UNIVERSITY OF MASSACHUSETTS DARTMOUTH
COLLEGE OF ENGINEERING
EGR 101 INTRODUCTION TO ENGINEERING I
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This assignment is due Thursday, February 21, 2013.
Use engineering papers.
Show all of your steps.
Present your work neatly and clearly.
Box your final answers.
Problem 1
What is the diameter in mm of a resistor made of gold, which is 0.1 m long and has a
resistance of 10m ? Use Table 1.3 for properties.
R
l
A
2
d2
d 
A   r2      
4
2
2
d
L
A

4
R
d
4 L

R
 4   2.443 106   cm 
10mm 
 1000mm 
  0.1 m  

 1cm 
 1m 
 10 103  
d  0.557mm
Problem 2
A 1.784-mm diameter wire that is 50 cm long has a resistor of 24.6m . What material
properties does this resistor match closely those listed in Table 1.3?
R
l
A
RA

l
d2
A
4
2

2  1cm  
 1.784  mm  
 
10mm  
3


 24.6 10   

4





50  cm 
  12.29 106   cm
Checking Table 1.3, the material is iron.
Problem 3
For comparison of two types of resistors it is advantageous to use RA   L instead of
L
R   . That is, you can compare one resistor of type 1 ( R1 A1  1L1 ) with another
A
R A
 L
resistor of type 2 ( R2 A2  2 L2 ) by expressing the ratio 2 2  2 2 .
R1 A1 1 L1
 Calculate the resistance ratio of a wire (type 2) to one whose diameter is tripled
and length is halved (type 1).
R2 A2  2 L2

R1 A1 1 L1
R2 A1  2 L2

R1 A2 1 L1
 d12 
2

4   d1 
A1


 
A2
 d 22   d 2 
 
 4 

d 2  3d1
A1 1

A2 9
L2 
L 1
1
L1  2 
2
L1 2
 2  1
R2 A1 L2 1 1 1



R1 A2 L1 9 2 18
R2 1

R1 18
 Find the resistance ratio of aluminum (type 2) to that of gold (type 1) for two
wires of the same length and diameter.
R2 A2  2 L2

R1 A1 1 L1
A1  A2
L1  L2
R2  2

R1 1
6
R2 2.825 10    cm 

 1.156
R1 2.443 106    cm 
R2
 1.156
R1
 Find the diameter ratio of aluminum (type 2) to that of gold (type 1) for the same
length and for an aluminum resistance that is one-half that of gold.
R2 A2  2 L2

R1 A1 1 L1
L1  L2
R1  2 R2
 d2 
A  
 4 
A2 R1  2 L2 2 R2  2



2 2
A1 R2 1 L1
R2 1
1
d 22
4  2 2
d2
1
 1
4

d2

2.825 10 6
 2 2  2
d1
1
2.443 10 6
d2
 1.52
d1
Problem 4
A battery provides a total charge of Q  50  C in 3ms .
 What is the current in mA?
Q  C  50 106  C 
I  A 

 16.67 103  A  16.67mA
3
t s
3 10  s 
I  16.67mA
 If the battery voltage is 240V, what is the energy generated in mJ, over this time
period?
V
E
Q
E  QV   50 106 C   240V 
E  12 103 J  12mJ
E  12mJ
 What is the power in mW?
P  VI
P   240V  16.67 103 A  4W
P  4W
Problem 5
A 10-Volt battery provides 3.5 J of energy to a load.
 How much charge in C does the battery provide?
E
Q
E 3.5 J
Q 
 0.35C
V 10V
Q  0.35C
V
What is the number of electrons in that charge?
e  # electrons
0.35C
e
 2.1875 1018 electrons
C
19
1.6 10
electron
e  2.1875 1018 electrons
 What is current in mA if the charge is delivered over a period of 10 s?
Q
t
0.35C
I
 0.035 A  35 10 3 A
10 s
I  35mA
I