2
CHAPTER
Electrical Units
LET’S LEARN ABOUT THE STANDARD UNITS THAT ENGINEERS USE IN DIRECT-CURRENT (DC) CIRCUITS. Many of these principles apply to common utility alternating-current (AC) systems as well.
The Volt
In Chap. 1, you learned about the volt, the standard unit of electromotive force (EMF), or potential difference. An accumulation of electrostatic charge, such as an excess or shortage of electrons, always occurs when we have a potential difference between two points or objects. A power plant, an electrochemical reaction, light rays striking a semiconductor chip, and other phenomena can also produce voltages. We can get an EMF when we move an electrical conductor through a fixed magnetic field, or when we surround a fixed electrical conductor with a fluctuating magnetic field.
A potential difference between two points, called poles, invariably produces an electric field, represented by electric lines of flux, as shown in Fig. 2-1. We call such a pair of electrically charged poles an electric dipole. One pole carries relatively positive charge, and the other pole carries relatively negative charge. The positive pole always has fewer electrons than the negative pole. Note that the electron numbers are relative, not absolute! An electric dipole can exist even if both poles carry surplus electrons, or if both poles suffer from electron deficiencies, relative to some external point of reference having an absolutely neutral charge.
2-1 Electric lines of flux always exist near poles of electric charge.
The abbreviation for volt (or volts) is V. Sometimes, engineers use smaller units. The millivolt (mV) equals 0.001 V. The microvolt (μV) equals 0.000001 V. Units larger than the volt also exist. One kilovolt (kV) represents 1000 V. One megavolt (MV) equals 1,000,000 V, or 1000 kV.
In an everyday dry cell, the poles maintain a potential difference somewhere between 1.2 and 1.7 V. In an automotive battery, it’s in the range of 12 V to 14 V. In household AC utility wiring, the potential difference alternates polarity and maintains an effective value of approximately 117 V for electric lights and most small appliances, and 234 V for washing machines, ovens, or other large appliances. In some high-power radio transmitters, the EMF can range in the thousands of volts. The largest potential differences on our planet—upwards of 1 MV—build up in thunderstorms, sandstorms, and violent erupting volcanoes.
The existence of a voltage always means that charge carriers, which are mostly electrons in a conventional circuit, will travel between the charge poles if we provide a decent path for them to follow. Voltage represents the driving force, or “pressure,” that impels charge carriers to move. If we hold all other factors constant, a high voltage will make the charge carriers flow in greater quantity per unit of time, thereby producing a larger electrical current than a low voltage. But that statement oversimplifies the situation in most practical systems, where “all other factors” rarely “hold constant”!
Current Flow
If we provide a conducting or semiconducting path between two poles having a potential difference, charge carriers flow in an attempt to equalize the charge between the poles. This current continues for as long as the path remains intact, and as long as a charge difference exists between the poles.
Sometimes the charge difference between two electric poles decreases to zero after a current has flowed for a while. This effect takes place in a lightning stroke, or when you touch a radiator after shuffling around on a carpet. In these instances, the charge between the poles equalizes in a fraction of a second. In other cases, the charge takes longer to dissipate. If you connect a piece of wire directly between the positive and negative poles of a dry cell, the cell “runs out of juice” after a few minutes. If you connect a light bulb across the cell to make a “flashlight,” the charge difference may take an hour or two to get all the way down to zero.
In household electric circuits, the charge difference never equalizes unless a power failure occurs. Of course, if you short-circuit an AC electrical outlet (don’t!), the fuse or breaker will blow or trip, and the charge difference will immediately drop to zero. But if you put a standard utility light bulb at the outlet, the charge difference will continue to exist at “full force” even as the current flows. The power plant can maintain a potential difference of 117 V across a lot of light bulbs indefinitely.
Have you heard that the deadly aspect of electricity results from current, not voltage? Literally, that’s true, but the statement plays on semantics. You could also say “It’s the heat, not the fire, that burns you.” Okay! But a deadly current can arise only in the presence of an EMF sufficient to drive a certain amount of current through your body. You don’t have to worry about deadly currents flowing between your hands when you handle a 1.5-V dry cell, even though, in theory, such a cell could produce currents strong enough to kill you if your body resistance were much lower. You’re safe when handling flashlight cells, but you’ve got good reason to fear for your life around household utility circuits. An EMF of 117 V can easily pump enough current through your body to electrocute you.
It all goes back to Ohm’s Law. In an electric circuit whose conductance (or resistance) never varies, the current is directly proportional to the applied voltage. If you double the voltage, you double the current. If you cut the voltage in half, the current goes down by half. Figure 2-2 shows this relationship as a graph in general terms. Here, we assume that the power supply can always provide as many charge carriers per unit of time as we need.
2-2 Relative current as a function of relative voltage for low, medium, and high resistances.
The Ampere
Current expresses the rate at which charge carriers flow past a fixed point per unit of time. The standard unit of current is the ampere, which represents one coulomb (6,240,000,000,000,000,000, or 6.24 × 1018) of charge carriers flowing past a given point every second.
An ampere is a comparatively large amount of current. The abbreviation is A. Often, you’ll want to express current in terms of milliamperes, abbreviated mA, where 1 mA = 0.001 A. You’ll also sometimes hear of microamperes (μA), where 1 μA = 0.000001 A or 0. 001 mA. You might even encounter nanoamperes (nA), where 1 nA = 0.000000001 A = 0.001 μA.
A current of a few milliamperes will give you a rude electrical shock. About 50 mA will jolt you severely, and 100 mA can kill you if it flows through your heart. An ordinary utility light bulb draws 0.5 A to 1 A of current in a household utility circuit. An electric iron draws approximately 10 A; an entire household normally uses between 10 A and 100 A, depending on the size of the house and the kinds of appliances it has, and also on the time of day, week, or year.
The amount of current that flows in an electrical circuit depends on the voltage, and also on the resistance. In some electrical systems, extremely large currents, say 1000 A, can flow. You’ll get a current like this if you place a metal bar directly across the output terminals of a massive electric generator. The bar has an extremely low resistance, and the generator can drive many coulombs of charge carriers through the bar every second. In some semiconductor electronic devices, a few nanoamperes will suffice to allow for complicated processes. Some electronic clocks draw so little current that their batteries last as long as they would if you left them on the shelf.
Resistance and the Ohm
Resistance quantifies the opposition that a circuit imposes against the flow of electric current. You can compare resistance to the reciprocal of the diameter of a garden hose (where conductance compares to the actual diameter). For metal wire, this analogy works pretty well. Small-diameter wire has higher resistance than large-diameter wire made of the same metal.
The standard unit of resistance is the ohm, sometimes symbolized as an upper-case Greek letter omega (Ω). You’ll also hear about kilohms (symbolized k or kΩ), where 1 k = 1000 ohms, or about 1 megohm (symbolized M or MΩ), where 1 M = 1,000,000 ohms or 1000 k. In this book, we’ll never use the omega symbol. Instead, we’ll always write out “ohm” or “ohms” in full.
Electric wire is sometimes rated for resistance per unit length. The standard unit for this purpose is the ohm per foot (ohm/ft) or the ohm per meter (ohm/m). You might also come across the unit ohm per kilometer (ohm/km). Table 2-1 shows the resistance per unit of length for various common sizes of solid copper wire at room temperature as a function of the wire size, as defined by a scheme known as the American Wire Gauge (AWG).
Table 2-1. Approximate resistance per unit of length in ohms per kilometer (ohms/km) at room temperature for solid copper wire as a function of the wire size in American Wire Gauge (AWG).
When we place a potential difference of 1 V across a component whose resistance equals 1 ohm, assuming that the power supply can deliver an unlimited number of charge carriers, we get a current of 1 A. If we double the resistance to 2 ohms, the current decreases to 0.5 A. If we cut the resistance by a factor of 5 to get only 0.2 ohms, the current increases by the same factor, from 1 A to 5 A. The current flow, for a constant voltage, varies in inverse proportion to the resistance. Figure 2-3 shows the current, through components of various resistances, given a constant potential difference of 1 V.
2-3 Current as a function of resistance through an electric device for a constant voltage of 1 V.
Whenever an electric current flows through a component, a potential difference appears across that component. If the component has been deliberately manufactured to exhibit a certain resistance, we call it a resistor. Figure 2-4 illustrates this effect. In general, the potential difference arises in direct proportion to the current through the resistance. Engineers take advantage of this effect when they design electronic circuits, as you’ll learn later in this book.
2-4 Whenever current passes through a component having resistance, a voltage exists across that component.
Electrical circuits always have some resistance. No such thing as a perfect conductor (an object with mathematically zero resistance) exists in the “real world.” When scientists cool certain metals down to temperatures near absolute zero, the substances lose practically all of their resistance, so that current can flow around and around for a long time. This phenomenon is called superconductivity. But nothing can ever become an absolutely perfect conductor.
Just as a perfectly resistance-free substance cannot exist in the real world, we’ll never encounter an absolutely infinite resistance, either. Even dry air conducts electric current to some extent, although the effect is usually so small that scientists and engineers can ignore it. In some electronic applications, engineers select materials based on how “nearly infinite” their resistance appears; but when they say that, they exploit a figure of speech. They really mean to say that the resistance is so gigantic that we can consider it “infinite” for all practical purposes.
In electronics, the resistance of a component often varies, depending on the conditions under which that component operates. A transistor, for example, might have high resistance some of the time, and low resistance at other times. High/low resistance variations can take place thousands, millions, or billions of times each second. In this way, oscillators, amplifiers, and digital devices function in radio receivers and transmitters, telephone networks, digital computers, and satellite links (to name just a few applications).
Conductance and the Siemens
Electricians and engineers sometimes talk about the conductance of a material, rather than about its resistance. The standard unit of conductance is the siemens, abbreviated S. When a component has a conductance of 1 S, its resistance equals 1 ohm. If we double the resistance of a component, its conductance drops to half the former value. If we halve the resistance, we double the conductance. Conductance in siemens always equals the reciprocal of resistance in ohms, as long as we confine our attention to one component or circuit at a time.
If we know the resistance of a component in ohms, we can get the conductance in siemens; we simply divide 1 by the resistance. If we know the conductance in siemens, we can get the resistance in ohms; we divide 1 by the conductance. In calculations and equations, engineers denote resistance by writing an italicized, uppercase letter R, and conductance by writing an italicized, uppercase letter G. If we express R in ohms and G in siemens, then
G = 1/R
and
R = 1/G
In “real-world” electrical and electronic circuits, you’ll often use units of conductance much smaller than the siemens. A resistance of 1 k represents a conductance of one millisiemens (1 mS). If we encounter a component whose resistance equals 1 M, its conductance is one microsiemens (1 μS). You’ll sometimes hear about kilosiemens (kS) or megasiemens (MS), representing resistances of 0.001 ohm and 0.000001 ohm, respectively. Short lengths of heavy wire have conductance values in the range of kilosiemens. A heavy, solid copper or silver rod might exhibit a conductance in the megasiemens range.
If a component has a resistance of 50 ohms, its conductance equals 1/50 S or 0.02 S. We can also call this quantity 20 mS. Now imagine a piece of wire with a conductance of 20 S. Its resistance equals 1/20 ohm or 0.05 ohm. You won’t often hear or read the term “milliohm” in technical conversations or papers, but you might say that an 0.05-ohm length of wire has a resistance of 50 milliohms, and you’d be technically correct.
When you want to determine the conductivity of a component, circuit, or system, you must exercise caution or you might end up calculating the wrong value. If wire has a resistance per unit length of 10 ohms/km, you can’t say that it has a conductivity of 1/10, or 0.1, S/km. A 1-km span of such wire does indeed have a conductance of 0.1 S, but a 2-km span of the same wire has a resistance of 20 ohms because you have twice as much wire. That’s not twice the conductance, but half the conductance, of the 1-km span. If you say that the conductivity of the wire is 0.1 S/km, then you might be tempted to say that 2 km of the wire has 0.2 S of conductance. That would be a mistake! Conductance decreases with increasing wire length.
Figure 2-5 illustrates the resistance and conductance values for various lengths of wire having a resistance per unit length of 10 ohms/km.
2-5 Resistance and conductance for various lengths of wire having a resistivity of 10 ohms/km.
Power and the Watt
Whenever we drive an electrical current through a resistive component, the temperature of that component rises. We can measure the intensity of the resulting heat in units called watts (symbolized W), representing power. (As a variable quantity in equations, we denote power by writing P.) Power can manifest itself in various forms such as mechanical motion, radio waves, visible light, or noise. But we’ll always find heat (in addition to any other form of power) in an electrical or electronic device, because no “real-world” system operates with 100-percent efficiency. Some power always goes to waste, and this waste shows up mainly as heat.
Look again at Fig. 2-4. A certain potential difference appears across the resistor, although the illustration does not reveal the actual voltage. A current flows through the resistor; again, the diagram doesn’t tell us the value. Suppose that we call the voltage across the resistor E and the current through the resistor I, expressed in volts (V) and amperes (A), respectively. If we let P represent the power in watts dissipated by the resistor, then
P = EI
If the voltage E across the resistance is caused by two flashlight cells in series, giving us 3 V, and if the current I through the resistance (a flashlight bulb, perhaps) equals 0.2 A, then E = 3 V and I = 0.2 A, and we can calculate the power P as
P = EI = 3 × 0.2 = 0.6 W
Suppose the voltage equals 220 V, giving rise to a current of 400 mA. To calculate the power, we must convert the current into amperes: 400 mA = 400/1000 A = 0.400 A. Then we have
P = EI = 220 × 0.400 = 88.0 W
You will often hear about milliwatts (mW), microwatts (μW), kilowatts (kW), and megawatts (MW). By now, you should know what these units represent when you see the prefixes. Otherwise, you can refer to Table 2-2, which lists the common prefix multipliers for physical units.
Table 2-2. Prefix multipliers from 0.000000000001 (trillionths, or units of 10-12) to 1,000,000,000,000 (trillions, or units of 1012).
Once in a while, you’ll want to take advantage of the power equation to find a current through a component or a voltage across a component. In that case, you can use the variant
I = P/E
to find current, or
E = P/I
to find the voltage. Always convert to standard units (volts, amperes, and watts) before performing calculations with any of these formulas. Otherwise, you risk getting an answer that’s too large or small by one or more orders of magnitude (powers of 10)!
A Word about Notation
Sometimes, symbols and abbreviations appear in italics, and sometimes they don’t. We’ll encounter subscripts often, and sometimes even the subscripts are italicized. Following are some rules that apply to notation in electricity and electronics.
• We never italicize the abbreviations for units such as volts (V), amperes (A), and watts (W).
• We never italicize the abbreviations for objects or components such as resistors (R), batteries (B), capacitors (C), and inductors (L).
• We never italicize the abbreviations for quantifying prefixes such as kilo- (k), micro- (μ), mega- (M), or nano- (n).
• Labeled points in drawings might or might not be italicized. It doesn’t matter, as long as a diagram remains consistent within itself. We might call a point either P or P, for example.
• We always italicize the symbols for mathematical constants and variables such as time (t), the speed of light in a vacuum (c), velocity (v), and acceleration (a).
• We always italicize the symbols for electrical quantities such as voltage (E or V), current (I), resistance (R), and power (P).
• We never italicize numeric subscripts. We might denote a certain resistor as R2, but never as R2; we might denote a certain amount of current as I4, but never as I4.
• For non-numeric subscripts, the same rules apply as for general symbols.
Once in a while we’ll see the same symbol italicized in one place and not in another—even within a single diagram or discussion! We might, for example, talk about “resistor number 3” (symbolized R3), and then later in the same paragraph talk about its value as “resistance number 3” (symbolized R3). Still later, we might talk about “the nth resistor in a combination of resistors” (Rn) and then “the nth resistance in a combination of resistances” (Rn).
Energy and the Watt-Hour
Have you heard the terms “power” and “energy” used interchangeably, as if they mean the same thing? Well, they don’t! The term energy expresses power dissipated over a certain period of time. Conversely, the term power expresses the instantaneous rate at which energy is expended at a particular moment in time.
Physicists measure energy in units called joules. One joule (1 J) technically equals a watt-second, the equivalent of 1 W of power dissipated for 1 s of time (1 W · s or 1 Ws). In electricity, you’ll more often encounter the watt-hour (symbolized W · h or Wh) or the kilowatt-hour (symbolized kW · h or kWh). As their names imply, a watt-hour represents the equivalent of 1 W dissipated for 1 h, and 1 kWh represents the equivalent of 1 kW of power dissipated for 1 h.
An energy quantity of 1 Wh can manifest itself in infinitely many ways. A light bulb rated at 60 W consumes 60 Wh in 1 h, the equivalent of a watt-hour per minute (1 Wh/min). A lamp rated at 100 W consumes 1 Wh in 1/100 h, or 36 s. Whenever we double the power, we halve the time required to consume 1 Wh of energy. But in “real-world” scenarios, the rate of power dissipation rarely remains constant. It can change from moment to moment in time.
Figure 2-6 illustrates two hypothetical devices that consume 1 Wh of energy. Device A uses its power at a constant rate of 60 W, so it consumes 1 Wh in 1 min. The power consumption rate of device B varies, starting at zero and ending up at more than 60 W. How do we know that this second device really consumes 1 Wh of energy? To figure that out, we must determine the area under the graph. In this case, the graph encloses a simple triangle. We recall from our basic geometry courses that the area of a triangle equals half the base length times the height. Device B receives power for 72 s, or 1.2 min; that’s 1.2/60 = 0.02 h. The area under the graph is therefore 1/2 × 100 × 0.02 = 1 Wh.
2-6 Two devices that consume 1 Wh of energy. Device A dissipates a constant amount of power as time passes. Device B dissipates an increasing amount of power as time passes.
When you calculate energy values, you must always keep in mind the units with which you work. In the example of Fig. 2-6, you would use the watt-hour, so you must multiply watts by hours. If you multiply watts by minutes or seconds, you’ll get the wrong kind of unit in your answer—and sometimes a “unit” that doesn’t have any technical definition at all!
Often, graphs of power versus time show up as complex curves, not “neat” figures, such as rectangles or triangles. Consider the graph of power consumption in your home, as a function of time, over the course of a hypothetical day. That graph might resemble the curve in Fig. 2-7. Obviously, you won’t find a simple formula to calculate the area under this curve, but you can use another scheme to determine the total energy consumed by your household over a period of time. You can employ a special meter that measures electrical energy in kilowatt-hours (kWh).
2-7 Graph showing the amount of power consumed by a hypothetical household, as a function of the time of day.
Every month, the power company sends a representative, or employs a wireless device, to record the number of kilowatt-hours that your meter displays. The power company then subtracts the reading taken the previous month from the current value. A few days later, you get a bill for the month’s energy usage. The “power meter” (a misnomer, because it’s really an energy meter) automatically keeps track of total consumed energy, without anybody having to go through high-level mathematical calculations to find the areas under irregular curves, such as the graph of Fig. 2-7.
Other Energy Units
The joule, while standard among scientists, isn’t the only energy unit that exists. You’ll occasionally encounter the erg, a tiny unit equivalent to 0.0000001 of a joule. Some scientists use the erg in laboratory experiments involving small amounts of expended energy.
Most folks have heard or read about the British thermal unit (Btu), equivalent to 1055 joules. People use the Btu to define the cooling or heating capacity of air-conditioning equipment. To cool your room from 85 to 78 degrees Fahrenheit, you need a certain amount of energy, perhaps best specified in Btu. If you plan to have an air conditioner or furnace installed in your home, an expert will determine the size of the unit that best suits your needs. That person might tell you how “powerful” the unit should be, in terms of its ability to heat or cool in British thermal units per hour (Btu/h).
Physicists also use, in addition to the joule, a unit of energy called the electron-volt (eV). It’s a minuscule unit indeed, equal to only 0.00000000000000000016 joule (you can count 18 zeroes after the decimal point and before the 1). Physicists represent this number as 1.6 × 10–19. A single electron in an electric field of 1 V gains 1 eV of energy. Atomic physicists rate particle accelerators (or, informally, “atom smashers”) in terms of megaelectron-volts (MeV, where 1 MeV = 1,000,000 eV), or gigaelectron-volts (GeV, where 1 GeV = 1000 MeV), or teraelectron-volts (TeV, where 1 TeV = 1000 GeV) of energy capacity.
Another energy unit, employed to denote mechanical work, is the foot-pound (ft-lb). It’s the amount of “labor” needed to elevate a weight of one pound (1 lb) straight upward by a distance of one foot (1 ft), not including any friction. One foot-pound equals 1.356 J.
Table 2-3 summarizes all of the energy units described here, along with conversion factors to help you change from any particular unit to joules or vice-versa. The table includes watt-hours and kilowatt-hours. In electricity and electronics, you’ll rarely need to concern yourself with any energy unit other than these two.
Table 2-3. Conversion Factors between Joules and Various Other Energy Units
Alternating Current and the Hertz
Direct current (DC) always flows in the same direction, but household utility current reverses direction at regular intervals. In the United States (and much of the world), the current direction reverses once every 1/120 second, completing one full cycle every 1/60 second. In some countries, the direction reverses every 1/100 second, taking 1/50 second to go through a complete cycle. When we encounter a periodically reversing current flow of this sort, we call it alternating current (AC).
Figure 2-8 shows a common “117-V” utility AC wave as a graph of voltage versus time. If you’re astute, you’ll see that the maximum positive and negative EMFs don’t equal 117 V. Instead, they come close to 165 V. The effective voltage for an AC wave usually differs from the instantaneous maximum, or peak, voltage. For the waveform shown in Fig. 2-8, the effective value is approximately 0.707 times the peak value (the theoretically exact multiplication factor equals the reciprocal of the square root of 2). Conversely, the peak value is approximately 1.414 times the effective value (theoretically the factor equals the square root of 2).
2-8 One cycle of utility alternating current (AC). The instantaneous voltage is the voltage at any particular instant in time. The peak voltages are approximately plus and minus 165 V.
The hertz (symbolized Hz) is the basic unit of AC frequency. One hertz represents one complete cycle per second. Because a typical utility AC cycle repeats itself every 1/60 second, we say that the wave has a frequency of 60 Hz. In the United States, 60 Hz is the standard frequency for AC. In much of the rest of the world, however, it’s 50 Hz.
In wireless communications, you’ll hear about kilohertz (kHz), megahertz (MHz), and gigahertz (GHz). These units relate to each other as follows:
• 1 kHz = 1000 Hz = 103 Hz
• 1 MHz = 1000 kHz = 1,000,000 Hz = 106 Hz
• 1 GHz = 1000 MHz = 1,000,000 kHz = 1,000,000,000 Hz = 109 Hz
Usually, but not always, the waves have shapes like the one shown in Fig. 2-8. Engineers and technicians call it a sine wave or a sinusoid.
Rectification and Pulsating Direct Current
Batteries and other sources of DC produce constant voltage, which we can graphically portray by plotting a straight, horizontal line on a coordinate grid, showing voltage as a function of time. Figure 2-9 shows a representation of DC. For pure DC, the peak voltage equals the effective voltage. In some systems that derive their power from sources other than batteries, the instantaneous DC voltage fluctuates rapidly with time. This situation exists, for example, if we pass the sinusoid of Fig. 2-8 through a rectifier circuit, which allows the current to flow in only one direction.
2-9 A representation of pure direct current (DC).
Rectification changes AC into DC. To obtain rectification, we can use a device called a diode. When we rectify an AC wave, we can either cut off or invert one half of the AC wave to get pulsating DC output. Figure 2-10 illustrates two different waveforms of pulsating DC. In the waveform at A, we simply remove the negative (bottom) half of the cycle. In the situation at B, we invert the negative portion of the wave, making it positive instead—a “mirror image” of its former self. Figure 2-10A shows half-wave rectification; it involves only half the waveform. Figure 2-10B illustrates full-wave rectification, in which both halves of the waveform contribute to the output. In the output of a full-wave rectifier, all of the original current still flows, even though it doesn’t alternate as the input does.
2-10 At A, half-wave rectification of common utility AC. At B, full-wave rectification of common utility AC. Effective voltages are shown by the dashed lines.
The effective value, compared with the peak value, for pulsating DC depends on whether we apply half-wave or full-wave rectification to an AC wave. In Figs. 2-10 A and B, the effective voltages appear as dashed lines, and the instantaneous voltages show up as solid curves. The instantaneous voltage changes from instant to instant in time (that’s where the term comes from).
In Fig. 2-10B, the effective voltage equals 2−1/2 (roughly 0.707) times the peak voltage, just as with ordinary AC. The direction of current flow, for many kinds of devices, doesn’t make any difference. But in Fig. 2-10A, half of the wave has been lost, cutting the effective value in half so that it’s only 2−1/2/2 (approximately 0.354) times the peak voltage.
In household “wall-outlet” AC for powering-up conventional appliances in the United States, we observe a peak EMF of about 165 V, and an effective EMF of about 117 V. If we subject this electricity to full-wave rectification, both the peak and the effective EMFs remain at these values. If we put such a wave through a half-wave rectifier, the peak EMF remains the same, but the effective output EMF drops to about 58.5 V.
Stay Safe!
For all “intents and purposes,” one rule applies concerning safety around electrical apparatus. Never forget it, even for a moment. One careless move can kill anyone.
Household electricity, with an effective EMF of about 117 V (but sometimes twice that for large appliances, such as electric ranges and laundry machines), is more than sufficient to kill you if it drives current through your chest cavity. Certain devices, such as spark coils, can produce lethal currents even from an automotive battery. Consult the American Red Cross or your electrician concerning what types of circuits, procedures, and devices are safe, and what kinds aren’t.
Magnetism
Whenever an electric current flows—that is, whenever charge carriers move—a magnetic field appears in the vicinity. In a straight wire that carries electrical current, magnetic lines of flux surround the wire in circles, with the wire at the center. (The lines of flux aren’t physical objects, but they offer a convenient way to represent the magnetic field.) You’ll sometimes hear or read about a certain number of flux lines per unit cross-sectional area, such as 100 lines per square centimeter. That terminology expresses, informally, the relative intensity of the magnetic field.
Magnetic fields arise whenever the atoms of certain materials align themselves. Iron is the most familiar element with this property. The atoms of iron in the earth’s core have become aligned to some extent, a complex phenomenon caused by the rotation of our planet and its motion with respect to the magnetic field of the sun. The magnetic field surrounding the earth gives rise to fascinating effects, such as the concentration of charged particles that you see as the aurora borealis during a “solar storm.”
When you wind a piece of wire into a tight coil, the resulting magnetic flux takes a shape similar to the flux field surrounding the earth. Two well-defined magnetic poles develop, as shown in Fig. 2-11. You can increase the intensity of such a field by placing a special core inside the coil. Iron, steel, or some other material that can be readily magnetized works very well for this purpose. We call such substances ferromagnetic materials.
2-11 Magnetic flux lines around a current-carrying coil of wire. The flux lines converge at the magnetic poles.
A ferromagnetic core doesn’t increase the total quantity of magnetism in and around a coil, but it can produce a more intense field. This is the principle by which an electromagnet works. It also facilitates the operation of electrical transformers for utility current. Technically, magnetic lines of flux emerge from north poles and converge toward south poles. Therefore, the magnetic field “flows” from the north end of a coil or bar magnet to the south end, following the lines of flux through the surrounding space.
Magnetic Units
We can express the overall quantity of a magnetic field in units called webers, abbreviated Wb. One weber is mathematically equivalent to one volt-second (1 V · s). For weaker magnetic fields, we can use a smaller unit called the maxwell (symbolized Mx). One maxwell equals 0.00000001 Wb, or 0.01 microvolt-second (0.01 μV · s).
We can express the flux density of a magnetic field in terms of webers or maxwells per square meter or per square centimeter. A flux density of one weber per square meter (1 Wb/m2) represents one tesla (1 T). One gauss (1 G) equals 0.0001 T, or one maxwell per square centimeter (1 Mx/cm2).
In general, as the electric current through a wire increases, so does the flux density near the wire. A coiled wire produces a greater flux density for a given electrical current than a single, straight wire. As we increase the number of turns in a coil of a specific diameter that carries a constant current, the flux density in and around the coil increases.
Sometimes, engineers specify magnetic field strength in ampere-turns (At). The ampere-turn quantifies a phenomenon called magnetomotive force. A one-turn wire loop, carrying 1 A of current, produces a magnetomotive force of 1 At. Doubling the number of turns with constant current doubles the magnetomotive force. Doubling the current for a constant number of turns also doubles the magnetomotive force. If you have 10 A flowing in a 10-turn coil, the magnetomotive force equals 10 × 10, or 100 At. If you have 100 mA flowing in a 200-turn coil, the magnetomotive force equals 0.1 × 200, or 20 At. (Remember that 100 mA = 0.1 A.)
Once in a while, you might hear or read about a unit of magnetomotive force called the gilbert (Gb). One gilbert equals approximately 0.796 At. Conversely, 1 At equals approximately 1.26 Gb.
Quiz
Refer to the text in this chapter if necessary. A good score is at least 18 correct answers. The answers are listed in the back of this book.
1. In an electric dipole of constant polarity, the positive charge center
(a) has more electrons than the negative charge center.
(b) has the same number of electrons as the negative charge center.
(c) has fewer electrons than the negative charge center.
(d) sometimes has more electrons than the negative charge center, sometimes has the same number, and sometimes has fewer.
2. If you touch two points that have DC voltage between them, one point with your left hand and the other point with your right hand, which of the following voltages would present the greatest electrocution hazard?
(a) 1.5 V
(b) 15 V
(c) 150 V
(d) All three voltages would present equal electrocution hazards because it’s the current that kills, not the voltage.
3. If you increase the DC voltage across a resistor by a factor of 100 but you also increase the resistance to keep the current constant, then (assuming the resistor doesn’t burn out) the resistor will dissipate
(a) 100 times as much power as it did before.
(b) 10 times as much power as it did before.
(c) the same amount of power as it did before.
(d) 1/10 as much power as it did before.
4. If a length of wire exhibits 500 mS of conductance, then it has a resistance of
(a) 0.02 ohm.
(b) 0.2 ohm.
(c) 2 ohms.
(d) an amount that depends on how much current the wire carries.
5. A 330-ohm resistor has a conductance of
(a) 0.303 mS.
(b) 3.03 mS.
(c) 30.3 mS.
(d) 303 mS.
6. A circuit breaker is rated for 15.0 A in a 13.8-V DC automotive system (with the alternator running). This breaker should cut off the current if you connect a set of devices that demand a total of more than
(a) 207 W.
(b) 20.7 W.
(c) 1.09 W.
(d) 920 mW.
7. A heater warms a space by 1,000,000 J over a period of time. This amount of energy represents
(a) 1055 Btu.
(b) 948 Btu.
(c) 10.55 Btu.
(d) None of the above. The British thermal unit quantifies power, not energy!
8. Suppose that a 6.00-V battery delivers 4.00 W of power to a light bulb. How much current flows through the bulb?
(a) 24.0 A
(b) 1.50 A
(c) 667 mA
(d) We must know the bulb’s resistance to calculate the current.
9. Imagine that a span of wire 200 m long has a conductance of 900 mS. A 600-m length of this wire would have a conductance of
(a) 8.10 S.
(b) 2.70 S.
(c) 300 mS.
(d) 100 mS.
10. Which of the following units quantifies energy?
(a) The erg
(b) The kilowatt-hour
(c) The joule
(d) All of the above
11. Suppose that an AC cycle repeats at a constant rate of one full cycle every 0.02 second. This wave has a frequency of
(a) 500 Hz.
(b) 200 Hz.
(c) 50 Hz.
(d) 20 Hz.
12. In many countries outside the United States, utility AC electricity has a frequency of
(a) 33 Hz.
(b) 50 Hz.
(c) 75 Hz.
(d) 100 Hz.
13. If we could see them, the magnetic flux contours near a straight, current-carrying wire would look like
(a) concentric circles with the wire at their centers.
(b) straight lines parallel to the wire.
(c) straight lines that all pass through the wire at right angles.
(d) spirals that originate on the wire and all lie in planes perpendicular to the wire.
14. A high DC voltage across a load (a component with DC resistance)
(a) gives rise to poor conductance.
(b) can exist even if the load has low resistance.
(c) invariably drives a lot of current through the load.
(d) All of the above
15. Suppose that DC flows through a wire coil. The magnetomotive force produced by this coil depends on
(a) the number of turns in the coil.
(b) the diameter of the coil.
(c) the resistance of the coil.
(d) the material around which the coil is wound.
16. Suppose that 3 A of current flows through a 100-turn, circular loop of wire wound around a powdered-iron rod. Then we remove the rod, leaving the coil with an air core. The magnetomotive force
(a) decreases.
(b) increases.
(c) stays the same.
(d) drops to zero.
17. Which, if any, of the following units can express magnetomotive force?
(a) The ampere-turn per square meter
(b) The weber per square meter
(c) The maxwell per square meter
(d) None of the above
18. Given a sine-wave AC input, the output of a full-wave rectifier
(a) has an average voltage equal to the peak voltage.
(b) comprises constant DC just like a battery produces.
(c) is pulsating DC.
(d) is also a sine wave.
19. Given a sine-wave AC input, the output of a half-wave rectifier
(a) has an average voltage equal to the peak voltage.
(b) comprises constant DC just like a battery produces.
(c) is pulsating DC.
(d) is also a sine wave.
20. Which of the following units can express overall magnetic-field quantity?
(a) The weber
(b) The coulomb
(c) The volt
(d) The watt