Download Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition

Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
Chapter 7: Electrical Engineering
7-1) For the circuit shown, find the voltage V.
15 A
Need: Voltage = _____ V
Know: I = 15 A through 100.Ω
resistor.
V
How: Ohm’s law, V = I R
Solve: V = 15 × 100. [V][Ω]
∴Voltage = 1500 V.
Copyright ©2010, Elsevier, Inc
100. Ω
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-2) For the circuit shown, find the current I.
217
I
Need: Current = ____ A
12 V
Know: V = 12 V across a 100. Ω
resistor
How: Ohm’ law, V = I R
Solve: I = V/R = 12/100. [V]/[Ω] = 0.12 A
Copyright ©2010, Elsevier, Inc
100. Ω
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-3) For the circuit shown, find the resistance R.
R
I = 1.5 A
12 V
Need: Resistance = ____ Ω
Know: I = 1.5 A through resistor driven by 12 V.
How: Ohm’ law, V = I R
Solve: R = V/I ∴resistance, R = 12/1.5 [V]/[A] = 8.0 Ω
Copyright ©2010, Elsevier, Inc
218
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-4) For the circuit shown, a power of 100. watts is dissipated in the resistor. Find the
current I.
I
R
12 V
Need: Current = ____ A
Know: P = 100. W; V = 12 V
How: Power law, P = I V
Solve: ∴ I = P/V = 100./12 [W]/[V] = 8.3 [V A]/[V] = 8.3 A
Copyright ©2010, Elsevier, Inc
219
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-5) For the circuit shown, a power of 100. watts is dissipated in the resistor. Find the
resistance, R.
R
I = 50. A
V
Need: Resistance = ____ Ω
Know: P = 100. W and I = 50. A
How: Power Law, P = I V, and Ohm’s law V = I R
Solve: Combine power law with Ohm’s law for the case in which the
voltage drop is not known a priori: then P = I2R or resistance, R = P/I2 =
100./(50.)2 [W]/[A]2 = 0.040 [V A]/[A]2 = 0.040 [V]/[A] = 0.040 Ω
Copyright ©2010, Elsevier, Inc
220
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-6) For the circuit shown, find the current I.
100.Ω
I
12 V
I1
I2
Need: Current, I = ____ A
Know: Each parallel resistor is 100. Ω with a 12 volt drop across it
How: For each parallel branch Ohm’s law, V = I1R = I2R with R = 100. Ω,
Solve: I1 = I2 = 12/100. [V]/[Ω] = 0.12 A. But by charge conservation, I =
I1 + I2 = 0.12 + 0.12 = 0.24 A.
Copyright ©2010, Elsevier, Inc
221
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-7) For the circuit shown, find the voltage V.
I = 4.0 A
V
I1
100.Ω
I2
Need: Voltage = ____ V
Know: Total current is 4.0 A so that I = 4.0 A = I1 + I2 through the two
parallel legs.
How: Current in each leg with resistance of 100. Ω is I1 = I2 = 2.0 A by
symmetry. Voltage drop across each leg by Ohm’s law, V = I R.
Solve: Voltage, V = 2.0 × 100. [A][Ω] = 200 [A][V/A] = 2.0 × 102 V.
Copyright ©2010, Elsevier, Inc
222
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-8) For the circuit shown, find the
resistance R.
Need: Resistance, R = ____ Ω
I = 0.10 A
9.0 V
I1
223
100.Ω
I2
Know: Total current = 0.10 A,
R1 = 100. Ω with voltage drop of 9.0 V.
How: Ohm’s law for leg 1 will give I1 = V/R1 and I = I1 + I2
Solve: Leg 1: I1 = V/R1 = 9.0/100. [V]/[Ω] = 0.090 A;
∴ Leg 2: I2 = I – I1 = 0.10 – 0.090 = 0.01 A. Hence resistance R = V/I2 =
9.0/ 0.01 [V]/[A] = 9. × 102 Ω.
Note: Only 1 sig. figure for I2 (since we are subtracting near equal currents
from each other and the difference is known to no better than one
significant figure).
Copyright ©2010, Elsevier, Inc
R
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
224
7-9) For the circuit shown, a power of 100. watts is dissipated in each resistor. Find the
current I and the voltage V.
60.Ω
I
V
I1
I2
Need: Current I = ____ A and V = ____ V.
Know: P = 100. W in each 60.Ω resistor.
How: I1 = I2 = ½I and P = I1V = I2V with unknown V given by Ohm’s
law, V = IR
Solve: In each resistor, P = I1 V = I12 R = I22 R (a useful general form for
current when the voltage drop is unknown.)
∴ I1= I2 = √(P/R) = √(100./60.) √{[W]/[Ω]} = 1.29 √{[A V]/[V/A]} =
1.29 A per leg. Total current, I = 2 × 1.29 = 2.6 A
The voltage drop is common to both legs: ∴V = I1R = I2R = 1.29 × 60. =
77 V.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
225
7-10) A circuit consists of a 3.0 V battery and two resistors connected in series with it. The first
resistor has a resistance of 10 Ω. The second has a resistance of 15 Ω. Find the current in the
circuit.
+
R1 = 10.Ω
R2 = 15 Ω
V = 3.0 V
V = 3.0 V
I
V = 0.V
Need: Current = _____ A
Know: Voltage drop overall = 3.0 V; series resistors of 10. and 15 Ω
respectively.
How: First draw the circuit sketch. Add small circles to the circuit
representing points where the local voltage potential is known. Then use
Ohm’
Solve: Current = I = V/Rtotal = 3.0/(10. + 15) [V]/[Ω] = 0.12 [V] [A/V] =
0.12 A.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
226
7-11) A circuit consists of a 12 V battery and a resistor connected in series with it. The
current is 105 A. Find the resistance.
V = 12 V
V= 12 V
+
R
I = 105 A
V = 0.V
Need: R = ____Ω
Know: Voltage drop is 12 V; current through resistor is 105 A.
How: First a sketch: then Ohm’s law, R = V/I
Solve: R = V/I = 12/105 [V]/[A] = 0.11 Ω.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
227
7-12) A circuit consists of a 9.0 volt battery and two parallel branches, one containing a
1500 Ω resistor and the other containing a 1.0 × 103 Ω resistor. Find the current drawn
from the battery.
I1
I2
1.0 × 103 Ω
I
9.0 V
1500Ω
V = 9.0 V
V = 0. V
Need: Current, I = _____ A.
Know: Parallel circuit with 1500 and 1.0 × 103 Ω resistors with 9.0 V
across each leg.
How: First a sketch: note where you know the potential. Then use Ohm’s
law for each leg, I = V/R; then add the branch currents.
Solve: For each leg, I = V/R. ∴ I1 = V/R1 = 9.0/1500 = 0.0060 [V][A/V] =
0.0060 A and likewise I2 = V/R2 = 9.0/1.0 × 103 = 0.0090 A
∴I = I1 + I2 = 0.015 A
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
228
7-13) A circuit consists of a 12 volt battery attached to a 1.0 × 102 Ω resistor, which is in
turn connected to two parallel branches, each containing a 1.0 × 103 Ω resistor. Find the
current drawn from the battery.
I
I2
1.0 × 103 Ω
I1
V?
1.0 × 103 Ω
V = 12 V
2
12 V 1.0 × 10 .Ω
V=0
Need: Current, I= _____A
Know: Parallel/series circuit with 2 by 1.0 × 103 Ω resistors in parallel
and a 1.0 × 102 Ω resistor in series. Total circuit voltage drop is 12 volts.
How: First a sketch: Mark the points where you know the potential and, in
this case, mark the unknown potential across the 1.0 × 103 Ω resistors.
Because these resistors are equal sizes with the same (unknown) voltage
drop across them, there is an equal current in each leg. I1 = I2 = ½ I. Then
Ohm’s law for each circuit, treated as series, with 12 volt drop in terms of
I.
Solve: Ohm’s law across the 100 Ω resistor: (12 – V) = I × 1.0 × 102.
Then across either of the 1,000 Ω resistors, (V – 0) = ½ I × 1.0 × 103.
These are two equations in two unknowns and solving by eliminating V
gives 12 = I × (1.0 × 102 + ½ × 1.0 × 103)
∴ I = 12/(1.0 × 102 + ½ × 1.0 × 103) [V][A/V] = 0.020 A
(Incidentally if you also needed V it is 10. volts. This is worth checking
since it is clear that most of the voltage drop is lost across the parallel
circuit with its high resistance. Only 2.0 V is dropped across the 100 Ω
resistor.)
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-14) An automobile’s 12 V battery is used to drive a starter motor, which for several
seconds draws a power of 3.0 kW from the battery. If the motor can be modeled by a
single resistor, what is the current while the motor is operating?
V = 12 V
I
R
V = 0. V
12 V
Need: Current = ____ A
Know: Voltage drop of 12 V drains 3.0 kW from battery.
How: First a sketch: Then power law P = I V with P and V known.
Solve: I = P/V = 3,000/12 [W]/[V] = 250 [VA]/[V] = 250 A.
Copyright ©2010, Elsevier, Inc
229
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
230
7-15) An automobile’s 12 V battery is used to light the automobile’s two headlights.
Each headlight can be modeled as a 1.00 Ω resistor. If the two headlights are hooked up
to the battery in series to form a circuit, what is the power produced in each headlight?
Why would you not wire car lights in series?
Need:
____ W
+
R1 = 1.00 Ω
R2 = 1.00 Ω
V = 12 V
V = 12 V
I
V = 0.V
Power =
Know: Series circuit with 2 × 1.00 Ω lights and a 12 V battery
How: First a sketch. Then note that series resistance sums to 2.00 Ω with
known voltage drop – use power law, P = V2/R form incorporating Ohm’s
law
Solve: P = V2/R = 122/2.00 [V]2[A/V] = 72 [V][A] = 72 W for two bulbs.
Hence 36 W/bulb.
If either bulb failed (“burned out”) then the circuit would has an infinite
resistance and then other bulb would draw no current and go off (think of
a switch opening). It’s hard to drive at night with both add italics)
headlights out!
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
231
7-16) An automobile’s 12 V battery is used to light the automobile’s two headlights.
Each headlight can be modeled as a 1.00 Ω resistor. If the two headlights are hooked up
to the battery in parallel to form a circuit, what is the power produced in each headlight?
Why is a parallel circuit preferred for this application?
V = 12 V
I
1.0 Ω
12 V
I1
1.0 Ω
I2
V = 0. V
Need: Power = ____ W
Know: Parallel circuit with 2 by 1.00 Ω lights and a 12 V battery
How: First a sketch. By symetry I1 = I2 = ½ I. Each leg has the same voltage drop
of 12 V. and I1 = I2 = V/R and the power/bulb = (I1 or I2) × V = V2/R.
Solve: Per bulb, P = V2/R = 122/1.00 [V]2[A/V] = 144[V][A] = 144 W.
Notice this is 4 × the previous power for the series solution. You certainly
want the reliability of parallel circuits because if one burns out the other
remains on.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
232
7-17) An automobile’s 12 V battery is used to light the automobile’s two headlights.
Each headlight can be modeled as a single resistor. If the two headlights are hooked up to
the battery in parallel to form a circuit, and each headlight is to produce a power of 100
W, what should the resistance of each headlight be?
V = 12 V
I
R
12 V
I1
R
I2
V = 0. V
Need: Resistance, R = ____Ω
Know: Same circuit as Problem 16, but different resistances for the bulbs.
Parallel circuit with two filaments each of R Ω and a 12 V battery.
How: First a sketch. P = I V and, since I unknown, use Ohm’s law for I so that P
= V2/R per bulb.
Solve: R = V2/P [V]2/[W] = 122/100. [V]2/[V-A] = 1.4 Ω per bulb.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
233
7-18) Suppose one of the headlights in problem 17 suddenly burns out. Will be the power
produced by the other headlight increase or decrease?
V = 12 V
I
12 V
RΩ
I1
R=∞Ω
I2
V = 0. V
Need: Is P > 100. W or < 100. W?
Know: Circuit is now modified because one leg is burned out. V = 12 volts still
and R1 = 1.4 Ω still.
How: First a sketch. In remaining leg, P = V2/R
Solve: Power = 100. W, the same power output as previously since
neither V nor R changed for the remaining bulb. Of course the current I
will halve since I = I1 + I2 = I1 only once the second bulb burns out.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
234
7-19) Suppose a car has headlights operating in parallel, but with different resistances,
one of 2.0 Ω and the other of 3.0 Ω. Suppose the headlight parallel circuit is connected in
series with a circuit for a car stereo that can be modeled by a 1.0 resistor. What is the
current being drawn from a 12 V battery when both the lights and the stereo are on?
1.0 Ω
V?
V = 12 V
12 V
I1
I
2.0 Ω
I2
3.0 Ω
V = 0. V
Need: Current, I = ____ A
Know: Asymmetric parallel circuit with 2.0 and 3.0 Ω resistor in series
with 1.0 Ω resistor all powered by a 12 V battery.
How: First a sketch; note the asymmetry of the parallel legs means that the
current through each is different through each, but their sum is still I. Use Ohm’s
law for each series circuit consisting of the stereo + the light.
Solve: For leg 1: V = 12 V = I × 1.0 + I1 × 2.0
or 2.0 I1 + 1.0I = 12
Eqn (1)
For leg 2: V = 12 V = I × 1.0 + I2 × 3.0
or 3.0 I2 + 1.0I = 12
Eqn (2)
Finally I = I1 + I2
Eqn (3)
These are 3 eqn in 3 unknowns, I, I1 , & I2. The solution is I1 = 3.3 A, I2 =
2.2 A and I = 5.5 A.
Can you immediately see why the current I1 is > I2? It’s because R1 < R2.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
235
7-20) Consider the circuit shown below. What are the currents Io through I6? (Hint: What
are the voltages Va and Vb? Note the symmetry between R1/R2 = 5.0/5.0 and R3/R4 =
10.0/10.0.)
12 V
I0
5.0 Ω
Va
12 V
5.0 Ω
I3
I1
10.0 Ω
5.0 Ω
Vb
I5
I4
I2
I6
10.0 Ω
0V
Need: I0, I1, I2, I3, I4, I5, I6, ______A
Know: The circuit has several levels of symmetry so that several of the
various currents are equal by inspection and on the conservation of charge
principle.
How: This is a relatively advanced problem. There are two ways to solve
this: the general solution and the intuitive solution.
The general solution: Assume all the currents are different. We have a
total of 6 unknown currents and will need a total of six equations for them.
There are also two unknown voltages, Va and Vb for a total of 8 unknowns.
We will also need Ohm’s Law to relate current and volts across the
resistors.
Solve: For currents write,
I0 = I1 + I3
I1 = I5 + I2
I4 = I5 + I3
I6 = I2 + I4
(1)
(2)
(3)
(4)
We need another 4 equations to solve our system of resistors. Note that
formulae such as 1/R = 1/R1 + 1/R2 cannot be easily applied, if at all. But
Ohm’s law is applicable to either branch, e.g.,
5.0 Ω × I1 = (12 – Va)
Copyright ©2010, Elsevier, Inc
(5)
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
5.0 Ω × I2 = Va
10.0 Ω × I3= (12 – Vb)
10.0 Ω × I4 = Vb
(6)
(7)
(8)
These 8 equations can be then be solved for each of the 6 currents and the
two voltages.
The most general circuit problem uses Kirchoff’s Law and you could rediscover the equations shown above - but there is a simpler solution.
The intuitive solution: Look at the symmetry of this particular problem.
What are the voltages Va and Vb? They must be equal since they are each
50% (i.e., either from R1/(R1 + R2) = 5.0/(5.0 + 5.0) or R3/(R3 + R4) =
10.0/(10.0 + 10.0) respectively) of the 12 V drop and so each is actually
6.0 V.
If this is true, what is the current I5? Ohm’s law tells you it is zero since I5
= (Va - Vb)/R = 0. Thus the resistor 5 carries no current and the circuit is
simply a parallel one!
∴I5 = 0.0
∴I1 = I2 = 12./10. [V][A/V] = 1.2 A
∴I3 = I4 = 12./20. [V][A/V] = 0.6 A
∴I0 = I6 = I4 + I2 = 1.8A
Copyright ©2010, Elsevier, Inc
236
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
237
7-21. Determine the output voltage, Vout, in Figure 7.5 if Vin = 90. V, R1 = 10. ohms, and
R2
= 50. ohms.
Vin
Vout
Need: Vout = ______ V.
Know: Vin = 90 V, R1 = 10. Ω, and R2 = 50. Ω.
How: Vout = V1n × R2/( R1 + R2)
Solve: By direct substitution, Vout = 90
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
238
7-22. Use the same RadioShack stock resistors in Example 7.6 to design the construction
of a voltage divider that will reduce the voltage by a factor of 30.
Vin
Vout
Need: Vout /V1n = 1/30.
Know: 5.0, 10., 15., and 20. ohm resistors in stock
How: Vout = V1n ×R2/( R1 + R2)
Solve: Pick smallest R2 & largest R1 resistors. We need about 150 Ω in the
denominator. Thus use 7 × 20. Ω resistors and 1 × 10. resistor. Then Vout/V1n =
5.0/(140. + 10.) = 1/30.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
239
7-23. Use Kirchhoff’s voltage and current laws to determine the voltage drop and current
in each of the three resistors below if R1 = 10. ohms, R2 = 20. ohms, and R3 = 30. ohms.
6.0
Need: V, I in each of R1, R2, & R3.
Know: R1 = 10. ohms, R2 = 20. ohms, and R3 = 30. ohms. And V = 6.0 V
How: Kirchoff’s Laws
Solve: Each loop has 6.0 V applied.
Thus I1 = 6.0/10. = 0.60 A
I2 = 6.0/20. = 0.30 A
I3 = 6.0/30. = 0.20 A
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
240
7-24. If the resistors in Exercise 23 are all equal to 10. ohms (R1 = R2 = R3 = 10. ohms),
what is the current supplied by the 6.0 volt battery?
Need: I = ______A
Know: V = 6.0V and R1 = R2 = R3 = 10. Ω
How: Kirchoff’s Laws – each circuit has 6.0 V across the resistor.
Solve: ∴I = 6.0/10. = 0.60 A in each loop.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
241
7-25. Determine the currents I1 and I2 in the current divider illustrated in Figure 7.7 if V =
50.3 volts, R1 = 125. ohms, and R2 = 375. ohms.
V
R1
Need: I1 = _____A, I2 =
R2
_______A
Know: Vin = 50.3 volts, R1 = 125. Ω, and R2 = 375.Ω
How: Kirchoff laws
Solve: Loop 1: V = 50.3 = I1R1; loop 2: V = 50.3 = I2R2
∴I1 = 50.3/125. = 0.402 A and I2 = 50.3/375 = 0.134 A.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-26. Repeat Example 7.7 with the resistor R1 changed to 100. Ω and all other values
remain unchanged.
C
R1
100. Ω
D
Need: I3 = ______ A
Know: R1 = 100. Ω, R2 = 20.0 Ω,
R3 = 40. Ω; V1 = 10. V, V2 = 20. V.
How: Kirchoff’s laws
Solve: Loop 1: V = 10. = 100. × I1 + 40. × I3 or 10. I1 + 4.0 I3 = 1.0
Loop 2: 20. × I2 + 40.× I3 = 20.
or I2 + 2.0 I3 = 2.0
Conservation of charge:
I1 + I2 = I3
Hence I3 = 2.1/3.4 = 0.62 A
Copyright ©2010, Elsevier, Inc
242
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-27. Repeat Example 7.7 with V1 increased to 100.volts and V2 reduced to 5.0 volts.
Need: I3 = ______ A
Know: R1 = 40. Ω, R2 = 20.0 Ω, R3 = 40. Ω; V1 = 100. V, V2 = 5.0
How: Kirchoff’s laws
Solve: Loop 1: V = 100. = 10. × I1 + 40. × I3
or I1 + 4.0 I3 = 10.
Loop 2: 20. × I2 + 40.× I3 = 5.0.
or 4.0 I2 + 8.0 I3 = 1.0
Conservation of charge: I1 + I2 = I3
Hence I3 = 28./5.0 = 5.6 A
Copyright ©2010, Elsevier, Inc
243
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
7-28. Determine the currents I1, I2, I3, and I4 in the figure below.
.
6.0 Ω
12. Ω
i=
5.0
9.0 Ω
9.0 Ω
12. V
Need: I1, I2, I3, and I4
Know - How: Kirchoff’s laws
Solve: Loop 1:
6.0 × I1 = 9.0 × I2
Loop2:
12. × I3 = 9.0 × I4
Overall Loop:
12. = 9.0 × I2 + 9.0 × I4
(or equally 12. = 6.0 I1 + 12. I3)
Conservation charge:
I1 + I2 = I3 + I4
I1 = 0.82A, I2 = 0.55A, I3 = 0.59A, I4 = 0.78A and I = 1.37 A
Check: Conservation of charge: 0.82 + 0.55 = (?) 0.59 + 0.78 or 1.37 = 1.37 …OK.
Copyright ©2010, Elsevier, Inc
244
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
245
7-29. It is the last semester of your senior year and you are anxious to get an exciting
electrical engineering position in a major company. You accept a position from company
A early in the recruiting process, but continue to interview hoping for a better offer.
Then your dream job offer comes along from company B. More salary, better company,
more options for advancement, it is just what you have been looking for. What do you
do?
a) Accept the offer from company B without telling company A (just don’t show
up for work).
b) Accept the offer from company B and advise company A that you have changed
your mind.
c) Write company A and ask them to release you from your agreement.
d) Write company B thanking them for their offer and explain that you have
already accepted an offer.
Summarize using an Engineering Ethics Matrix.
1) Apply the Fundamental Canons: Engineers, in the fulfillment of their professional
duties, shall:
1) Hold paramount the safety, health and welfare of the public. Does not apply
2) Perform services only in areas of their competence. Does not apply
3) Issue public statements only in an objective and truthful manner. This rules
out option a).
4) Act for each employer or client as faithful agents or trustees. Does not apply company is not yet your employer.
5) Avoid deceptive acts. Option a) would be deceptive
6) Conduct themselves honorably, responsibly, ethically, and lawfully so as to
enhance the honor, reputation, and usefulness of the profession. Option b)
possibly violates this canon—accepting a job offer and then changing
your mind is neither honorable nor responsible.
2) Apply the Engineering Ethics Matrix
Options a) Accept B
without telling
A
Canons
Does not
Hold
paramount the apply
safety, health
and welfare of
the public.
Does not
Perform
apply
services only
in the area of
your
competence
b) Accept B,
notify A
c) Ask A for
release
d) Refuse B,
stick with A
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
Issue public
statements
only in an
objective and
truthful
manner
Act for each
employer or
client as
faithful agents
or trustees
Avoid
deceptive acts
Conduct
themselves
honorably …
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
No
Yes
Yes
Yes
No
No
Yes
Yes
Solution: The canons narrow down the list of acceptable options to c) and d).
Copyright ©2010, Elsevier, Inc
246
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
247
7-30. A female student in your class mentions to you that she is being sexually harassed
by another student. What do you do?
a)
b)
c)
d)
Do nothing; it is none of your business.
Ask her to report the harassment to the course instructor.
Confront the student accused of harassment and get their side of the issue.
Talk to the course instructor or the college human resource director on your
own.
Summarize using an Engineering Ethics Matrix.
1) Apply the Fundamental Canons: Engineers, in the fulfillment of their professional
duties, shall:
1) Hold paramount the safety, health and welfare of the public. Just because she
is a colleague does not mean she is not one of the “public” so you cannot
do a).
2) Perform services only in areas of their competence. You are probably not
competent to make a judgment as to the truth of the student’s claims
3) Issue public statements only in an objective and truthful manner. Does not
apply
4) Act for each employer or client as faithful agents or trustees. Does not apply
5) Avoid deceptive acts. Does not apply.
6) Conduct themselves honorably, responsibly, ethically, and lawfully so as to
enhance the honor, reputation, and usefulness of the profession. Honorable
and responsible conduct here would consist of first becoming aware of the
procedures the college has established for dealing with harassment issues,
and helping the student make use of these procedures.
2) Apply the Engineering Ethics Matrix
Options a) Do nothing
Canons
Hold
paramount the
safety, health
and welfare of
the public.
Perform
services only
in the area of
your
competence
Issue public
statements
b) Ask her to
report
d) Talk to
course
instructor
Does not
apply
Does not
apply
Does not
apply
c) Confront
student
accused
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
only in an
objective and
truthful
manner
Act for each
employer or
client as
faithful agents
or trustees
Avoid
deceptive acts
Conduct
themselves
honorably …
248
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
No
Does not
apply
No
Does not
apply
Yes
Does not
apply
Yes
Solution: None of the proposed options meet the need brought out by canon 6. A
better answer than any of the given options might be the following: “Assist the
student to initiate the college’s procedures for dealing with a potential harassment
issue”. This suggests that students have an ethical obligation to know institutional
policies. This is a dimension of behaving honorably and responsibly in an
organizational contest.
Copyright ©2010, Elsevier, Inc