48510 Introduction to Electrical and Electronic Engineering Final Exam Pack PDF

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F.1 Final Exam Pack Contents Introduction ..................................................................................................... F.2 23.1 Final Exam Topics ................................................................................... F.3 23.1.1 DC Resistive Circuit Analysis ...................................................... F.3 23.1.2 AC Circuit Analysis...................................................................... F.3 23.1.3 Electromagnetics .......................................................................... F.4 23.1.4 Electronic Circuits ........................................................................ F.4 23.2 Exam Technique ...................................................................................... F.5 23.3 The Final Exam as a Communication Tool ............................................. F.5 23.4 Sample Final Exam with Worked Solutions ........................................... F.6 23.5 Practice Final Exam with Answers........................................................ F.23 23.6 Writing Final Exam Solutions ............................................................... F.28 23.6.1 Do’s ............................................................................................ F.28 23.6.2 Don’ts ......................................................................................... F.29 23.7 Example Final Exam Scripts ................................................................. F.30

F.2 Introduction Proficiency in a discipline entails the internalisation, through memory and understanding, of the body of knowledge currently active in the discipline. A bachelor’s degree qualification is indicative of a broad grasp of a discipline such as would enable its application in the working world. The bachelor’s degree should also provide a sufficiently broad knowledge base to prepare for specialisation and to sustain the directions in which research might lead if undertaking a higher degree. For efficient exercise of a discipline in the workplace, the knowledge of which it is composed needs to be lodged in the memory of the individual. Such knowledge need not be complete, but it needs to be considerable, otherwise there is little to distinguish the educated from the uneducated. A good examination functions as a quality-control measure for this important characteristic. The final exam is the final assessment task in many subjects for this reason. A final exam is used for assessment purposes because: 

it assures and calibrates the quality of the graduate



it indicates an individual’s educational attainment



it ensures there are no external sources of assistance

A good examination requires of the student a novel application and interpretation of factual, memorised material which ensures rote learning without understanding is minimally rewarded. A good examination does not test what you know, but: what you can do with what you know, and how well you do it. Recall the quote by James Clerk Maxwell from the preface: In this class, I hope you will learn not merely results, or formulae applicable to cases that may possibly occur in our practice afterwards, but the principles on which those formulae depend, and without which the formulae are mere mental rubbish. A good examination will therefore present you with challenges and problems that you have never seen before.

F.3 23.1 Final Exam Topics Broadly, the final exam will cover the following areas: 23.1.1 DC Resistive Circuit Analysis The summaries at the end of each of Topic Notes 1 to 6 are a good guide on what you should know. For nodal and mesh analysis, you may be asked to solve equations symbolically or numerically up to and including “rank 3” matrix equations, i.e. G or R is a 3 x 3 real matrix, and there are 3 variables to solve. You should be able to: 

Model (i.e., approximate) a real system using an ideal circuit.



Apply a variety of appropriate circuit analysis techniques to determine all voltages and currents.



Find the power dissipated or delivered in various branches of the circuit.



Determine the maximum power deliverable to a load resistor, and the conditions under which this occurs.

23.1.2 AC Circuit Analysis The summaries at the end of each of Topic Notes 20 to 22 are a good guide on what you should know. For nodal and mesh analysis, you may be asked to solve equations symbolically or numerically up to and including “rank 2” matrix equations, i.e. Y or Z is a 2 x 2 complex matrix, and there are 2 variables to solve. You should be able to: 

Transform a circuit from the time-domain into the frequency-domain, and vice-versa.



Perform phasor analysis (i.e., using quantities such as V, I and Z).



Manipulate impedances to arrive at Zeq or Yeq.



Apply the concepts of power factor, complex power, real power and reactive power to a “simple” circuit.



Draw a power triangle.



Sketch time-domain sinusoids based on their phasor representation.

F.4 23.1.3 Electromagnetics The summaries at the end of each of Topic Notes 9 to 14 are a good guide on what you should know. You may be asked to perform the integrations involved in Gauss’ Law and Ampère's Law for “simple” geometries. You should be able to: 

Sketch a picture of a field using flux lines (and equipotentials for electric fields), including illustrating the concept of a flux tube.



Apply Gauss’ Law to the ideal: point charge; line charge; and parallel plate capacitor.



Determine D, E, Ψ , Q, C and V for simple geometrical arrangements of conductors.



Apply Ampère's Law to the ideal: infinitely long conductor; toroid; solenoid.



Determine B, H,  ,  , L and I for simple geometrical arrangements of conductors and magnetic material.

23.1.4 Electronic Circuits The summaries at the end of each of Topic Notes 7-8 and 15 are a good guide on what you should know. You should be able to: 

Analyse a DC resistive circuit containing up to 3 ideal diodes.



Sketch waveforms for circuits that contain 1 diode and 1 AC source.



Draw up a digital logic truth table based on English statements.



Determine Boolean functions that describe a truth table.



Apply Boolean algebra to reduce a Boolean function to a minimum number of literals.



Apply De Morgan’s Theorem.



Draw digital logic circuits using any combination of AND, OR , NOT, NAND and NOR gates.

F.5 23.2 Exam Technique Planning your time is vital. Read the entire question paper before embarking on ANY solution. Take your time to prioritise the questions, which can be attempted in ANY order. Start with the question you are most confident with, and leave questions that require more thought to last. Make a rough estimate of the time you would like to work on a problem, taking into account your knowledge of the material and the expectations of the question. Your time may not necessarily be divided evenly between questions. Stick to your allotted time – there is nothing worse than “knowing how to do something” and not having the time to write it down to demonstrate it.

23.3 The Final Exam as a Communication Tool There is a logical sequence to the format of a solution:          

Sketching out, drawing, or simplifying the problem Labelling quantities and defining variables English description of what the next step in the solution process is Mathematics English description of what the next step in the solution process is Mathematics English description of what the next step in the solution process is Mathematics … Making sense of the solution (does it make intuitive sense, are the numbers reasonable?, do the parts “line up”?)

Understand the above sequence, and practice it. The following pages show 2 typical final exams. The first exam has fully worked solutions. See if you can attempt the exam ON YOUR OWN before looking at the worked solutions (this will be good practice). The second exam just has questions and answers. You should check and compare your solutions with a study partner – perhaps get them to mark your exam paper, and receive feedback on your solutions and written style!

F.6 23.4 Sample Final Exam with Worked Solutions

QUESTION 1 The Sydney Light Rail is supplied with power by substations which are located 10 km apart. Each of these substations maintains a constant voltage between trolley wire and track of 3.3 kV at the substations. The trolley wire has a resistance of 180 mΩ/km and the rails have an equivalent resistance of 30 mΩ/km. Two trams are in the section between two substations A and B as shown in the figure:

trolley wire A

I

II

B

rails One tram is 2.5 km from substation A and draws a current of 185 A. The other tram is 3.5 km from substation B and draws a current of 315 A. 3 marks

(i)

Show that the current drawn from each substation is:

I A  249 A

I B  251 A

Determine the voltage across each tram.

2 marks

(ii)

1 mark

(iii) Determine the resistance of each tram.

2 marks

(iv) Calculate the total power supplied by the two substations.

2 marks

(v)

Calculate the total power delivered to the two trams.

F.7 Answer 3 marks (i)

Calculating the rail and overhead line resistances, and using current sources to represent the trams, we can draw the equivalent circuit:

450 m

3.3 kV

720 m

630 m

i3

i2

i1 75 m

3.3 kV

315 A

185 A 120 m

105 m

Combining resistors in series, this can be reduced to:

3.3 kV

525 m

720 m

735 m

i1

i2

i3

185 A supermesh

3.3 kV

315 A

120 m

Applying KVL about the supermesh:

 3300  0.525i 1  0.72i 2  0.735i 3  3300  0.12i 2  0 0.525i 1  0.84i 2  0.735i 3  0 For the independent current sources, we relate the source current to the mesh currents: i1  i2  185

and:

i 2  i3  315 Rewriting these last three equations in matrix form, we have:

0.525 0.84 0.735 i1   0   1 0  i2   185 1      1  0  1  i3  315

F.8 Applying Cramer’s rule for i1 :

0 0.84 0.735 185 1 0 315 1 1 0.84  185  0.735185  315 522.9 i1     249 A 0.525 0.84 0.735 0.525  0.84  0.735 2.1 1 0 1 0 1 1 The other mesh currents are: i2  64 A and i3   251 A. The substation currents are therefore: I A  i1  249 A

and:

I B  i 3  251 A 2 marks (ii)

KVL around the left mesh gives:

 3300  0.525i 1 V I  0 V I  3300  0.525  249  3169 V KVL around the right mesh gives:

 V II  0.735i 3  3300  0

V II  3300  0.735   251  3116 V 1 mark

(iii) Applying Ohm’s Law to each tram:

RI 

VI 3169   17.13  185 185

R II 

VII 3116   9.891 315 315

2 marks (iv)

The power supplied by each substation is: PA  VA I A  3300  249  821.7 kW

PB  VB I B  3300 251  828.3 kW

The total power supplied by the substations is then: PT  PA  PB  821.7  828.3 1650 kW

F.9

2 marks (v)

The power supplied to each tram is: PI  VI I I  3169 185  586.3 kW

PII  V II I II  3116 315 981.5 kW

The total power delivered to the two trams is then: PT  PI  PII  586.3  981.5 1568 kW

The difference between the power generated by the substations and the power delivered to the trams is the power dissipated in the trolley wire and rails.

F.10 QUESTION 2 A generator, Vg  100 V RMS, supplies power to a load which is connected between terminals A and B in the circuit below.

0.4

j0.8 A 2

Vg 100 0° V

0.5 load

j5 

-j1 B

4 marks (i)

Show that the current I L in the load, expressed as a phasor, is:

I L  96.3321.02  A 2 marks (ii)

Show that the load voltage between terminals A and B is given by the phasor:

VL  107.7  42.41 V 2 marks (iii)

Calculate the real and reactive power in the load.

2 marks (iv)

Determine the required resistance and reactance of the load for maximum power transfer.

F.11 Answer 4 marks (i)

Simplify the circuit be applying Thévenin’s Theorem to terminals A and B:

ZS

A

VS

B The Thévenin voltage VS is the open-circuit voltage. Using the voltage divider rule we get:

VS 

2  j5 5.385 68.20  100 0  85.79 0.678 V  100 0  2.4  j5.8 6.277 67.52

Setting the independent voltage source to zero, looking into terminals A and B we see two impedances in parallel: Z S  0.4  j 0.8 || 2  j5  

 3.2  j 3.6 4.817131.6  2.4  j 5.8 6.27767.52

 0.767364.08   0.3350  j0.6904 

The equivalent circuit is now:

ZS A 0.5 load

VS -j1 B The load current is then given by Ohm’s Law:

IL 

85.790.678  85.790.678  VS  96.3321.02  A   Z S  Z L 0.3350  j0.6904  0.5  j 0.8905  20.34

F.12 (ii) 2 marks

The voltage between terminals A and B is given by Ohms’ Law: V L  Z L I L  0.5  j  96.33 21.02  1.118  63.43  96.3321.02   107.7  42.41 V  79.52  j 72.64 V 2 marks (iii)

The complex power is:

S L  V LI *L  107.7  42.41  96.33  21.02  10.38  63.43 kVA  P  jQ  4.640  j 9.280 kVA Therefore: P  4.640 kW Q  -9.280 kvar 2 marks (iv)

For maximum power transfer, we have: Z L max  Z *S  0.3350 j 0.6904 

Thus: RL max  335 m

X L max  690.4 m

F.13 QUESTION 3 2 marks (a)

(b)

State Gauss’ Law and illustrate it by a simple example.

A cylindrical co-axial cable has an inner conductor of radius a and an outer conductor of radius b. The space between the conductors is filled with a dielectric of relative permittivity  r  1 . Assume a uniform charge density of  Cm -1 on the inner conductor. Assume the cable is so long that end effects can be neglected.

2 marks

(i)

2 marks

(ii)

Sketch the electric field distribution in the space between the conductors. Use Gauss’ Law to find an expression for the electric flux density (D) and electric field intensity (E) at a radius r between a and b.

2 marks

(iii) Find an expression for the capacitance per unit length of cable, and explain your reasoning.

2 marks

(iv) How would flux density, field intensity, and capacitance be affected if the space between the conductors were filled with a dielectric of relative permittivity

r  2 ?

F.14 Answer 2 marks (a)

Gauss’ Law in integral form is

 D  dA  q

enclosed

.

A simple example is an isolated point charge:

D

D dA r q

"Gaussian surface" D

D We start by stating Gauss’ Law:

 D  dA  q

enclosed

Now since D and dA, for any point on the surface of the sphere, point radially outward, the dot product in the integral reduces to a simple expression. That is, we use the fact that:

D  dA  DdA cos 0  DdA to rewrite the integral as:

 DdA  q Invoking a spherical symmetry argument, the magnitude (not the direction) of the flux density, D, must be a constant on the surface of the sphere. If D is a constant, we can “bring it out the front” of our integral: D  dA  q

F.15 Now

 dA

indicates that we wish to find the sum of all the infinitesimally small

areas dA over the whole of A, i.e. over the entire surface of the sphere. This must be just the surface area of the sphere:

 dA  A  4r

2

Gauss’ Law has now been reduced to:

DA  D4r 2  q Rearranging, we now have a formula for the magnitude of the electric flux density: D

q 4r 2

Since we already know that the direction of D is radially outward, we can add back the direction information to get a vector formula for the electric flux density: D

q rˆ 4r 2

Cm -2

(b) 2 marks

(i) Since the conductor is assumed to be infinitely long, there are no “end effects”. Since the conductor has a uniform linear charge density, the field is uniform in the axial direction. The field distribution will have radially directed electric flux lines:

radially directed E field a

b

relative permittivity r =1

F.16 (ii) 2 marks

A view of a section of length l of the coaxial cable is shown below: Gaussian surface S (a cylinder)

uniform charge density



r l

Due to cylindrical symmetry of the electric field, we choose a cylindrical Gaussian surface S, of radius r, and length l. From Gauss' Law, the electric flux leaving the Gaussian surface S is: Ψ   D  dA  qenclosed

The charge enclosed by the Gaussian surface is:

q enclosed  l The “closed” integral of the flux density over the entire Gaussian surface is composed of 3 parts: the curved part of the cylinder (area A1 ), and the two ends of the cylinder (areas A2 and A3 ):

 D  dA  

A1

D  dA 1   D  dA 2   D dA 3 A2

A3

Since the direction of D is parallel to the ends of the cylinder, then we must have



A2

D  dA 2   DdA2 cos 90  0 and A2



A3

D  dA 3   DdA3 cos 90  0 . A3

Since the direction of D is radial we must have



A1

D  dA1   DdA1 cos 0   DdA1 . A1

A1

Also, due to cylindrical symmetry, D must have a constant magnitude at a certain radial distance r. Then:

 D  dA  D

A1

dA1  D 2 rl

F.17 Therefore, Gauss’ Law gives the magnitude of D as:

 D dA  q

enclosed

D2rl  l D

 2r

The electric flux density is then: D

 ˆ r

2r

and the electric field intensity is: E

2 marks

 2 0 r

a r b

rˆ

(iii) The voltage between the inner and outer conductor is: a

a

Vab   E  dl    Edl b b where the path l is from outer ( b) to inner (a) conductor which is against the direction of the E field, so the dot product introduces a negative sign (cancelling the one outside the integral). Now since a little bit of path length dl is opposite to a little bit of radius dr, we have dl  dr and so: a

a

Vab    Edl    Edr b

b


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