UNIVERSITY OF MASSACHUSETTS DARTMOUTH
COLLEGE OF ENGINEERING
EGR 101 INTRODUCTION TO ENGINEERING THROUGH APPLIED
SCIENCE I





This assignment is due Thursday, September 26, 2013.
Use engineering papers.
Show all of your steps.
Present your work neatly and clearly.
Box your final answers.
Problem 1
Problem #3 on page 73 of the textbook.
360kΩ+/- 5% = 360kΩ+/-18kΩ
360kΩ - 18kΩ = 342kΩ
360kΩ + 18kΩ = 398kΩ
The resistor values range from 342kΩ - 398kΩ.
Problem 2
Problem #6 on page 73 of the textbook.
Band 1
Band 2
Band 3
Value
Red
Orange
Red
White
Brown
Blue
Black
Black
Red
Brown
Blue
Gray
Brown
Silver
Black
Green
Blue
Silver
200Ω
0.3Ω
22Ω
9.1MΩ
16MΩ
0.68Ω
Band 2
Orange
Red
Red
Black
Band 3
Orange
Red
Yellow
Blue
Problem 3
Problem #7, parts a, d, e, f.
Value
33kΩ
2.2kΩ
120kΩ
10MΩ
Band 1
Orange
Red
Brown
Brown
Problem 4
Consider e-t/RC, where R is resistance in Ohms, C is capacitance in Farads, and t
is time in s
a) What are the units associated with the product R*C
The exponent –t/RC must be dimensionless, therefore, RC must have units of
seconds.
b) Look up the definition of Farad. Determine how you could get the units from
part a) using R*C. Hint: you may have to use Ohm's law to find a suitable form
for the resistance.
RC     F  V C  C  CC  s
I V
I
s
c) If R = 1 MΩ and C = 1 µF , use excel to plot e-t/RC , for the interval 0 ≤ t ≤
10 sec
t
exp(t/RC)
0
1
2
3
4
5
6
7
8
9
10
1
0.36788
0.13534
0.04979
0.01832
0.00674
0.00248
0.00091
0.00034
0.00012
4.5E-05
Problem 5
Find the % increase or decrease in resistance of a wire whose diameter is halved
Let R1 
l
A1
be the original resistor with diameter = d1
2
d2
d 
A1  r12    1    1
4
2
d
l
Let R2 
be the resistor with diameter d2= 1 .
A2
2
A2 
d12
16
l
d1 2
R2
A
A
 2  1  42 4
l A2
R1
d
 1
A1
4
4
R2  4 R1  400% R1
Find the ratio of the resistance of copper to that of aluminum for two wires of the
same length and diameter.
Let R1 
Let R2 
 1l
1l
A
 2l
A
be the resistance of the Copper wire.
be the resistance of the Aluminum wire.


R1
1.723 X 10 6
 A  1  Cu 
 0.61
R2  2 l  2  Al 2.825 X 10 6
A
Find the diameter of a copper wire that has the same length and resistance of an
aluminum wire. The aluminum wire has a diameter of 0.1 mm.
Let R1 be the resistance of the Aluminum wire and R2 be the resistance of
the Copper wire.
R1  R2
 1l
A1

 2l
A2
A2  A1
d 22

2
 A1 Cu  0.61A1
1
 Al
 0.61
d12
4
4
d 2  0.078mm