University of Waterloo
Midterm Examination
Department of Electrical and Computer Engineering
Term: Spring
Year: 2010
Course Abbreviation and Number:
MTE 120
Course Title:
Circuits
Instructor:
Akrem El-Ghazal
Date of Exam:
Time Period:
Duration of Exam:
Number of Exam Pages:
June 16, 2010
Start time: 7:00 PM
2 hours
15
End time: 9:00 PM
Student Name:
UW Student ID Number:
Instructions:
1. Answer all FOUR questions.
2. The exam is a closed book examination.
3. All the questions carry equal marks.
4. Electronic calculators are allowed.
5. A formula sheet is attached at the end of the examination paper.
6. Clearly show all steps used in the solution process. No marks will be given for
numerical results unless accompanied by a correct solution method.
7. Use correct SI units in all your answers.
MARKING SCHEME:
Question
Q1
Q2
Q3
Q4
TOTAL
Mark
Marker
Circuits (MTE 120)
Question (1)
Midterm Exam (Solution)(Spring-2010)
2/11
(25 Marks)
Four charges are located at the corners of a square of side a=2 m as shown in the Figure
below.
If Q1= Q2= Q3=10 µC and Q4= -10 µC.
at the center of the square, at point (0, 0).
(a) Find the electric field (b) If a test charge Q=2 µC is placed at the square center, pinot (0, 0), find the electric
force on this test charge due the other four charges.
(c) If the four charges do not move, determine the potential difference VA- VB.
(d) What are the conditions for the magnitudes and the signs of Q1, Q2, Q3 and Q4 that
make the net electric field at the center of the square, at point (0, 0), equal to zero?
Circuits (MTE 120)
Midterm Exam (Solution)(Spring-2010)
Solution
(a)
The electric field due to cancels the electric field due to |E| | | | | | | | | | | | |
√1 1 √2
(b)
8.99 x10
|E| 10 10 10 89900
2
!
"# $45'
|F| |E|. q 89900 2x 10 0.1798 N
", $45'
(c)
3/11
Circuits (MTE 120)
Midterm Exam (Solution)(Spring-2010)
-
. .
The voltage due to cancels the voltage due to Then,
-/ . . . .
2x 8.99 x10 10 10 179800V
1
1
. . . .
-1 2
2
and
2 √1 4 √5
-1 8.99 x10 3
.10 10 .
.10 10 .
4 80409 √5
√5
-/ $ -1 179800 $ 80409 99391 (d) The conditions are
and 4/11
Circuits (MTE 120)
Midterm Exam (Solution)(Spring-2010)
Question (2)
5/11
(25 Marks)
A square wire loop with 4 m sides is perpendicular to a uniform magnetic field with half
the area of the loop in the field as shown below. The loop contains 20 V battery. If the
magnitude of the filed varies with time according to B=(0.042 -0.87 t ) [Tesla].
(a) What is the magnitude of the net emf in the circuit?
(b) What are the amount and the direction of the net current in the loop if the
resistance of the loop is 4 Ω?
Figure 2
Solution
Φ 78
dΦ
dB A
dB
|9:;|<=>?@A> B B B
BA B B
dt
dt
dt
A
B
4 x4
8 m
2
dB
B 0.870
dt
|9:;|<=>?@A> 8 x 0.870 6.96 Since the magnetite field is decreasing the induced B is coming out of the page and hence
the direction of induced current is counterclockwise and the direction of the |9:;|<=>?@A> is
as indicated below:
Circuits (MTE 120)
Midterm Exam (Solution)(Spring-2010)
|9:;|=AI 20 $ 6.96 13.03 13.03
3.26 8
4
(Clockwise)
J=AI 6/11
Circuits (MTE 120)
Midterm Exam (Solution)(Spring-2010)
7/11
Question (3)
(25 Marks)
For the circuit shown below:
(a) Use Kirchhoff’s law and Ohm’s law to find the voltage (Vx) across the
dependent source.
(b) Find the power supplied or absorbed by the 4 V voltage source.
4V
1Ω
1Ω
-
+
IB
VX
0.5 IB
+
2Ω
12 V
-
+
Figure 3
Solution
4V
1Ω
1Ω
A
-
+
IB
i
VX
+
0.5 IB
Loop#
2
2Ω
Loop#
1
+
-
12 V
Circuits (MTE 120)
KCL at A
Midterm Exam (Solution)(Spring-2010)
K J1 $ 0.5J1 0.5J1
KVL around loop#1
$12 1 J1 0.5J1 2 0
J1 6 8
K .5 6 38
KVL around loop#2
$K 2 4 1 0.5 J1 $ -L 0
-L 1
MN - .5 J1 4 .5 6 12 O PQRSQ9T
8/11
Circuits (MTE 120)
Midterm Exam (Solution)(Spring-2010)
9/11
Question (4)
(25 Marks)
For the circuit shown below, use the NODAL ANALYSIS to find the node voltages and the
power of the 4 Iy dependent voltage source.
Figure 4
Solution:
Node voltages
KCL at node #1
- $20 - … … 1
- 4 JV … … . 2
- $ - - 20 -
0
2
2
4
$2- 5- $40 … … 3
Describe the extra variable JV by using node voltages:
JV Substitute JV in equation 2
- $ - 2
- 4 - $ - 2
$ - 2- 0 … … 4
Circuits (MTE 120)
Midterm Exam (Solution)(Spring-2010)
Solving Eq.(3) and Eq.(4)
From Eq.(4)
- 2-
Substitute - in Eq.(43)
$22- 5- $40
- $40
Then
- $80
The power of the 4 Iy dependent voltage source
KCL at node #1
K JV $ 2-L 0
K $JV 2-L
JV - $ - $80 40
$20 8
2
2
-L - $ - $40 20 $20 K 20 2$20 $20 8
M W4JV X K 4 $20 $20 1600O
The source absorbs power of 1600W
10/11
Circuits (MTE 120)
Midterm Exam (Solution)(Spring-2010)
11/11
Formula Sheet
Constants:
The force between two parallel long wires
carrying currents:
[C]
e =-1.6 x 10
|K
|
!
.c .€ ]Z n  K‚
YZ 8.85 x 10 [
\
2_T
. :
]Z^ 4_ x10`
a
[
8
\
:
1
8.99 x 10 [ . \
b
4_YZ
Electric Charge
d d
[ \ !SefeS:g R ;Sb9 fPh
ca
Electric Field ( c
[ \
d !
For point charge:
a [ \
!
Electric Potential (V)
-a
[-\ i9S jSfkPl9 Pk ∞
Magnetic Field (7
Magnetic force on a moving particle:
[ \
c d jx 7
Magnetic force on current carrying wire:
x 7
[ \
c K n
Ampere’s Law
.ƒ 7
. ]Z K
. TR
Magnetic Flux
. T8
Φ „7
Faraday’s Law
emfvoltage $N
Circuit
Ohm’s Law
2
x r
]Z K TR
[p\
4_
For infinit long wire:
]Z K
[p\
7
2_
For a wire of length L:
]Z K
Q
Q$n
7
z{
$ {
} [p\
4_ √Q |Q $ n For circular arc:
.7 . ]Z K ~
4_ 2
[p\
dΦ
dt
[-\
J
Resistors in series:
2Aˆ 2 2 2 ‰ 2=
Resistors in parallel:
Š‹Œ
Š
Voltage-divider rule:
- Current-divider rule:
J J
Š{
Biot–Savart Law
.
T7 . [OQ\
ŠŽ
‰
2
2 2
2
2 2
Š