1. Concepts of Alternating Current
All of your study thus far has been with direct current (dc), that is, current which does not change direction. However, as you saw in module 1 and will see later in this module, a coil rotating in a magnetic field actually generates a current which regularly changes direction. This current is called ALTERNATING CURRENT or ac.
2. AC and DC
Alternating current is current which constantly changes in amplitude, and which reverses direction at regular intervals. You learned previously that direct current flows only in one direction, and that the amplitude of current is determined by the number of electrons flowing past a point in a circuit in one second. If, for example, a coulomb of electrons moves past a point in a wire in one second and all of the electrons are moving in the same direction, the amplitude of direct current in the wire is one ampere. Similarly, if half a coulomb of electrons moves in one direction past a point in the wire in half a second, then reverses direction and moves past the same point in the opposite direction during the next half-second, a total of one coulomb of electrons passes the point in one second. The amplitude of the alternating current is one ampere. The preceding comparison of dc and ac as illustrated. Notice that one white arrow plus one striped arrow comprise one coulomb.
3. Disadvantages of DC Compared to AC
When commercial use of electricity became wide-spread in the United States, certain disadvantages in using direct current in the home became apparent. If a commercial direct-current system is used, the voltage must be generated at the level (amplitude or value) required by the load. To properly light a 240- volt lamp, for example, the dc generator must deliver 240 volts. If a 120-volt lamp is to be supplied power from the 240-volt generator, a resistor or another 120-volt lamp must be placed in series with the 120-volt lamp to drop the extra 120 volts. When the resistor is used to reduce the voltage, an amount of power equal to that consumed by the lamp is wasted.
Another disadvantage of the direct-current system becomes evident when the direct current (I) from the generating station must be transmitted a long distance over wires to the consumer. When this happens, a large amount of power is lost due to the resistance (R) of the wire. The power loss is equal to I2R. However, this loss can be greatly reduced if the power is transmitted over the lines at a very high voltage level and a low current level. This is not a practical solution to the power loss in the dc system since the load would then have to be operated at a dangerously high voltage. Because of the disadvantages related to transmitting and using direct current, practically all modern commercial electric power companies generate and distribute alternating current (ac).
Unlike direct voltages, alternating voltages can be stepped up or down in amplitude by a device called a TRANSFORMER. (The transformer will be explained later in this module.) Use of the transformer permits efficient transmission of electrical power over long-distance lines. At the electrical power station, the transformer output power is at high voltage and low current levels. At the consumer end of the transmission lines, the voltage is stepped down by a transformer to the value required by the load. Due to its inherent advantages and versatility, alternating current has replaced direct current in all but a few commercial power distribution systems.
O4. What disadvantage of a direct current is due to the resistance of the transmission wires?
4. Voltage Waveforms
You now know that there are two types of current and voltage, that is, direct current and voltage and alternating current and voltage. If a graph is constructed showing the amplitude of a dc voltage across the terminals of a battery with respect to time, it will appear in Figure 1 view A. The dc voltage is shown to have a constant amplitude. Some voltages go through periodic changes in amplitude like those shown in Figure 1 view B. The pattern which results when these changes in amplitude with respect to time are plotted on graph paper is known as a WAVEFORM. Figure 1 view B shows some of the common electrical waveforms. Of those illustrated, the sine wave will be dealt with most often.
5. Basic AC Generation
In a previous chapter, you learned that a current-carrying conductor produces a magnetic field around itself. You also learned how a changing magnetic field produces an emf in a conductor. That is, if a conductor is placed in a magnetic field, and either the field or the conductor moves, an emf is induced in the conductor. This effect is called electromagnetic induction.
The sine wave illustrated in Figure 1 view B is a plot of a current which changes amplitude and direction. Although there are several ways of producing this current, the method based on the principles of electromagnetic induction is by far the easiest and most common method in use.
5.1. Cycle
Figure 2 and Figure 3 show a suspended loop of wire (conductor) being rotated (moved) in a clockwise direction through the magnetic field between the poles of a permanent magnet. For ease of explanation, the loop has been divided into a dark half and light half. Notice in (A) of the figure that the dark half is moving along (parallel to) the lines of force. Consequently, it is cutting NO lines of force. The same is true of the light half, which is moving in the opposite direction. Since the conductors are cutting no lines of force, no emf is induced. As the loop rotates toward the position shown in (B), it cuts more and more lines of force per second (inducing an ever-increasing voltage) because it is cutting more directly across the field (lines of force). At (B), the conductor is shown completing one-quarter of a complete revolution, or 90° , of a complete circle. Because the conductor is now cutting directly across the field, the voltage induced in the conductor is maximum. When the value of induced voltage at various points during the rotation from (A) to (B) is plotted on a graph (and the points connected), a curve appears as shown below.
As the loop continues to be rotated toward the position shown below in (C), it cuts fewer and fewer lines of force. The induced voltage decreases from its peak value. Eventually, the loop is once again moving in a plane parallel to the magnetic field, and no emf is induced in the conductor.
The loop has now been rotated through half a circle (one alternation or 180° ). If the preceding quarter-cycle is plotted, it appears as shown below.
When the same procedure is applied to the second half of rotation (180° through 360° ), the curve appears as shown below. Notice the only difference is in the polarity of the induced voltage. Where previously the polarity was positive, it is now negative.
The sine curve shows the value of induced voltage at each instant of time during rotation of the loop. Notice that this curve contains 360° , or two alternations. TWO ALTERNATIONS represent ONE complete CYCLE of rotation.
Assuming a closed path is provided across the ends of the conductor loop, you can determine the direction of current in the loop by using the LEFT-HAND RULE FOR GENERATORS. Refer to Figure 4. The left-hand rule is applied as follows: First, place your left hand on the illustration with the fingers as shown. Your THUMB will now point in the direction of rotation (relative movement of the wire to the magnetic field); your FOREFINGER will point in the direction of magnetic flux (north to south); and your MIDDLE FINGER (pointing out of the paper) will point in the direction of electron current flow.
By applying the left-hand rule to the dark half of the loop in (B) in Figure 3, you will find that the current flows in the direction indicated by the heavy arrow. Similarly, by using the left-hand rule on the light half of the loop, you will find that current therein flows in the opposite direction. The two induced voltages in the loop add together to form one total emf. It is this emf which causes the current in the loop.
When the loop rotates to the position shown in (D) of Figure 3, the action reverses. The dark half is moving up instead of down, and the light half is moving down instead of up. By applying the left-hand rule once again, you will see that the total induced emf and its resulting current have reversed direction. The voltage builds up to maximum in this new direction, as shown by the sine curve in Figure 3. The loop finally returns to its original position (E), at which point voltage is again zero. The sine curve represents one complete cycle of voltage generated by the rotating loop. All the illustrations used in this chapter show the wire loop moving in a clockwise direction. In actual practice, the loop can be moved clockwise or counterclockwise. Regardless of the direction of movement, the left-hand rule applies.
If the loop is rotated through 360° at a steady rate, and if the strength of the magnetic field is uniform, the voltage produced is a sine wave of voltage, as indicated in Figure 4. Continuous rotation of the loop will produce a series of sine-wave voltage cycles or, in other words, an ac voltage. As mentioned previously, the cycle consists of two complete alternations in a period of time. Recently the HERTZ (Hz) has been designated to indicate one cycle per second. If ONE CYCLE PER SECOND is ONE HERTZ, then 100 cycles per second are equal to 100 hertz, and so on. Throughout the NEETS, the term cycle is used when no specific time element is involved, and the term hertz (Hz) is used when the time element is measured in seconds.
QI7. One cycle is equal to how many degrees of rotation of a conductor in a magnetic field?
5.2. Frequency
If the loop in the figure 1-8 (A) makes one complete revolution each second, the generator produces one complete cycle of ac during each second (1 Hz). Increasing the number of revolutions to two per second will produce two complete cycles of ac per second (2 Hz). The number of complete cycles of alternating current or voltage completed each second is referred to as the FREQUENCY. Frequency is always measured and expressed in hertz.
Alternating-current frequency is an important term to understand since most ac electrical equipments require a specific frequency for proper operation.
5.3. Period
An individual cycle of any sine wave represents a definite amount of TIME. Notice that Figure 5 shows 2 cycles of a sine wave which has a frequency of 2 hertz (Hz). Since 2 cycles occur each second, | cycle must require one-half second of time. The time required to complete one cycle of a waveform is called the PERIOD of the wave. In Figure 5, the period is one-half second. The relationship between time (t) and frequency (f) is indicated by the formulas
where t = period mn seconds and t = frequency ti hertz
Each cycle of the sine wave shown in figure 1-10 consists of two identically shaped variations in voltage. The variation which occurs during the time the voltage is positive is called the POSITIVE ALTERNATION. The variation which occurs during the time the voltage is negative is called the NEGATIVE ALTERNATION. In a sine wave, these two alternations are identical in size and shape, but opposite in polarity.
The distance from zero to the maximum value of each alternation is called the AMPLITUDE. The amplitude of the positive alternation and the amplitude of the negative alternation are the same.
5.4. Wavelength
The time it takes for a sine wave to complete one cycle is defined as the period of the waveform. The distance traveled by the sine wave during this period is referred to as WAVELENGTH. Wavelength, indicated by the symbol λ (Greek lambda), is the distance along the waveform from one point to the same point on the next cycle. You can observe this relationship by examining Figure 6. The point on the waveform that measurement of wavelength begins is not important as long as the distance is measured to the same point on the next cycle (see Figure 7).
6. Alternating Current Values
In discussing alternating current and voltage, you will often find it necessary to express the current and voltage in terms of MAXIMUM or PEAK values, PEAK-to-PEAK values, EFFECTIVE values, AVERAGE values, or INSTANTANEOUS values. Each of these values has a different meaning and is used to describe a different amount of current or voltage.
6.1. Peak and Peak-To-Peak Values
Refer to Figure 8. Notice it shows the positive alternation of a sine wave (a half-cycle of ac) and a dc waveform that occur simultaneously. Note that the de starts and stops at the same moment as does the positive alternation, and that both waveforms rise to the same maximum value. However, the de values are greater than the corresponding ac values at all points except the point at which the positive alternation passes through its maximum value. At this point the dc and ac values are equal. This point on the sine wave is referred to as the maximum or peak value.
During each complete cycle of ac there are always two maximum or peak values, one for the positive half-cycle and the other for the negative half-cycle. The difference between the peak positive value and the peak negative value is called the peak-to-peak value of the sine wave. This value is twice the maximum or peak value of the sine wave and is sometimes used for measurement of ac voltages. Note the difference between peak and peak-to-peak values in Figure 9. Usually alternating voltage and current are expressed in EFFECTIVE VALUES (a term you will study later) rather than in peak-to-peak values.
6.2. Instantaneous Value
The INSTANTANEOUS value of an alternating voltage or current is the value of voltage or current at one particular instant. The value may be zero if the particular instant is the time in the cycle at which the polarity of the voltage is changing. It may also be the same as the peak value, if the selected instant is the time in the cycle at which the voltage or current stops increasing and starts decreasing. There are actually an infinite number of instantaneous values between zero and the peak value.
6.3. Average Value
The AVERAGE value of an alternating current or voltage is the average of ALL the INSTANTANEOUS values during ONE alternation. Since the voltage increases from zero to peak value and decreases back to zero during one alternation, the average value must be some value between those two limits. You could determine the average value by adding together a series of instantaneous values of the alternation (between 0° and 180° ), and then dividing the sum by the number of instantaneous values used. The computation would show that one alternation of a sine wave has an average value equal to 0.636 times the peak value. The formula for average voltage is
Eavg = 0.636 x Emax
where Eavg is the average voltage of one alternation, and Emax 18 the maximum or peak voltage. Similarly, the formula for average current is
lavg = 0.636 X Imax
where Iavg is the average current in one alternation, and I… 18 the maximum or peak current.
Do not confuse the above definition of an average value with that of the average value of a complete cycle. Because the voltage is positive during one alternation and negative during the other alternation, the average value of the voltage values occurring during the complete cycle is zero.
O30. If Emax, ts 115 volts, what is Eavg? O31. If lavg is 1.272 ampere, what is Imax?
6.4. Effective Value of a Sine Wave
Emax, Eavg, Imax, and Lavg are values used in ac measurements. Another value used is the EFFECTIVE value of ac This is the value of alternating voltage or current that will have the same effect on a resistance as a comparable value of direct voltage or current will have on the same resistance.
In an earlier discussion you were told that when current flows in a resistance, heat is produced. When direct current flows in a resistance, the amount of electrical power converted into heat equals IR watts. However, since an alternating current having a maximum value of 1 ampere does not maintain a constant value, the alternating current will not produce as much heat in the resistance as will a direct current of 1 ampere.
Figure 10 compares the heating effect of 1 ampere of dc to the heating effect of 1 ampere of ac.
Examine views A and B of Figure 10 and notice that the heat (70.7° C) produced by 1 ampere of alternating current (that is, an ac with a maximum value of 1 ampere) is only 70.7 percent of the heat (100° C) produced by | ampere of direct current. Mathematically,
Therefore, for effective value of ac (Ieff) = 0.707 X Imax.
The rate at which heat is produced in a resistance forms a convenient basis for establishing an effective value of alternating current, and is known as the "heating effect" method. An alternating current is said to have an effective value of one ampere when it produces heat in a given resistance at the same rate as does one ampere of direct current.
You can compute the effective value of a sine wave of current to a fair degree of accuracy by taking equally-spaced instantaneous values of current along the curve and extracting the square root of the average of the sum of the squared values.
For this reason, the effective value is often called the "root-mean-square" (rms) value. Thus,
Stated another way, the effective or rms value (Ieff) of a sine wave of current is 0.707 times the maximum value of current (Imax). Thus, Ieff = 0.707 X Imax. When Ieff known, you can find I, by using the formula Imax = 1.414 X Ieff. You might wonder where the constant 1.414 comes from. To find out, examine figure 1-15 again and read the following explanation. Assume that the dc in figure 1-15(A) is maintained at | ampere and the resistor temperature at 100° C. Also assume that the ac in figure 1-15(B) is increased until the temperature of the resistor is 100° C. At this point it is found that a maximum ac value of 1.414 amperes is required in order to have the same heating effect as direct current. Therefore, in the ac circuit the maximum current required is 1.414 times the effective current. It is important for you to remember the above relationship and that the effective value (Ieff) of any sine wave of current is always 0.707 times the maximum value (Imax).
Since alternating current is caused by an alternating voltage, the ratio of the effective value of voltage to the maximum value of voltage is the same as the ratio of the effective value of current to the maximum value of current. Stated another way, the effective or rms value (Eeff) of a sine-wave of voltage is 0.707 times the maximum value of voltage (Emax),
When an alternating current or voltage value is specified in a book or on a diagram, the value is an effective value unless there is a definite statement to the contrary. Remember that all meters, unless marked to the contrary, are calibrated to indicate effective values of current and voltage.
Problem: A circuit is known to have an alternating voltage of 120 volts and a peak or maximum current of 30 amperes. What are the peak voltage and effective current values?
Figure 11 shows the relationship between the various values used to indicate sine-wave amplitude. Review the values in the figure to ensure you understand what each value indicates.
O34. What is the formula for finding the effective value of an alternating current? O35. If the peak value of a sine wave is 1,000 volts, what is the effective (Eeff) value? O36. If Ieff = 4.25 ampere, what is Imax?
6.5. Sine Waves in Phase
When a sine wave of voltage is applied to a resistance, the resulting current is also a sine wave. This follows Ohm’s law which states that current is directly proportional to the applied voltage. Now examine Figure 12. Notice that the sine wave of voltage and the resulting sine wave of current are superimposed on the same time axis. Notice also that as the voltage increases in a positive direction, the current increases along with it, and that when the voltage reverses direction, the current also reverses direction. When two sine waves, such as those represented by Figure 12, are precisely in step with one another, they are said to be IN PHASE. To be in phase, the two sine waves must go through their maximum and minimum points at the same time and in the same direction.
In some circuits, several sine waves can be in phase with each other. Thus, it is possible to have two or more voltage drops in phase with each other and also be in phase with the circuit current.
6.6. Sine Waves Out of Phase
Figure 13 shows voltage wave E1 which is considered to start at 0° (time one). As voltage wave E1 reaches its positive peak, voltage wave E2 starts its rise (time two). Since these voltage waves do not go through their maximum and minimum points at the same instant of time, a PHASE DIFFERENCE exists between the two waves. The two waves are said to be OUT OF PHASE. For the two waves in Figure 13 the phase difference is 90°.
To further describe the phase relationship between two sine waves, the terms LEAD and LAG are used. The amount by which one sine wave leads or lags another sine wave is measured in degrees. Refer again to figure 1-18. Observe that wave E; starts 90° later in time than does wave E;. You can also describe this relationship by saying that wave E, leads wave E> by 90°, or that wave E> lags wave E, by 90° . (Either statement is correct; it is the phase relationship between the two sine waves that is important.)
It is possible for one sine wave to lead or lag another sine wave by any number of degrees, except 0° or 360° . When the latter condition exists, the two waves are said to be in phase. Thus, two sine waves that differ in phase by 45° are actually out of phase with each other, whereas two sine waves that differ in phase by 360° are considered to be in phase with each other.
A phase relationship that is quite common is shown in Figure 14. Notice that the two waves illustrated differ in phase by 180° . Notice also that although the waves pass through their maximum and minimum values at the same time, their instantaneous voltages are always of opposite polarity. If two such waves exist across the same component, and the waves are of equal amplitude, they cancel each other. When they have different amplitudes, the resultant wave has the same polarity as the larger wave and has an amplitude equal to the difference between the amplitudes of the two waves.
To determine the phase difference between two sine waves, locate the points on the time axis where the two waves cross the time axis traveling in the same direction. The number of degrees between the crossing points is the phase difference. The wave that crosses the axis at the later time (to the right on the time axis) is said to lag the other wave.
O39. What is the phase relationship between two voltage waves that differ in phase by 360°?
7. Ohm’s Law in AC Circuits
Many ac circuits contain resistance only. The rules for these circuits are the same rules that apply to dc circuits. Resistors, lamps, and heating elements are examples of resistive elements. When an ac circuit contains only resistance, Ohm’s Law, Kirchhoff’s Law, and the various rules that apply to voltage, current, and power in a dc circuit also apply to the ac circuit. The Ohm’s Law formula for an ac circuit can be stated as
Remember, unless otherwise stated, all ac voltage and current values are given as effective values. The formula for Ohm’s Law can also be stated as
The important thing to keep in mind is: Do Not mix ac values. When you solve for effective values, all values you use in the formula must be effective values. Similarly, when you solve for average values, all values you use must be average values. This point should be clearer after you work the following problem: A series circuit consists of two resistors (R1 = 5 ohms and R2 = 15 ohms) and an alternating voltage source of 120 volts. What is Iavg?
Given:
Solution: First sobre for total resistance RT.
The alternating voltage is assumed to be an effective value (since it is not specified to be otherwise). Apply the Ohm’s Law formula.
The problem, however, asked for the average value of current (I 4,2). To convert the effective value of current to the average value of current, you must first determine the peak or maximum value of current, Imax.
You can now find Iavg. Just substitute 8.484 amperes in the Iavg formula and solve for Iavg.
Remember, you can use the Ohm’s Law formulas to solve any purely resistive ac circuit problem. Use the formulas in the same manner as you would to solve a de circuit problem.
O41. A series circuit consists of three resistors (RI = 10Q, R2 = 209 R3 = 15Q)) and an alternating voltage source of 100 volts. What is the effective value of current in the circuit? O42. If the alternating source in Q41 is changed to 200 volts peak-to-peak, what is Iavg? O43. If Eeff 130 volts and Ieff 3 amperes, what is the total resistance (RT) in the circuit?