Description:
UNIT 3
ASSIGNMENT 26: Simultaneous Equations Read: Section 61.
Answer: SelfTest questions 1 through 9. Then refer to page 176 to check your answers.
Quick Review: From studying Algebra you are familiar solving equations with one unknown (for example, 3 + x = 7). Some equations, however, have more than one unknown (for example, x+ y= 3 and x + y+ z = 10). Depending upon the number of unknowns they contain, such equations cannot be solved unless you can determine a value for one or more of the unknowns. You can do this if you have been given
additional equations involving the same unknowns. The total number of equations you need is equal to the number of unknown variables in the equations. That is, if you have one variable, you can find its value with one equation. If you have two unknowns, you will need two equations. If you have three unknowns, you will have three equations, and so on. These groups or sets of equations are called simultaneous equations. To solve these equations you can use the variableelimination method, which involves three operations: multiplying by a constant, adding equations, and substituting numerical value for variables. Another method uses determinants. Regardless of which method is used, the solution should be checked by substituting values into the original equations.
Supplementary Example:
Given the following three equations, solve for the three unknowns A, B, and C:
(1)7A — 2B + 3C 
= 
46 
(2)3A + 5B — 4C 
= 
20 
(3)4A + 11B — 2C 
= 
162 
Step 1: Eliminate 1 of your unknowns or variables.
Pick any 2 equations to combine in order to eliminate a variable. For purposes of example, we will eliminate variable C using equations (2) and (3). To do this, we need to multiply equation 3 by a constant.
We get this number by dividing the term with the variable C in equation (2) by the term with the variable C in equation (3) and multiplying by 1. This gives us
x 1 =  2
2C}
 I
I
$
Multiplying equation (3) by 2 gives us the new equation
Adding this equation to equation (2) gives us equation (4) 
4A + 11B – 2C = 162 x z + 3A + 5B – 4C = 20 11A – 17B = 304 
Step 2: Pick a different combination of equations to eliminate the same variable as in Step 1. This time we will combine equation (1) and equation (2) to eliminate variable C.
3C
x 1 = 0.75
4C
Multiplying equation (2) by 0.75 yields
3A+ 5B – 4C = 20
x .75
2.25A + 3.75B – 3C = 15 Adding this to (1) + 7A – 2B + 3C = 46 gives us
equation (5) 9.25A + 1.75B = 31
Eliminate decimals by x 4
multiplying by 4: (5) 37A + 7B = 124
Step 3: Solve equations (4) and (5) by eliminating 1 of the 2 remaining variables.
For our example we will eliminate variable A.
37A37
x 1 =
11A  11
37
Multiplying equation (4) by _— yields
11

^{629}B ^{11248} 37A + – 


11 11 

Adding this equation to equation (5): 
37A + 7B = 124 



yields 
629 _{\}_{0} + 
11248 
124 

11 
11 



Multiplying both sides of this equation by 11 yields
706B = 9884
Dividing both sides by 706 yields
B = 14
So we now know the value of one of our unknowns.
Step 4: Substitute the known value into equations to find the value of the other unknowns. Substituting B = 14 into equation (5) yields
37A + 7(B) = 124
37A + 7(14) = 124
37A + 98 = 124
37A = 222
—222
A— =  6
37
Substituting A = 6 and B = 14 into equation (1) yields
7(6) — 2(14) + 3C = 46
4228+3C=46
70+3C=46
Adding 70 to each side: Dividing each side by 3: So now we have solved for all the unknowns. 
3C = 24 
Step 5: Finally, check your solution by substituting the values into the original equations. If any of the equations do not hold true then you know there is an error in the calculations.
Equation (1): 7(6) 
— 2(14) 
+ 3(8) 
= 
42 
— 28 
+ 24 
= 
46 
O.K. 
Equation (2): 3(6) 
+ 5(14) 
— 4(8) 
= 
18 
+ 70 
— 32 
= 
20 
O.K. 
Equation (3): 4(6) +11(14) — 2(8) = 24 +154 — 16 = 162
ASSIGNMENT 27: Loop Equations for Complex Circuits
Read: Section 62, pages 142146 on single source circuits. The information on
pages 146150 on multiple source circuits is optional material.
Answer SelfTest questions 10 through 14. Then refer to page 176 to check your
answers.
Quick Review: When some or all of the components have no direct series or parallel relationship to other components, then circuit analysis requires solving a set of simultaneous equations.
By using Kirchoff's voltage law applied to a set of current paths, a set of loop equations can be generated involving unknown current values. After solving the set of simultaneous loop equations. the current through each component can be determined. This method works as long as each resistor is included in at least one loop.
Supplementary Material: When some resistors are directly in parallel, combine them into an equivalent resistance before applying loop analysis.
Section 63 is considered optional material.
ASSIGNMENT 28: Superposition Theorem
Read: Section 64, pages 156 and 157. The information on page 158, from Three
Loop Circuits, through page 159 is optional material.
Answer: SelfTest questions 27 through 29. Then refer to page 176 to check your answers.
Quick Review: The current through each resistor in a circuit is the sum of the currents caused by each source working independently. To determine the net effect of an individual source, all other sources are replaced by their internal resistance. Since most sources have very low internal resistances (less than 1% of the total circuit resistance) they are replaced with a wire.
ASSIGNMENT 29: Voltage Sources and Thevenin's Theorem
Read: Sections 65 and 66. pages 160163. The information on pages 164166
on complex circuits is optional.
Answer: SelfTest questions 32 through 34. Then refer to page 176 to check your answers.
Quick Review: A constant voltage source provides the same terminal voltage regardless of the load. All real sources have some internal resistance (Rs). so the terminal voltage drops when current is drawn. However, when Rs is less than 1% of the total circuit resistance, its effect can be ignored.
A power source can be modeled as a constant voltage source with an open circuit voltage ( Voc) in series with a source resistance (Rs) to form an equivalent voltage source.
Using Thevenin's theorem, a two terminal network may be converted to an equivalent voltage source. The current and voltage associated with a resistor in a complex network
is easily determined after converting the rest of the circuit (with respect to the resistor's terminals) to an equivalent voltage source.
Supplementary Material: If the physical circuit is available then the equivalent voltage source values with respect to two terminals may be found as follows: taking a voltmeter reading at the terminals provides Voc; taking an ammeter reading at the terminals provides isc (short circuit current).
The internal resistance Rs is equal to ______ .
ASSIGNMENT 30: Current Sources and Norton's Theorem Read: Sections 67 through 69.
Answer: SelfTest questions 39 through 41. Then refer to page 176 to check your answers.
Quick Review: A constant current source provides the same current through any load. A real source has some internal resistance (Rs), so the current supplied to a load is decreased when the load resistance is large.
Using Norton's theorem, a two terminal network can be converted to an equivalent current source. The voltage and current associated with a resistor in a complex circuit is easily determined after converting the rest of the circuit (with respect to the resistor's terminals) into an equivalent current source.
Any source may be treated as an equivalent voltage source with a constant voltage source ( Voc) in series with the internal resistance (Rs), or an equivalent current source with constant current (/sc) in parallel with Rs. Sources are easily converted from one type to the other since Voc = /sc x Rs.
Comparison of Techniques: To determine the voltage and current associated with a changing resistor Rx, source transformation using either Thevenin's theorem or Norton's theorem should be used. The other techniques would require the entire network to be analyzed for each value of Rx.
If finding the voltage and current of one resistor makes it a simple matter to analyze the rest of the circuit, then source transformation is very useful. When the short circuit
current through that resistor is easily determined, then Norton's theorem is usually more useful then Thevenin's theorem.
The superposition theorem is often useful when many sources are involved. When some of the components have no direct series or parallel relationship, then loop analysis is required. Sometimes a combination of analysis techniques is needed.
CHAPTER SUMMARY and REVIEW Study the chapter summary and related formulas on page 173. Answer the chapter review questions on page 174. Then refer to the Answer Key at the back of this study guide to check your answers. Chapter review problems are optional. 
ASSIGNMENT 31: Magnets Read: Sections 71 and 72.
Answer: SelfTest questions 1 through 10. Then refer to page 204 to check your answers.
Quick Review: A magnetic field is an invisible force field found around a magnet. It consists of continuous loops of magnetic flux. Flux lines leave the magnets north pole and enter the south pole. The flux is most dense at the poles.
Like poles repel. while unlike poles attract. The force of repulsion (or attraction) is inversely proportional to the square of the distance between the poles.
When two opposite poles touch, the flux joins together and the two magnets act as a single magnet.
ASSIGNMENT 32: Electromagnetism Read: Section 73.
Answer: SelfTest questions 11 through 17. Then refer to page 204 to check your answers.
Quick Review: A straight current carrying conductor has a circular magnetic field with a strength directly proportional to the amount of current. The field has no poles, but the flux direction can be determined from the direction of the current using the lefthand rule. Two parallel currentcarrying conductors attract each other when their current is flowing in the same direction, otherwise they are repelled.
Forming a coil out of a conductor creates an electromagnet, with poles which may be determined using the lefthand rule. The space between turns of a coil should be minimized since this space weakens the field produced at the poles.
ASSIGNMENT 33: Theory of Magnetism Read: Sections 74 and 75.
Answer: SelfTest questions 18 through 24. Then refer to page 204 to check your answers.
Quick Review: Magnetic materials are attracted by magnetic fields and can be magnetized. Magnetic materials have molecules with more electron spin in one direction than the other, so a small magnetic field is produced. These molecules form small clusters called domains which share a magnetic field. The alignment of the domains is randomly arranged, so the overall magnetic field is neutral.
When placed in a magnetic field, domains are aligned and the material is magnetized. If most of the domains remain aligned after the magnetic field is removed, the material has been permanently magnetized.
Pure iron, ferrite, and silicon steel make good temporary magnets. They are used as a core material for electromagnets, transformers, motors, and other applications where a magnetic field must be electrically controlled.
ASSIGNMENT 34: Magnetomotive Force, Saturation, Residual Magnetism, and Reluctance
Read: Sections 76 through 711.
Answer: SelfTest questions 25 through 32. Then refer to page 204 to check your answers.
Quick Review: Increasing the current or number of turns in a coil increases the magnetomotive force (mmf). Continually increasing the mmf in a coil will increase the flux, until the core material is saturated.
A permanent magnet may be demagnetized by heating it to a high temperature, or by placing it in a changing magnetic field and slowly removing it.
Residual magnetism refers to the flux remaining in a material when removed from a magnetic field.
Magnetic flux lines try to take both the shortest path between poles and the path of least reluctance. Since both of these conditions are improved when the material is moved closer to the poles, magnetic materials are drawn to a magnet. Low reluctance materials can guide magnetic flux around an object to act as a magnetic shield.
ASSIGNMENT 35: Generators, Transformers, Magnetic Quantities and Units. Read: Sections 712 and 713.
Answer: SelfTest questions 33 through 44. Then refer to page 204 to check your answers.
Quick Review: A changing magnetic field will induce a voltage in a conductor. The voltage is directly related to the amount and rate of flux change. Moving a conductor across lines of flux is called generator action. Using one coil to generate a changing magnetic field which induces a voltage in another coil is called transformer action.
The ampereturn is the base unit of magnetomotive force (mmf). The weber (Wb) is the base unit of magnetic flux (0). A flux change of 1 Wb in 1 second will induce 1 volt per turn in a coil.
The ampereturn per meter _{(}A t ) is the base unit for magnetic field strength (H ), and
111
refers to the mmf per meter available in a magnetic circuit.
Flux density (8 ) is the flux per unit crosssectional area of a material, and has as its base unit the tesla (T). One tesla is one weber per square meter.
Permeability is the ratio of flux density to magnetic field strength (p = —^{B}), and has a base unit of weber per meter per ampereturn t Whi
.• t • in).
Relative permeability is the ratio of a material's permeability to that of air. As flux density increases in a magnetic material. its permeability decreases.
ASSIGNMENT 36: Electromagnets, dc Motors, Solenoids, and Relays Read: Sections 714 through 718.
Answer: SelfTest questions 45 through 55. Then refer to page 204 to check your answers.
Quick Review: An electromagnet is strongest when the core has low reluctance and high permeability, and the coil has high mmf.
Torque is generated in a dc motor due to the attraction and repulsion between the armature and the field poles. A commutator and brush arrangement is used to change the poles of a rotating armature with respect to the constant field poles.
A solenoid is an electromechanical device which pulls a magnetic plunger into a coil. The pull is determined by the same factors as any electromagnet.
A relay is an electromagnetically activated switch. Relays have many different switch arrangements such as DPDT, DPST NO, etc. Both coil and contacts have voltage and current ratings. Both relays and solenoids come in ac and dc varieties.
FUNDAMENTALS OF ELECTRICITY Examination 3Continued
Student's Name________________________________ Student Number___________________
For each of the following circle T if the statement is true, F if it is false.
22 T F There is only one proper selection of current loops in a loop analysis of a single source circuit.
Answer the questions on the following pages in the space provided. Write your final answer on the blank lines provided and show your main calculations below. To receive full credit for an answer you MUST show your main calculations. Partial credit can also be given for incorrect answers that show some understanding of the concepts involved in solving a problem if the main calculations are provided.
Answer: A = _____________________________________ bind with (2) to eliminate vari
able C, and solve for variable A.
B = ________
C=________
Examination 3Continued
(loop 1) : 
12 V = 
150 Q 
(1_{k}) 
+ 
0 
(1,) 
+ 100 Q 
(1_{3}) 
12 V = 

(/_{i}) 
+ 

(1_{2}) 
+ 
(1_{3}) 

12 V= 


+ 

(/_{2}) 
+ 
(1_{3}) 









(loop 1): V_{T} = V_{R} V_{R 2}
12 V = 100 Q + /_{3}) + 50 Q (1,) = 150 Q (1_{k}) + 100 Q (1_{3})
(loop 2): V_{T} = V_{R3} + V_{R4}
(loop 3): V_{T} =
(More)
FUNDAMENTALS OF ELECTRICITY Examination 3—Continued
Student's Name________________________________ Student Number___________________
RT = 6Q, /R1 = 1A, /R2 = 1A, /R3 = 2A)

R2 R, 4f) 
R3 
_L__ _{B2} 12V 




4 S 2 










Examination 3Continued
(a) _____________ I_{N}
(More)
FUNDAMENTALS OF ELECTRICITY Examination 3Continued Student's Name____________________________________________ Student Number___________________
(b) ____________ RN
FILL IN YOUR NAME AND THE OTHER REQUIRED INFORMATION ON EACH PAGE OF THE EXAMINATION AND MAIL THE EXAMINATION TO THE SCHOOL.
UNIT 4
ASSIGNMENT 37: AC Waveforms
Read: Sections 81 through 83.
Answer: SelfTest questions 1 through 6. Then refer to page 230 to check your answers.
Quick Review: An ac voltage produces an alternating current across a load. Alternating current periodically changes direction.
An electrical waveform is often shown as a graph. with voltage, current, or power shown on the vertical axis and time shown on the horizontal axis.
AC waveforms include sine, square, and sawtooth waves. The sinusoidal waveform is commonly used. It has a continuously changing value.
A square wave periodically alternates between two constant values. A sawtooth wave has a magnitude that is continuously changing, but the rate of change is usually constant.
ASSIGNMENT 38: AC Terms
Read: Section 84.
Answer: SelfTest questions 7 through 17. Then refer to page 230 to check your answers.
Quick Review: A cycle of a periodic waveform is the largest part that does not repeat itself. It is composed of a negative alteration and a positive alteration (1 cycle = 2 alterations).
The period (T) of a waveform is the time required to complete one cycle. The frequency (f) is the rate at which cycles are produced, and has units of hertz (Hz). One hertz is one cycle per second. The frequency and period of a waveform are reciprocals of each
other
f T
The voltage of a sinewave may be expressed as peak voltage (Vp), peaktopeak voltage (Vp  p), average voltage ( Vav), and effective or rms voltage ( Vms). These values have the following relationships :
= 2V„ V 0.637V„ V 0.707V,
Whorl an ac voltage or current value is specified, the value is assumed to be the rms value unless otherwise indicated. This rms value supplies the same power to a resistive
load as the equivalent dc value. Thus the rms value of ac voltage or current is the effective value for calculating power.
ASSIGNMENT 39: AC Sinewave Generators Read: Sections 85 through 87.
Answer: SelfTest questions 18 through 32. Then refer to page 230 to check your answers.
Quick Review: When a coil is rotating in a perfect circle, at a constant speed, in a uniform magnetic field, a perfect sinusoidal voltage is produced. One cycle of the coil produces 360 electrical degrees. The magnitude of the output voltage is dependent on the field flux density, the number of turns in the coil, and the speed of rotation.
Most generators have more than one pair of field poles, so each revolution produces more than one output cycle. The electrical degrees are a multiple of the mechanical degrees.
The frequency of the voltage waveform produced by an ac generator is dependent on the number of field coils, and the rotational speed of the coils as follows :
r/min
60____ x (11 of pairs of poles)
There are many advantages to using ac rather than dc for commercial power. AC voltages are easily converted using transformers. The constant frequency is used to control the speed of clocks and motors. AC generators and motors are simpler and more reliable than their dc counterparts.
ASSIGNMENT 40: ThreePhase AC
Read: Section 88.
Textbook correction: On page 217 in Figure 820(a), change the bottom line to Line 3.
Answer: SelfTest questions 33 through 46. Then refer to page 230 to check your answers.
Quick Review: Electricity is usually produced and transmitted from a power plant as threephase sinusoidal alternating current. Threephase power is created using three separate windings on the armature of a generator. The threephase coils produce sinusoidal voltages that are 120 electrical degrees out of phase from each other. At any instance, the sum of the three coil voltages (or currents) is zero.
The coils of a threephase generator may be connected in a delta or wye configuration. In a delta connection, the phase windings are connected in a continuous loop. Three transmission lines are connected to the phase coil terminals. The voltage across any
two lines is equal to the phase coil voltages. However, the line currents are 1.732 times the phase coil currents.
In a wye connection, the phase coils are connected to a neutral central node (star point), and three transmission lines are connected to the other coil terminals. The line currents are equal to the phase coil currents. However, the voltage between two lines is 1.732 times the phase coil voltage.
Most threephase loads are balanced loads since they draw the same power from each pair of conductors. For balanced loads, three transmission lines are sufficient. In order to power a nonbalanced load, a wye connection is used with a fourth line connected to the center node (star point). There are many advantages to using a threephase system of power generation and distribution over a singlephase system.
CHAPTER SUMMARY and REVIEW Study the chapter summary and related formulas on page 227. Answer the chapter review questions and complete the chapter review problems on pages 228 and 229. Then refer to the Answer Key at the back of this study guide to check your answers. 
ASSIGNMENT 41: Power in AC Circuits
Read: Sections 91 and 92. (Also read the two following pages for a quick review
of the basics of trigonometry of the right triangle, which is the basis of some of the material in Section 92.)
Answer: SelfTest questions 1 through 27. Then refer to pages 248 and 249 to check your answers.
Quick Review: Reactance is the opposition to current caused by capacitors and inductors. Capacitive reactance causes the current to lead the voltage by 90°, while inductive reactance causes the current to lag the voltage by 90". A purely reactive load uses no power.
A QUICK REVIEW OF TRIGONOMETRY OF THE RIGHT TRIANGLE
Trigonometry deals with the relationships among the 3 sides and 3 angles of triangles. It is based on the concept of similar triangles. If all the corresponding angles of two
triangles are the same, then the two triangles are said to be similar. The lengths of their sides are not necessarily the same, but they will be proportional. For example, take the two triangles illustrated below.
r. 
In these triangles Lb = /e and La = Zd, Therefore the two triangles are similar. In addition, although it is obvious by looking at the two triangles that the length of their sides are not equal, the length of their sides are proportional, that is the ratio —^{A }= ^{B C} D E F 

A 

The length of side A is .5 inches, the length of side D is 1 inch, the length of side B is .5 inches. therefore the length of side E is 1 inch. (Use a ruler and try out the measurements for yourself.)
This concept of similar triangles is particularly useful when applied to right triangles. A right triangle is a triangle which has as one of its angles an angle which is 90°. Our two triangles above are right triangles. The box in the corner indicates the 90° angle. Thus, in a right triangle two sides are perpendicular to each other and the third side, called the hypotenuse, is the longest side.
As the 3 angles in any triangle total 180°, in a right triangle the 2 acute angles (the two angles which are not 90°) must together total 90°. So if you know the value of one acute angle you can determine the value of the other and if the acute angle of one triangle equals the acute angle of another, the two triangles are similar. And if the two triangles are similar, their sides are proportional. Each of the possible proportions or ratios of the sides has been given a name and is defined in terms of the angles, illustrated and described below.
Opposite side
Adjacent side
In your textbook the only angle which matters is the angle the textbook calls angle 6, illustrated above (6 is the Greek letter theta). Therefore, we will define the possible relationships, called trigonometric functions, in terms of this angle:
length of opposite side
sine 0 (usually written sin, and pronounced "sign") = _______________________
length of hypotenuse
cosine 8 (usually written cos) length of adjacent side
length of hypotenuse
tangent 6 (usually written tan) length of opposite side
For each possible degree of the angle 6 there is a specific value, which you do not have to compute. They are given in tables of trigonometric functions such as that in Appendix G at the back of your textbook. Using the values from this table and knowing the length of one of the sides, you can use the appropriate ratio above and solve for the length of either of the other two sides.
In a mixed load (reactance and resistance), only the resistive component uses power. The phase shift is between 0° and 90°, and is dependent on the relative amounts of resistance and reactance.
The power used by a load at any instant is equal to the product of the voltage and current at that instant. The instantaneous power used by a mixed load will be negative at some portion of each cycle, but the net power will be a positive value.
The power used by a mixed load can be easily calculated as the product of the voltage and current components which are in phase. This is done using phasors and right triangle relationships.
A phasor represents the magnitude and phase angle of an electrical quantity at an instant in time. A phasor is like a vector, only it changes with time, rotating in a counterclockwise direction. A phasor can be broken into a component at 0° and another at 90° or 270° using triangle relationships:





If C and 6 represent the magnitude and phase angle of a phasor, then the component at 0° is side A, which equals C x cosh. The component at 90° is side B, which equals C x sin 6.
If 6 is the phase difference between a load current and the applied voltage, then the total current, resistive current, and reactive current could be represented by sides C, A, and B respectively. The power used by the load is the voltage times the current represented by side A. If 8 is the phase difference between the voltage ( V) and current (/ ) through a load, then the power may be calculated as follows :
Power = 1 • V • cos q
Supplementary Material: Another property of the above triangle is that A' + B^{2} = 0^{2}. This equation is known as the Pythagorean Theorem. Thus if values for two sides are known, the third side is easily determined. For example. if the total load current is 5 amperes. and the resistive current is 4 amperes, then the reactive current is 3 amperes. (Solving for B you get 4^{2} + B^{2} = 25; B^{2} = 25  16 = 9; B^{2}= 9: B = 3.
ASSIGNMENT 42:Tnie Power, Apparent Power, and Power Factor
Read: Sections 93 and 94.
Answer: SelfTest questions 28 through 43. Then refer to page 249 to check your answers.
Quick Review: The true power of an ac circuit can be measured with a wattmeter. Calculating power from separate readings of circuit voltage
( V) and current (I) results in a power value that ignores the phase difference. Power calculated as V/ is called apparent power, and has base unit voltampere (VA).
The power factor (PF) of a circuit is the ratio of its apparent power (Papp) to its true power (P).
Since p=vxIx cosq then Pr = — =cosq
app
The power factor can range from 0 (in a purely reactive load) to 1 (in a resistive load).
When the current and voltage are out of phase. more current is required to provide the same power. Thus it is more difficult for electric utilities to supply power at a low power factor.
Section 95 is optional material.
CHAPTER SUMMARY and REVIEW Study the chapter summary and related formulas on page 246. Answer the chapter review questions and complete the chapter review problems on pages 246 and 247. Then refer to the Answer Key at the back of this study guide to check your answers. Problems 910 and 911 are optional. 
FUNDAMENTALS OF ELECTRICITY Examination 4Continued
Student's Name__________________________ Student Number______________
Answer the following questions in the space provided. Write your final answer on the blank lines provided and show your main calculations below. To receive full credit for an answer you MUST show your main calculations. Partial credit can also be given for incorrect answers that show some understanding of the concepts involved in solving a problem if the main calculations are provided.
(c) 

Vey 

(Over)
(a)___________ Resistive current
(b)___________ Reactive current
(c)_____________ Power used by the load
(More)
FUNDAMENTALS OF ELECTRICITY Examination 4Continued
Student's Name_________________________ Student Number_________________
(a)___________ Apparent power
(a)____________ Apparent power (Hint: P = Papp • PF)
(b)____________ The rms value of the current (Hint: Papp = I • V)
(Over)
Examination 4Continued
(a)________ Total current (Hint: Use the Pythagorean Theorem.)
(b)_______________ Phase angle (9 ) (Hint: tan B = ^{°}PP ) adj 


12A 
FILL IN YOUR NAME AND THE OTHER REQUIRED INFORMATION ON EACH PAGE OF THE EXAMINATION AND MAIL THE EXAMINATION TO THE SCHOOL.
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