UNIVERSITY OF MASSACHUSETTS DARTMOUTH
COLLEGE OF ENGINEERING
EGR 101 INTRODUCTION TO ENGINEERING I
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This assignment is due Tuesday, October 2, 2012.
Use engineering papers.
Show all of your steps.
Present your work neatly and clearly.
Box your final answers.
# 2; # 5 and # 14 on page 41 of your textbook.
Problem 1
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
Since e must be unitless, t/RC must also be unitless and therefore the
quantity RC must have the units of time
b) Look up the definition of Farad. Determine how you could get the units from part a) from
R*C. Hint: you may have to use Ohm's law to find a suitable form for the resistance.
R=V/I by ohms law C = Q/V from definition of C so RC = VQ/IV or Q/I.
However, the definition of I is Q/t so the result becomes t
c) If R = 1 MΩ and C = 1 µf , use excel to plot e-t/RC , for the interval 0 ≤ t ≤ 10 sec
Problem 2
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Find the % increase or decrease in resistance of a wire whose diameter is halved
% change = 100*(original- new)/original
R1 =ρ*L1/A1 R2= ρ*L1/A2 assuming the length and material are the same
(R1-R2)/R1 =A1*(1/A1-1/A2) and using πd2/4 for the cross sectional area of
the wire,
and that d2 = d1/2 we get % change = 300%
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Find the ratio of the resistance of copper to that of aluminum for two wires of the same
length and diameter.
Ratio of ρ's or 10.37/17 = 0.61
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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.
Rearranging the formulae for R we can get dCU2 = 0.61*dAL2
So that dCU = 0.078 mm
Problem 3
A battery provides 16,000 J of energy by virtue of the internal chemical reaction, which is
capable of moving 5 x 1020 e- through a wire.
 What is the charge Q in Coulomb (C)?
Q = N* Qe = 5 x 1020* 1.6 x 10-19 C = 80 C
 What is the voltage of the battery in Volts (V)?
V = E/Q = 16.0 kJ / 80 C = 200 V
 What is the current in (μA) if the charges take 3 months to pass through a section of the
wire?
t= 3 mo * 30 day/mo * 24 hrs/day * 3600 sec/hr = 7.776 x 106 sec
I = Q/t = 10.3 µA