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Department of Electrical and Computer Engineering
E&CE 100 FUNDAMENTALS OF ELECTRICAL ENGINEERING
First Midterm Exam
June 2, 2006
Instructors: Ayman El-Hag, Hatem Zeineldin
Time Allowed: 1.5 Hours
Name:
I.D.#:
Program (Circle):
ELECTRICAL
COMPUTER
Instructions:
1. Answer all questions.
2. The exam is a closed book examination.
3. Electronic calculators are allowed.
4. A formula sheet is attached at the end of the examination paper.
5. Clearly show all steps used in the solution process. No marks will be given
for numerical results unless accompanied by a correct solution method.
6. Use correct SI units in all your answers.
Question
Marks
1
2
3
4
TOTAL
ECE 100 Fundamental of Electrical Engineering
2/10
1. If all the capacitors in the following figure are 6 F, find C eq .
June 2, 2006
[10 marks]
Solution:
Ceq’ = (C3 in series with C5) in parallel with C6 = 6x6/(6+6) +6 = 9 F
Ceq’’=C2 parallel with C4 =6+6=12 F
Ceq’’’= C7 parallel with C8 =6+6 = 12 F
(2)
(2)
(2)
Then,
Ceq’’’’ = Ceq’ is series with C9 = 9x6/(9+6) = 3.6 F
(1)
Then,
Ceq’’’’’= Ceq’’’’ parallel with Ceq’’’ = 3.6+12 = 15.6 F
(1)
Lastly,
Ceq = Ceq’’’’’ in series with Ceq’’ in series with C1
1/Ceq = 1/15.6 +1/12 +1/6
Ceq = 3.18 F
(2)
ECE 100 Fundamental of Electrical Engineering
3/10
June 2, 2006
The Figure shown below shows a cross sectional area of a very long conductor wire with
radius = R carrying a uniform current I and with a uniform charge distribution on its
surface (  q ) and  is the change per unit length. Find the following:
[15 marks]
a. The magnetic field B at distances M and N from the center of the wire (using
either biot-Savart or Ampere law).
b. The electric field E at distances M and N from the center of the wire (using
either coulomb or gauss law).
c. The potential difference V21 between points 1 and 2 , where N is greater than L .
+
+
+
+
2
+
N
+
+
+
L
+
1
+
R
M
+
+
+
+
+
+
Solution:
a. Using Amper’s Law
 B  dl   I
o enc
At any point inside the wire
B 2Πr =
B
o I
 r2
R 2
o I
r , and thus at distance M,
2R 2
B
o I
M
2R 2
(3)
For a point outside the wire
B 2Πr = o I
B
o I
and thus at a distance N ,
2 r
B
o I
2 N
(3)
ECE 100 Fundamental of Electrical Engineering
4/10
June 2, 2006
b. Using Gauss Law
 E  dS 
qenc
o
For any pint inside the wire, since the charge enclosed is equal to zero,
E = 0 and thus E at distance M is equal to zero
(3)
For a point outside the wire:
 E  dS 
qenc
o
E  2 rL 
q
o
q


 o 2 rL  o 2 r
Thus the electric field at distance N is
q

 E 

 o 2 rL  o 2 N
d.
The potential difference between 1 and 2
 E 


L
dr 
ln( )
 2 r
 o 2
N
L o
(3)
N
V2  V1   
(2)
ECE 100 Fundamental of Electrical Engineering
5/10
June 2, 2006
Two particle charges, Q1 and Q2 are 4 cm apart from each other and located along the xaxis, Q1 = +5 C and Q2 = +4 C it is required to find the following:
a. The electric field at point P .
b. The electric potential at point P .
c. A point along the x-axis at which the electric field = 0
d. A point along the x-axis at which the electric potential = 0
[15 marks]
P
5 cm
3 cm
Y
4 cm
Q1
Q2
X
Solution:
q
= 5x10-6/(4Πx8.854x10-12x0.032) =49931902.43 j
4o r 2
E (due to Q2) = 4x10-6/(4Πx8.854x10-12x0.052) = 14380387.9 <143.13
The net E at P = 49931902.43 j +14380387.9 <143.13 = 11504294.9i+58560155.72j=59679482.57<101.11
b- The potential at P
q1
q2
Vp =
+
= 5x10-6/(4Πx8.854x10-12x0.03)+ 4x10-6/(4Πx8.854x102
4o r 4o r
12x0.05)=2216976.468 V
a- E (due to Q1) =
(1)
(2)
(1)
(3)
c- Point along X-axis where E is zero (point will be between the two charges). Taking
Q1 to be at the origin then
(2)
ECE 100 Fundamental of Electrical Engineering
q1
6/10
June 2, 2006
q2
4o x
4o (4  x) 2
4
5
=
2
(4  x) 2
x
2
=
x= 2.11 cm and theus the point is at a distance 2.11 cm from Q1
(3)
d- Point at which potential is zero
Since both changes are positive, thus the potential will not equal zero
(3)
ECE 100 Fundamental of Electrical Engineering
7/10
June 2, 2006
4. A current of I = 2 A flows around a wire shaped like an equilateral triangle. Each side of
the triangle has a length of L = 1m. A constant magnetic field B = 1.5 T is directed
horizontally as shown in the figure. Find the force on each segment of the wire and the net
force on the entire loop.
[10 marks]
B
60o
I
I
60o
L =1m
60o
I
Solution:
F = I Lx B
For the horizontal side, the force on it will be zero since the angel between B and L is zero.
(3)
For the left side of the triangle
F = I Lx B = 2 [ 1cos30(-j )+ 1sin30 (-i)]x ( 1.5i) = 2.598k
(3)
For the right side of the triangle
F = I Lx B = 2 [ 1cos30(j )+ 1sin30 (-i)]x ( 1.5i) = -2.598k
(3)
The net force on the Loop = zero
(1)
ECE 100 Fundamental of Electrical Engineering
8/10
June 2, 2006
ECE 100 Fundamental of Electrical Engineering
9/10
June 2, 2006
FORMULA SHEET
Electrostatics:
 Force:

Field:
Energy 
1
CV 2
2
Current and Resistance:

Voltage:

Capacitance:
ECE 100 Fundamental of Electrical Engineering
Magnetic Fields:
 Bio-Savart's law:

Ampere's Law:

Force:
10/10
June 2, 2006