E, = 30 volts A, = 1082 E = 150 volts Rz + Ę =? 100 Rz E, = ?
300 60,2 FIGURE 8-58. Voltage drops in a circuit. FIGURE 8-59. Applying Ohm's law.
15 ohms, resistors are added (10 V + 20 V), a value equal to the applied voltage, 30 volts, is obtained. This = - 159 60 V - = 4 amperes, confirms the basic formula for series circuits:
Er = E1 + Er.
and if voltage is reduced to half its original value, with resistance constant, current will decrease to half its original value.
In any d.c. series circuit, a missing quantity such as voltage, resistance, or current can be calculated by using Ohm's law if any two of the 1 = R E = 15 g = 1 ampere. 15 V Thus, if resistance remains constant, and voltage increases, current must also increase. If voltage decreases, current decreases also.
It is important to distinguish between the terms "voltage" and "voltage drop" in discussing quantities are known.
Figure 8-59 is a series circuit containing three known values of resistance and an applied voltage of 150 volts. Using these values, the unknown circuit quantities can be determined by applying Ohm's law as follows:
R1 = 30 g series circuits. Voltage drop refers to the loss in R2 = 60 9 electrical pressure caused by forcing electrons through a resistance. In figure 8-58, the applied R3 = 10 Q voltage (the battery) is 30 volts and is labeled Er. Rr = Since there are two resistances in the circuit, there will be two separate voltage drops. These Ir = two voltage drops will be the loss in electrical ERI = pressure used to force electrons through the resistances. The amount of electrical pressure ER, = required to force a given number of electrons ER: = through a resistance is proportional to the size of the resistance. Thus, the voltage drop across Ri Total resistances:
will be twice that across Ra since R, has two times RT = R1 + Rz + R3 the resistance value of Rz. The drop across R, is labeled E1, and that across R, is E2. The current, = 30 + 60 + 10 I, is the same throughout the circuit. = 100 R.
Using: Total Current:
E = IR E1 = IR1 E2 = 1R2 Ez = 2a X 5 Ix = RI Er E1 = 2a × 10 Ez = 10 V. 150 V E1 = 20 V = 100 If the voltage drops (used) across the two = 1.5 amperes.
Voltage Drops:
E = IR ER, = IT X R1 = 1.5 amps X 30 = 45 V.
ERr = IT X Rz = 1.5 amps X 60 = 90 V.
ERS = IT X R3 = 1.5 amps X 10 = 15 V.
Do these values for voltage drops equal the applied voltage?
ET = ER: + ERZ + ERS Er = 150 volts 150 V = 45 V + 90 V + 15 V.
The sum of the voltage drops equals the applied voltage.
Kirchhoff Laws In 1847, a German physicist, G. R. Kirchhoff, elaborated on Ohm's law and developed two statements that are known as Kirchhoff's laws for current and voltage. An understanding of these laws enables the aircraft technician to gain a better understanding of the behavior of electriity. Using Kirchhoff's laws, it is possible to find:
(1) The current in each branch of a network circuit when both the resistance and the electromotive force in each branch are known, or (2) the electromotive force in each branch when both the resistance of, and the current in, each branch are known. These laws are stated as follows:
Current Law-The algebraie sum of the.
currents at any junction of conductors in a circuit is zero. This means that the amount of current flowing away from a point in a circuit is equal to the amount flowing to that point.
Voltage Law- The algebraic sum of the applied voltage and the voltage drop around any closed circuit is zero, which means that the voltage drop around any closed circuit is equal to the applied voltage.
When applying Kirchhoff's laws, use the follow- 50 102 + 5 v 10 v 30 v 15 v < 15 g FIGURE 8-60. Polarity of voltage drops.
ing procedures to simplify the work:
I. When the direction of current is not apparent, assume a direction of flow.
If the assumption is wrong, the answer will be numerically correct but preceded by a negative sign.
2. Place polarity markings (plus and minus signs) on all resistors and batteries in the circuit being solved. The assumed direction of current flow will not affect the polarities of the batteries, but it will affect the polarity of the voltage drop on resistors.
Therefore, the voltage drop should be marked so that the end of a resistor into which the current is assumed to flow is negative, and the end from which it leaves is positive.
In the statements of Kirchhoff's laws, the term algebraic sum was used. An algebraic sum differs from an arithmetic sum in that both the magnitude and the sign of each number must be considered. In electrical circuits a voltage drop occurs when current flows through a resistor.
The magnitude of the voltage is determined by the size of the resistor and the amount of current flow. The polarity (sign) of the voltage drop is determined by the direction of the current flow.
For example, observe the polarities of the applied electromotive force (e.m.f.) and the voltage drop as shown in figure 8-60. The applied e.m.f. causes electrons to flow through the opposition offered by the resistances. The voltage drop across each resistance is therefore opposite in polarity to that of the applied e.m.f. Note that the side of each resistor where the current enters is labeled the negative side.
2 = P = 1 amp d 150 E, = 30 volts B FIGURE: 8-61. Circuit demonstrating Kirchhoff's laws, (A) current law and (B) voltage law.
A portion of a circuit which illustrates Kirchhoff's current law is shown in figure 8-61.
The current flowing through resistor Ri has an intensity of four amperes. The current flowing through resistor R, has a magnitude of one ampere and is flowing into the same junction as the current through Ri. Using Kirchhoff's current law, it is possible to determine how much current is flowing through R2 and whether it is flowing toward or away from the common junction. This is expressed in equation form as:
11 + 12 + Is = 0.
Substituting the current values in the equation gives 4 + 12+ (-1) =0 12=1+4 12 = 5
- 4 + (-1) +5 = 0.
Kirchhoff's current law finds a wider application in more complex parallel or series-parallel circuits.
Figure 8-61 (B) is a series d.c. circuit which is used to demonstrate Kirchhoff's voltage law.
The total resistance is the sum of R1, R2, and Rs which is 30 ohms. Since the applied voltage is 30 volts, the current flowing in the circuit is 1 ampere. Therefore, the voltage drops across R1, R2, and R, are 5 volts, 10 volts, and 15 volts, respectively. The sum of the voltage drops is equal to the applied voltage of 30 volts.
This circuit may also be solved by using the polarities of the voltages and showing that the algebraic sum of the voltages is zero. Consider voltages positive if the (+) sign is met first and negative if the (-) sign is met first when tracing the current flow. By starting at the battery and going in the direction of current flow (as indicated by the arrows) the following equation may be set up:
Total voltage (El) = + 30 - 5 - 10 - 15 E. = O.
The point to start around the circuit and the polarity to use are arbitrary and are a matter of choice for each circuit.
PARALLEL D. C. CIRCUITS A circuit in which two or more electrical resistances, or loads, are connected across the same voltage source is a parallel circuit. The parallel circuit differs from the series circuit in that more than one path is provided for current flowthe more paths added in parallel, the less opposition to flow of electrons from the source. In a series circuit the addition of resistance increases the opposition to current flow. The minimum requirements for a parallel circuit are the following:
(1) a power source.
(2) conductors.
(3) a resistance or load for each current path.
(4) two or more paths for current flow.
Figure 8-62 shows a parallel circuit with three paths for current flow. Points A, B, C, and D are connected to the same conductor and are at the same electrical potential. In a similar manner, points E, F, G, and H are at the same potential.
Since the applied voltage appears between points A and E, the same voltage is applied between points B and F, points C and G, and between points D and H. Thus, when resistors are connected in parallel across a voltage source, each resistor has the same applied voltage, although the currents through the resistors may differ de- 290
The total current, Ir, is Ir = I1 + I2 + 13 Ir = 4 amps + .24 amps + .5 amps Ir = 1.14 amps.
In a parallel circuit, Ir = I1 + I2 + Is. By Ohm's law the following relationships can be obtained:
FIGURE 8-62. A parallel circuit.
pending on the values of resistance. The voltage in a parallel circuit may be expressed as follows:
Er = E1 = E2 = E3.
Where Er is the applied voltage, E, is the voltage across R1, Ez is the voltage across R2, and Es is the voltage across R3 (figure 8-62).
The current in a parallel circuit divides among the various branches in a manner depending on the resistance of each branch (see figure 8-63).
A branch containing a small value of resistance will have a greater current flow than a branch containing a high resistance. Kirchhoff's current law states that the current flowing toward a point is equal to the current flowing away from that point. Thus, the current flow in a circuit may be expressed mathematically as follows:
IT = 11 + 12 + 13 where IT is the total current and I1, I2, and Is are the currents through R1, R2, and R3, respectively.
Kirchhoff's and Ohm's law can be applied to find the total current flow in the circuit shown in figure 8-63.
The current flow through the branch containing resistance R1 is I1 = Ri E = 15 6 = 4 amps.
The current through Re is Iz = Rz E = 25 6 = .24 amps.
The current through R3 is Із = = 12 -_ = .5 amps.
Rz İT = Substituting these values in the equation for total current, In a parallel circuit Er = E1 = E2 = E}.
Therefore, Dividing through by E gives, This equation is the reciprocal formula for finding the total or equivalent resistance of a parallel circuit. Another form of the equation may be derived by solving for Rr.
RT = An analysis of the equation for total resistance in a parallel circuit shows that Rr is always less R 6V {15 & 25 gl | 1212 FIGURE 8-63. Current flow in a parallel circuit.
A, = 200 R, = 100 { R, = 100 FICURE 8-65. A series-parallel circuit.
FIGURE 8-64. A series-parallel circuit.
than the smallest resistance in a parallel circuit.
Thus a 10-ohm, a 20-ohm, and a 40-ohm resistor connected in parallel have a total resistance of less than 10 ohms.
If there are only two resistors in a parallel circuit, the reciprocal formula is Simplified, this becomes:
Ry Rz Rr = Re + R2 This simplified, shorter formula can be used when two resistances are in parallel. Another method can be used for any number of resistors in parallel if they are of equal resistance. The resistance value of one resistor is divided by the number of resistors in parallel to determine the total resistance. Expressed mathematically this becomes:
Rr = Where Rr is the total resistance, R is the resistance of one resistor, and N is the number of resistors.
SERIES-PARALLEL D. C. CIRCUITS Most circuits in electrical equipment are not series or parallel circuits. They are usually seriesparallel circuits, which are combinations of series and parallel circuits. A series-parallel circuit consists of groups of parallel resistors connected in series with other resistors. An example of a series-parallel circuit is shown in figure 8-64.
The requirements for a series-parallel circuit are as follows:
(1) power source (battery).
(2) conductors (wires).
(3) load (resistances).
(4) more than one path for current flow.
(5) a control (switch).
(6) safety device (fuse).
While series-parallel circuits may appear extremely complex, the same rule used for series and parallel circuits can be applied to simplify and solve them.
The easiest method of handling series-parallel circuits is to break them apart and redraw them as equivalent circuits. The circuit in figure 8-65 is an example of a simple series-parallel circuit that can be redrawn to illustrate this procedure.
In this circuit the same voltage is applied to Re and Re; thus they are in parallel. The equivalent resistance of these two resistors is equal to the value of one resistor divided by the number of resistors in parallel. This is true only when the parallel resistors have the same ohmic value.
If this rule is applied, the circuit can be redrawn as shown in figure 8-66.
This has converted the original series-parallel circuit into a simple series circuit containing two resistances. To further simplify the circuit, the two series resistances can be added, and the circuit can be redrawn as shown in figure 8-67.
Although the last redrawing of the circuit could have been omitted and the calculations done mentally, this circuit illustrates clearly that one 25-ohm resistor is the resistive equiv- R, = 202 52 2 R. and R equivalent FIGURE 8-66. A redrawn series-parallel circuit.
R, 602 ww R, 25g equivalent 100g 30 2 R; 602 m Re 602 FIGURE 8-67. An equivalent series-parallel circuit alent of the three resistors of the original circuit.
Figure 8-68 contains a more complex seriesparallel circuit.
The first step in simplifying this circuit is to reduce each group of parallel resistors to a single equivalent resistor. The first group is the parallel combination of Re and R3. Since these resistors have unequal values of resistance, the formula for two parallel resistances is used:
Rz Rz - 120·10 - 4800 Ra = Rr + R3 = 30 Q.
160 Then the parallel combination of R, and R3 can be replaced with a single 30 I resistor, as shown in figure 8-69.
Next, the equivalent resistance of the parallel combination of R4, R5, and Re can be determined using the formula Ri = R/N: where, R& is the equivalent resistance of R4, R5, and R6, R is the value of one of the resistors, and N is the number of resistors in parallel, A =Ñ = 60 - = 20 Q.
The parallel combination of R4, R5, and Re can 100 22 120g R, 60 g m R, 60 g
80.2 m 100 g FIGURE 8-69. Series-parallel circuit with one equivalent resistance.
now be redrawn as a single 20 I resistor, as shown in figure 8-70.
The original series-parallel circuit has now been replaced with its equivalent series circuit. This circuit could be redrawn again to replace the five resistors in series with one 330-ohm resistor.
This can be proved by using the total resistance formula for series circuits:
Rr = R: + Ra + Ro + Rr + Res = 100 + 30 + 20 + 80 + 100 = 330 ohms.
The first series-parallel circuit used is redrawn to discuss the behavior of current flow (figure 8-71).
Unlike the parallel circuit, the branch currents, I1 and 12, cannot be established using the applied voltage. Since R, is in series with the parallel combination of R2 and R3, a portion of the applied voltage is dropped across Ri. In order to find the branch currents, total resistance and total current must be found first. Since R, and R3 are 4052 Re 600 1000 3022 2022 R 802 - 100g Re 10052 FIGURE 8-68. A more complex series-parallel circuit. FIGURE 8-70. Series-parallel equivalent circuit.
The voltage drops across parallel resistors are 210 always equal. It should also be remembered that, when the voltage is held constant and the resistance of any resistor in a series-parallel circuit 28V 31480 is increased, the total current will decrease. This should not be confused as adding another parallel 140 resistor to a parallel combination, which could reduce total resistance and increase total current FICURE 8-71. Current flow in a series-parallel VOLTAGE DIVIDERS circuit.
Voltage dividers are devices which make it equal resistance, possible to obtain more than one voltage from a single power source.
Roger - N= 2 = 70. A voltage divider usually consists of a resistor, or resistors connected in series, with fixed or Total resistance is movable contacts and two fixed terminal contacts. As current flows through the resistor, Rr = Ri + Requiv. different voltages can be obtained between the = 210+70 contacts. A typical voltage divider is shown in figure 8-72.
= 28 0. A load is any device which draws current. A Using Ohm's law, total current is Is = Rr Er = 28 h = 1 ampere. 28 V The total current, 1 ampere, flows through Ri and divides at point A, with part of the current flowing through R2, and the other part through Re. Since R, and R3 are of equal size, it is obvious that half of the total current, or .5 amps, will flow through each branch.
The voltage drops in the circuit are determined by Ohm's law:
E = IR.
ER, = ITRI heavy load means a heavy current drain. In addition to the current drawn by the various loads, there is a certain amount drawn by the voltage divider itself. This is known as bleeder To understand how a voltage divider works, examine the illustration in figure 8-73 carefully and observe the following:
Each load draws a given amount of current: I1, I2, I3. In addition to the load currents, some bleeder current (Is) flows. The current I, is drawn from the power source and is equal to the sum of all currents.
The voltage at each point is measured with respect to a common point. Note that the common = 1 × 21 = 21 Volts.
E = IR.
ER2 = 11R2 = .5 × 14 = 7 Volts.
E = IR.
ERs = 12 R$ = .5 × 14 = 7 Volts. FIGURE 8-72. A voltage divider circuit.
200 v 90 +50V + 30V - 20V 150 v 90 v 100V ‚oad Load FIGURE 8-75. Current flow through a voltage divider.
FIGURE 8-73. A typical voltage divider.
point is the point at which the total current (I) divides into separate currents (I1, I2, I3).
Each part of the voltage divider has a different current flowing in it. The current distribution is as follows:
Through R:-bleeder current (Ia) Through R2Is plus 11 Through R3-Is plus 11, plus 12.
The voltage across each resistor of the voltage divider is:
90 volts across R1 60 volts across Rz 50 volts across Rz.
The voltage divider circuit discussed up to this point has had one side of the power supply (battery) at ground potential. In figure 8-74 the common reference point (ground symbol) has been moved to a different point on the voltage divider.
The voltage drop across R; is 20 volts; however, since tap A is connected to a point in the circuit that is at the same potential as the negative › C A O 50V 30V 20V R, = 1008 R, = 602 R, = 4052 100V FIGURE 8-74. Positive and negative voltage on a voltage divider.
side of the battery, the voltage between tap A and the reference point is a negative (-) 20 volts. Since resistors R2 and Ra are connected to the positive side of the battery, the voltages between the reference point and tap B or C are positive.
A simple method of determining negative and positive voltages is provided by the following rules: (1) If current enters a resistance flowing away from the reference point, the voltage drop across that resistance is positive in respect to the reference point; (2) if current flows out of a resistance toward the reference point, the voltage drop across that resistance is negative in respect to the reference point. It is the location of the reference point that determines whether a voltage is negative or positive.
Tracing the current flow provides a means for determining the voltage polarity. Figure 8-75 shows the same circuit with the polarities of the voltage drops and the direction of current flow indicated.
The current flows from the negative side of the battery to R1. Tap A is at the same potential as the negative terminal of the battery since the slight voltage drop caused by the resistance of the conductor is disregarded; however, 20 volts of the source voltage are required to force the current through Ri and this 20-volt drop has the polarity indicated. Stated another way, there are only 80 volts of electrical pressure left in the circuit on the ground side of R1.
When the current reaches tap B, 30 more volts have been used to move the electrons through R2, and in a similar manner the remaining 50 volts are used for R3. But the voltages across R2 and Ra are positive voltages, since they are above ground potential.
Figure 8-76 shows the voltage divider used 295
AQ 50V 20V R, 100V FIGURE 8-76. Voltage divider with changed ground.
previously. The voltage drops across the resistances are the same; however, the reference point (ground) has been changed. The voltage between ground and tap A is now a negative 100 volts, or the applied voltage. The voltage between ground and tap B is a negative 80 volts, and the voltage between ground and tap C is a negative 50 volts.
RHEOSTATS AND POTENTIOMETERS The voltage dividers discussed thus far have consisted of resistors of various sizes across which a variety of voltage drops were developed.
Rheostats and potentiometers are variable resistors which are sometimes used in connection with voltage dividers.
A rheostat is a variable resistor used to vary the amount of current flowing in a circuit.
The rheostat is represented schematically as a two-terminal resistance with a sliding arm contact. Figure 8-77 shows a rheostat connected in series with an ordinary resistance in a series circuit. As the slider arm moves from point A to B, the amount of rheostat resistance (AB) is increased. Since the rheostat resistance and the fixed resistance are in series, the total resistance in the circuit also increases, and the current in the circuit decreases. On the other hand, if the slider arm is moved toward point A, the total resistance decreases and the current in the circuit increases.
The potentiometer is a variable resistor which has three terminals. Two ends and a slider arm are connected in a circuit. A potentiometer is used to vary the amount of voltage in a circuit and is one of the most common controls used in electrical and electronic equipment. Some examples are the volume control in radio receivers and the brightness control in television receivers.
In A of figure 8-78 a potentiometer is used to obtain a variable voltage from a fixed voltage source to apply to an electrical load. The voltage applied to the load is the voltage between points B and C. When the slider arm is moved to point A, the entire voltage is applied to the electrical device (load); when the arm is moved to point C, the voltage applied to the load is zero. The potentiometer makes possible the application of any voltage between zero and full voltage to the load.
The current flowing through the circuit of figure 8-78 leaves the negative terminal of the battery and divides, one part flowing through the lower portion of the potentiometer (points C to B) and the other part through the load. Both parts combine at point B and flow through the upper portion of the potentiometer (points B to A) back to the positive terminal of the battery.
In B of figure 8-78, a potentiometer and its schematic symbol are shown.
In choosing a potentiometer resistance, consideration should be given to the amount of current drawn by the load as well as the current flow through the potentiometer at all settings of the slider arm. The energy of the current through the potentiometer is dissipated in the form of B Movable Contact B FIGURE 8-77. Rheostat.
tenance. Small distances, for example, can usually be measured in inches, but larger distances are more meaningfully expressed in feet, yards, or miles. Since electrical values often vary from numbers that are a millionth part of a basic unit of measurement to very large values, it is often necessary to use a wide range of numbers to represent the values of such units as volts, amperes, or ohms. A series of prefixes which appear with the name of the unit have been devised for the various multiples or submultiples of the basic units. There are 12 of these prefixes, which are also known as conversion factors. Six of the most commonly used prefixes with a short definition of each are as follows:
Shaft Mega means one million (1,000,000).
Kilo means one thousand (1,000).
Centi means one-hundredth (1/100).
FIGURER 8-78. Potentiometer.
heat. It is important to keep this wasted current as small as possible by making the resistance of the potentiometer as large as practicable. In most cases, the resistance of the potentiometer can be several times the resistance of the load.
Rheostats and potentiometers are constructed of a circular resistance material over which a sliding contact moves. The resistance may be distributed in many ways, and the method used determines the classification as either linear or tapered. The linear type provides a resistance evenly distributed over its entire length, while the tapered has more resistance per unit length at one end than at the other. As an example, a one-half turn of a linear rheostat places one half of the total resistance between either end and the slider, while a one-half turn of a tapered rheostat places one-tenth (or any desired fraction) of the total resistance between one end and the slider.
Prefixes In any system of measurements, a single set of units is usually not sufficient for all the computations involved in electrical repair and main- Milli means one-thousandth (1/1000).
Micro means one-millionth (1/1,000,000).
Micro micro means one-millionth-millionth (1/1,000,000,000,000).
One of the most extensively used conversion factors, kilo, can be used to explain the use of prefixes with basic units of measurement. Kilo means 1,000, and when used with volts is expressed as kilovolt, meaning 1,000 volts. The symbol for kilo is the letter "K". Thus, 1,000 volts is one kilovolt or 1KV. Conversely, one volt would equal one-thousandth of a KV, or 1/1000 KV.
This could also be written 0.001 KV.
Similarly, the word "milli" means one-thousandth, and thus, I millivolt equals one-thousandth (1/1000) of a volt.
These prefixes may be used with all electrical units. They provide a convenient method for writing extremely large or small values. Most electrical formulas require the use of values expressed in basic units; therefore, all values must usually be converted before computation can be made. Figure 8-79 contains a conversion table which lists a number of the most commonly used electrical values.
1 ampere = 1,000,000 microamperes.
1 ampere = 1,000 milliamperes.
I farad = 1,000,000,000,000 micromicrofarade.
1 farad = 1,000,000 microfarads.
I farad = 1,000 millifarads.
1 henry = 1,000,000 microhenrys.
the lodestone is the only natural magnet. All other magnets are produced artifically.
From the earliest times a great deal was known about the elementary behavior of magnets. For 1 henry = 1,000 millihenrys. example, it was known that the property of 1 kilovolt = 1,000 volte. magnetism could be induced in an iron bar by 1 kilowatt = 1,000 watts. stroking it with a lodestone. In addition, it was I megohm = 1,000,000 ohms.
1 microampere = .000001 ampere. known that if the north-seeking end of a suspended 1 microfarad = .000001 farad. magnet was brought near the north-seeking end of 1 microhm = .000001 ohm. another, the magnets would repel each other. On I microvolt = .000001 volt. the other hand, they found that a north-seeking 1 microwatt = .000001 watt.
1 micromicrofarad = .000000000001 farad. and a south-seeking end would attract each other.
1 milliampere = .001 ampere. Magnetism is defined as the property of an 1 millihenry = .001 henry. object to attract certain metallic substances. In 1 millimho = .001 mho. general, these substances are ferrous materials;
1 milliohm = .001 ohm. that is, materials composed of iron or iron alloys, 1 millivolt = .001 volt.
1 milliwatt = .001 watt. such as soft iron, steel, and alnico. These materials, 1 volt = 1,000,000 microvolts. sometimes called magnetic materials, today 1 volt = 1,000 millivolts. include at least three nonferrous materials:
1 watt = 1,000 milliwatts. nickel, cobalt and gadolinium, which are magnetic 1 watt = .001 kilowatt. to a limited degree. All other substances are considered nonmagnetic, and a few of these non- FIGURE 8-79. Conversion table. magnetic substances can be classified as diamagnetic since they are repelled by both poles of a Figure 8-80 contains a complete list of the Magnetism is an invisible force, the ultimate multiples used to express electrical quantities, nature of which has not been fully determined.
together with the prefixes and symbols used to represent each number.
Number Prefix Symbol MAGNETISM 1,000,000,000,000 tera Magnetism is so closely allied with electricity 1,000,000,000 giga g in the modern industrial world it can be safely 1,000,000 mega m stated that without magnetism the electrical 1,000 kilo k world would not be possible. Knowledge of magnetism has existed for many centuries, but it 100 hecto h was not until the eighteenth century that this 10 deka dk stream of knowledge was joined with that of 0.1 deci d electricity by the discoveries of science.
0.01 centi The earliest known magnetism was the lode- 0.001 mili stone, a natural mineral found in Asia Minor. 0.000,001 micro Today this substance is called magnetite or magnetic oxide of iron. When a piece of this ore is 0.000,000,001 nano 11 suspended horizontally by a thread or floated 0.000,000,000,001 pico P on wood in undisturbed water, it will align itself in a north-south direction. This characteristic led to its use as a compass and the name lodestone, FIGURE 8-80. Prefixes and symbols for multiples meaning leading stone. Other than the earth itself, of basic quantities.
NORTH POLE SOUTH POLE (A) Unmagnetized
N N S N N FIGURE 8-81. One end of magnetized strip N IN S IN S IN points to the magnetic north pole.
N S N S It can best be described by the effects it produces.
Examination of a simple bar magnet similar to that illustrated in figure 8-81 discloses some basic characteristics of all magnets. If the magnet is suspended to swing freely, it will align itself with the earth's magnetic poles. One end is labelled "N," meaning the north-seeking end or pole of the magnet. If the "N" end of a compass or magnet is referred to as north-seeking rather than north, there will be no conflict in referring to the pole it seeks, which is the north magnetic Lines of Force FIGURE 8-82. Magnetic field around magnets.
(B) Magnetized FIGURE 8-83. Arrangement of molecules in a piece of magnetic material.
pole. The opposite end of the magnet, marked "S," is the south-seeking end and points to the south magnetic pole. Since the earth is a giant magnet, its poles attract the ends of the magnet.
These poles are not located at the geographic poles.
The somewhat mysterious and completely invisible force of a magnet depends on a magnetic field that surrounds the magnet as illustrated in figure 8-82. This field always exists between the poles of a magnet, and will arrange itself to conform to the shape of any magnet.
The theory that explains the action of a magnet holds that each molecule making up the iron bar is itself a tiny magnet, with both north and south poles as illustrated in A of figure 8-83. These molecular magnets each possess a magnetic field, but in an unmagnetized state the molecules are arranged at random throughout the iron bar. If a magnetizing force, such as stroking with a lodestone, is applied to the unmagnetized bar, the molecular magnets rearrange themselves in line with the magnetic field of the lodestone, with all north ends of the magnets pointing in one direction and all south ends in the opposite direction. This is illustrated in B of figure 8-83. In such a configuration, the 299