# UNIVERSITY OF MASSACHUSETTS DARTMOUTH COLLEGE OF ENGINEERING ```UNIVERSITY OF MASSACHUSETTS DARTMOUTH
COLLEGE OF ENGINEERING
EGR 101 INTRODUCTION TO ENGINEERING I
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This assignment is due Thursday, October 11, 2012.
Use engineering papers.
Present your work neatly and clearly.
Problem 1
Resistors are rated not only in terms of resistance, but also in terms of their ability to
dissipate power without burning. For example, if you use a resistor rated to dissipate 2 W
in a circuit where the voltage and current conditions require that the wire dissipates 3 W
then the wire will burn, and the circuit may be damaged. The ratings of resistors are
given in section 2.3.5. on p. 56.
Chapter 3: solve problem 15 on p. 93, and refer to section 2.3.5 on p. 56 in order to select
the required power rating.
I = V/R = 10V/10kΩ = 1 mA P = V*I = 10V*1mA = 10 mW = 0.010 W so 1/8W OK
Problem 2
Chapter 3: problem 27 on p. 94.
The range of possible current values is found by calculating the minimum and maximum
V
values of the current. If you consider Ohm’s law in the form of I  , then you can
R
infer that for fixed voltage, as R increases, I decreases, and vice versa.
Use the 4-band color code of the resistor to find its nominal value and tolerance. From
this information find the minimum and maximum values of resistance and the
corresponding currents.
Red Red Red Gold = 2200 Ω 5% tolerance
resistance range = 2090-2310 Ω
Imax = 8V/Rmin = 8V/2090Ω = 3.83 mA
Imin = 8V/Rmax = 8V/2310Ω = 3.46 mA
Problem 4
Refer to Figure 3.2 (a) on p. 77. What is the % setting of the potentiometer if its
resistance rating is 2.5k ?
R = V/I = 10V / 100 mA = 100 Ω 100/1500x100 = 6.67%
Problem 5
Chapter 4: problem 5 on p. 122.
R3 = 53k -12k -33k = 8kΩ 8k/10k x 100 = 80%
Problem 6
Chapter 4: problem 9 on p. 122.
Rt = 51 + 39 + 10 = 100 Ω I = V/Rt = 1V / 100 Ω = 10 mA
Problem 7
Chapter 4: problem 13 on p. 122.
Rt = V/I = 12V / 560 &micro;A = 21,429 Ω.
3629/5000 x 100 = 72.58%
21429 - 6800 -11000 = Rpot = 3629 Ω
Problem 8
Chapter 4: problem 17 on p. 123.
Rt = 51 + 39 + 20 = 110 Ω Vs = I*R = 50 mA * 110 Ω = 5.5 V
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