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UNIT 3

ASSIGNMENT 26: Simultaneous Equations Read:       Section 6-1.

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 variable-elimination 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 8A – 22B + 4C = -324 +  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

 629B                            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

-42-28+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 C = 8

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 6-2, pages 142-146 on single source circuits. The information on

pages 146-150 on multiple source circuits is optional material.

Answer Self-Test questions 10 through 14. Then refer to page 176 to check your

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 6-3 is considered optional material.

ASSIGNMENT 28: Superposition Theorem

Read:            Section 6-4, pages 156 and 157. The information on page 158, from Three-

Loop Circuits, through page 159 is optional material.

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 6-5 and 6-6. pages 160-163. The information on pages 164-166

on complex circuits is optional.

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 6-7 through 6-9.

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 7-1 and 7-2.

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 7-3.

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 left-hand rule. Two parallel current-carrying 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 left-hand 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 7-4 and 7-5.

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

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 7-12 and 7-13.

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 ampere-turn 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 ampere-turn 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 cross-sectional 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 ampere-turn 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 7-14 through 7-18.

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 electro-mechanical 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 3--Continued

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.

1. T F A constant current source provides a constant voltage at its terminals under all load conditions.
2. T F9 If an electrical current were flowing straight at you through a straight conductor, the direction of the flux would be counterclockwise.
3. T CF Two parallel conductors carrying current in the same direction produce a force of
4. CT-. F Temporary magnet material has low residual magnetism.
5. (I' F When a conductor is moved across lines of flux, a voltage is induced.
6. F DC motors operate on the principle of transformer action.
7. T (1= Solenoids and relays operate on the principle of the Hall-effect.

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.

1. Solve for A, B, and C when : (1) -2A + B — 3C = -9 Hint: Combining (1) and (3)
• 2A — C = -1           will eliminate variable This
• A — B + 4C = 11 new equation can then be corn-

Answer: A =  _____________________________________ bind with (2) to eliminate vari-

able C, and solve for variable A.

B = ________

C=________

Examination 3--Continued

1. For a loop analysis of the circuit illustrated below, assume passes through R1 and R2, I, passes through R3 and R4, and 13 passes through RI, R5, and R4. Add the resistor values to complete the loop equations below. Show your work in the space below. The first one is done for you. (Hint: See solution for Figure 6-3.)
 (loop 1) : (loop 2) : (loop 3) : 12 V = 150 Q (1k) + 0 (1,) + 100 Q (13) 12 V = (/i) + (12) + (13) 12 V= + (/2) + (13)

(loop 1): VT = VR VR 2

12 V = 100 Q + /3) + 50 Q (1,) = 150 Q (1k) + 100 Q (13)

(loop 2): VT = VR3 + VR4

(loop 3): VT =

(More)

FUNDAMENTALS OF ELECTRICITY Examination 3—Continued

Student's Name________________________________ Student Number___________________

1. In the circuit on the preceding page, /I= 0.08332 A, /2= 0.18653 A, and /3= -0.005 A. Determine the following values.
• ___________ 1R4
• ___________ V1
1. Use the superposition theorem to analyze the circuit diagrammed below to determine the resistor currents. (Hint: See solution for Figure 6-17. Shorting out Bi gives

RT = 6Q, /R1 = -1A, /R2 = 1A, /R3 = 2A)

 R2 R,           4f) . 412 _ R3 _L__ B2  12V 4 S 2

•  IR I
•  /R2
•  1R3

Examination 3--Continued

1. Use Thevenin's theorem to analyze the circuit on the left in order to determine the values of the Thevenin equivalent circuit on the right. (Hint: See Figure 6-23.)

• __________
• __________
1. Use Norton's theorem to analyze the circuit in question 34 in order to determine the values of an equivalent current source. (Hint: See Figure 6-30.)

(a) _____________  IN

(More)

FUNDAMENTALS OF ELECTRICITY Examination 3--Continued Student's Name____________________________________________ Student Number___________________

(b) ____________  RN

1. A magnetic circuit has a 20 cm length core with a permeability of 0.005 Wb/(A • t • m). The coil has 250 turns, and draws 0.4 ampere. Determine the following:
• ________ the mmf of the coil
• __________ the magnetic field strength
• __________ the flux density (Hint: B =RH )

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

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

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), peak-to-peak 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 8-5 through 8-7.

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: Three-Phase AC

Textbook correction: On page 217 in Figure 8-20(a), change the bottom line to Line 3.

Quick Review: Electricity is usually produced and transmitted from a power plant as three-phase sinusoidal alternating current. Three-phase power is created using three separate windings on the armature of a generator. The three-phase 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 three-phase 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 three-phase 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 non-balanced load, a wye connection is used with a fourth line connected to the center node (star point). There are many advantages to using a three-phase system of power generation and distribution over a single-phase 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 9-1 and 9-2. (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 9-2.)

Answer: Self-Test 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

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:

 sinq = cosq = tang =

 8

 A

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' + B2 = 02. 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 42 + B2 = 25; B2 = 25 - 16 = 9; B2= 9: B = 3.

ASSIGNMENT 42:-Tnie Power, Apparent Power, and Power Factor

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 volt-ampere (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 9-5 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 9-10 and 9-11 are optional.

FUNDAMENTALS OF ELECTRICITY Examination 4--Continued

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.

1. A standard electrical outlet provides a 120-V, 60-Hz sinewave. Determine the following:
• Time required for one cycle
• Vp p (Hint: V. = 120 V, find Vp first)
 (c) Vey

1.  If a 600042 resistor is connected to the outlet described in question 29,
how much power would be used?
2.  What is the frequency produced by a 4-pole generator rotating at
1800 r/min ?

(Over)

1.  In a three-phase system, what is the instantaneous voltage of phase 2 when
phase 1 is -85 volts and phase 3 is 116 volts? (Hint: See page 227 #18.)
2.  A generator connected in a four-wire system provides 208-V three-phase
What is the phase voltage? (Hint: See Figure 8-22.)
3. A load draws 10 A of current from a 120-V source. The voltage leads the current by 36.9°. Determine the following:

(a)___________ Resistive current

(b)___________ Reactive current

(c)_____________ Power used by the load

(More)

FUNDAMENTALS OF ELECTRICITY Examination 4--Continued

Student's Name_________________________ Student Number_________________

1. The measured power, current, and voltage delivered to a load are 100 W, I A, and 120 V respectively. Determine the following:

(a)___________ Apparent power

• Power factor
• Phase angle (0 (Hint: PF = cos e)
• Reactive current
1. A manufacturing plant uses 2.1 MW of 220-V single phase power at a power factor of 0.8. Determine the following:

(a)____________ Apparent power (Hint: P = Papp • PF)

(b)____________ The rms value of the current (Hint: Papp = I • V)

(Over)

Examination 4--Continued

1. The resistive and reactive currents in a load are 12 A, and 5 A respectively. Determine the following:

(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.

Electricity Engineering Pricniples

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