Alternating Current

CHAPTER 13








Alternating Current


Direct current (dc) can be expressed in terms of two variables: the polarity
(or direction) and the amplitude. Alternating current (ac) is more compli- cated. There are additional variables: the period (and its reciprocal, the frequency), the waveform, and the phase.




Definition of Alternating Current

Direct current has a polarity, or direction, that stays the same over a long period of time. Although the amplitude can vary—the number of amperes, volts, or watts can fluctuate—the charge carriers always flow in the same direction through the circuit. In ac, the polarity reverses repeatedly.


PERIOD

In a periodic ac wave, the kind discussed in this chapter, the mathematical function of amplitude versus time repeats precisely and indefinitely; the same pattern recurs countless times. The period is the length of time between one repetition of the pattern, or one wave cycle, and the next. This
is illustrated in Fig. 13-1 for a simple ac wave.
The period of a wave, in theory, can be anywhere from a minuscule frac- tion of a second to many centuries. Some electromagnetic (EM) fields have periods measured in quadrillionths of a second or smaller. The charged par- ticles held captive by the magnetic field of the Sun reverse their direction

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Period Period




Time









Amplitude
Fig. 13-1. A sine wave. The period is the length of time required for one cycle to be completed.

over periods measured in years. Period, when measured in seconds, is
symbolized T.


FREQUENCY

The frequency, denoted f, of a wave is the reciprocal of the period. That is,
f 1/T, and T 1/f. In the olden days (prior to the 1970s), frequency was specified in cycles per second, abbreviated cps. High frequencies were expressed in kilocycles, megacycles, or gigacycles, representing thousands, millions, or billions of cycles per second. Nowadays, the standard unit of frequency is known as the hertz, abbreviated Hz. Thus 1 Hz 1 cps, 10 Hz
10 cps, and so on.
Higher frequencies are given in kilohertz (kHz), megahertz (MHz),
gigahertz (GHz), and terahertz (THz). The relationships are
1 kHz 1,000 Hz 103 Hz
1 MHz 1,000 kHz 106 Hz
1 GHz
1,000 MHz
109 Hz
1 THz
1,000 GHz
1012 Hz
CHAPTER 13 Alternating Current 325


PROBLEM 13-1
The period of an ac wave is 5.000 10 6 s. What is the frequency in hertz? In kilohertz? In megahertz?

SOLUTION 13-1
First, find the frequency fHz in hertz by taking the reciprocal of the period in seconds:
fHz 1/(5.000 10
3
6 ) 2.000 105 Hz
Next, divide fHz by 1,000 or 10
to get the frequency fkHz in kilohertz:
3fkHz fHz/10
2.000 105/103
3
200.0 kHz
Finally, divide fkHz by 1,000 or 10
3
to get the frequency fMHz in megahertz:
3
fMHz fkHz/10
200.0/10
0.2000 MHz



Waveforms

If you graph the instantaneous current or voltage in an ac system as a function of time, you get a waveform. Alternating currents can manifest themselves in an infinite variety of waveforms. Here are some of the simplest ones.


SINE WAVE

In its purest form, alternating current has a sine-wave, or sinusoidal, nature. The waveform in Fig. 13-1 is a sine wave. Any ac wave that consists of a single frequency has a perfect sine-wave shape. Any perfect sine-wave cur- rent contains one, and only one, component frequency.
In practice, a wave can be so close to a sine wave that it looks exactly like the sine function on an oscilloscope when in reality there are traces of other frequencies present. Imperfections are often too small to see. Utility ac in the United States has an almost perfect sine-wave shape, with a fre- quency of 60 Hz. However, there are slight aberrations.


SQUARE WAVE

On an oscilloscope, a theoretically perfect square wave would look like a pair of parallel dotted lines, one having positive polarity and the other hav- ing negative polarity (Fig. 13-2a). In real life, the transitions often can be seen as vertical lines (see Fig. 13-2b).

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(a)



Time




(b)




Amplitude
Fig. 13-2. (a) A theoretically perfect square wave. (b) The more common rendition.


A square wave might have equal negative and positive peaks. Then the
absolute amplitude of the wave is constant at a certain voltage, current, or power level. Half the time the amplitude is x, and the other half it is
x volts, amperes, or watts.
Some square waves are asymmetrical, with the positive and negative magnitudes differing. If the length of time for which the amplitude is positive differs from the length of time for which the amplitude is negative, the wave is not truly square but is described by the more general term rectangular wave.


SAWTOOTH WAVES

Some ac waves reverse their polarity at constant but not instantaneous rates. The slope of the amplitude-versus-time line indicates how fast the magnitude is changing. Such waves are called sawtooth waves because of their appearance.
In Fig. 13-3, one form of sawtooth wave is shown. The positive-going slope (rise) is extremely steep, as with a square wave, but the negative-
CHAPTER 13 Alternating Current 327











Time









Amplitude
Fig. 13-3. A fast-rise, slow-decay sawtooth wave.

going slope (fall or decay) is gradual. The period of the wave is the time
between points at identical positions on two successive pulses.
Another form of sawtooth wave is just the opposite, with a gradual positive-going slope and a vertical negative-going transition. This type of wave is sometimes called a ramp (Fig. 13-4). This waveform is used for scanning in cathode-ray-tube (CRT) television sets and oscilloscopes.
Sawtooth waves can have rise and decay slopes in an infinite number of different combinations. One example is shown in Fig. 13-5. In this case, the positive-going slope is the same as the negative-going slope. This is a triangular wave.

PROBLEM 13-2
Suppose that each horizontal division in Fig. 13-5 represents 1.0 microsecond
(1.0 s or 1.0 10 6 s). What is the period of this triangular wave? What is the frequency?

SOLUTION 13-2
The easiest way to look at this is to evaluate the wave from a point where it
crosses the time axis going upward and then find the next point (to the right or left) where the wave crosses the time axis going upward. This is four hori- zontal divisions, at least within the limit of our ability to tell by looking at it. The period T is therefore 4.0 s or 4.0 10 6 s. The frequency is the reciprocal
of this: f 1/T 1/(4.0 10 6) 2.5 105 Hz.

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Time









Amplitude
Fig. 13-4. A slow-rise, fast-decay sawtooth wave, also called a ramp wave.











Time









Amplitude
Fig. 13-5. A triangular wave.
CHAPTER 13 Alternating Current 329


Fractions of a Cycle

Scientists and engineers break the ac cycle down into small parts for analysis and reference. One complete cycle can be likened to a single revolution around a circle.


SINE WAVES AS CIRCULAR MOTION

Suppose that you swing a glowing ball around and around at the end of a string at a rate of one revolution per second. The ball thus describes a circle
in space (Fig. 13-6a). Imagine that you swing the ball around so that it is always at the same level; it takes a path that lies in a horizontal plane. Imagine that you do this in a pitch-dark gymnasium. If a friend stands some distance away with his or her eyes in the plane of the ball’s path, what does your friend see? Only the glowing ball, oscillating back and forth. The ball seems to move toward the right, slow down, and then reverse its direction,


Top view



Ball



String




(a)



Side view
Ball



(b)

Fig. 13-6. Swinging ball and string. (a) as seen from above; (b) as seen from some distance away in the plane of the ball’s circular path.

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going back toward the left (see Fig. 13-6b). Then it moves faster and faster and
then slower again, reaching its left-most point, at which it turns around again. This goes on and on, with a frequency of 1 Hz, or a complete cycle per sec- ond, because you are swinging the ball around at one revolution per second.
If you graph the position of the ball as seen by your friend with respect to time, the result will be a sine wave (Fig. 13-7). This wave has the same char- acteristic shape as all sine waves. The standard, or basic, sine wave is described by the mathematical function y sin x in the (x, y) coordinate plane. The general form is y a sin bx, where a and b are real-number constants.

Time






Ball












Left Right

Position of ball
Fig. 13-7. Position of ball as seen edge-on as a function of time.

DEGREES

One method of specifying fractions of an ac cycle is to divide it into 360 equal increments called degrees, symbolized ° or deg (but it’s okay to write out the whole word). The value 0° is assigned to the point in the cycle where the magnitude is zero and positive-going. The same point on the next cycle
is given the value 360°. Halfway through the cycle is 180°; a quarter cycle is 90°; an eighth cycle is 45°. This is illustrated in Fig. 13-8.
CHAPTER 13 Alternating Current 331







180°




0° 90° 270°
360°
Time








Amplitude
Fig. 13-8. A wave cycle can be divided into 360 degrees.


RADIANS

The other method of specifying fractions of an ac cycle is to divide it into exactly 2 , or approximately 6.2832, equal parts. This is the number of radii of a circle that can be laid end to end around the circumference. One radian, symbolized rad (although you can write out the whole word), is equal to about 57.296°. Physicists use the radian more often than the degree when talking about fractional parts of an ac cycle.
Sometimes the frequency of an ac wave is measured in radians per second
(rad/s) rather than in hertz (cycles per second). Because there are 2 radians
in a complete cycle of 360°, the angular frequency of a wave, in radians per second, is equal to 2 times the frequency in hertz. Angular frequency is symbolized by the lowercase italicized Greek letter omega ( ).

PROBLEM 13-3
What is the angular frequency of household ac? Assume that the frequency
of utility ac is 60.0 Hz.

SOLUTION 13-3
Multiply the frequency in hertz by 2 . If this value is taken as 6.2832, then the
angular frequency is

6.2832 60.0 376.992 rad/s

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This should be rounded off to 377 rad/s because our input data are given only
to three significant figures.

PROBLEM 13-4
A certain wave has an angular frequency of 3.8865 105 rad/s. What is the frequency in kilohertz? Express the answer to three significant figures.

SOLUTION 13-4
To solve this, first find the frequency in hertz. This requires that the angular
frequency, in radians per second, be divided by 2 , which is approximately
6.2832. The frequency fHz is therefore
fHz (3.8865 105)/6.2832
6.1855 104 Hz

To obtain the frequency in kilohertz, divide by 103, and then round off to three significant figures:

fkHz 6.1855 104/103
61.855 kHz » 61.9 kHz




Amplitude

Amplitude also can be called magnitude, level, strength, or intensity. Depending on the quantity being measured, the amplitude of an ac wave can be specified in amperes (for current), volts (for voltage), or watts (for power).


INSTANTANEOUS AMPLITUDE

The instantaneous amplitude of an ac wave is the voltage, current, or power
at some precise moment in time. This constantly changes. The manner in which it varies depends on the waveform. Instantaneous amplitudes are represented by individual points on the wave curves.


AVERAGE AMPLITUDE

The average amplitude of an ac wave is the mathematical average (or mean) instantaneous voltage, current, or power evaluated over exactly one wave cycle or any exact whole number of wave cycles. A pure ac sine wave always has an average amplitude of zero. The same is true of a pure ac square wave or triangular wave. It is not generally the case for sawtooth
CHAPTER 13 Alternating Current 333


waves. You can get an idea of why these things are true by carefully looking
at the waveforms illustrated by Figs. 13-1 through 13-5. If you know calculus, you know that the average amplitude is the integral of the waveform eval- uated over one full cycle.


PEAK AMPLITUDE

The peak amplitude of an ac wave is the maximum extent, either positive or negative, that the instantaneous amplitude attains. In many waves, the positive and negative peak amplitudes are the same. Sometimes they differ, however. Figure 13-9 is an example of a wave in which the positive peak amplitude
is the same as the negative peak amplitude. Figure 13-10 is an illustration of a wave that has different positive and negative peak amplitudes.


PEAK-TO-PEAK AMPLITUDE

The peak-to-peak (pk-pk) amplitude of a wave is the net difference between the positive peak amplitude and the negative peak amplitude (Fig. 13-11). Another way of saying this is that the pk-pk amplitude is equal to the positive peak amplitude plus the absolute value of the negative peak amplitude.





Positive
Peak
Amplitude


Time


Negative
Peak
Amplitude




Amplitude
Fig. 13-9. Positive and negative peak amplitudes. In this case, they are equal.

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Positive
Peak
Amplitude


Time


Negative
Peak
Amplitude




Amplitude
Fig. 13-10. A wave in which the positive and negative peak amplitudes differ.












Peak-
to- peak
amplitude
Time








Amplitude
Fig. 13-11. Peak-to-peak amplitude.
CHAPTER 13 Alternating Current 335


Peak to peak is a way of expressing how much the wave level “swings”
during the cycle.
In many waves, the pk-pk amplitude is twice the peak amplitude. This is the case when the positive and negative peak amplitudes are the same.


ROOT-MEAN-SQUARE AMPLITUDE

Often it is necessary to express the effective amplitude of an ac wave. This
is the voltage, current, or power that a dc source would produce to have the same general effect in a real circuit or system. When you say a wall outlet has 117 V, you mean 117 effective volts. The most common figure for effective ac levels is called the root-mean-square, or rms, value.
The expression root mean square means that the waveform is mathemat- ically “operated on” by taking the square root of the mean of the square of all its instantaneous values. The rms amplitude is not the same thing as the average amplitude. For a perfect sine wave, the rms value is equal to 0.707 times the peak value, or 0.354 times the pk-pk value. Conversely, the peak value is 1.414 times the rms value, and the pk-pk value is 2.828 times the rms value. The rms figures often are quoted for perfect sine-wave sources of voltage, such as the utility voltage or the effective voltage of a radio signal. For a perfect square wave, the rms value is the same as the peak value,
and the pk-pk value is twice the rms value and twice the peak value. For sawtooth and irregular waves, the relationship between the rms value and the peak value depends on the exact shape of the wave. The rms value is never more than the peak value for any waveshape.


SUPERIMPOSED DC

Sometimes a wave can have components of both ac and dc. The simplest example of an ac/dc combination is illustrated by the connection of a dc source, such as a battery, in series with an ac source, such as the utility main. Any ac wave can have a dc component along with it. If the dc compo-
nent exceeds the peak value of the ac wave, then fluctuating or pulsating dc will result. This would happen, for example, if a 200-V dc source were connected in series with the utility output. Pulsating dc would appear, with an average value of 200 V but with instantaneous values much higher and lower. The waveshape in this case is illustrated by Fig. 13-12.

PROBLEM 13-5
An ac sine wave measures 60 V pk-pk. There is no dc component. What is
the peak voltage?

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Voltage
500


400
ac Component


300
dc
Component


200


100


0

Fig. 13-12. Composite ac/dc wave resulting from 117-V rms ac in series with 200-V dc.
Time


SOLUTION 13-5
In this case, the peak voltage is exactly half the peak-to-peak value, or 30 V pk.
Half the peaks are 30 V; half are 30 V.

PROBLEM 13-6
Suppose that a dc component of 10 V is superimposed on the sine wave
described in Problem 13-5. What is the peak voltage?

SOLUTION 13-6
This can’t be answered simply, because the absolute values of the positive
peak and negative peak voltages differ. In the case of Problem 13-5, the pos- itive peak is 30 V and the negative peak is 30 V, so their absolute values are the same. However, when a dc component of 10 V is superimposed on the wave, both the positive peak and the negative peak voltages change by
10 V. The positive peak voltage thus becomes 40 V, and the negative peak voltage becomes 20 V.




Phase Angle

Phase angle is an expression of the displacement between two waves having identical frequencies. There are various ways of defining this. Phase angles are usually expressed as values such that 0° 360°. In radians, this

CHAPTER 13
Alternating Current
337

range is 0

2 . Once in awhile you will hear about phase angles

specified over a range of 180° 180°. In radians, this range is
. Phase angle figures can be defined only for pairs of waves whose frequencies are the same. If the frequencies differ, the phase changes from moment to moment and cannot be denoted as a specific number.


PHASE COINCIDENCE

Phase coincidence means that two waves begin at exactly the same moment. They are “lined up.” This is shown in Fig. 13-13 for two waves having different amplitudes. (If the amplitudes were the same, you would see only one wave.) The phase difference in this case is 0°.
If two sine waves are in phase coincidence, the peak amplitude of the resulting wave, which also will be a sine wave, is equal to the sum of the peak amplitudes of the two composite waves. The phase of the resultant is the same as that of the composite waves.


PHASE OPPOSITION

When two sine waves begin exactly one-half cycle, or 180°, apart, they are said to be in phase opposition. This is illustrated by the drawing of Fig. 13-14.







180°




0° 90° 270°
360°
Time








Amplitude
Fig. 13-13. Two sine waves in phase coincidence.

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0° 90° 270°
360°
Time








Amplitude
Fig. 13-14. Two sine waves in phase opposition.


If two sine waves have the same amplitude and are in phase opposition,
they cancel each other out because the instantaneous amplitudes of the two waves are equal and opposite at every moment in time.
If two sine waves have different amplitudes and are in phase opposition, the peak value of the resulting wave, which is a sine wave, is equal to the difference between the peak values of the two composite waves. The phase of the resultant is the same as the phase of the stronger of the two composite waves.


LEADING PHASE

Suppose that there are two sine waves, wave X and wave Y, with identical frequencies. If wave X begins a fraction of a cycle earlier than wave Y, then wave X is said to be leading wave Y in phase. For this to be true, X must begin its cycle less than 180° before Y. Figure 13-15 shows wave X leading wave Y by 90°. The difference can be anything greater than 0°, up to but not including 180°.
Leading phase is sometimes expressed as a phase angle such that 0°
180°. In radians, this is 0 . If we say that wave X has a phase of /2 rad relative to wave Y, we mean that wave X leads wave Y by /2 rad.
CHAPTER 13 Alternating Current 339







180°




0° 90° 270°
360°
Time




Wave X
(Leading)

Wave Y

Amplitude
Fig. 13-15. Wave X leads wave Y by 90°.


LAGGING PHASE

Suppose that wave X begins its cycle more than 180° but less than 360° ahead of wave Y. In this situation, it is easier to imagine that wave X starts its cycle later than wave Y by some value between but not including 0° and
180°. Then wave X is lagging wave Y. Figure 13-16 shows wave X lagging wave Y by 90°. The difference can be anything between but not including
0° and 180°.
Lagging phase is sometimes expressed as a negative angle such that
180° 0°. In radians, this is 0. If we say that wave X has a phase of 45° relative to wave Y, we mean that wave X lags wave Y by 45°.


VECTOR REPRESENTATIONS OF PHASE

If a sine wave X is leading a sine wave Y by x degrees, then the two waves can be drawn as vectors, with vector X oriented x degrees counterclockwise from vector Y. If wave X lags Y by y degrees, then X is oriented y degrees clockwise from Y. If two waves are in phase, their vectors overlap (line up).
If they are in phase opposition, they point in exactly opposite directions.

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Wave Y Wave X
(Lagging)








0° 90°


270°
360°
Time

180°





Amplitude
Fig. 13-16. Wave X lags wave Y by 90°.

Figure 13-17 shows four phase relationships between waves X and Y.
Wave X always has twice the amplitude of wave Y, so vector X is always twice as long as vector Y. In part a, wave X is in phase with wave Y. In part b, wave X leads wave Y by 90°. In part c, waves X and Y are 180° opposite in phase. In part d, wave X lags wave Y by 90°.
In all cases, the vectors rotate counterclockwise at the rate of one complete circle per wave cycle. Mathematically, a sine wave is a vector that goes around and around, just like the ball goes around and around your head when you put it on a string and whirl it.
In a sine wave, the vector magnitude stays the same at all times. If the waveform is not sinusoidal, the vector magnitude is greater in some direc- tions than in others. As you can guess, there exist an infinite number of variations on this theme, and some of them can get complicated.

PROBLEM 13-7
Suppose that there are three waves, called X, Y, and Z. Wave X leads wave
Y by 0.5000 rad; wave Y leads wave Z by precisely one-eighth cycle. By how many degrees does wave X lead or lag wave Z?

SOLUTION 13-7
To solve this, let’s convert all phase-angle measures to degrees. One radian
is approximately equal to 57.296°; therefore, 0.5000 rad 57.296° 0.5000
CHAPTER 13 Alternating Current 341


90 90
X


X,Y Y
180
0 180 0




(a)
270
(b)
270





90 90



X
180
Y Y
0 180 0




(c)
270
(d)
X
270
Fig. 13-17. Vector representations of phase. (a) waves X and Y are in phase;
(b) wave X leads wave Y by 90 degrees; (c) waves X and Y are in phase opposition; (d) wave X lags wave Y by 90 degrees.


28.65° (to four significant figures). One-eighth of a cycle is equal to 45.00°
(that is 360°/8.000). The phase angles therefore add up, so wave X leads wave Z by 28.65° 45.00°, or 73.65°.

PROBLEM 13-8
Suppose that there are three waves X, Y, and Z. Wave X leads wave Y by
0.5000 rad; wave Y lags wave Z by precisely one-eighth cycle. By how many degrees does wave X lead or lag wave Z?

SOLUTION 13-8
The difference in phase between X and Y in this problem is the same as that
in the preceding problem, namely, 28.65°. The difference between Y and Z is also the same, but in the opposite sense. Wave Y lags wave Z by 45.00°. This is the same as saying that wave Y leads wave Z by 45.00°. Thus wave X

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leads wave Z by 28.65° ( 45.00°), which is equivalent to 28.65° 45.00°,
or 16.35°. It is better in this case to say that wave X lags wave Z by 16.35°
or that wave Z leads wave X by 16.35°.

As you can see, phase relationships can get confusing. It’s the same sort of thing that happens when you talk about negative numbers. Which number
is larger than which? It depends on point of view. If it helps you to draw pictures of waves when thinking about phase, then by all means go ahead.






Quiz


Refer to the text in this chapter if necessary. A good score is eight correct. Answers are in the back of the book.
1. Approximately how many radians are in a quarter of a cycle?
(a) 0.7854
(b) 1.571
(c) 3.142
(d) 6.284
2. Refer to Fig. 13-18. Suppose that each horizontal division represents 1.0 ns
(1.0 10 9 s) and that each vertical division represents 1 mV (1.0 10 3 V). What is the approximate rms voltage? Assume the wave is sinusoidal.
(a) 4.8 mV
(b) 9.6 mV
(c) 3.4 mV
(d) 6.8 mV
3. In the wave illustrated by Fig. 13-18, given the same specifications as those for the previous problem, what is the approximate frequency of this wave?
(a) 330 MHz
(b) 660 MHz
(c) 4.1 109 rad/s
(d) It cannot be determined from this information.
4. In the wave illustrated by Fig. 13-18, what fraction of a cycle, in degrees, is represented by one horizontal division?
(a) 60
(b) 90
CHAPTER 13 Alternating Current 343











Time









Voltage
Fig. 13-18. Illustration for quiz questions 2, 3, and 4.


(c) 120
(d) 180
5. The maximum instantaneous current in a fluctuating dc wave is 543 mA over several cycles. The minimum instantaneous current is 105 mA, also over sev- eral cycles. What is the peak-to-peak current in this wave?
(a) 438 mA
(b) 648 mA
(c) 543 mA
(d) It cannot be calculated from this information.
6. The pk-pk voltage in a square wave is 5.50 V. The wave is ac, but it has a dc component of 1.00 V. What is the instantaneous voltage?
(a) More information is needed to answer this question.
(b) 3.25 V
(c) 1.25 V
(d) 1.00 V
7. Given the situation in the preceding question, what is the average voltage?
(a) More information is needed to answer this question.
(b) 3.25 V
(c) 1.25 V
(d) 1.00 V

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8. Suppose that there are two sine waves having identical frequency and that their
vector representations are at right angles to each other. What is the difference in phase?
(a) More information is needed to answer this question.
(b) 90°
(c) 180°
(d) 2 rad
9. A square wave is a special form of
(a) sine wave.
(b) sawtooth wave.
(c) ramp wave.
(d) rectangular wave.
10. An ac wave has a constant frequency f. Its peak voltage Vpk is doubled. What happens to the period T?
(a) It doubles to 2T.
(b) It is reduced to T/2.
(c) It is reduced to 0.707T.
(d) It remains at T.