform the independent
voltage source to a current
source. Define the voltage
across the dependent current
source as v1.
4/3 Ω
+
3.75 A
4Ω
2Ω
v1
i1
3vx
_
+
_
v1
_
4/3 Ω
4
 v1  5  i1  5  4vx
3
Also, from the original circuit
v1  15  vx
5V
+
i1  3vx
 vx  2 V
3vx
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
Q3. In the circuit in Figure 3, the op-amp is ideal.
(a) Find the output voltage v0. (7 pts)
(b) Find the current i0 (output current of the op-amp). (8 pts)
5 kΩ
Figure 3
i0
v0
2 mA
2 kΩ
SOLUTION:
5 kΩ
if
i=0 A
i0
v0
v-=0 V
iL
2 mA
(a)
i f  2 mA
(b)
KCL at the output node:
i0  i f  iL


2 kΩ
v0  5 k  i f  10 V
iL 
v0
 5 mA
2 k

i0  7 mA
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
Q4. The switch in the circuit in Figure 4 has been closed for a long time. It is opened at t = 0.
(a) Find the initial value of the capacitor voltage v(0+) just after the switch is opened. (5 pts)
(b) Find the capacitor voltage v(t) for t > 0. (10 pts)
(c) Determine the time instant t1 > 0 at which the capacitor voltage becomes equal to 10 V. (5 pts)
t=0
Figure 4
+
10 Ω
9A
8Ω
v(t)
0.25 F
SOLUTION:
(a)
For t < 0
i
By current division,
+
10 Ω
9A
8Ω
v(t)
i
10
9  5 A
10  8

(b)
v (0 )  8i  40 V
v(0 )  v(0 )  40 V
For t > 0
+
v(t )  V0 e t /
8Ω
v(t)
0.25 F
(c)

V0  40 V,   RC  8  0.25  2 s
 v(t )  40e0.5 t V
v(t1 )  40e 0.5t1  10  e 0.5t1  0.25  t1  2 ln(0.25)  2.7726 s
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
Q5. In the circuit in Figure 5,
v(t )  100 e  200 t V ,
i (t )  I 0 e  200 t A
(a) Find L and I0. (5 pts)
(b) Determine the time constant . (5 pts)
(c) Find the energy dissipated in the resistor
until
t  5 ms (i.e. in the time
interval 0  t  5 ms ).
for t  0 .
i
Figure 5
+
L
(5 pts)
v
5Ω
-
SOLUTION:
(a)
v(t )  R i(t )  100  5I 0  I 0  20 A
Also,
v (t )   L
di (t )
dt
since L does not satisfy the passive sign convention
 100   LI 0 ( 200)  L 
100
 25 mH
200  20
(b)

1
 5 ms
200
or

L 25 mH

 5 ms
R
5
( c) The energy dissipated in the resistor from t = 0 s to t = 5 ms is equal to the change in the energy
stored in the inductor between those instants.
1 2 1 2
Li1  Li0
i0  i (0)  20 A, i1  i (5 ms)  20 e 2000.005  20 e 1  7.3576 A
2
2
1
1
WL  L(202  7.35762 )   0.025  345.87  4.3233 J
2
2
WL 
OR the energy dissipated may be found by integrating the power
5ms
WR 
5ms
 R i (t ) dt  RI  e
2
0
2
0
0
400 t
 1  400t 5ms
  5(1  e 4000.005 )  4.3233 J
dt  2000  
 e
0
400


EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
Q6. In the circuit in Figure 6, u(t ) is the unit step function.
(a) Sketch graphs of the current sources i1 (t )  2[1  u(t )] A and i2 (t )  5u(t ) A. (4 pts)
(b) Find and sketch the capacitor voltage v0 (t ) for all time. (16 pts)
+ v 0(t)
Figure 6
0.1 F
10 Ω
2 [1  u(t) ] A
10 Ω
5 u(t) A
SOLUTION:
(a)
i1
i2
5A
2A
t
t
(b) For t < 0 under dc conditions:
_
+ v 0(0 )
10 Ω
2A
v0 (0 )  2  10  20 V
10 Ω
For t > 0
Under dc conditions (as t 
+ v 0(t)
Req
+ v 0()
0.1 F
10 Ω
10 Ω
5A
10 Ω
10 Ω
10 Ω
5A
10 Ω
v0 (0 )  v0 (0 )  20 V, v0 ()  50 V, Req  20     ReqC  20  0.1  2 s
v0 (t )  v0 ( )   v0 (0)  v0 ( )  e  t /  50  (20  50) e  t /2
 50  30 e
 t /2
v0 (V)
50
V
20
t