Sabtu, 21 November 2009

Direct Current

CHAPTER 12




Direct Current


You now have a solid grasp of physics math, and you know the basics of classical physics. It is time to delve into the workings of things that can’t be observed directly. These include particles, and forces among them, that make it possible for you to light your home, communicate instantly with people on the other side of the world, and in general do things that would have been considered magical a few generations ago.




What Does Electricity Do?

When I took physics in middle school, they used 16-millimeter celluloid film projectors. Our teacher showed us several films made by a well-known professor. I’ll never forget the end of one of these lectures, in which the professor said, “We evaluate electricity not by knowing what it is, but by scrutinizing what it does.” This was a great statement. It really expresses the whole philosophy of modern physics, not only for electricity but also for all phenomena that aren’t directly tangible. Let’s look at some of the things electricity does.


CONDUCTORS

In some materials, electrons move easily from atom to atom. In others, the electrons move with difficulty. And in some materials, it is almost impos- sible to get them to move. An electrical conductor is a substance in which the electrons are highly mobile.
The best conductor, at least among common materials, at room tempera- ture is pure elemental silver. Copper and aluminum are also excellent electrical

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conductors. Iron, steel, and various other metals are fair to good conductors of
electricity. Some liquids are good conductors. Mercury is one example. Salt water is a fair conductor. Gases are, in general, poor conductors because the atoms or molecules are too far apart to allow a free exchange of electrons. However, if a gas becomes ionized, it can be a fair conductor of electricity.
Electrons in a conductor do not move in a steady stream like molecules of water through a garden hose. They pass from atom to atom (Fig. 12-1). This happens to countless atoms all the time. As a result, trillions of elec- trons pass a given point each second in a typical electric circuit.


Outer electron shell












Outer electron shell
Electron
Fig. 12-1. In an electrical conductor, electrons pass easily from atom to atom. This drawing is greatly simplified.


Imagine a long line of people, each one constantly passing a ball to his or her neighbor on the right. If there are plenty of balls all along the line, and if everyone keeps passing balls along as they come, the result is a steady stream of balls moving along the line. This represents a good conductor. If the peo- ple become tired or lazy and do not feel much like passing the balls along, the rate of flow decreases. The conductor is no longer very good.


INSULATORS

If the people refuse to pass balls along the line in the preceding example, the line represents an electrical insulator. Such substances prevent electric cur- rents from flowing, except in very small amounts under certain circumstances.
CHAPTER 12 Direct Current 299


Most gases are good electrical insulators (because they are poor con-
ductors). Glass, dry wood, paper, and plastics are other examples. Pure water is a good electrical insulator, although it conducts some current when minerals are dissolved in it. Metal oxides can be good insulators, even though the metal in pure form is a good conductor.
An insulating material is sometimes called a dielectric. This term arises from the fact that it keeps electric charges apart, preventing the flow of electrons that would equalize a charge difference between two places. Excellent insulating materials can be used to advantage in certain electrical components such as capacitors, where it is important that electrons not be able to flow steadily. When there are two separate regions of electric charge having opposite polarity (called plus and minus, positive and negative, or and ) that are close to each other but kept apart by an insulating mate- rial, that pair of charges is called an electric dipole.


RESISTORS

Some substances, such as carbon, conduct electricity fairly well but not very well. The conductivity can be changed by adding impurities such as clay to a carbon paste. Electrical components made in this way are called resistors. They are important in electronic circuits because they allow for the control of current flow. The better a resistor conducts, the lower is its resistance; the worse it conducts, the higher is the resistance.
Electrical resistance is measured in ohms, sometimes symbolized by the uppercase Greek letter omega (W). In this book we’ll sometimes use the symbol W and sometimes spell out the word ohm or ohms, so that you’ll get used to both expressions. The higher the value in ohms, the greater is the resistance, and the more difficult it is for current to flow. In an electrical system, it is usually desirable to have as low a resistance, or ohmic value,
as possible because resistance converts electrical energy into heat. This heat is called resistance loss and in most cases represents energy wasted. Thick wires and high voltages reduce the resistance loss in long-distance electrical lines. This is why gigantic towers, with dangerous voltages, are employed in large utility systems.


CURRENT

Whenever there is movement of charge carriers in a substance, there is an elec- tric current. Current is measured in terms of the number of charge carriers, or

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particles containing a unit electric charge, passing a single point in 1
second.
Charge carriers come in two main forms: electrons, which have a unit negative charge, and holes, which are electron absences within atoms and which carry a unit positive charge. Ions can act as charge carriers, and in some cases, atomic nuclei can too. These types of particles carry whole- number multiples of a unit electric charge. Ions can be positive or negative
in polarity, but atomic nuclei are always positive.
Usually, a great many charge carriers go past any given point in 1 sec- ond, even if the current is small. In a household electric circuit, a 100-W light bulb draws a current of about 6 quintillion (6 1018) charge carriers per second. Even the smallest minibulb carries a huge number of charge carriers every second. It is ridiculous to speak of a current in terms of charge carriers per second, so usually it is measured in coulombs per sec- ond instead. A coulomb (symbolized C) is equal to approximately 6.24
1018 electrons or holes. A current of 1 coulomb per second (1 C/s) is called an ampere (symbolized A), and this is the standard unit of electric current.
A 60-W bulb in a common table lamp draws about 0.5 A of current.
When a current flows through a resistance—and this is always the case, because even the best conductors have resistance—heat is generated. Sometimes visible light and other forms of energy are emitted as well. A light bulb is deliberately designed so that the resistance causes visible light
to be generated. However, even the best incandescent lamp is inefficient, creating more heat than light energy. Fluorescent lamps are better; they produce more light for a given amount of current. To put this another way, they need less current to give off a certain amount of light.
In physics, electric current is theoretically considered to flow from the positive to the negative pole. This is known as conventional current. If you connect a light bulb to a battery, therefore, the conventional current flows out of the positive terminal and into the negative terminal. However, the electrons, which are the primary type of charge carrier in the wire and the bulb, flow in the opposite direction, from negative to positive. This is the way engineers usually think about current.


STATIC ELECTRICITY

Charge carriers, particularly electrons, can build up or become deficient on objects without flowing anywhere. You’ve experienced this when walking on a carpeted floor during the winter or in a place where the humidity is
CHAPTER 12 Direct Current 301


low. An excess or shortage of electrons is created on and in your body. You
acquire a charge of static electricity. It’s called static because it doesn’t go anywhere. You don’t feel this until you touch some metallic object that is connected to an electrical ground or to some large fixture, but then there is
a discharge, accompanied by a spark and a small electric shock. It is the current, during this discharge, that causes the sensation.
If you were to become much more charged, your hair would stand on end because every hair would repel every other one. Objects that carry the same electric charge, caused by either an excess or a deficiency of electrons, repel each other. If you were massively charged, the spark might jump several cen- timeters. Such a charge is dangerous. Static electric (also called electrostatic) charge buildup of this magnitude does not happen with ordinary carpet and shoes, fortunately. However, a device called a Van de Graaff generator, found
in some high-school physics labs, can cause a spark this large. You have to be careful when using this device for physics experiments.
On the grand scale of the Earth’s atmosphere, lightning occurs between clouds and between clouds and the surface. This spark is a greatly magni- fied version of the little spark you get after shuffling around on a carpet. Until the spark occurs, there is an electrostatic charge in the clouds, between different clouds, or between parts of a cloud and the ground. In Fig. 12-2, four types of lightning are shown. The discharge can occur with-
in a single cloud (intracloud lightning, part a), between two different clouds (intercloud lightning, part b), or from a cloud to the surface (cloud-
to-ground lightning, part c), or from the surface to a cloud (ground-to-cloud lightning, part d). The direction of the current flow in these cases is con- sidered to be the same as the direction in which the electrons move. In cloud-to-ground or ground-to-cloud lightning, the charge on the Earth’s surface follows along beneath the thunderstorm cloud like a shadow as the storm is blown along by the prevailing winds.
The current in a lightning stroke can approach 1 million A. However, it takes place only for a fraction of a second. Still, many coulombs of charge are displaced in a single bolt of lightning.


ELECTROMOTIVE FORCE

Current can flow only if it gets a “push.” This push can be provided by a buildup of electrostatic charges, as in the case of a lightning stroke. When the charge builds up, with positive polarity (shortage of electrons) in one place and negative polarity (excess of electrons) in another place, a powerful

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+ +
+ + +
+ + +
+
+ + A













_ _ _
_
B

_ _ _ _
_ _ + +
_ _ _ _ _
+ + +
+

C D
+ + +
+ + + +
+ + +
_ _ _ _ _ _




Surface

Fig. 12-2. (a) Lightning can occur within a single cloud (intracloud),
(b) between clouds (intercloud), or between a cloud and the surface
(c) cloud to ground or (d) ground to cloud.


electromotive force (emf) exists. This effect, also known as voltage or electri-
cal potential, is measured in volts (symbolized V).
Ordinary household electricity has an effective voltage of between 110
and 130 V; usually it is about 117 V. A car battery has an emf of 12 V (6 V
in some older systems). The static charge that you acquire when walking on a carpet with hard-soled shoes can be several thousand volts. Before a discharge of lightning, millions of volts exist.
An emf of 1 V, across a resistance of 1 W, will cause a current of 1 A to
flow. This is a classic relationship in electricity and is stated generally as
CHAPTER 12 Direct Current 303


Ohm’s law. If the emf is doubled, the current is doubled. If the resistance is
doubled, the current is cut in half. This law of electricity will be covered in detail a little later.
It is possible to have an emf without having current flow. This is the case just before a lightning bolt occurs and before you touch a metallic object after walking on the carpet. It is also true between the two prongs of a lamp plug when the lamp switch is turned off. It is true of a dry cell when there
is nothing connected to it. There is no current, but a current can flow if there is a conductive path between the two points.
Even a large emf might not drive much current through a conductor or resistance. A good example is your body after walking around on the carpet. Although the voltage seems deadly in terms of numbers (thousands), not many coulombs of charge normally can accumulate on an object the size of your body. Therefore, not many electrons flow through your finger, in relative terms, when you touch the metallic object. Thus you don’t get a severe shock. Conversely, if there are plenty of coulombs available, a moderate voltage,
such as 117 V (or even less), can result in a lethal flow of current. This is why
it is dangerous to repair an electrical device with the power on. The utility power source can pump an unlimited number of coulombs of charge through your body if you are foolish enough to get caught in this kind of situation.




Electrical Diagrams

To understand how electric circuits work, you should be able to read electri- cal wiring diagrams, called schematic diagrams. These diagrams use schematic symbols. Here are the basic symbols. Think of them as something like an alphabet in a language such as Chinese or Japanese, where things are represented by little pictures. However, before you get intimidated by this comparison, rest assured that it will be easier for you to learn schematic sym- bology than it would be to learn Chinese (unless you already know Chinese!).


BASIC SYMBOLS

The simplest schematic symbol is the one representing a wire or electrical conductor: a straight solid line. Sometimes dashed lines are used to repre- sent conductors, but usually, broken lines are drawn to partition diagrams into constituent circuits or to indicate that certain components interact with

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each other or operate in step with each other. Conductor lines are almost
always drawn either horizontally across or vertically up and down the page so that the imaginary charge carriers are forced to march in formation like soldiers. This keeps the diagram neat and easy to read.
When two conductor lines cross, they are not connected at the crossing point unless a heavy black dot is placed where the two lines meet. The dot always should be clearly visible wherever conductors are to be connected, no matter how many of them meet at the junction.
A resistor is indicated by a zigzaggy line. A variable resistor, such as a rheostat or potentiometer, is indicated by a zigzaggy line with an arrow through it or by a zigzaggy line with an arrow pointing at it. These symbols are shown in Fig. 12-3.











(a) (b) (c)

Fig. 12-3. (a) A fixed resistor.
(b) A two-terminal variable resistor.
(c) A three-terminal potentiometer.



An electrochemical cell is shown by two parallel lines, one longer than the other. The longer line represents the positive terminal. A battery, or combination of cells in series, is indicated by an alternating sequence of parallel lines, long-short-long-short. The symbols for a cell and a battery are shown in Fig. 12-4.


SOME MORE SYMBOLS

Meters are indicated as circles. Sometimes the circle has an arrow inside it, and the meter type, such as mA (milliammeter) or V (voltmeter), is written alongside the circle, as shown in Fig. 12-5a. Sometimes the meter type is indicated inside the circle, and there is no arrow (see Fig. 12-5b). It doesn’t
CHAPTER 12 Direct Current 305



_

+





(a) (b)

Fig. 12-4. (a) An electrochemical cell. (b) A battery.





mA

mA







(a) (b)

Fig. 12-5. Meter symbols: (a) designator outside; (b) designator inside.


matter which way it’s done as long as you are consistent everywhere in a
given diagram.
Some other common symbols include the lamp, the capacitor, the air-core coil, the iron-core coil, the chassis ground, the earth ground, the alternating- current (AC) source, the set of terminals, and the black box (which can stand for almost anything), a rectangle with the designator written inside. These are shown in Fig. 12-6.




Voltage/Current/Resistance Circuits

Most direct current (dc) circuits can be boiled down ultimately to three major components: a voltage source, a set of conductors, and a resistance. This is shown in the schematic diagram of Fig. 12-7. The voltage of the emf source

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(a) (b) (c)



(d) (e) (f)









(g) (h) (i)
Fig. 12-6. More common schematic symbols:
(a) incandescent lamp; (b) fixed-value capacitor;
(c) air-core coil; (d) iron-core coil; (e) chassis ground;
(f) earth ground; (g) ac source; (h) terminals;
and (i) and black box.



is called E (or sometimes V); the current in the conductor is called I; the
resistance is called R. The standard units for these components are the volt
(V), the ampere (A), and the ohm (W), respectively. Note which characters here are italicized and which are not. Italicized characters represent mathe- matical variables; nonitalicized characters represent symbols for units.
You already know that there is a relationship among these three quanti- ties. If one of them changes, then one or both of the others also will change.
If you make the resistance smaller, the current will get larger. If you make
CHAPTER 12 Direct Current 307


R



I


E

Fig. 12-7. A simple dc circuit. The voltage is E, the current is I,
and the resistance is R.


the emf source smaller, the current will decrease. If the current in the cir-
cuit increases, the voltage across the resistor will increase. There is a sim- ple arithmetic relationship between these three quantities.


OHM’S LAW

The interdependence among current, voltage, and resistance in dc circuits
is called Ohm’s law, named after the scientist who supposedly first expressed it. Three formulas denote this law:

E IR I E/R R E/I
You need only remember the first of these formulas to be able to derive the others. The easiest way to remember it is to learn the abbreviations E for emf, I for current, and R for resistance; then remember that they appear in alphabetical order with the equals sign after the E. Thus E IR.
It is important to remember that you must use units of volts, amperes, and ohms in order for Ohm’s law to work right. If you use volts, mil- liamperes (mA), and ohms or kilovolts (kV), microamperes ( A), and megohms (MW), you cannot expect to get the right answers. If the initial quantities are given in units other than volts, amperes, and ohms, you must convert to these units and then calculate. After that, you can convert the units back again to whatever you like. For example, if you get 13.5 million ohms as a calculated resistance, you might prefer to say that it is 13.5 megohms. However, in the calculation, you should use the number 13.5
million (or 1.35 107) and stick to ohms for the units.

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CURRENT CALCULATIONS

The first way to use Ohm’s law is to find current values in dc circuits. In order to find the current, you must know the voltage and the resistance or be able to deduce them.
Refer to the schematic diagram of Fig. 12-8. It consists of a variable dc generator, a voltmeter, some wire, an ammeter, and a calibrated wide-range potentiometer. Actual component values are not shown here, but they can be assigned for the purpose of creating sample Ohm’s law problems. While calculating the current in the following problems, it is necessary to mentally
“cover up” the meter.

PROBLEM 12-1
Suppose that the dc generator (see Fig. 12-8) produces 10 V and that the potentiometer is set to a value of 10 W. What is the current?

SOLUTION 12-1
This is solved easily by the formula I E/R. Plug in the values for E and
R; they are both 10, because the units are given in volts and ohms. Then
I 10/10 1.0 A.

PROBLEM 12-2
The dc generator (see Fig. 12-8) produces 100 V, and the potentiometer is set to 10.0 kW. What is the current?



Current
=I

A








dc generator

V


Voltage
=E
Resistance
=R



Fig. 12-8. Circuit for working Ohm’s law problems.
CHAPTER 12 Direct Current 309


SOLUTION 12-2
First, convert the resistance to ohms: 10.0 kW 10,000 W. Then plug the val-
ues in: I 100/10,000 0.0100 A.


VOLTAGE CALCULATIONS

The second use of Ohm’s law is to find unknown voltages when the current and the resistance are known. For the following problems, uncover the ammeter and cover the voltmeter scale in your mind.

PROBLEM 12-3
Suppose that the potentiometer (see Fig. 12-8) is set to 100 W, and the meas-
ured current is 10.0 mA. What is the dc voltage?

SOLUTION 12-3
Use the formula E IR. First, convert the current to amperes: 10.0 mA
0.0100 A. Then multiply: E 0.0100 100 1.00 V. This is a low, safe volt- age, a little less than what is produced by a flashlight cell.


RESISTANCE CALCULATIONS

Ohm’s law can be used to find a resistance between two points in a dc cir- cuit when the voltage and the current are known. For the following prob- lems, imagine that both the voltmeter and ammeter scales in Fig. 12-8 are visible but that the potentiometer is uncalibrated.

PROBLEM 12-4
If the voltmeter reads 24 V and the ammeter shows 3.0 A, what is the value
of the potentiometer?

SOLUTION 12-4
Use the formula R E/I, and plug in the values directly because they are expressed in volts and amperes: R 24/3.0 8.0 W.


POWER CALCULATIONS

You can calculate the power P (in watts, symbolized W) in a dc circuit such as that shown in Fig. 12-8 using the following formula:

P EI

where E is the voltage in volts and I is the current in amperes. You may not be given the voltage directly, but you can calculate it if you know the cur- rent and the resistance.

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Remember the Ohm’s law formula for obtaining voltage: E IR. If you
know I and R but don’t know E, you can get the power P by means of this formula:
P (IR) I I 2R

That is, take the current in amperes, multiply this figure by itself, and then multiply the result by the resistance in ohms.
You also can get the power if you aren’t given the current directly. Suppose that you’re given only the voltage and the resistance. Remember the Ohm’s law formula for obtaining current: I E/R. Therefore, you can calculate power using this formula:
P E (E/R) E 2/R

That is, take the voltage, multiply it by itself, and divide by the resistance. Stated all together, these power formulas are
P EI I 2R E 2/R

Now we are all ready to do power calculations. Refer once again to
Fig. 12-8.

PROBLEM 12-5
Suppose that the voltmeter reads 12 V and the ammeter shows 50 mA. What
is the power dissipated by the potentiometer?

SOLUTION 12-5
Use the formula P EI. First, convert the current to amperes, getting I
0.050 A. Then P EI 12 0.050 0.60 W.


How Resistances Combine

When electrical components or devices containing dc resistance are con- nected together, their resistances combine according to specific rules. Sometimes the combined resistance is more than that of any of the compo- nents or devices alone. In other cases the combined resistance is less than that of any of the components or devices all by itself.


RESISTANCES IN SERIES

When you place resistances in series, their ohmic values add up to get the total resistance. This is intuitively simple, and it’s easy to remember.
CHAPTER 12 Direct Current 311


PROBLEM 12-6
Suppose that the following resistances are hooked up in series with each
other: 112 ohms, 470 ohms, and 680 ohms (Fig. 12-9). What is the total resistance of the series combination?


112 470 680




R



Fig. 12-9. An example of three specific resistances in series.


SOLUTION 12-6
Just add the values, getting a total of 112 470 680 1,262 ohms. You
can round this off to 1,260 ohms. It depends on the tolerances of the compo- nents—how much their actual values are allowed to vary, as a result of man- ufacturing processes, from the values specified by the vendor. Tolerance is more of an engineering concern than a physics concern, so we won’t worry about that here.


RESISTANCES IN PARALLEL
When resistances are placed in parallel, they behave differently than they do in series. In general, if you have a resistor of a certain value and you place other resistors in parallel with it, the overall resistance decreases. Mathematically, the rule is straightforward, but it can get a little messy.
One way to evaluate resistances in parallel is to consider them as con- ductances instead. Conductance is measured in units called siemens, some- times symbolized S. (The word siemens serves both in the singular and the plural sense). In older documents, the word mho (ohm spelled backwards)
is used instead. In parallel, conductances add up in the same way as resist- ances add in series. If you change all the ohmic values to siemens, you can add these figures up and convert the final answer back to ohms.
The symbol for conductance is G. Conductance in siemens is the recip- rocal of resistance in ohms. This can be expressed neatly in the following two formulas. It is assumed that neither R nor G is ever equal to zero:

G 1/R R 1/G

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PROBLEM 12-7
Consider five resistors in parallel. Call them R1 through R5, and call the total resistance R, as shown in the diagram of Fig. 12-10. Let R1 100 ohms, R2
200 ohms, R3 300 ohms, R4 400 ohms, and R5 500 ohms, respec- tively. What is the total resistance R of this parallel combination?





R R 1 R 2 R 3 R 4 R 5



Fig. 12-10. Five general resistances in parallel.


SOLUTION 12-7
Converting the resistances to conductance values, you get G1 1/100
0.0100 siemens, G2 1/200 0.00500 siemens, G3 1/300 0.00333
siemens, G4 1/400 0.00250 siemens, and G5 1/500 0.00200
siemens. Adding these gives G 0.0100 0.00500 0.00333 0.00250
0.00200 0.02283 siemens. The total resistance is therefore R 1/G
1/0.02283 43.80 ohms. Because we’re given the input numbers to only three significant figures, we should round this off to 43.8 ohms.

When you have resistances in parallel and their values are all equal, the total resistance is equal to the resistance of any one component divided by the num- ber of components. In a more general sense, the resistances in Fig. 12-10 combine like this:

R 1/(1/R1 1/R2 1/R3 1/R4 1/R5) If you prefer to use exponents, the formula looks like this:
1
R
1
R
1
1
5R (R1 2
3 R4
R 1) 1
These resistance formulas are cumbersome for some people to work with, but mathematically they represent the same thing we just did in Prob- lem 12-7.


CURRENT THROUGH SERIES RESISTANCES

Have you ever used those tiny holiday lights that come in strings? If one bulb burns out, the whole set of bulbs goes dark. Then you have to find out which bulb is bad and replace it to get the lights working again. Each bulb
CHAPTER 12 Direct Current 313


works with something like 10 V, and there are about a dozen bulbs in the
string. You plug in the whole bunch, and the 120-V utility mains drive just the right amount of current through each bulb.
In a series circuit such as a string of light bulbs, the current at any given point is the same as the current at any other point. An ammeter can be connected in series at any point in the circuit, and it will always show the same reading. This is true in any series dc circuit, no matter what the components actually are and regardless of whether or not they all have the same resistance.
If the bulbs in a string are of different resistances, some of them will consume more power than others. In case one of the bulbs burns out and its socket is shorted out instead of filled with a replacement bulb, the current through the whole chain will increase because the overall resistance of the string will go down. This will force each of the remaining bulbs to carry too much current. Another bulb will burn out before long as a result of this excess current. If it, too, is replaced by a short circuit, the current will be increased still further. A third bulb will blow out almost right away. At this point it would be wise to buy some new bulbs!


VOLTAGES ACROSS SERIES RESISTANCES

In a series circuit, the voltage is divided up among the components. The sum total of the potential differences across each resistance is equal to the dc power-supply or battery voltage. This is always true, no matter how large or how small the resistances and whether or not they’re all the same value.
If you think about this for a moment, it’s easy to see why this is true. Look at the schematic diagram of Fig. 12-11. Each resistor carries the same current. Each resistor Rn has a potential difference En across it equal to the product of the current and the resistance of that particular resistor. These En values are in series, like cells in a battery, so they add together. What if the En values across all the resistors added up to something more or less than the supply voltage E? Then there would be a “phantom emf” someplace, adding or taking away voltage. However, there can be no such thing. An emf cannot come out of nowhere.
Look at this another way. The voltmeter V in Fig. 12-11 shows the volt- age E of the battery because the meter is hooked up across the battery. The meter V also shows the sum of the En values across the set of resistors sim- ply because the meter is connected across the set of resistors. The meter

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E n



R n


V








E
Fig. 12-11. Analysis of voltage in a series dc circuit. See text for discussion.


says the same thing whether you think of it as measuring the battery volt-
age E or as measuring the sum of the En values across the series combina- tion of resistors. Therefore, E is equal to the sum of the En values.
This is a fundamental rule in series dc circuits. It also holds for common utility ac circuits almost all the time.
How do you find the voltage across any particular resistor Rn in a cir- cuit like the one in Fig. 12-11? Remember Ohm’s law for finding voltage:
E IR. The voltage is equal to the product of the current and the resist- ance. Remember, too, that you must use volts, ohms, and amperes when making calculations. In order to find the current in the circuit I, you need to know the total resistance and the supply voltage. Then I E/R. First find the current in the whole circuit; then find the voltage across any par- ticular resistor.

PROBLEM 12-8
In Fig. 12-11, suppose that there are 10 resistors. Five of them have values
of 10 ohms, and the other 5 have values of 20 ohms. The power source is 15
V dc. What is the voltage across one of the 10-ohm resistors? Across one of the 20-ohm resistors?

SOLUTION 12-8
First, find the total resistance: R (10 5) (20 5) 50 100 150
ohms. Then find the current: I E/R 15/150 0.10 A. This is the current
through each of the resistors in the circuit. If Rn 10 ohms, then
CHAPTER 12 Direct Current 315


En I (Rn) 0.10 10 1.0 V
If Rn 20 ohms, then
En I (Rn) 0.10 20 2.0 V
You can check to see whether all these voltages add up to the supply volt- age. There are 5 resistors with 1.0 V across each, for a total of 5.0 V; there are also 5 resistors with 2.0 V across each, for a total of 10 V. Thus the sum of the voltages across the 10 resistors is 5.0 V 10 V 15 V.


VOLTAGE ACROSS PARALLEL RESISTANCES

Imagine now a set of ornamental light bulbs connected in parallel. This is the method used for outdoor holiday lighting or for bright indoor lighting. You know that it’s much easier to fix a parallel-wired string of holiday lights if one bulb should burn out than it is to fix a series-wired string. The failure of one bulb does not cause catastrophic system failure. In fact, it might be awhile before you notice that the bulb is dark because all the other ones will stay lit, and their brightness will not change.
In a parallel circuit, the voltage across each component is always the same and is always equal to the supply or battery voltage. The current drawn by each component depends only on the resistance of that particular device. In this sense, the components in a parallel-wired circuit work inde- pendently, as opposed to the series-wired circuit, in which they all interact.
If any branch of a parallel circuit is taken away, the conditions in the other branches remain the same. If new branches are added, assuming that the power supply can handle the load, conditions in previously existing branches are not affected.


CURRENTS THROUGH PARALLEL RESISTANCES

Refer to the schematic diagram of Fig. 12-12. The total parallel resistance
in the circuit is R. The battery voltage is E. The current in branch n, con- taining resistance Rn, is measured by ammeter A and is called In.
The sum of all the In values in the circuit is equal to the total current I
drawn from the source. That is, the current is divided up in the parallel cir-
cuit, similarly to the way that voltage is divided up in a series circuit.

PROBLEM 12-9
Suppose that the battery in Fig. 12-12 delivers 12 V. Further suppose that
there are 12 resistors, each with a value of 120 ohms in the parallel circuit.

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A
In





R n











E
Fig. 12-12. Analysis of current in a parallel dc circuit. See text for discussion.


What is the total current I drawn from the battery?

SOLUTION 12-9
First, find the total resistance. This is easy because all the resistors have the
same value. Divide Rn 120 by 12 to get R 10 ohms. Then the current I
is found by Ohm’s law:

I E/R 12/10 1.2 A


PROBLEM 12-10
In the circuit of Fig. 12-12, what does the ammeter A say, given the same
component values as exist in the scenario of the preceding problem?

SOLUTION 12-10
This involves finding the current in any given branch. The voltage is 12 V
across every branch; Rn 120. Therefore, In, the ammeter reading, is found by Ohm’s law:

In E/Rn 12/120 0.10 A
Let’s check to be sure all the In values add to get the total current I. There are
12 identical branches, each carrying 0.10 A; therefore, the sum is 0.10 12
1.2 A. It checks out.
CHAPTER 12 Direct Current 317


POWER DISTRIBUTION IN SERIES CIRCUITS

n nLet’s switch back now to series circuits. When calculating the power in a circuit containing resistors in series, all you need to do is find out the cur- rent I, in amperes, that the circuit is carrying. Then it’s easy to calculate the power Pn, in watts, dissipated by any particular resistor of value Rn , in ohms, based on the formula P I 2R .
The total power dissipated in a series circuit is equal to the sum of the wattages dissipated in each resistor. In this way, the distribution of power
in a series circuit is like the distribution of the voltage.

PROBLEM 12-11
Suppose that we have a series circuit with a supply of 150 V and three resis-
tors: R1 330 ohms, R2 680 ohms, and R3 910 ohms. What is the power dissipated by R2?

SOLUTION 12-11
Find the current in the circuit. To do this, calculate the total resistance first.
Because the resistors are in series, the total is resistance is R 330 680
910 1920 ohms. Therefore, the current is I 150/1920 0.07813 A
78.1 mA. The power dissipated by R2 is
2 2P I 2R 0.07813 0.07813 680 4.151 W

We must round this off to three significant figures, getting 4.15 W.


POWER DISTRIBUTION IN PARALLEL CIRCUITS

n n nWhen resistances are wired in parallel, they each consume power accord- ing to the same formula, P I2R. However, the current is not the same in each resistance. An easier method to find the power Pn dissipated by resis- tor of value R is by using the formula P E2/R , where E is the voltage
of the supply. This voltage is the same across every resistor.
In a parallel circuit, the total power consumed is equal to the sum of the wattages dissipated by the individual resistances. This is, in fact, true for any dc circuit containing resistances. Power cannot come out of nowhere, nor can it vanish.

PROBLEM 12-12
A circuit contains three resistances R1 22 ohms, R2 47 ohms, and R3
68 ohms, all in parallel across a voltage E 3.0 V. Find the power dissipated
by each resistor.

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SOLUTION 12-12
First find E 2, the square of the supply voltage: E 2 3.0 3.0 9.0. Then
P1 9.0/22 0.4091 W, P2 9.0/47 0.1915 W, and P3 9.0/68 0.1324
W. These should be rounded off to P1 0.41 W, P2 0.19 W, and P3 0.13 W,
respectively.




Kirchhoff’s Laws

The physicist Gustav Robert Kirchhoff (1824–1887) was a researcher and experimentalist in electricity, back in the time before radio, before electric lighting, and before much was understood about how electric cur- rents flow.


KIRCHHOFF’S CURRENT LAW

Kirchhoff reasoned that current must work something like water in a net- work of pipes and that the current going into any point has to be the same as the current going out. This is true for any point in a circuit, no matter how many branches lead into or out of the point (Fig. 12-13).
In a network of water pipes that does not leak and into which no water
is added along the way, the total number of cubic meters going in has to be the same as the total volume going out. Water cannot form from nothing, nor can it disappear, inside a closed system of pipes. Charge carriers, thought Kirchhoff, must act the same way in an electric circuit.

PROBLEM 12-13
In Fig. 12-13, suppose that each of the two resistors below point Z has a value
of 100 ohms and that all three resistors above Z have values of 10.0 ohms. The current through each 100-ohm resistor is 500 mA (0.500 A). What is the current through any of the 10.0-ohm resistors, assuming that the current is equally distributed? What is the voltage, then, across any of the 10.0-ohm resistors?

SOLUTION 12-13
The total current into Z is 500 mA 500 mA 1.00 A. This must be divided
three ways equally among the 10-ohm resistors. Therefore, the current through any one of them is 1.00/3 A 0.333 A 333 mA. The voltage across any one of the 10.0-ohm resistors is found by Ohm’s law: E IR 0.333
10.0 3.33 V.
CHAPTER 12 Direct Current 319


I
4


3I I
5






Z






I I
1 2
Fig. 12-13. Kirchhoff’s current law. The current enter- ing point Z is equal to the current leaving point Z. In this case, I1 I2 I3 I4 I5.


KIRCHHOFF’S VOLTAGE LAW

The sum of all the voltages, as you go around a circuit from some fixed point and return there from the opposite direction, and taking polarity into account, is always zero. At first thought, some people find this strange. Certainly there is voltage in your electric hair dryer, radio, or computer! Yes, there is—between different points in the circuit. However, no single point can have an electrical potential with respect to itself. This is so sim- ple that it’s trivial. A point in a circuit is always shorted out to itself.
What Kirchhoff was saying when he wrote his voltage law is that volt- age cannot appear out of nowhere, nor can it vanish. All the potential dif- ferences must balance out in any circuit, no matter how complicated and no matter how many branches there are.
Consider the rule you’ve already learned about series circuits: The volt- ages across all the resistors add up to the supply voltage. However, the polarities of the emfs across the resistors are opposite to that of the battery. This is shown in Fig. 12-14. It is a subtle thing, but it becomes clear when

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a series circuit is drawn with all the components, including the battery or
other emf source, in line with each other, as in Fig. 12-14.


E E
+ 2 _
+ 3 _







_ + _ +
E E
1 E 4
Fig. 12-14. Kirchhoff’s voltage law. The sum of the voltages
E E1 E2 E3 E4 0, taking polarity into account.


PROBLEM 12-14
Refer to the diagram of Fig. 12-14. Suppose that the four resistors have val-
ues of 50, 60, 70, and 80 ohms and that the current through them is 500 mA
(0.500 A). What is the supply voltage E?

SOLUTION 12-14
Find the voltages E1, E2, E3, and E4 across each of the resistors. This is done using Ohm’s law. In the case of E1, say, with the 50-ohm resistor, calculate E1 0.500 50 25 V. In the same way, you can calculate E2 30 V, E3 35 V, and E4 40 V. The supply voltage is the sum E1 E2 E3 E4
25 30 35 40 V 130 V.





Quiz


Refer to the text in this chapter if necessary. A good score is eight correct. Answers are in the back of the book.
1. Suppose that 5.00 1017 electrical charge carriers flow past a point in 1.00 s. What is the electrical voltage?
(a) 0.080 V
(b) 12.5 V
CHAPTER 12 Direct Current 321


(c) 5.00 V
(d) It cannot be calculated from this information.
2. An ampere also can be regarded as
(a) an ohm per volt.
(b) an ohm per watt.
(c) a volt per ohm.
(d) a volt-ohm.
3. Suppose that there are two resistances in a series circuit. One of the resistors has a value of 33 kW (that is, 33,000 or 3.3 104 ohms). The value of the other resistor is not known. The power dissipated by the 33-kW resistor is 3.3 W. What is the current through the unknown resistor?
(a) 0.11 A
(b) 10 mA
(c) 0.33 mA
(d) It cannot be calculated from this information.
4. If the voltage across a resistor is E (in volts) and the current through that resis- tor is I (in milliamperes), then the power P (in watts) is given by the following formula:
(a) P EI.
(b) P EI 103.
(c) P EI 10 3.
(d) P E/I.
5. Suppose that you have a set of five 0.5-W flashlight bulbs connected in paral- lel across a dc source of 3.0 V. If one of the bulbs is removed or blows out, what will happen to the current through the other four bulbs?
(a) It will remain the same.
(b) It will increase.
(c) It will decrease.
(d) It will drop to zero.
6. A good dielectric is characterized by
(a) excellent conductivity.
(b) fair conductivity.
(c) poor conductivity.
(d) variable conductivity.
7. Suppose that there are two resistances in a parallel circuit. One of the resistors has a value of 100 ohms. The value of the other resistor is not known. The power dissipated by the 100-ohm resistor is 500 mW (that is, 0.500 W). What
is the current through the unknown resistor?
(a) 71 mA
(b) 25 A
(c) 200 A
(d) It cannot be calculated from this information.

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8. Conventional current flows
(a) from the positive pole to the negative pole.
(b) from the negative pole to the positive pole.
(c) in either direction; it doesn’t matter.
(d) nowhere; current does not flow.
9. Suppose that a circuit contains 620 ohms of resistance and that the current in the circuit is 50.0 mA. What is the voltage across this resistance?
(a) 12.4 kV
(b) 31.0 V
(c) 8.06 10 5 V
(d) It cannot be calculated from this information.
10. Which of the following cannot be an electric charge carrier?
(a) A neutron
(b) An electron
(c) A hole
(d) An ion

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