Alternating Current by Robert Cecci CONTENTS

Study Unit Alternating Current By Robert Cecci CONTENTS

INTRODUCTION

SECTION 1: INTRODUCTION TO ALTERNATING CURRENT 1 Uses of Alternating Current 2 Alternating Current Versus Direct Current 2 Generation of Alternating Current 4 Elementary Alternator 6 Generation of Voltage Cycle 7 Sine Wave 8 A Cycle of Alternating Current 8 Frequency 9 Time Measured in Degrees 10

SECTION 2: CHARACTERISTIC VALUES OF THE AC CYCLE 14 Types of Characteristic Values 14 Instantaneous Values 15 Maximum and Peak-to-Peak Values 16 Average Value 17 Effective Value 17 Importance of Effective Value 20

SECTION 3: SINGLE-PHASE ALTERNATING CURRENT 23 Opposition to Alternating Current 23 Phase Angle 23 Phase Angle in Resistive AC Circuits 25 Phase Angle in Reactive AC Circuits 26 Power in Resistive AC Circuits 27 Power in Purely Reactive AC Circuits 30 Power in Partially Reactive AC Circuits 32 Apparent Power in an AC Circuit 34 Real Power in AC Circuits 37 Reactive Power in AC Circuits 38 Application of Power Formula 38 Power Factors in Industry 42

SECTION 4: POLYPHASE ALTERNATING CURRENT 45 Polyphase Systems 45 Single-Phase System 46 Three-Phase Circuits 46 Star- or Y-Connected Three-Phase Systems 47 Delta-Connected Three-Phase Systems 48 Power in Three-Phase Systems 49

SELF-CHECK ANSWERS 51

This study unit covers the most common form of electric power used in homes, businesses, and industry: AC current. AC current is used in industry to power computers, control systems, ovens, motors, and there are many more applications.

We’ll begin with a study of the basic characteristics of alternating current and the values used to describe AC cycles. You’ll then be presented with information on single-, split-, and three-phase AC current.

When you complete this study unit, you’ll be able to QQ Draw a graph of an AC voltage and describe how AC voltage is created

QQ Explain what an AC cycle is using the terms alternation, peak, positive, and negative

QQ Express the time period of an AC cycle in degrees

QQ List the characteristic values of an AC cycle and describe the relationship between the values

QQ Define phase angle and describe how it relates to reactive circuits

QQ Calculate power for single-phase and three-phase circuits

QQ Describe how a 220 VAC, single-phase circuit operates

QQ Calculate the phase and line voltages of multiphase wave forms

QQ Determine real power by reading a power factor meter

QQ Describe delta-connected and wye-connected three-phase circuit connections

USES OF ALTERNATING CURRENT

Our modern way of life depends on electricity: we use electrically operated machines and devices in our work and recreation, and nearly all manufactured products are produced with the aid of electricity.

Most electric devices and machines are powered by alternating current (AC). Alternating current is very different from direct current (DC). In a DC circuit, there’s a steady flow of electrons from the negative terminal of the generator, power supply, or battery. These electrons flowing through the circuit or load are attracted to the positive terminal of the supply.

In an AC circuit, the flow of electrons reverses periodically. You may have heard of 60 cycle current delivered by your local utility to your plant or home. 60 cycles means the current begins as a positive cycle and then reverses to a negative cycle 60 times per second.

AC is widely used to provide electricity for lighting, for the majority of household appliances, and for most industrial motors, controllers, ovens, and process systems. Because of the widespread use of AC electric energy, it’s something you’ll want to know about and be able to work with. Before you study the applications of alternating current, though, you should be familiar with its general characteristics.

ALTERNATING CURRENT VERSUS DIRECT CURRENT

AC power can be produced and transmitted at a lower cost than DC power. Also, AC can be distributed more conveniently than DC. For these reasons, AC is a widely used choice.

An alternator, or AC generator, is a machine that produces or generates AC voltage. It can be designed and built to produce higher voltages and to have higher power capacity than any DC generator of the same size. Also, the larger an alternator is, the more efficient it is.

It’s relatively simple to generate a high alternating voltage and, using transformers, step it up to a still higher voltage and, finally, transmit it through power lines over long distances to the place where it will be used. For instance, electric power systems can generate voltage at about 1300 volts (V); step this power up to 500,000 V or more for transmission;

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 2 Section 1 and, finally, step it down to 480, 240, or 120 volts alternating current (VAC) at the points of industrial use. Very high voltage is desirable in a transmission line because it makes it possible to reduce the power loss in that line.

A low current at a high voltage and a high current at a low voltage transmit the same amount of power. During any power transmission, losses occur mostly due to the heating of conductors due to the conductor’s resistance. Power lost due to the heating of conductors can be determined by using the following formula:

P 5 I 2R

In this formula, the letter P stands for the power loss in watts, the letter I stands for the current in amperes, and the letter R stands for the resistance of the conductors in ohms.

Resistance in a transmission line is constant, but we can change the current. As you can see in the formula, decreasing the current will substantially lower the power loss. Therefore, the current in transmission lines should be kept as low as possible. Sending the desired electric power through the line at very high voltage will lower the current and reduce the overall power level.

Because step-up transformers can easily raise the voltage in AC circuits to high values, and in this way lower the power loss, AC is much more economical than DC for transmission of electric power.

Transformers allow for the convenient distribution of alternating current within an industrial plant. The high transmission voltage is first stepped down by transformers. Then, the transformed electric power is applied wherever it’s needed, at a safer and more convenient low voltage such as the common 480, 240, or 120 VAC used for general-purpose industrial circuits.

In the early days of electric power generation and distribution, such an aggressive competition raged over how our electric power was to be distributed that it was referred to as a “war.” The Edison Electric Light Company wanted to distribute low voltage DC current to homes and businesses. On the other side of the distribution competition, the Westinghouse Electric Company wanted to distribute electric power as alternating current. At that time, alternating current was widely believed to be dangerous. This idea came from the highly publicized electrocution of relatively few people by the high transmission voltages present on the electricity distribution poles found in cities with AC power systems. The competition went so far that the Edison company publicized the fact that AC power was used in the electric chair.

Thankfully, the advantages of AC power generation and transmission gradually won out and today we rely on an AC power distribution system. An AC transformer, an extraor- dinarily simple and efficient device, makes it easy to increase or decrease voltages to any required value. Meanwhile, changing voltage level in a DC system usually requires a number of electronic devices connected in a circuit to convert DC to AC. Once the DC is converted to AC, it’s regulated to the desired voltage level then rectified, or converted back to DC.

Most of the alternating current used in home and industry is generated by alternators. Alternators operate on the principle of electromagnetic induction. The generating action of electromagnetic induction occurs whenever there’s a relative motion between a conductor and a magnetic field. As the result of electromagnetic induction, a voltage is induced in the conductor and, if the conductor is a part of a closed circuit, a current will flow through the conductor.

The direction of the conventional current (current flow from positive to negative) induced in a conductor by electromagnetic induction can be found by Fleming’s right-hand, or generator, rule. Do you know that rule? It’s illustrated in Figure 1. Here’s how you can apply it. First, using your right hand, aim your index, or pointing, finger in the direction of the magnetic field. At the same time, aim your thumb in the direction in which the conductor is moving through the magnetic field. Your thumb and index finger will now be at a right angle to each other. Now, point your middle finger so that it’s at a right angle to both your thumb and your index finger. Look again at Figure 1 to see how to hold your fingers. If you follow directions, your middle finger will point in the direction of the voltage induced in the conductor. That’s the same direction as that of conventional current flow through the conductor.

Conductor Direction

Current

Magnetic Field

S N

Magnetic Field

Current

FIGURE 1—The right-hand rule is illustrated here. The direction of the magnetic field flowing from the North Pole (N) to the South Pole is indicated by the index finger. The upward motion of the conductor is indicated by the thumb. The direction of the conventional current is toward you, as indicated by the middle finger.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 4 Section 1 In an alternator, a group of conductors revolves in a magnetic field. Because of the continuously changing position of the conductors relative to the lines of force, the induced voltage is continuously changing, or alternating, and the current output of the alternator is an alternating current.

The value of the current at any instant is its instantaneous value. The instantaneous values of any alternating current change in direction and in magnitude at regular intervals. The current rises from zero to a maximum value, maximum positive for example, then decreases to zero. The current then increases again, but in the opposite direction, to a maximum negative value, then returns to zero again. Consider one set of such changes, or reversals, to be one change from zero to maximum positive to zero and another change from zero to maximum negative to zero. This set of changes is a cycle.

In an AC circuit, the current doesn’t stop at just one cycle. The cycles are repeated over and over because they’re generated continually by the alternator. Thus AC voltages and currents consist of many cycles. One such cycle is shown in Figure 2. This picture, or graph, displays not only AC current but also the resulting AC voltage.

Maximum Positive

Rise Decrease

Maximum Negative

One Cycle

FIGURE 2—Here’s one cycle of AC power. Note that this waveform reaches a positive maximum then crosses zero and reaches a negative maximum then returns to zero.

Understanding the simultaneous creation of AC current and voltage is very important. DC current and voltage in a given circuit are linked; increasing voltage may be linked to a decrease or increase in current, depending on the circumstance, but the change in voltage and current levels happens at the same time. Meanwhile, changes in AC current and voltage levels can occur on a different time schedule. It’s useful to think of this scheduling difference between changes in AC current and AC voltage levels as a time shift. This AC-current-versus-AC-voltage time shift can occur in any circuit containing capacitors and/or inductors.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 5 Section 1 Remember, a capacitor’s plates must first build a difference in charge before the electrical energy stored in the capacitor is released. The time it takes to build this charge results in a time delay between the steady current flow through the capacitor and the built-up voltage leaving the capacitor. It’s important for you to remember that in an AC circuit incorporating capacitors, voltage change occurs after current change. Conversely, in an AC circuit with an inductor, through which voltage easily passes but current is delayed, voltage change occurs before current change. As you progress through this study unit, you’ll learn much more about the shifting relationship between voltage and current in AC circuits.

ELEMENTARY ALTERNATOR

Let’s see how a cycle of an AC voltage is generated. To keep the discussion simple, we’ll use an elementary alternator. This elementary alternator has only one conductor coil and uses a permanent magnet to provide the field poles. Commercial alternators have many coils and use electromagnets, rather than permanent magnets, for the field.

In the elementary alternator shown in Figure 3A, the magnetic field is provided by a stationary horseshoe magnet. The direction of the magnetic lines of force is indicated by vertical arrows, leading from the north pole (N) of the magnet down to its south pole (S).

A one-turn coil is connected to two slip rings. The coil and the rings are mounted on a shaft (not shown), which is made to rotate by an outside source of mechanical power, such as a motor driven by gasoline or diesel fuel or by a turbine powered by water flow or steam. As the shaft rotates, the sides of the coil cut the lines of force. The carbon brushes are held stationary against the slip rings, connecting the outside circuit to the coil.

Permanent Terminals Magnet Brushes To Outside Circuit ltage N Time Vo 3 3’ 2 4 Axis 1 5 78910110 0 6 11 7 021 3 4 10 8 Slip Rings 9 9’ One-Turn Coil S

Elementary Alternator One Cycle of Induced Voltage (A) (B)

FIGURE 3—One revolution of the coil in the elementary alternator in 3A generates the varying voltage shown in 3B. The wavy line in 3B represents the alternating voltage in the outside circuit.

An illustration of how an AC voltage changes during one complete cycle and an end view of the alternator are shown in Figure 3B. The positions of the alternator conductor as it rotates are numbered from 0 to 11 at the left in Figure 3B. The lines of force are indicated by the lines between N and S.

The instantaneous values of the voltage during one cycle are shown on the time line, or axis, at the right in Figure 3B. The axis is divided into the same number of points as the conductor has positions. This numbered axis—0 through 11, and back to 0—represents the time it takes the conductor to complete one cycle. The vertical distances above and below the time axis represent the values of the voltage or current induced at the corresponding positions of the conductor.

When the conductor is moving past position 0 in Figure 3B, its motion is parallel to the lines of force, so no voltage or current is induced in it, and voltage and current are zero at point 0 on the time axis. As the conductor passes position 1, cutting a number of lines of force, a corresponding voltage and current are induced. A horizontal broken line is drawn parallel to the time axis from conductor position 1. A vertical line from point 1 on the time axis to the horizontal broken line represents the induced voltage and current at that instant. In position 2, the conductor cuts more lines of force than in position 1. Therefore, the vertical distance at point 2 on the time axis, representing the induced voltage and current, is higher than at point 1.

As the conductor passes position 3, its motion is at right angles to the lines of force, and it cuts the largest number of lines. The induced voltage and current shown at the positive peak, or 3’, has, therefore, the maximum possible value. At position 4, fewer lines are cut. The number of lines is the same as at position 2. Therefore, the induced voltages and currents at points 2 and 4 are of the same value.

At position 5, there’s a further decrease in lines cut, and the induced voltage and current are the same as those generated at position 1.

When the conductor reaches position 6, no lines of force are cut, and the induced voltage and current are zero as they were at the start of the cycle. Voltages and currents in all positions from 0 through 6 are generated in the same direction. All are represented by distances above the time axis which means they’re positive.

After the conductor has passed position 6, the voltages and currents induced are in the opposite direction. These voltages and currents are represented by distances below the time axis which means they’re negative. The voltages and currents have increasing negative values as the conductor moves past positions 7 through 9. The maximum negative voltage and current are reached at 9, as indicated by the negative peak, or 9, shown below the time axis. The values then decrease through positions 10 and 11, until zero voltage and current are induced at the end of one full revolution of the conductor.

The wavy line, or curve, that connects all points reached by the voltage or the current during a given period of time is the sine wave. It’s a graph showing the changes in AC voltage or current. The sine wave is very helpful in a discussion of alternating voltages and currents. In Figure 3B, only one set of reversals is shown. Such a set is a cycle. The sine wave, though, contains many cycles. They repeat themselves, again and again, as long as the voltage or current is generated.

Both the voltage and the current in an AC circuit vary in the form of sine waves. In Figure 3B, a voltage curve is shown, but the changes in the current flow correspond to the changes in the voltage.

A CYCLE OF ALTERNATING CURRENT

A graph of one cycle of alternating current is shown in Figure 4. The horizontal axis indi- 1 cates the time in seconds. In this example, the duration of one cycle is ⁄60 second (s). Thus, in one second there are 60 such cycles, which gives us the term 60 hertz (Hz). One hertz is one complete cycle consisting of a positive and negative alternation. The word hertz is used in honor of the scientist Heinrich R. Hertz who’s famous for experimenting with electricity in the nineteenth century.

Cycle Alternation Alternation 100 4 75

Positive 50 3 1 25 13 1 240 240 240 60 0 2 1 Time, In Seconds 25 Current, in Amperes Current, in 50

Negative 75

100 5 One Cycle of Alternating Current

1 FIGURE 4—One full cycle of a 60 Hz sine wave takes ⁄60 second. Try to imagine just how fast the variations are: 60 sets of changes completed every second!

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 8 Section 1 In Figure 4, observe that zero values of the current occur at the beginning of the cycle, 1 or at point 0; then again ⁄ 20 second later, at point 1; and finally, at the end of the cycle, at point 2. The amount of current changes every instant. The consecutive values of current are represented by the distances between the points on the sine wave and the horizontal time axis. The distances, measured in amperes (A), are indicated on the vertical scale through the point 0. Values above the horizontal axis are positive; values below are negative.

The current starts from point 0. At point 3, it has reached a value of 50 A in the positive direction. The current reaches the highest point in the positive direction (100 A) at 1 point 4 which occurs ⁄240 second after the beginning of the cycle. After that point, the current starts to decrease in value, until it reaches zero amperes at point 1. Here the current reverses its direction, and starts to increase in the opposite direction until, at point 5, it 3 reaches the maximum negative value (100 A negative). That occurs ⁄240 second after the beginning of the cycle. In the last fourth of the cycle, the current is still negative, but it decreases gradually to zero amperes at point 2.

One half of a cycle, the change of current or voltage from one zero point to the next, is an alternation. There are two alternations in one cycle. The highest value a current or a voltage may reach in a cycle is the maximum value, or peak value, of the cycle.

FREQUENCY

As previously mentioned, the direction of the flow of alternating current changes continually. One complete change of alternating current is a cycle, and the frequency is the number of cycles completed in one second. Measurements of frequency are given in hertz.

A frequency of 60 Hz is used almost exclusively for lighting and power applications in the United States and Canada. In South America, Europe, and Asia, AC power of 50 Hz is widely used.

Engineers now design most modern electrical equipment—such as relays or contac- tors, transformers, and motors—to operate on 50 Hz or 60 Hz, AC power. Some slight excess heating may result from operating a transformer, relay, or motor on 60 Hz power if it’s designed for 50 Hz power. Also, motors that are designed for use on either speed will rotate faster on 60 Hz power. Normally, the two speeds will be listed on the motor’s nameplate.

While you can use a 60-Hz power source to operate equipment designed for 50-Hz power, some equipment intended for 60-Hz power won’t operate correctly at 50 Hz. For instance, the magnetic core of electromagnetic equipment rated for 60 Hz will saturate much more quickly when operated with a 50-Hz power source. When operated with an incorrect 50-Hz power supply, the magnetic material will heat up faster and may be damaged by over-heating. One good example of this potential problem is an electrical solenoid intended to operate at 60 Hz. If exposed to 50-Hz power, the solenoid will heat up and be destroyed.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 9 Section 1 If you must operate 60 Hz electromagnetic equipment in a location served by a 50-Hz power source, it’s best to use a frequency inverter, which will convert the 50-Hz AC power to 60 Hz.

TIME MEASURED IN DEGREES

Changes in current or voltage happen in fractions of a second. However, the time within a cycle isn’t usually measured in fractions of a second, as it was in Figure 4. Instead, it’s much more practical to measure the time in degrees, as shown in Figure 5.

+ lts Vo

210 240 270 300 330 360 0 30 60 90 120 150180 Time, in Degrees

Cycle Divided in Degrees

FIGURE 5—Each of the twelve time intervals shown in Figure 3B corresponds to an interval of 30° shown in this cycle. Thus, the time of 360° in a full cycle corresponds to one full revolution of the coil.

The time needed for completion of one cycle depends on the rate of change, or frequency, if it’s measured in seconds. However, without knowing the duration of one cycle in fractional parts of a second, it’s convenient to represent the time interval in degrees. In Figure 5, the sine wave of the voltage is drawn to a suitable scale, and the part of the time axis between the beginning and the end of one cycle of the sine wave is divided into 360 equal parts, or degrees. Don’t confuse these degrees with those that measure temperature. These degrees are measuring time. How large the scale is, or how much time the interval actually represents, isn’t important because 360° (degrees) always represents the interval needed for one cycle.

One alternation takes 180° or half the time necessary for a complete cycle. Zero values of voltage occur at 0°, 180°, and 360°, as shown in Figure 5. The positive maximum is reached at 90° and the negative maximum at 270°. For the value at any other instant, the corresponding number of degrees can be determined on the horizontal time axis.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 10 Section 1 Timing an AC cycle in degrees rather than fractions of a second is very important in electrical and electronics work. As you’ll see as you progress through this study unit, most AC-voltage or AC-current calculations deal with an AC signal expressed in degrees. Also, when studying how semiconductor devices are triggered during the AC cycle, an understanding of AC cycles in degrees is important.

Before we continue, let’s examine a typical AC waveform as it appears on an oscilloscope. Figure 6 shows a 60-Hz AC waveform on the display screen of an oscilloscope. Note that more than one cycle is shown and that the waveform looks like a typical sine wave with equal portions of voltage and current above and below the zero amplitude, or horizontally positioned, centerline.

FIGURE 6—This oscilloscope display shows a typical AC sine wave.

Now, take a few moments to review what you’ve learned by completing Self-Check 1.

At the end of each section of Alternating Current, you’ll be asked to pause and check your understanding of what you’ve just read by completing a “Self-Check” exercise. Answering these questions will help you review what you’ve studied so far. Please complete Self-Check 1 now.

1. Which of the following may be used as a backup source of AC voltage for emergency lighting in your plant? a. Transformer c. Motor b. Alternator d. Insulator

2. Which of the following may be used to lower or raise an AC voltage? a. Transformer c. Motor b. Alternator d. Insulator

3. Which one of these figures shows one cycle of anAC voltage?

a. c.

b. d.

(Continued)

4. Which one of these figures shows one alternation of anAC voltage?

a. c.

b. d.

5. If an AC-voltage sine wave started from 0° and reached the point P indicated on the figure below, how many degrees has it completed?

6. Which of the following would be used to provide the magnetic field in a commercial alternator? a. Electromagnet c. Industrial transformers b. Permanent magnet d. Fleming’s generator

Check your answers with those in the back of this study unit.

TYPES OF CHARACTERISTIC VALUES

If a DC circuit carries 5 A, a current of 5 A is continually flowing through the circuit, but that’s not true of an AC circuit. The values of AC current and AC voltage vary for every instant in time. Therefore, it’s necessary to establish one characteristic amount of amperes and volts to be used in most calculations on AC signals. We’ll identify that one amount a little later in this study unit.

First, however, we’ll take a look at the several values of current and voltage to be identified in every AC cycle. These five values are 1. Instantaneous values 2. Maximum, or peak, values 3. Peak-to-peak value 4. Average value 5. Effective, or rms, value

Each of these values for voltage will be discussed in greater detail with the aid of the voltage cycle in Figure 7.

1.0 Epk 0.707 E rms E = 2.0 pk-pk 0.636 Ea

090 180 270 360 0 ltage Vo

Characteristic Values

FIGURE 7—Any point in the sine wave represents one instantaneous value of the voltage. Other characteristic values are effective, or rms, value (Erms ), maximum, or peak, value (Epk ), average value (Ea ), and peak-to-peak value (Epk – pk ).

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 14 Section 2 Assume the peak voltage (Epk) in Figure 7 is 1 V. The other characteristic values are all relative to Epk. Thus, effective, or rms, voltage (Erms) is 0.707 V, average voltage (Ea) is

0.636 V, and peak-to-peak voltage (Epk – pk) is 2.0 V. The numerical relations of the values are

= 3 = 3 Erms Epk 0.707 Ea 1.11 = 3 = 3 Epk Erms 1.414 Ea 1.572 = 3 = 3 Ea Epk 0.636 Erms 0.9 = 3 Epk – pk Epk 2.0 INSTANTANEOUS VALUES

The instantaneous value is the actual value of voltage or current at any particular time. In the field of electronics, it’s sometimes necessary to consider instantaneous values. In Figure 8, instantaneous values are represented by vertical distances between the points of the sine wave and the horizontal time axis. They’re read from the scale on the vertical axis. Instantaneous values of an alternating current are different at every instant. They aren’t used in the calculations in this lesson, although instantaneous values are used sometimes to help explain the conditions in a circuit at a given instant.

100 3 75

50 5 6 25 90 180 270 360 0 1 Time, in Degrees 2 -25 Current, in Amperes Current, in -50

-75

-100 4 Instantaneous Values

FIGURE 8—At points 0, 1, and 2, the instantaneous value of the current is zero amperes. At point 3, the instantaneous value is the peak positive value (100 A); at point 4, it’s the peak negative value (also 100 A, but in the opposite direction). The instantaneous values at points 5 and 6 are both 50 A.

The maximum value, or peak value, is the highest instantaneous value of voltage or current reached in either the positive or negative direction during one cycle. Therefore, there are two peak values in each cycle: a positive peak and a negative peak.

The peak current value in an AC circuit is important for determining the size of conductor sufficient to carry the current. The peak voltage value in a circuit should be considered when the conductor insulation is selected.

The peak-to-peak value is the change of voltage or current from the positive peak value to the negative peak value, as indicated in Figure 9. For example, if the peak current is 100 A, the peak-to-peak current value is 200 A. For a sine wave voltage or current, the peak-to-peak value is twice the peak value. You may also find the peak-to-peak value by adding the two peak values (ignoring that one is positive and the other is negative). The peak-to-peak value of a voltage wave can be measured conveniently by using an instrument called an oscilloscope. Many industrial engineers and electronic technicians deal continually with peak-to-peak values.

100 3 75

25 90 180 270 360 0 2 1 Time, In Degrees -25 Current, in Amperes Current, in -50 Peak-to-Peak -75

-100 4

FIGURE 9—The two peak values of this wave are shown at points 3 and 4. The peak-to-peak value is the total range between the two peak values, which are particular instantaneous values.

The average value of an alternating current or voltage is obtained by calculating the average of all instantaneous values during one alternation. The average isn’t taken for the full cycle because the sum of all positive values cancels out the sum of all negative values, and we would be left with an average of zero.

The calculation of average value results in the following two formulas:

= 3 Ia 0.636 Ipk = 3 Ea 0.636 Epk

In the first formula, Ia stands for the average value of current and Ipk stands for the peak value of current. Both values are expressed in amperes. In the second formula, Ea stands for the average value of voltage and Epk stands for the peak value of voltage. Both of these values are expressed in volts.

For example, for the cycle in Figure 9, the peak value of the current (Ipk) is shown as 100 3 A. The average value of the current (Ia) is 0.636 100, or 63.6 A.

A voltage measurement of an AC waveform performed on a standard oscilloscope deter- mines the waveform’s peak-to-peak voltage. If you used the same standard oscilloscope to measure the 120 VAC line voltage delivered by a convenience receptacle in your home, it would read almost 170 volts! Of course, if you then checked this result with a digital multimeter, it would read the expected value of 120 VAC. What causes this difference? The standard oscilloscope and multimeter indicate different voltage readings because instead of providing a peak-to-peak value, most multimeters measure the effective or rms value of an alternating current. You’ll learn much more about this difference in the next part of this study unit.

EFFECTIVE VALUE

The effective, or rms value of an alternating current or voltage is the value that will have the same heating effect as a known direct current or voltage. For example, if an unknown alternating current has the same heating effect as a direct current of 10 A, the alternating current is known to have an rms value of 10 A.

The term rms is an abbreviation for root mean square. This value is created by mathe- matically converting the peak value into a number of instantaneous values all taken from one segment of the AC waveform. Each value is then multiplied by itself or squared. An average, or mean, is then taken of all of the squared values. Finally, the square root of this mean is calculated and equals the rms value.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 17 Section 2 While you’re not likely to manually calculate it, the rms value is very important to those working as an electrician or technician. As you’ve already read, digital multimeters usually measure rms rather than the peak or peak-to-peak values. Specifications provided with a multimeter will often indicate it provides true RMS measurements. In fact, such a label may even appear on the meter’s face. As you’ll soon learn, most AC-circuit power calculations are based on rms voltage and current.

If the peak value is known, the rms value can be determined by the following formulas:

= 3 Irms 0.707 Ipk = 3 Erms 0.707 Epk

Use the first formula to find rms current (I ). In this formula, Ipk stands for the peak value of current and both I and Ipk are expressed in amperes. Use the second formula to find the rms value of voltage (Erms). Epk, stands for the peak value of voltage and both Erms and Epk are expressed in volts.

It’s the rms current that’s used in the calculation of power. When the heat loss in the transmission line discussed earlier was determined by using the formula P = I 2R, it was the rms value of the current that was used.

The rms value wasn’t shown in Figure 8 or 9, because it isn’t one of the instantaneous values. For the current shown in Figure 9, the rms value is equal to 0.707 3 100, or 70.7 A which is about 70% of the peak value. The rms value of the voltage in Figure 7 is 0.707 3 1, or 0.707 V.

When you already know the rms values of current and voltage, determine the peak values by using these formulas:

I I 551.414 3I pk 0.707 rms E E 551.414 3 E pk 0.707 rms

These formulas show that the peak values are 1.414 times the rms values. The two illus- trations shown in Figure 10 display the rms values of current and voltage compared to the peak values.

Irms = 70.7 A

Effective (rms) Current (A)

Epk = 1.0 V

Erms = .707 V

Effective (rms) Voltage (B)

FIGURE 10—As shown in 10A, an rms current of 70.7 A will be produced by a 100 A, AC supply. In 10B, an rms voltage of 0.707 volts is produced by a 1.0 VAC signal.

When direct current is measured, a constant amount of current flows through the measuring instrument. That amount is shown on the scale of the meter. That method of measurement can’t be applied to alternating current. Since alternating current varies all the time, its amount can’t be measured directly; it can only be measured by its effects.

The most reliable effect is the heat produced by the current, because the amount of heat depends on the amount of current and isn’t influenced by the direction of flow. Therefore, if an alternating current produces the same amount of heat as a given direct current, we know that it has an rms value equal to that of the direct current.

Figure 11 shows a sine wave representing an alternating current with a peak value of 20 A. The rms value of this current is 0.707 3 20, or 14.14 A. The alternating current in this circuit will produce the same amount of heat in the conductor as the steady direct current of 14.14 A, shown by the horizontal line. There are instants when the alternating current is higher than 14.14 A, and other instants when no current is flowing at all, but the overall heating effect of the changing current is the same as that produced by the steady flow of direct current.

30 Maximum Value of Alternating Current

20 Direct Current 14.14 10

0 Amperes

-10 -14.14

-20

-30 RMS, or Effective, Value

FIGURE 11—A steady flow of direct current is represented by the horizontal line drawn at a distance of A from the time axis. This steady flow of direct current will cause the same heat as the alternating current, represented by the figure’s sine wave, which has a peak value of 20 A.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 20 Section 2 The rms values of current and voltage are used in calculations concerning AC circuits. It’s an accepted practice that, unless otherwise specified, rms values are meant when any values are mentioned in AC applications. For example, if an AC circuit is said to carry 20 A at 240 V, we know that the current in the circuit has an rms value of 20 A and that the voltage has an rms value of 240 V. Most AC measuring instruments, such as the pop- ular digital multimeter, are designed so that they indicate the rms value of the electrical quantity they measure.

Many instruments that measure AC voltages and currents will have printed on their scale plates RMS VOLTAGE or RMS CURRENT. Remember that effective value equals rms value.

Here are some sample problems to give you practice in using the formulas for finding AC values. Read each problem carefully and study the solution until you understand it thoroughly. 1. An alternating current has the peak value of 15 A. What is the average value of the current?

= 3 Ia 0.636 Ipeak Write the formula for average current. = 3 Ia 0.636 15 A Substitute the value for Ipk. Multiply (0.636 3 15 = 9.54).

= Ia 9.54 A Answer: The average value of the current is 9.54 amperes. 2. The electric power supplied to an electric drill is usually at 110 V. What is the peak value of the AC voltage?

= 3 Epk 1.414 Erms Write the formula for peak voltage. = 3 Epk 1.414 110 V Substitute the value for E. (The given voltage is the rms voltage.) Multiply (1.414 3 110 = 155.54).

= Epk 155.54 V Answer: The peak value of the AC voltage is 155.54 volts. 3. An alternator provides an rms current of 10 A when connected to a load of 10 W. (Greek letter omega symbolizing ohm). What is the heating power delivered by the generator?

P = I 2R Write the formula for power loss.

P = (10 A)2 3 (10 θ) Substitute the rms value for I and the value for R.

P = 10 3 10 3 10 Multiply (10 3 10 3 10 = 1000).

P = 1000 watts (W) Answer: The heating power delivered by the generator is 1000 watts.

Now, take a few moments to review what you’ve learned by completing Self-Check 2.

1. The instantaneous value is one of the characteristic values in an AC cycle. Name the four other characteristic values. ______

2. How many maximum, or peak, values are found in one AC cycle? a. One c. Three b. Two d. Four

3. If you read that an AC circuit carries 30 A at 230 V, what characteristic value of the current can you assume is being measured by the 30 A? a. Rms current c. Peak-to-peak current b. Average current d. Instantaneous current

4. Which of the following characteristic AC values can be either positive or negative? a. Average value c. Peak value b. Peak-to-peak value d. Rms value

Check your answers with those in the back of this study unit.

OPPOSITION TO ALTERNATING CURRENT

Simple AC circuits with one voltage are single-phase AC circuits. If two, three, or more AC voltages are applied to interconnected circuits, the system is a polyphase AC circuit. The prefix poly means many. We’ll discuss single-phase AC circuits first.

In DC circuits, the only opposition to the flow of current is the resistance. InAC circuits, however, in addition to resistance, there’s another type of opposition to the current. This other type of opposition is reactance. Reactance is either inductive or capacitive, depending upon whether an inductor or a capacitor is in the circuit.

The total opposition to an alternating current is impedance. Impedance represents the combined effect of the resistance, the inductive reactance (if any), and the capacitive reactance (if any). The impedance of an AC circuit may contain no reactances at all, only one kind of reactance, or both kinds.

In practice, there’s always some resistance present in any AC circuit. However, if a circuit, in addition to resistance, contains any reactances at all, it’s called a reactive circuit.

PHASE ANGLE

In every AC circuit, the current waveform, the voltage waveform, and their relation to each other must be considered. It very seldom happens that the voltage and the current go through their maximum positive, zero, and maximum negative values at the same point in time. If both the voltage and the current rise and fall together at the same time throughout an AC cycle, the voltage and current are in phase with each other. However, it seldom happens that the voltage wave is in phase with the current wave. There’s usually a delay of either the voltage wave or the current wave due to conditions in the circuit. When such a delay exists, the voltage and current are out of phase.

In Figure 12A, the current wave is shown leading the voltage wave in the circuit. The distance between their zero points is measured in degrees. This distance represents their phase difference, or phase angle. The usual symbol for the phase angle is θ (Greek letter theta). The phase angle in Figure 12A is 90°, leading.

0 Time, in Degrees 90 180 270 360 Amperes or Volts

Current Wave

Leading Current (A)

Voltage Wave

Current Wave

Time, 0 in 90 180 270 Degrees Amperes or Volts

Lagging Current (B)

FIGURE 12—The voltage wave in 12A starts at the maximum point of the current wave, which means the current wave leads the voltage. The voltage wave in 12B is already at its maximum point when the current wave starts, which means the current wave lags the voltage.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 24 Section 3 Let’s examine one cycle of the current wave shown in Figure 12A. At 0°, the current starts to rise. It reaches its maximum positive value at 90° when the voltage wave has a zero value. At 180°, the current has dropped to its zero value and the voltage wave has reached its maximum value. At 270°, the current has reached its maximum negative value and the voltage has zero value. At 360°, the current has reached its zero value, completing its cycle, and the voltage has reached its maximum negative value. This sequence continues, with the current wave in this circuit always leading the voltage wave by 90°, or one fourth of a cycle.

The waves shown in Figure 12B represent the opposite situation, with the current wave lagging 90° behind the voltage wave. Thus, the phase angle θ, which equals 90°, is lagging. The current wave starts when the voltage wave has already reached its maximum positive value.

The phase angle in actual circuits is usually small; the phase angles in Figures 12A and B represent extreme phase differences.

The phase angle plays a very important part in calculating the power of a circuit. It deter- mines the power factor of the circuit. The greater the phase angle, the lower the power factor which governs the active power produced by the circuit. This relationship will be explained later in the study unit.

PHASE ANGLE IN RESISTIVE AC CIRCUITS

When an AC circuit contains only a resistive load, such as an incandescent lamp, a heater, or a resistor, the only opposition to the current is the resistance. The circuit has no reactances. Such a circuit is called a resistive circuit. The rms current, the rms voltage, and the resistance of an AC circuit are related to one another by Ohm’s law in exactly the same way as in a DC circuit. Ohm’s law states E 5 I 3 R. Voltage is equal to current times resistance.

In a purely resistive load, the current and voltage waves are in phase, as shown in Figure 13.

Both waves start from zero, reach their maximum positive values at the same time, pass through zero together, reach their maximum negative values at the same time, and return to zero together. The two waves are therefore exactly in phase. There is no phase difference between the voltage and the current waves in a purely resistive circuit.

Current Wave

0 360 Time, in 0 = 0˚ Degrees or Voltage, in Volts or Voltage, Current, in Amperes Current, in

Current and Voltage in Phase

FIGURE 13—Since the voltage and current waves of a resistive circuit are in phase, the difference in phase, or phase angle θ, equals zero.

PHASE ANGLE IN REACTIVE AC CIRCUITS

You’re already familiar with the principle of electromagnetic induction. Electric devices that work on the induction principle are inductors. Electric devices designed to accumu- late and hold a charge of electricity are capacitors. Circuits using inductors are inductive circuits and those using capacitors are capacitive circuits. Both capacitors and inductors offer opposition to alternating current.

The opposition from a capacitor is the capacitive reactance and the opposition from an inductor is the inductive reactance. Any AC circuit containing reactance is a reactive circuit. Some AC circuits contain resistance plus reactance from a capacitor or an inductor or both. This combination of resistance and reactance is impedance.

For now, we’ll learn a little more about the effects of reactance on AC voltage and current waves. There’ll be more on inductive reactance and capacitive reactance in other study units.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 26 Section 3 In a reactive circuit, phase differences between voltage and current are caused by inductance or capacitance or a combination of both. The current in an AC circuit lags the voltage if the circuit contains mostly inductive reactance; the current leads the voltage if the circuit contains mostly capacitive reactance. Either way, the current and voltage in a reactive circuit are always out of phase.

The sine waves in Figure 12A represent the phase conditions found in a capacitive circuit, while the sine waves in Figure 12B represent the phase conditions found in an inductive circuit.

In a capacitive circuit, the current leads the voltage and thus the voltage lags the current. In an inductive circuit, the current lags the voltage and the voltage leads the current.

While you may find this confusing at first, you can easily remember these relations by using the following phrase:

ELI the ICE man

In this phrase, E represents the voltage and I represents the current. The type of circuit is shown by the middle letters. L represents inductance or an inductive circuit and C represents capacitance or a capacitive circuit.

For example, the letter E is at the beginning of the symbol ELI and the letter I is at the end. Therefore, I follows E. That is, the current I lags the voltage E in an inductive circuit L. In the same way, the letter I is at the beginning of the symbol ICE. Therefore, the current I leads the voltage E in a capacitive circuit C.

In practice, we usually say what the current is doing, leading or lagging, with regard to the voltage. Sometimes the voltage and current aren’t even mentioned. For example, if it’s said that the phase angle is 45° leading, you can understand this to mean that the current is leading the voltage by 45°.

POWER IN RESISTIVE AC CIRCUITS

The same basic method for calculating power is used for an AC circuit as for a DC circuit. But for AC circuits, voltages and currents vary and sometimes aren’t in phase with each other. You should take these varying conditions and phase relations into account when calculating AC power.

Let’s briefly review how to calculate power in a DC circuit. Power delivered to a DC circuit is equal to the applied voltage multiplied by the total current flowing into the circuit.

This relationship is described in the following formula:

P = E 3 I

In this formula, P stands for the power delivered to the circuit in watts, E stands for the applied voltage in volts, and I stands for the total current in amperes.

In an AC circuit, the power is also equal to the applied voltage multiplied by the total current flowing into the circuit. However, in an AC circuit, the voltage and current are changing continually and, therefore, the power changes from one instant to the next. The instantaneous power, at any point in time, is equal to the instantaneous voltage multiplied by the instantaneous current all measured at the same point in time. In Figure 14, you see a sine wave of voltage with a positive peak value of +3 V and a sine wave of current with a positive peak value of +2 A. Because the voltage is in phase with the current, this can be recognized as a resistive circuit.

The drawing of the instantaneous power wave in Figure 14 shows graphically how the power in this circuit will vary with time. The maximum value of the instantaneous power wave is 6 W (3 V 3 2 A) and the minimum value of the instantaneous power wave is 0 W. Note that all of the instantaneous power curve is above the horizontal line. Although the power varies, its instantaneous value measured at any point in time will always be positive. We say that all power delivered to this circuit can produce work and cause heat, or is real power. Figure 14 shows that all power is real power in a purely resistive circuit.

To find the value of the real power in a purely resistive AC circuit, multiply the rms values of the voltage and current. Use the same formula as was used for calculating the power in a DC circuit except now we’ll let E and I stand for rms values:

= 3 Preal Erms Irms

For the purely resistive circuit, which produces waveforms like those in Figure 14, you can find the real power by following these three steps: 1. Find the rms value of the voltage.

= 3 Erms 0.707 Epk Write the formula for calculating the rms value of the voltage.

= 3 Erms 0.707 3 Substitute the positive value for Epk in volts.

= Erms 2.121 V Multiply. Answer: The rms value of the voltage is 2.121 volts. 2. Find the rms value of the current.

= 3 Irms 0.707 Ipk Write the formula for calculating the rms value of the current.

= 3 Irms 0.707 2.0 Substitute the value for Ipk in amperes. = Irms 1.414 A Multiply. Answer: The rms value of the current is 1.414 amperes (rounded).

5 Instantaneous Power Wave

0 90˚ 270˚

-1 Instantaneous Current Wave

-2

Instantaneous Voltage Wave -3

-4

-5 Power in a Resistive Circuit

-6

FIGURE 14—The instantaneous power wave in a resistive circuit is in phase with both the voltage and current waves. Because the instantaneous power wave has only positive values, all of the power is active, or real, power.

= 3 Preal Erms Irms Write the formula for calculating the real power in a resistive circuit.

= 3 Preal 2.121 1.414 Substitute the values for Erms and Irms. = Preal 2.999 W Multiply. Answer: The real power in this resistive circuit is 3 watts (rounded).

Another method for finding the real power in an AC circuit is to find the average value of the instantaneous power using a graph of the power wave. We can find this average value for the circuit producing the power wave in Figure 14. All we have to do is men- tally shift the instantaneous power wave down the vertical axis until the horizontal axis is exactly between the power wave’s positive and negative peaks. You’ll have to shift the power wave down 3 increments along the vertical axis to position the peaks equally distant from the horizontal axis. If you really had shifted each point of the wave down by 3 increments, the newly positioned wave would have a positive peak value of +3 W and a negative peak value of –3 W. The number of increments that you would shift the power wave down is the average value of the instantaneous power and, expressed as watts, is equivalent to the real power of the circuit.

Thus, we can see by the amount of shift required that the real power in this circuit is 3 W. This is the same amount we had previously calculated using the rms values of the voltage and current.

POWER IN PURELY REACTIVE AC CIRCUITS

Because every circuit contains at least a small amount of resistance, there’s no such thing as a purely reactive circuit. However, in order to discuss reactance, it helps to assume that such circuits exist. The waveforms that would exist in a purely reactive circuit are shown in Figure 15.

Figure 15’s waveforms resemble what would be found in an inductive circuit in that the current lags the voltage by 90°. The voltage and current waves, like those in Figure 14, have positive peak values of +3 V and +2 A respectively, but the current’s 90° lag keeps the voltage and current out of phase. The instantaneous power still equals the instantaneous voltage multiplied by the instantaneous current at every point in time. Note that not all of the instantaneous power is above the zero line as it was in a purely resistive AC circuit.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 30 Section 3 We found the real power in a purely resistive AC circuit by multiplying the rms voltage by the rms current. In any reactive circuit, which would be any circuit containing an inductor or a capacitor, we need a third factor to find the real power. This third factor is the power factor and will be described later in the study unit.

However, we can also find the real power for a purely reactive circuit as we did before with the waveform from a purely resistive circuit by finding the average value of the instantaneous power. The average value of the instantaneous power is the amount of vertical shift required to make the wave’s positive and negative peaks equally distant from the horizontal axis. In Figure 15, we can see that the instantaneous power wave has a positive peak value of +3 W and a negative peak value of –3 W. Because both instantaneous power peaks are already equally distant from the horizontal axis, no shift is required and the real power is thus 0 W.

90˚

Instantaneous 1 Power Wave

-1 Instantaneous Current Wave

-2

Instantaneous Voltage Wave -3

Power in Purely Reactive Circuit

FIGURE 15—The current wave lags the voltage wave by 90° in what would be a purely reactive circuit. Note that the instantaneous power wave is positive half of the time and negative half of the time.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 31 Section 3 Remember that, in practice, there’s no such thing as a purely reactive circuit. There’s always some resistance. If a purely reactive circuit did exist, however, its real power would be 0 W. No work can be done nor can any heat be generated with a purely reactive AC circuit.

POWER IN PARTIALLY REACTIVE AC CIRCUITS

Figure 16 shows the relation of instantaneous voltage, current, and power waves in a partially reactive AC circuit. The phase shift between the current and the voltage of such a circuit would be less than 90°. For the circuit current that produced the waves depicted in Figure 16, the current leads the voltage by 45°. This is the phase shift for a capacitive circuit with the capacitive reactance equal to the resistance. As in Figures 14 and 15, the voltage and current waves also have positive peak values of +3 V and +2 A respectively. The 45° phase shift between the voltage and current waves can be clearly seen.

The instantaneous power is still equal to the instantaneous voltage multiplied by the instantaneous current at every point in time. Again, note that not all of the instantaneous power is above the zero line as it was in a purely resistive AC circuit.

When shifting the instantaneous power wave in Figure 16 to find the average value of the instantaneous power, the real power will also be revealed. The peak positive value of the instantaneous power is +5.12 W while its peak negative value is –0.88 W. If you shift each point of the instantaneous power wave down the vertical axis by 2.12 increments, the positive peak value for the repositioned wave will be +3 W while the negative peak value will be –3 watts. Therefore, the real power in this circuit is 2.12 W.

4 Instantaneous Power Wave 3

Instantaneous 0 Current Wave

-0.88 Instantaneous Voltage Wave -2

-3

-4

Power in a Partially Reactive Circuit -5

FIGURE 16—The voltage and current waves are shifted 45° out of phase in this partially reactive circuit. Some of the instantaneous power is positive and some is negative.

Multiplying the rms voltage by the rms current in an AC circuit won’t always give you the real power. The product of the applied rms voltage and the total rms current in a circuit is the apparent power.

As you’ve learned, once an AC circuit contains a capacitor or inductor it’s considered to be a reactive circuit. As you’ve also learned, the presence of these components means there is a time or phase shift between the circuit’s current and the voltage levels; the point in time at which the circuit voltage reaches its peak will not be the same as the time when the current level reaches its maximum. This phase shift results in a power loss. In practical terms, this loss means that not all of the power applied to a reactive circuit is converted into useful work. The concept of a power loss is easier to understand by refer- ring to the graphical explanation shown in Figure 17.

Reactive Power (VAR)

Apparent Power (VA)

Ø Phase Angle in Degrees

Real or Active Power in Watts (W)

FIGURE 17—This is a typical power triangle for an AC circuit containing reactive components.

You can better understand a few characteristics of a reactive AC circuit by viewing the triangle in Figure 17. Notice the bottom horizontal axis contains the label Real or Active Power. This represents the amount of electric energy actually used by the supplied load to create useful work. The vertical axis is labeled Reactive Power, which is power that doesn’t perform any useful work. Reactive power is what causes the phase shift represented in Figure 17 by the angle θ. Finally, the triangle’s hypotenuse, or its sloped side, is labeled Apparent Power. The real or active power is always less than the apparent power. Also, the larger the value of reactive power, the larger the triangle’s phase angle shift and the greater the amount of apparent power compared to real power.

As you might expect, the inverse relationship is also true. Less capacitance or inductance present in a circuit reduces the triangle’s vertical axis, which produces a smaller phase angle. This condition means the level of apparent power in the circuit will be closer to the amount of real or active power.

= 3 Papparent Erms Irms

In this formula, Papparent stands for the apparent power in volt-amperes (VA), Erms stands for the rms voltage in volts, and Irms stands for the rms current in amperes. Note that this formula is the same as the real power formula for a purely resistive circuit. Therefore, for a purely resistive circuit, the real power in watts is equal to the apparent power in volt-amperes.

Only the real power, measured in watts, performs useful work. As shown in Figure 17, the real power may never be greater than the apparent power in an AC circuit.

Apparent power must be considered, however, when determining conductor sizes and insulation requirements. Reactive power is power in the circuit that doesn’t contribute useful electrical work. Reactive power is measured in volt-amperes reactive (VAR).

Apparent power in volt-amperes (VA), real power in watts (W), and reactive power in volt-amperes reactive (VAR) are related as shown in the following equation:

Apparent power51() real power22reactive power

In terms of the units of measurement:

VA 51W 22VAR

The symbol ( ) means you take the square root of whatever quantity is enclosed by the symbol. For example, 36 means take the square root of 36 which is 6 because 6 3 6 is 36.

Let’s find the apparent power for the circuit in Figure 16. Assume that the reactive power is 2.12 VAR.

VA 51W 22VAR Write the formula for apparent power.

VA 5121..222212 Substitute the values for real power and reactive power.

VA 5121..222212 Keystrokes Display Description

2.12 [x2] 4.4944 Calculate the value of the real power squared. [+] 2.12 [x2] [=] 8.9888 Add the value of the reactive power squared.

[ x ] 2.998132752 Calculate the square root. Answer: Apparent power equals 3.00 volt-amperes (rounded).

In the circuits in a plant, it’s common to use larger power units. For example, kilovolts and kilowatts may be used instead of volts and watts. Kilo means 1000 so 1 kilowatt (kW) = 1000 watts (W); 1 kilovolt-ampere (kVA) = 1000 volt-amperes (VA); and 1 kilovolt-ampere reactive (kVAR 1000 volt-amperes reactive (VAR).

The volt-ampere reactive meter shown in Figure 18 is part of the switchgear at a modern industrial plant. This meter reports on all characteristics of the power entering the switchgear as well as the phase shift and associated power values resulting from the inductive and/or capacitive loads inside the plant.

FIGURE 18—This device, located on an industrial plant’s main switchgear, serves as a VAR meter as well as providing many other measurement functions.

To determine the real power in AC circuits, you won’t have to plot out the instantaneous power wave as was done in Figures 14, 15, and 16. The real power delivered to any AC circuit can be found by using the formula for apparent power and multiplying it by a power factor:

= 3 3 Preal Erms Irms PF

In this formula Preal is the real power in watts, Erms is the rms value of the voltage, Irms is the rms value of the current and PF is the power factor of the circuit.

The power factor is a number between 0 and 1, such as 0.1, 0.25, 0.7 or 0.85.

Normally you’ll simply read the power factor for an AC circuit or system from a power factor meter as shown in Figure 19. If a meter isn’t readily available, you may also find the value of your power factor on a table or in a chart. To understand how the power factor is related to the circuit’s phase shift and that it’s actually a trigonometric function, you should practice calculating it.

FIGURE 19—This power-monitoring center is now configured to serve as a power factor meter.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 37 Section 3 The power factor is equal to the cosine of the angle theta (cos q). The angle q is the angle by which the current lags or leads the voltage. The cosine for any angle can easily be looked up on a table or computed using a calculator. For a phase angle of 0°, the power factor (cos 0°) is 1. For a phase angle of 45°, the power factor (cos 45°) is 0.707. For a phase angle of 90°, the power factor (cos 90°) is 0. If you know the phase angle, the following would be a practical expression of the formula for real power:

= 3 3 q Preal Erms Irms (cos )

The value of the power factor can also be calculated as a ratio of the resistance (R), to the impedance (Z): R PF 5 Z

Another way to express the power factor is real power PF 5 apparent power

Do you recall that impedance is made up of combined resistance and reactance? Thus, if a circuit contains any reactance at all, impedance must be larger than resistance, so the power factor must be a decimal between 0 and 1. Similarly, because in a practical AC system apparent power is always greater than real power, their ratio, which is the power factor, must always be less than 1. The power factor may equal 1 and the reactive power would be 0 only if, as in a resistive circuit, the impedance equals the resistance.

REACTIVE POWER IN AC CIRCUITS

The reactive power in an AC circuit can be calculated using a formula similar to the one used to calculate real power. Instead of the trigonometric term, cos q, the formula for reactive power uses the sine of the angle theta (sin q):

= 3 q Preactive Erms Irms (sin )

The values of sin 0 for the three angles of phase shift depicted in Figures 14, 15, and 16 are sin 0° = 0, sin 45° = 0.707, and sin 90° = 1.

APPLICATION OF POWER FORMULA

= 3 3 As you just learned, real power can be found by using the formula Preal Erms Irms PF. To use the formula, you only need to determine the cosine of the phase angle and multiply. Here are some examples that apply to the conditions in Figures 14, 15, and 16.

= 3 3 Preal Erms Irms PF Write the formula for real power. PF = Cos q Write the formula for the power factor.

Let’s use the scientific calculator to calculate the power factor.

Keystrokes Display Description 0 0 Enter the value of the phase angle.

[cos] 1. Press the cosine button to determine the cosine of 0°. Answer: The power factor is l.

= 3 3 Preal 2.121 1.414 1 Substitute the rms values of the voltage and current and the value of the power factor in the formula.

= 3 Preal 2.999 1 Multiply 2.121 by 1.414. = Preal 2.999 Answer. The real power is 3 watts (rounded).

This is the value that, before we knew the term power factor, we had found by shifting the instantaneous power down the vertical axis on the graph shown in Figure 14.

What would be the reactive power? Reactive power is determined by the formula = 3 q Preactive Erms Irms (sin ). = 3 q Preactive Erms Irms (sin ) Write the formula for reactive power.

Use the scientific calculator to determine the sine of q.

Keystroke Display Description 0 0 Enter the value of the phase angle θ. [sin] 0. Calculate the sine of 0°. Answer: The sine of θ is 0.

Because the sine of q is 0, you’ll be multiplying by 0 in the equation. You know that any value multiplied by 0 equals 0. So, for the wave form shown in Figure 13, the reactive power is 0 VAR.

In Figure 15, the values of the voltage and current are the same as in Figure 14, but the angle of phase shift, q, between the voltage and current is 90°. The power factor, cos q, is 0 and, when multiplied through the formula for real power, it gives the purely reactive circuit a value of 0 for real power. Once more, we had confirmed this value for the real power when we tried to shift the instantaneous power wave down the vertical axis on the graph.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 39 Section 3 = 3 The reactive power for a purely reactive circuit is given by the formula Preactive Erms Irms (sin q). For a 90° phase angle, the sine of q is 1. Therefore, reactive power will simply be a product of the rms values for current and voltage: 2.121 3 1.414 = 3 VAR (rounded). This is the result you would expect because, in a purely reactive circuit, all power would be reactive power.

In Figure 16, the voltage and current again have the same values as before, but this time they have a phase shift angle of only 45°. What is the real power displayed in Figure 16?

= 3 3 Preal Erms Irms PF Write the formula for real power.

PF = cos q Write the formula for the power factor.

Using a scientific calculator, determine the power factor. Keystroke Display Description 45 45 Enter the value of the phase angle θ. [cos] 0.707106781 Press the cosine button to determine the cosine of 45°. Answer: The power factor is 0.707 (rounded).

= 3 3 Preal 2.121 1.414 0.707 Substitute the rms values of the voltage and current and the value of the power factor in the formula.

= 3 Preal 2.999094 0.707 Multiply 2.121 by 1.414. = Preal 2.120 (rounded) Multiply 2.999094 by 0.707. Answer: The real power is 2.120 watts (rounded).

We had found this same value when we shifted the instantaneous power down the vertical axis on the graph.

= 3 The sine of 45° is also 0.707. Thus, the reactive power in the circuit—Preactive Erms Irms 3 sin q—is 2.121 3 1.414 3 0.707, or 2.12 VAR.

Be careful. Just because the sine of 45° and the cosine of 45° both equal .707, don’t assume that for partially reactive circuits, the reactive power and real power will always be the same. For example, if the phase angle in Figure 16 were 60° instead of 45°, what would the real power be?

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 40 Section 3 Keystroke Display Description 60 60 Enter the value of the phase angle θ. [cos] 0.5 Press the cosine button to determine the cosine of 60°. Answer: The power factor is 0.5.

= 3 3 Preal 2.121 1.414 0.5 Substitute the rms values of the voltage and current and the value of the power factor in the formula.

= 3 Preal 2.9999 0.5 Multiply 2.121 by 1.414. = Preal 1.4999 Multiply 2.9999 by 0.5. Answer: The real power is 1.5 watts (rounded).

Now find the reactive power.

= 3 3 q Preactive Erms Irms sin Write the formula for reactive power.

Use a scientific calculator to determine the sine of the phase angle.

Keystroke Display Description 60 60 Enter the value of the phase angle θ. [sin] 0.866025404 Press the sine button to determine the sine of 60°.

= 3 3 Preactive 2.121 1.414 .866 Substitute the rms values of the voltage and current and the value of the sine of θ (rounded).

= 3 Preactive 2.999 .866 Multiply 2.121 by 1.414. = Preactive 2.597 Multiply 2.999 by .866. Answer: The reactive power is 2.60 volt-ampere reactive (rounded).

So, by changing the phase angle by 15° in the partially reactive circuit, the real power decreased to 1.5 W and the reactive power increased to 2.6 W.

In any AC circuit, the power factor is the ratio of the real power to the apparent power. This is the same ratio as the resistance to the impedance of the circuit. The greater the phase difference between the voltage and current, the smaller the power factor.

The power factor is also very often expressed as a percentage. For example, a 0.707 power factor may be expressed as a power factor of 70.7% (percent). A 1.0 power factor is the same as a power factor of 100%.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 41 Section 3 Whether the current lags or leads the voltage doesn’t affect the value of the power factor or the value of the real power. However, if the current lags the voltage, as shown in Figure 15, we call the power factor a lagging power factor. A circuit with a lagging power factor would be primarily inductive. If the current in the circuit leads the voltage, as shown in Figure 16, its power factor is a leading power factor and it would be primarily capacitive.

In addition to the size of the phase angle, the value of the power factor depends on the frequency of the applied voltage and the resulting current. No matter what the frequency, though, the formula used to calculate the power factor would stay the same.

In summary, here are the most common terms used to describe power:

QQ Instantaneous power—The product of the instantaneous values of the voltage and current measured at any one instant in time. Its units are watts (W).

QQ Apparent power—The product of the rms, or effective, values of the voltage and current. Its units are volt-amperes (VA) or kilo-volt amperes (kVA).

QQ Real power—The apparent power multiplied by the power factor. Its units are watts (W). Other names for real power are average power, true power, and active power.

QQ Reactive power—The apparent power multiplied by the sine of q where q is the phase angle between the voltage and current. Its units are volt-amperes reactive (VAR).

Power factor—The cosine of q where q is the phase angle between the voltage and current. It may also be calculated as the ratio resulting from dividing the resistance by the impedance. The power factor is always between 0 and 1 although it’s frequently expressed as a percentage between 0 and 100%. Because it’s only a ratio between two values, it has no units of measurement.

POWER FACTORS IN INDUSTRY

When the current and voltage are in phase in a circuit, all the current is effective in producing work or heat, the phase angle is 0, the power factor is 1, and all of the apparent power is real power. In a typical industrial system, however, a power factor of 1 would be rare. In an industrial system, the wide use of motors, transformers, and other inductive loads will prevent a circuit from being purely resistive.

If the current and voltage were 90° out of phase, there would be only reactance in the circuit. The power factor would then be 0 and all apparent power would be reactive power. A power factor of 0 and purely reactive circuits aren’t possible in practice.

The power factors of commercial power distribution circuits commonly range from 0.75 to 1, or 75% to 100% lagging.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 42 Section 3 Values as low as 0.5 are sometimes found, but a value below that is rare in commercial settings. Low power factors are not desirable. The lower the power factor, the greater the current needed to transmit a given amount of power. The capacitance of long transmission and distribution lines can cause a leading power factor, which can cancel the lagging power factor caused by the inductive reactance of the loads.

The extra reactance in a circuit with a low power factor merely increases the losses of the circuit and requires larger conductors to carry the necessary increase in current.

The useful power of alternators and of the lines they serve may often be increased by connecting to the lines certain devices that correct the power factor. These devices make adjustments to bring the power factor closer to 1, or unity. If the line has many motors causing inductive reactance, the power factor can be corrected by connecting capacitors to the line. If the power factor in a system is leading, shunt reactors can be used to correct the power factor.

Power companies monitor the condition of the AC power they supply. When lagging current and a high power factor are identified, the power company corrects both by installing capacitors on the supply lines. In residential locations, you can often see line voltage capacitors mounted on a power pole leading into a development. These are typically small capacitors. In an industrial park, large capacitor banks are often found where the power lines enter the park. Sometimes, when a specific facility displays a very low power factor, the utility requires the industrial facility to install its own capacitor bank at the location where the electrical power enters the building.

Now, take a few moments to review what you’ve learned by completing Self-Check 3.

1. In AC circuits, the current wave is usually either leading or lagging the voltage wave. Because of this, the current and voltage are considered out of ______.

2. In the circuit shown below, the current wave reaches its maximum value at 90°, and the voltage wave reaches its maximum value at 135°. What is the phase angle?

a. 45°, leading c. 90°, leading b. 45°, lagging d. 90°, lagging

3. The phrase ELI the ICE man helps you to remember that the current (I) leads the voltage (E) in ______circuits, (inductive or capacitive)

4. In a partially reactive AC circuit, some power is positive and some negative. What is the difference between the positive and negative power? a. The apparent power c. The phase power b. The reactive power d. The active power

5. Suppose you’re working on a circuit with capacitive reactance. The phase shift is 45°, the power factor is 0.707, the rms current is 0.707 A, and the rms voltage is 1.414 V. What is the real power? a. 0 W c. 1.414 W b. 0.707 W d. 4.0 W

6. Which of the following will always have a decimal value between 0 and 1? a. The rms voltage c. The impedance b. The total circuit current d. The power factor

Check your answers with those in the back of this study unit.

POLYPHASE SYSTEMS

Alternating voltages and currents that are out of phase with each other must be considered in polyphase AC systems. In the electrical field, any system of more than one phase is considered a polyphase system. A two-phase system uses, at the same time, two circuits connected to a two-phase alternator. Likewise, a three-phase system utilizes three circuits produced by a three-phase alternator. It’s very likely that in your work as a plant maintenance electrician or electronic technician you’ll be working on equipment requiring multiphase power.

A two-phase alternator, as shown in Figure 20A, has two groups of coils, 1 and 2. Each coil group generates an alternating voltage and current. Since the two groups of coils are physically arranged at an angle of 90°, the two generated voltages are not in phase but are 90° out of phase with each other. That means that the phase angle f (Greek letter phi) between the two voltages is 90°. That phase relation is also shown in Figure 20A: the

1 2 1

2 ltage 0 Time, Vo in Degrees = 90˚

Two-Phase Alternator and Voltages Generated (A)

1 = 120˚ 2 3

2 1 3 ltage 0 Time, Vo in Degrees = 120˚ = 120˚

Three-Phase Alternator and Voltages Generated (B)

FIGURE 20—The coils of a two-phase alternator are at right angles, so their waves are 90° apart. The three-phase alternator coils, spaced equally apart from each other, produce waves 120° apart.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 45 Section 4 two voltages have the same frequency and the same maximum value, but their maximum and minimum and other corresponding instantaneous values don’t occur at the same time.

A three-phase alternator has three groups of coils (1, 2, and 3) arranged 120° apart, as shown in Figure 20B. In this alternator, three separate alternating voltages are generated, and the phase difference is (f)120°, as shown.

SINGLE-PHASE SYSTEM

Single-phase, or split-phase, systems use an AC power that is similar to the output of the simple alternator such as in Figure 3. Typically, with an AC voltmeter, you’ll measure either 220 VAC or 480 VAC across the terminals in a single-phase system.

In our homes, single-phase, 220 VAC is used to power our electric heat, electric clothes dryers, and electric water heaters. This voltage is used to lower the current drawn by these high current appliances. In industry, single-phase circuits are much less common. However, electric heaters, some smaller machinery, and some smaller motors may be powered by 220 VAC, split-phase power. This split-phase power is normally developed by using two of the three power feeds from a three-phase circuit.

THREE-PHASE CIRCUITS

The most commonly used polyphase system in industry is a three-phase system. A three-phase system or circuit receives three simultaneous voltages from a three-phase alternator. The voltages are generated in phase windings connected internally. The windings may be connected in a star, or Y (wye) shape, or in a D (Greek letter delta) shape. The three generated voltages in the system differ in phase by 120°. Since the maximum values of the three voltages occur at different times, the voltages overlap and so provide a smoother output power than single-phase, or split-phase, systems.

The current is carried to the load by line wires. Three-phase systems are classified as three-wire or four-wire systems according to the number of line wires.

The load on a system is the total current drawn by the individual loads connected to the system. These loads, such as motors, are generally designed for a three-phase supply of current. However, there may also be individual single-phase loads connected to the lines in such a way that the load on the lines is evenly distributed, or balanced. When we discuss three-phase systems, we usually assume that the system is a balanced system, with equal voltages, currents, and power factors in all the lines. Unbalanced loads are uneconomical and should be avoided.

The name star-connected is given to a three-phase system because of the starlike connection of the phase windings, as shown in Figures 21A and B. This connection also resembles the letter Y (wye), which accounts for the name Y-connected, the one more commonly used. The three alternator phase windings (1, 2, and 3) in Figure 21A are connected to a common star point (0). Each phase winding has an end, or terminal. The terminals (a, b, and c) are connected to the line wires (A, B, and C). The loads (4, 5, and 6) are symmetrically connected, or evenly distributed, between the lines.

The voltage across any phase winding is the phase voltage. The voltage across any two wires in a circuit, such as across A and B or B and C or A and C, is the line voltage for those two wires. The line voltages for the pairs of wires in Figure 21A would be referred to as Eab, Ebc, and Eca.

The following are two rules for a balanced Y-connected system. The first rule is the line voltage equals 1.732 times the phase voltage of a single alternator winding. The second rule is the line current equals the phase current. The currents are equal because each line wire is in series with a phase winding.

The system shown in Figure 21A is a three-wire system. In a four-wire Y-connected system, such as the one shown in Figure 21B, a fourth wire (N) is connected to the common star point (0). Three-phase loads, such as the three-phase motors (7 and 8) can be connected to the three line wires (A, B, and C) in the same way as in three-wire systems. In addition, it’s possible to connect single-phase loads, such as the lamps (9, 10, and 11) between one line wire and the neutral wire. The voltages between A and N, B and N, and C and N are each equal to the phase voltages of the system.

7 8 c C c C E E 3 bc ca 4 3 11 1 a A 1 a 0 5 0 A 9 E 6 N 2 ab 2 B 10 b b B Three-Wire Wye-Connected System Four-Wire Wye-Connected System (A) (B)

FIGURE 21—Two three-phase Y-connected systems are shown here. If the line voltages of the two systems are 208 V, then loads 4, 5, and 6 are 208 V single-phase loads; loads 7 and 8 are 208 V three-phase loads; and loads 9, 10, and 11 are 120 V single-phase loads.

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 47 Section 4 If, for example, the line voltage in the system in Figure 21B is 208 V, then three-phase motors rated at 208 V can be connected to the line 208 wires. The phase voltage is 208 ÷ 1,732, or 120 V (rounded). Therefore, single-phase loads such as lamps designed for 120 V operation may be connected between the neutral wire and any line wire.

DELTA-CONNECTED THREE-PHASE SYSTEMS

The three-phase windings of a delta system are connected in a closed loop, similar to D (Greek letter delta), which gives the system its name. In Figures 22A and B, the phase windings (1, 2, and 3) don’t have a common point. Their terminals (a, b, and c) are connected to lines (A, B, and C ). Here are two rules for delta-connected systems. Rule one is the line voltage equals the phase voltage. Rule two is the line current equals 1.732 times the current in any phase winding (1, 2, or 3).

In the three-wire system in Figure 22A, the loads, or load phases, are numbered 4, 5, and 6. They are connected symmetrically and in A to the lines (A, B, and C). Each voltage across 4, 5, and 6 is the same as the line voltage. If the current in each load phase is 50 A, each line current in A, B, and C is 1.732 3 50, or 86.6 A.

87 c C c C 2 3 4 5 23 A 1 b 6 b A 1 a B 0 a B 910 11 Danger N

Three-Wire Delta-Connected System Four-Wire Delta-Connected System (A) (B)

FIGURE 22—If the line voltage of two three-phase delta-connected systems is 240 V, then loads 4, 5, and 6 are 240 V single phase loads; loads 7 and 8 are 240 V three-phase loads; and loads 9 and 10 are 120 V single-phase loads. Load 11 should not be used because the voltage would be 208 V which is higher than 120 V, the maximum safe voltage.

If a lower voltage is desired from a system, a fourth wire (N) is connected to the center tap (0) of winding 1, as shown in Figure 22B. In such a four-wire delta system, the voltage between N and A, and between N and B, is one half of the line voltage. Therefore, single-phase loads, such as 9 and 10, may be connected to the system. If the line voltage

© PENN FOSTER, INC. 2017 ALTERNATING CURRENT PAGE 48 Section 4 is 240 V, loads 9 and 10 may be designed to operate safely at 120 V. Care should be taken not to connect a single-phase 120 V load, such as 11, between lines N and C. The voltage across N and C is 208 V and could damage the device so connected.

POWER IN THREE-PHASE SYSTEMS

For a balanced three-phase system, in either a Y or a D shape, the total real power is found by using the following equation:

= 3 3 3 Preal 1.732 Erms Irms PF

In this formula, Preal stands for the total real power, Erms stands for rms value of the line voltage, Irms stands for the rms value of line current, and PF stands for the power factor for the system.

Here is a sample problem to give you practice in using this formula.

A balanced Y-connected three-phase system has each phase voltage equal to 7970 V, and each line carries a current of 500 A. What’s the total real power delivered by the system if it operates at a power factor of 0.8?

= 3 = Line voltage (Erms) 1.732 7970 V 13,804.04 V

First find the rms value of line voltage. For a balanced Y system, line voltage is equal to 1.732 times the phase voltage.

= 3 3 3 Preal 1.732 Erms Irms PF Next, write the formula for real power. = 3 3 3 Preal 1.732 13,800 V (rounded) 500 A 0.8

Substitute the values for Erms, Irms, and PF. You

can round off the value of Erms to the nearest tens place.

= Preal 9,560,000 W Multiply. Answer: The real power is 9,560,000 watts (rounded).

9,560,640 W 4 1000 = 9560 kW Divide this answer by 1000 to express it in kilowatts. Answer: The total real power delivered by the system is 9560 kilowatts.

Now, take a few moments to review what you’ve learned by completing Self-Check 4.

1. The two voltages shown in the figure below would be generated by which one of the following?

a. Single-phase alternator c. Three-phase alternator b. Two-phase alternator d. Six-phase alternator

2. If you’re examining a three-phase alternator, you can expect to find ______groups of coils arranged ______° apart.

3. In order for a three-phase system to be operated economically, the load on the three lines must be ______.

4. Which one of the following represents a Y-, or star-connected, four-wire system?

a. c.

b. d.

5. In a balanced Y-connected system, if the phase voltage of a single winding is 240 V, what is the rms value of the line voltage? a. 120 V c. 240 V b. 208 V d. 416 V

Check your answers with those in the back of this study unit.

Self-Check 1

1. b 2. a 3. c 4. d 5. 270° 6. a

Self-Check 2

1. The four values are maximum, or peak, values; peak-to-peak value; average value; and effective, or rms, value. 2. b 3. a 4. c

Self-Check 3

1. phase 2. a 3. capacitive 4. d 5. b 6 d

1. b 2. three, 120 3. balanced (or evenly distributed) 4. c 5. d