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# Chapter 21. Electricity and circuits

Where the telescope ends, the microscope begins. Which of the two has the grander view? -- Victor Hugo

His father died during his mother's pregnancy. Rejected by her as a boy, he was packed off to boarding school when she remarried. He himself never married, but in middle age he formed an intense relationship with a much younger man, a relationship that he terminated when he underwent a psychotic break. Following his early scientific successes, he spent the rest of his professional life mostly in frustration over his inability to unlock the secrets of alchemy.

The man being described is Isaac Newton, but not the triumphant Newton of the standard textbook hagiography. Why dwell on the sad side of his life? To the modern science educator, Newton's lifelong obsession with alchemy may seem an embarrassment, a distraction from his main achievement, the creation the modern science of mechanics. To Newton, however, his alchemical researches were naturally related to his investigations of force and motion. What was radical about Newton's analysis of motion was its universality: it succeeded in describing both the heavens and the earth with the same equations, whereas previously it had been assumed that the sun, moon, stars, and planets were fundamentally different from earthly objects. But Newton realized that if science was to describe all of nature in a unified way, it was not enough to unite the human scale with the scale of the universe: he would not be satisfied until he fit the microscopic universe into the picture as well.

It should not surprise us that Newton failed. Although he was a firm believer in the existence of atoms, there was no more experimental evidence for their existence than there had been when the ancient Greeks first posited them on purely philosophical grounds. Alchemy labored under a tradition of secrecy and mysticism. Newton had already almost single-handedly transformed the fuzzyheaded field of “natural philosophy” into something we would recognize as the modern science of physics, and it would be unjust to criticize him for failing to change alchemy into modern chemistry as well. The time was not ripe. The microscope was a new invention, and it was cutting-edge science when Newton's contemporary Hooke discovered that living things were made out of cells.

## 21.1 The quest for the atomic force

a / Four pieces of tape are prepared, 1, as described in the text. Depending on which combination is tested, the interaction can be either repulsive, 2, or attractive, 3.

Newton was not the first of the age of reason. He was the last of the magicians. -- John Maynard Keynes

Nevertheless it will be instructive to pick up Newton's train of thought and see where it leads us with the benefit of modern hindsight. In uniting the human and cosmic scales of existence, he had reimagined both as stages on which the actors were objects (trees and houses, planets and stars) that interacted through attractions and repulsions. He was already convinced that the objects inhabiting the microworld were atoms, so it remained only to determine what kinds of forces they exerted on each other.

His next insight was no less brilliant for his inability to bring it to fruition. He realized that the many human-scale forces --- friction, sticky forces, the normal forces that keep objects from occupying the same space, and so on --- must all simply be expressions of a more fundamental force acting between atoms. Tape sticks to paper because the atoms in the tape attract the atoms in the paper. My house doesn't fall to the center of the earth because its atoms repel the atoms of the dirt under it.

Here he got stuck. It was tempting to think that the atomic force was a form of gravity, which he knew to be universal, fundamental, and mathematically simple. Gravity, however, is always attractive, so how could he use it to explain the existence of both attractive and repulsive atomic forces? The gravitational force between objects of ordinary size is also extremely small, which is why we never notice cars and houses attracting us gravitationally. It would be hard to understand how gravity could be responsible for anything as vigorous as the beating of a heart or the explosion of gunpowder. Newton went on to write a million words of alchemical notes filled with speculation about some other force, perhaps a “divine force” or “vegetative force” that would for example be carried by the sperm to the egg.

Luckily, we now know enough to investigate a different suspect as a candidate for the atomic force: electricity. Electric forces are often observed between objects that have been prepared by rubbing (or other surface interactions), for instance when clothes rub against each other in the dryer. A useful example is shown in figure a/1: stick two pieces of tape on a tabletop, and then put two more pieces on top of them. Lift each pair from the table, and then separate them. The two top pieces will then repel each other, a/2, as will the two bottom pieces. A bottom piece will attract a top piece, however, a/3. Electrical forces like these are similar in certain ways to gravity, the other force that we already know to be fundamental:

• Electrical forces are universal. Although some substances, such as fur, rubber, and plastic, respond more strongly to electrical preparation than others, all matter participates in electrical forces to some degree. There is no such thing as a “nonelectric” substance. Matter is both inherently gravitational and inherently electrical.
• Experiments show that the electrical force, like the gravitational force, is an inverse square force. That is, the electrical force between two spheres is proportional to $$1/r^2$$, where $$r$$ is the center-to-center distance between them.

Furthermore, electrical forces make more sense than gravity as candidates for the fundamental force between atoms, because we have observed that they can be either attractive or repulsive.

## 21.2 Electrical forces

b / A charged piece of tape attracts uncharged pieces of paper from a distance, and they leap up to it.

c / The paper has zero total charge, but it does have charged particles in it that can move.

### Charge

“Charge” is the technical term used to indicate that an object has been prepared so as to participate in electrical forces. This is to be distinguished from the common usage, in which the term is used indiscriminately for anything electrical. For example, although we speak colloquially of “charging” a battery, you may easily verify that a battery has no charge in the technical sense, e.g., it does not exert any electrical force on a piece of tape that has been prepared as described in the previous section.

#### Two types of charge

We can easily collect reams of data on electrical forces between different substances that have been charged in different ways. We find for example that cat fur prepared by rubbing against rabbit fur will attract glass that has been rubbed on silk. How can we make any sense of all this information? A vast simplification is achieved by noting that there are really only two types of charge. Suppose we pick cat fur rubbed on rabbit fur as a representative of type A, and glass rubbed on silk for type B. We will now find that there is no “type C.” Any object electrified by any method is either A-like, attracting things A attracts and repelling those it repels, or B-like, displaying the same attractions and repulsions as B. The two types, A and B, always display opposite interactions. If A displays an attraction with some charged object, then B is guaranteed to undergo repulsion with it, and vice-versa.

#### The coulomb

Although there are only two types of charge, each type can come in different amounts. The metric unit of charge is the coulomb (rhymes with “drool on”), defined as follows:

One Coulomb (C) is the amount of charge such that a force of $$9.0\times10^9$$ N occurs between two pointlike objects with charges of 1 C separated by a distance of 1 m.

The notation for an amount of charge is $$q$$. The numerical factor in the definition is historical in origin, and is not worth memorizing. The definition is stated for pointlike, i.e., very small, objects, because otherwise different parts of them would be at different distances from each other.

#### A model of two types of charged particles

Experiments show that all the methods of rubbing or otherwise charging objects involve two objects, and both of them end up getting charged. If one object acquires a certain amount of one type of charge, then the other ends up with an equal amount of the other type. Various interpretations of this are possible, but the simplest is that the basic building blocks of matter come in two flavors, one with each type of charge. Rubbing objects together results in the transfer of some of these particles from one object to the other. In this model, an object that has not been electrically prepared may actually possesses a great deal of both types of charge, but the amounts are equal and they are distributed in the same way throughout it. Since type A repels anything that type B attracts, and vice versa, the object will make a total force of zero on any other object. The rest of this chapter fleshes out this model and discusses how these mysterious particles can be understood as being internal parts of atoms.

#### Use of positive and negative signs for charge

Because the two types of charge tend to cancel out each other's forces, it makes sense to label them using positive and negative signs, and to discuss the total charge of an object. It is entirely arbitrary which type of charge to call negative and which to call positive. Benjamin Franklin decided to describe the one we've been calling “A” as negative, but it really doesn't matter as long as everyone is consistent with everyone else. An object with a total charge of zero (equal amounts of both types) is referred to as electrically $$neutral$$.

self-check:

Criticize the following statement: “There are two types of charge, attractive and repulsive.”

#### Coulomb's law

A large body of experimental observations can be summarized as follows:

Coulomb's law: The magnitude of the force acting between pointlike charged objects at a center-to-center distance $$r$$ is given by the equation

$\begin{equation*} |\mathbf{F}| = k\frac{|q_1||q_2|}{r^2} , \end{equation*}$

where the constant $$k$$ equals $$9.0\times10^9\ \text{N}\!\cdot\!\text{m}^2/\text{C}^2$$. The force is attractive if the charges are of different signs, and repulsive if they have the same sign.

Clever modern techniques have allowed the $$1/r^2$$ form of Coulomb's law to be tested to incredible accuracy, showing that the exponent is in the range from 1.9999999999999998 to 2.0000000000000002.

Note that Coulomb's law is closely analogous to Newton's law of gravity, where the magnitude of the force is $$Gm_1m_2/r^2$$, except that there is only one type of mass, not two, and gravitational forces are never repulsive. Because of this close analogy between the two types of forces, we can recycle a great deal of our knowledge of gravitational forces. For instance, there is an electrical equivalent of the shell theorem: the electrical forces exerted externally by a uniformly charged spherical shell are the same as if all the charge was concentrated at its center, and the forces exerted internally are zero.

### Conservation of charge

An even more fundamental reason for using positive and negative signs for electrical charge is that experiments show that charge is conserved according to this definition: in any closed system, the total amount of charge is a constant. This is why we observe that rubbing initially uncharged substances together always has the result that one gains a certain amount of one type of charge, while the other acquires an equal amount of the other type. Conservation of charge seems natural in our model in which matter is made of positive and negative particles. If the charge on each particle is a fixed property of that type of particle, and if the particles themselves can be neither created nor destroyed, then conservation of charge is inevitable.

### Electrical forces involving neutral objects

As shown in figure b, an electrically charged object can attract objects that are uncharged. How is this possible? The key is that even though each piece of paper has a total charge of zero, it has at least some charged particles in it that have some freedom to move. Suppose that the tape is positively charged, c. Mobile particles in the paper will respond to the tape's forces, causing one end of the paper to become negatively charged and the other to become positive. The attraction between the paper and the tape is now stronger than the repulsion, because the negatively charged end is closer to the tape.

self-check:

What would have happened if the tape was negatively charged?

##### Discussion Questions

If the electrical attraction between two pointlike objects at a distance of 1 m is $$9\times10^9$$ N, why can't we infer that their charges are $$+1$$ and $$-1$$ C? What further observations would we need to do in order to prove this?

An electrically charged piece of tape will be attracted to your hand. Does that allow us to tell whether the mobile charged particles in your hand are positive or negative, or both?

## 21.3 Current

d / Michael Faraday (1791-1867) was the son of a poor blacksmith.

e / Gymnotus carapo, a knifefish, uses electrical signals to sense its environment and to communicate with others of its species.

f / André Marie Ampère (1775-1836).

### Unity of all types of electricity

We are surrounded by things we have been told are “electrical,” but it's far from obvious what they have in common to justify being grouped together. What relationship is there between the way socks cling together and the way a battery lights a lightbulb? We have been told that both an electric eel and our own brains are somehow electrical in nature, but what do they have in common?

British physicist Michael Faraday (1791-1867) set out to address this problem. He investigated electricity from a variety of sources --- including electric eels! --- to see whether they could all produce the same effects, such as shocks and sparks, attraction and repulsion. “Heating” refers, for example, to the way a lightbulb filament gets hot enough to glow and emit light. Magnetic induction is an effect discovered by Faraday himself that connects electricity and magnetism. We will not study this effect, which is the basis for the electric generator, in detail until later in the book.

 shocks attraction and repulsion heating rubbing × × × × battery × × × × animal × × (×) × magnetically induced × × × ×

The table shows a summary of some of Faraday's results. Check marks indicate that Faraday or his close contemporaries were able to verify that a particular source of electricity was capable of producing a certain effect. (They evidently failed to demonstrate attraction and repulsion between objects charged by electric eels, although modern workers have studied these species in detail and been able to understand all their electrical characteristics on the same footing as other forms of electricity.)

Faraday's results indicate that there is nothing fundamentally different about the types of electricity supplied by the various sources. They are all able to produce a wide variety of identical effects. Wrote Faraday, “The general conclusion which must be drawn from this collection of facts is that electricity, whatever may be its source, is identical in its nature.”

If the types of electricity are the same thing, what thing is that? The answer is provided by the fact that all the sources of electricity can cause objects to repel or attract each other. We use the word “charge” to describe the property of an object that allows it to participate in such electrical forces, and we have learned that charge is present in matter in the form of nuclei and electrons. Evidently all these electrical phenomena boil down to the motion of charged particles in matter.

### Electric current

If the fundamental phenomenon is the motion of charged particles, then how can we define a useful numerical measurement of it? We might describe the flow of a river simply by the velocity of the water, but velocity will not be appropriate for electrical purposes because we need to take into account how much charge the moving particles have, and in any case there are no practical devices sold at Radio Shack that can tell us the velocity of charged particles. Experiments show that the intensity of various electrical effects is related to a different quantity: the number of coulombs of charge that pass by a certain point per second. By analogy with the flow of water, this quantity is called the electric current, $$I$$. Its units of coulombs/second are more conveniently abbreviated as amperes,
[3] 1 A=1 C/s. (In informal speech, one usually says “amps.”)

The main subtlety involved in this definition is how to account for the two types of charge. The stream of water coming from a hose is made of atoms containing charged particles, but it produces none of the effects we associate with electric currents. For example, you do not get an electrical shock when you are sprayed by a hose. This type of experiment shows that the effect created by the motion of one type of charged particle can be canceled out by the motion of the opposite type of charge in the same direction. In water, every oxygen atom with a charge of $$+8e$$ is surrounded by eight electrons with charges of $$-e$$, and likewise for the hydrogen atoms.

We therefore refine our definition of current as follows:

##### definition of electric current

When charged particles are exchanged between regions of space A and B, the electric current flowing from A to B is

$\begin{equation*} I = \frac{\Delta q}{\Delta t} , \end{equation*}$

where $$\Delta q$$ is the change in region B's total charge occurring over a period of time $$\Delta t$$.

In the garden hose example, your body picks up equal amounts of positive and negative charge, resulting in no change in your total charge, so the electrical current flowing into you is zero.

##### Example 1: Interpretation of $$\Delta q/\Delta t$$
$$\triangleright$$ How should the expression $$\Delta q/\Delta t$$ be interpreted when the current isn't constant?

$$\triangleright$$ You've seen lots of equations of this form before: $$v=\Delta x/\Delta t$$, $$F=\Delta p/\Delta t$$, etc. These are all descriptions of rates of change, and they all require that the rate of change be constant. If the rate of change isn't constant, you instead have to use the slope of the tangent line on a graph. The slope of a tangent line is equivalent to a derivative in calculus; applications of calculus are discussed in section 21.7.

##### Example 2: Ions moving across a cell membrane
$$\triangleright$$ Figure g shows ions, labeled with their charges, moving in or out through the membranes of four cells. If the ions all cross the membranes during the same interval of time, how would the currents into the cells compare with each other?

$$\triangleright$$ Cell A has positive current going into it because its charge is increased, i.e., has a positive value of $$\Delta q$$.

Cell B has the same current as cell A, because by losing one unit of negative charge it also ends up increasing its own total charge by one unit.

Cell C's total charge is reduced by three units, so it has a large negative current going into it.

Cell D loses one unit of charge, so it has a small negative current into it.

g / Example 2

It may seem strange to say that a negatively charged particle going one way creates a current going the other way, but this is quite ordinary. As we will see, currents flow through metal wires via the motion of electrons, which are negatively charged, so the direction of motion of the electrons in a circuit is always opposite to the direction of the current. Of course it would have been convenient of Benjamin Franklin had defined the positive and negative signs of charge the opposite way, since so many electrical devices are based on metal wires.

##### Example 3: Number of electrons flowing through a lightbulb

$$\triangleright$$ If a lightbulb has 1.0 A flowing through it, how many electrons will pass through the filament in 1.0 s?

$$\triangleright$$ We are only calculating the number of electrons that flow, so we can ignore the positive and negative signs. Solving for $$\Delta q= I \Delta t$$ gives a charge of 1.0 C flowing in this time interval. The number of electrons is

\begin{align*} \text{number of electrons} &= \text{coulombs}\times\frac{\text{electrons}}{\text{coulomb}} \\ &= \text{coulombs}/\frac{\text{coulombs}}{\text{electron}} \\ &= 1.0\ \text{C} / e \\ &= 6.2\times10^{18} \end{align*}

## 21.4 Circuits

h / 1. Static electricity runs out quickly. 2. A practical circuit. 3. An open circuit. 4. How an ammeter works. 5. Measuring the current with an ammeter.

How can we put electric currents to work? The only method of controlling electric charge we have studied so far is to charge different substances, e.g., rubber and fur, by rubbing them against each other. Figure h/1 shows an attempt to use this technique to light a lightbulb. This method is unsatisfactory. True, current will flow through the bulb, since electrons can move through metal wires, and the excess electrons on the rubber rod will therefore come through the wires and bulb due to the attraction of the positively charged fur and the repulsion of the other electrons. The problem is that after a zillionth of a second of current, the rod and fur will both have run out of charge. No more current will flow, and the lightbulb will go out.

Figure h/2 shows a setup that works. The battery pushes charge through the circuit, and recycles it over and over again. (We will have more to say later in this chapter about how batteries work.) This is called a complete circuit. Today, the electrical use of the word “circuit” is the only one that springs to mind for most people, but the original meaning was to travel around and make a round trip, as when a circuit court judge would ride around the boondocks, dispensing justice in each town on a certain date.

Note that an example like h/3 does not work. The wire will quickly begin acquiring a net charge, because it has no way to get rid of the charge flowing into it. The repulsion of this charge will make it more and more difficult to send any more charge in, and soon the electrical forces exerted by the battery will be canceled out completely. The whole process would be over so quickly that the filament would not even have enough time to get hot and glow. This is known as an open circuit. Exactly the same thing would happen if the complete circuit of figure h/2 was cut somewhere with a pair of scissors, and in fact that is essentially how an ordinary light switch works: by opening up a gap in the circuit.

The definition of electric current we have developed has the great virtue that it is easy to measure. In practical electrical work, one almost always measures current, not charge. The instrument used to measure current is called an ammeter. A simplified ammeter, h/4, simply consists of a coiled-wire magnet whose force twists an iron needle against the resistance of a spring. The greater the current, the greater the force. Although the construction of ammeters may differ, their use is always the same. We break into the path of the electric current and interpose the meter like a tollbooth on a road, h/5. There is still a complete circuit, and as far as the battery and bulb are concerned, the ammeter is just another segment of wire.

Does it matter where in the circuit we place the ammeter? Could we, for instance, have put it in the left side of the circuit instead of the right? Conservation of charge tells us that this can make no difference. Charge is not destroyed or “used up” by the lightbulb, so we will get the same current reading on either side of it. What is “used up” is energy stored in the battery, which is being converted into heat and light energy.

## 21.5 Voltage

i / Alessandro Volta (1745-1827).

j / Example 4.

### The volt unit

Electrical circuits can be used for sending signals, storing information, or doing calculations, but their most common purpose by far is to manipulate energy, as in the battery-and-bulb example of the previous section. We know that lightbulbs are rated in units of watts, i.e., how many joules per second of energy they can convert into heat and light, but how would this relate to the flow of charge as measured in amperes? By way of analogy, suppose your friend, who didn't take physics, can't find any job better than pitching bales of hay. The number of calories he burns per hour will certainly depend on how many bales he pitches per minute, but it will also be proportional to how much mechanical work he has to do on each bale. If his job is to toss them up into a hayloft, he will get tired a lot more quickly than someone who merely tips bales off a loading dock into trucks. In metric units,

$\begin{equation*} \frac{\text{joules}}{\text{second}} = \frac{\text{haybales}}{\text{second}} \times \frac{\text{joules}}{\text{haybale}} . \end{equation*}$

Similarly, the rate of energy transformation by a battery will not just depend on how many coulombs per second it pushes through a circuit but also on how much mechanical work it has to do on each coulomb of charge:

$\begin{equation*} \frac{\text{joules}}{\text{second}} = \frac{\text{coulombs}}{\text{second}} \times \frac{\text{joules}}{\text{coulomb}} \end{equation*}$

or

$\begin{equation*} \text{power} = \text{current} \times \text{work per unit charge} . \end{equation*}$

Units of joules per coulomb are abbreviated as volts, 1 V=1 J/C, named after the Italian physicist Alessandro Volta. Everyone knows that batteries are rated in units of volts, but the voltage concept is more general than that; it turns out that voltage is a property of every point in space. To gain more insight, let's think more carefully about what goes on in the battery and bulb circuit.

### The voltage concept in general

To do work on a charged particle, the battery apparently must be exerting forces on it. How does it do this? Well, the only thing that can exert an electrical force on a charged particle is another charged particle. It's as though the haybales were pushing and pulling each other into the hayloft! This is potentially a horribly complicated situation. Even if we knew how much excess positive or negative charge there was at every point in the circuit (which realistically we don't) we would have to calculate zillions of forces using Coulomb's law, perform all the vector additions, and finally calculate how much work was being done on the charges as they moved along. To make things even more scary, there is more than one type of charged particle that moves: electrons are what move in the wires and the bulb's filament, but ions are the moving charge carriers inside the battery. Luckily, there are two ways in which we can simplify things:

The situation is unchanging. Unlike the imaginary setup in which we attempted to light a bulb using a rubber rod and a piece of fur, this circuit maintains itself in a steady state (after perhaps a microsecond-long period of settling down after the circuit is first assembled). The current is steady, and as charge flows out of any area of the circuit it is replaced by the same amount of charge flowing in. The amount of excess positive or negative charge in any part of the circuit therefore stays constant. Similarly, when we watch a river flowing, the water goes by but the river doesn't disappear.

Force depends only on position. Since the charge distribution is not changing, the total electrical force on a charged particle depends only on its own charge and on its location. If another charged particle of the same type visits the same location later on, it will feel exactly the same force.

The second observation tells us that there is nothing all that different about the experience of one charged particle as compared to another's. If we single out one particle to pay attention to, and figure out the amount of work done on it by electrical forces as it goes from point A to point $$B$$ along a certain path, then this is the same amount of work that will be done on any other charged particles of the same type as it follows the same path. For the sake of visualization, let's think about the path that starts at one terminal of the battery, goes through the light bulb's filament, and ends at the other terminal. When an object experiences a force that depends only on its position (and when certain other, technical conditions are satisfied), we can define an electrical energy associated with the position of that object. The amount of work done on the particle by electrical forces as it moves from A to B equals the drop in electrical energy between A and B. This electrical energy is what is being converted into other forms of energy such as heat and light. We therefore define voltage in general as electrical energy per unit charge:

##### definition of voltage difference

The difference in voltage between two points in space is defined as

$\begin{equation*} \Delta V=\Delta PE_{elec}/q , \end{equation*}$

where $$\Delta PE_{elec}$$ is the change in the electrical energy of a particle with charge $$q$$ as it moves from the initial point to the final point.

The amount of power dissipated (i.e., rate at which energy is transformed by the flow of electricity) is then given by the equation

$\begin{equation*} P = I \Delta V . \end{equation*}$

##### Example 4: Energy stored in a battery
$$\triangleright$$ The 1.2 V rechargeable battery in figure j is labeled 1800 milliamp-hours. What is the maximum amount of energy the battery can store?

$$\triangleright$$ An ampere-hour is a unit of current multiplied by a unit of time. Current is charge per unit time, so an ampere-hour is in fact a funny unit of charge:

\begin{align*} \text{(1 A)(1 hour)} &= \text{(1 C/s)(3600 s)} \\ &= \text{3600 C} \end{align*}

1800 milliamp-hours is therefore $$1800\times10^{-3}\times 3600\ \text{C}=6.5\times10^3\ \text{C}$$. That's a huge number of charged particles, but the total loss of electrical energy will just be their total charge multiplied by the voltage difference across which they move:

\begin{align*} \Delta PE_{elec} &= q \Delta V \\ &= (6.5\times10^3\ \text{C})(1.2\ \text{V}) \\ &= 7.8\ \text{kJ} \end{align*}

##### Example 5: Units of volt-amps

$$\triangleright$$ Doorbells are often rated in volt-amps. What does this combination of units mean?

$$\triangleright$$ Current times voltage gives units of power, $$P= I\Delta V$$, so volt-amps are really just a nonstandard way of writing watts. They are telling you how much power the doorbell requires.

##### Example 6: Power dissipated by a battery and bulb

$$\triangleright$$ If a 9.0-volt battery causes 1.0 A to flow through a lightbulb, how much power is dissipated?

$$\triangleright$$ The voltage rating of a battery tells us what voltage difference $$\Delta V$$ it is designed to maintain between its terminals.

\begin{align*} P &= I \ \Delta \text{V} \\ &= 9.0 \ \text{A}\cdot\text{V} \\ &= 9.0 \ \frac{\text{C}}{\text{s}}\cdot\frac{\text{J}}{\text{C}} \\ &= 9.0 \ \text{J/s} \\ &= 9.0 \ \text{W} \end{align*}

The only nontrivial thing in this problem was dealing with the units. One quickly gets used to translating common combinations like $$\text{A}\cdot\text{V}$$ into simpler terms.

{}Question: OK, so what is voltage, really?
{}Answer: A device like a battery has positive and negative charges inside it that push other charges around the outside circuit. A higher-voltage battery has denser charges in it, which will do more work on each charged particle that moves through the outside circuit.

To use a gravitational analogy, we can put a paddlewheel at the bottom of either a tall waterfall or a short one, but a kg of water that falls through the greater gravitational energy difference will have more energy to give up to the paddlewheel at the bottom.

{}Question: Why do we define voltage as electrical energy divided by charge, instead of just defining it as electrical energy?
{}Answer: One answer is that it's the only definition that makes the equation $$P=I \Delta V$$ work. A more general answer is that we want to be able to define a voltage difference between any two points in space without having to know in advance how much charge the particles moving between them will have. If you put a nine-volt battery on your tongue, then the charged particles that move across your tongue and give you that tingly sensation are not electrons but ions, which may have charges of $$+e$$, $$-2e$$, or practically anything. The manufacturer probably expected the battery to be used mostly in circuits with metal wires, where the charged particles that flowed would be electrons with charges of $$-e$$. If the ones flowing across your tongue happen to have charges of $$-2e$$, the electrical energy difference for them will be twice as much, but dividing by their charge of $$-2e$$ in the definition of voltage will still give a result of 9 $$V$$.

{}Question: Are there two separate roles for the charged particles in the circuit, a type that sits still and exerts the forces, and another that moves under the influence of those forces?
{}Answer: No. Every charged particle simultaneously plays both roles. Newton's third law says that any particle that has an electrical force acting on it must also be exerting an electrical force back on the other particle. There are no “designated movers” or “designated force-makers.”

{}Question: Why does the definition of voltage only refer to voltage differences?
{}Answer: It's perfectly OK to define voltage as $$V=PE_{elec}/q$$. But recall that it is only differences in interaction energy, $$U$$, that have direct physical meaning in physics. Similarly, voltage differences are really more useful than absolute voltages. A voltmeter measures voltage differences, not absolute voltages.

##### Discussion Questions

A roller coaster is sort of like an electric circuit, but it uses gravitational forces on the cars instead of electric ones. What would a high-voltage roller coaster be like? What would a high-current roller coaster be like?

Criticize the following statements:

• “He touched the wire, and 10000 volts went through him.”
• “That battery has a charge of 9 volts.”
• “You used up the charge of the battery.”

When you touch a 9-volt battery to your tongue, both positive and negative ions move through your saliva. Which ions go which way?

I once touched a piece of physics apparatus that had been wired incorrectly, and got a several-thousand-volt voltage difference across my hand. I was not injured. For what possible reason would the shock have had insufficient power to hurt me?

## 21.6 Resistance

k / Georg Simon Ohm (1787-1854).

l / Four objects made of the same substance have different resistances.

m / A medical MRI scanner, which uses superconductors.

o / The voltmeter doesn't care which of these setups you use.

p / Example 8. In 1 and 2, charges that are visible on the front surface of the conductor are shown as solid dots; the others would have to be seen through the conductor, which we imagine is semi-transparent.

q / Short-circuiting a battery. Warning: you can burn yourself this way or start a fire! If you want to try this, try making the connection only very briefly, use a low-voltage battery, and avoid touching the battery or the wire, both of which will get hot.

Resistors.

r / The symbol used in schematics to represent a resistor.

s / An example of a resistor with a color code.

 color meaning black 0 brown 1 red 2 orange 3 yellow 4 green 5 blue 6 violet 7 gray 8 white 9 silver ±10% gold ±5%

Color codes used on resistors.

u / 1. A simplified diagram of how a voltmeter works. 2. Measuring the voltage difference across a lightbulb. 3. The same setup drawn in schematic form. 4. The setup for measuring current is different.

### Resistance

So far we have simply presented it as an observed fact that a battery-and-bulb circuit quickly settles down to a steady flow, but why should it? Newton's second law, $$a=F/m$$, would seem to predict that the steady forces on the charged particles should make them whip around the circuit faster and faster. The answer is that as charged particles move through matter, there are always forces, analogous to frictional forces, that resist the motion. These forces need to be included in Newton's second law, which is really $$a=F_{total}/m$$, not $$a=F/m$$. If, by analogy, you push a crate across the floor at constant speed, i.e., with zero acceleration, the total force on it must be zero. After you get the crate going, the floor's frictional force is exactly canceling out your force. The chemical energy stored in your body is being transformed into heat in the crate and the floor, and no longer into an increase in the crate's kinetic energy. Similarly, the battery's internal chemical energy is converted into heat, not into perpetually increasing the charged particles' kinetic energy. Changing energy into heat may be a nuisance in some circuits, such as a computer chip, but it is vital in a lightbulb, which must get hot enough to glow. Whether we like it or not, this kind of heating effect is going to occur any time charged particles move through matter.

What determines the amount of heating? One flashlight bulb designed to work with a 9-volt battery might be labeled 1.0 watts, another 5.0. How does this work? Even without knowing the details of this type of friction at the atomic level, we can relate the heat dissipation to the amount of current that flows via the equation $$P=I\Delta$$V. If the two flashlight bulbs can have two different values of $$P$$ when used with a battery that maintains the same $$\Delta V$$, it must be that the 5.0-watt bulb allows five times more current to flow through it.

For many substances, including the tungsten from which lightbulb filaments are made, experiments show that the amount of current that will flow through it is directly proportional to the voltage difference placed across it. For an object made of such a substance, we define its electrical resistance as follows:

##### definition of resistance
If an object inserted in a circuit displays a current flow proportional to the voltage difference across it, then we define its resistance as the constant ratio
$\begin{equation*} R = \Delta V / I \end{equation*}$

The units of resistance are volts/ampere, usually abbreviated as ohms, symbolized with the capital Greek letter omega, $$\Omega$$.

##### Example 7: Resistance of a lightbulb

$$\triangleright$$ A flashlight bulb powered by a 9-volt battery has a resistance of 10 $$\Omega$$. How much current will it draw?

$$\triangleright$$ Solving the definition of resistance for $$I$$, we find

\begin{align*} I &= \Delta V/ R \\ &= 0.9\ \text{V}/\Omega \\ &= 0.9\ \text{V}/(\text{V}/\text{A}) \\ &= \text{0.9 A} \end{align*}

Ohm's law states that many substances, including many solids and some liquids, display this kind of behavior, at least for voltages that are not too large. The fact that Ohm's law is called a “law” should not be taken to mean that all materials obey it, or that it has the same fundamental importance as Newton's laws, for example. Materials are called ohmic or nonohmic, depending on whether they obey Ohm's law. Although we will concentrate on ohmic materials in this book, it's important to keep in mind that a great many materials are nonohmic, and devices made from them are often very important. For instance, a transistor is a nonohmic device that can be used to amplify a signal (as in a guitar amplifier) or to store and manipulate the ones and zeroes in a computer chip.

If objects of the same size and shape made from two different ohmic materials have different resistances, we can say that one material is more resistive than the other, or equivalently that it is less conductive. Materials, such as metals, that are very conductive are said to be good conductors. Those that are extremely poor conductors, for example wood or rubber, are classified as insulators. There is no sharp distinction between the two classes of materials. Some, such as silicon, lie midway between the two extremes, and are called semiconductors.

On an intuitive level, we can understand the idea of resistance by making the sounds “hhhhhh” and “ffffff.” To make air flow out of your mouth, you use your diaphragm to compress the air in your chest. The pressure difference between your chest and the air outside your mouth is analogous to a voltage difference. When you make the “h” sound, you form your mouth and throat in a way that allows air to flow easily. The large flow of air is like a large current. Dividing by a large current in the definition of resistance means that we get a small resistance. We say that the small resistance of your mouth and throat allows a large current to flow. When you make the “f” sound, you increase the resistance and cause a smaller current to flow.

Note that although the resistance of an object depends on the substance it is made of, we cannot speak simply of the “resistance of gold” or the “resistance of wood.” Figure l shows four examples of objects that have had wires attached at the ends as electrical connections. If they were made of the same substance, they would all nevertheless have different resistances because of their different sizes and shapes. A more detailed discussion will be more natural in the context of the following chapter, but it should not be too surprising that the resistance of l/2 will be greater than that of l/1 --- the image of water flowing through a pipe, however incorrect, gives us the right intuition. Object l/3 will have a smaller resistance than l/1 because the charged particles have less of it to get through.

### Superconductors

All materials display some variation in resistance according to temperature (a fact that is used in thermostats to make a thermometer that can be easily interfaced to an electric circuit). More spectacularly, most metals have been found to exhibit a sudden change to zero resistance when cooled to a certain critical temperature. They are then said to be superconductors. Currently, the most important practical application of superconductivity is in medical MRI (magnetic resonance imaging) scanners. The mechanism of MRI is explained on p. 465, but the important point for now is that when your body is inserted into one of these devices, you are being immersed in an extremely strong magnetic field produced by electric currents flowing through the coiled wires of an electromagnet. If these wires were not superconducting, they would instantly burn up because of the heat generated by their resistance.

There are many other potential applications for superconductors, but most of these, such as power transmission, are not currently economically feasible because of the extremely low temperatures required for superconductivity to occur.

However, it was discovered in 1986 that certain ceramics are superconductors at less extreme temperatures. The technological barrier is now in finding practical methods for making wire out of these brittle materials. Wall Street is currently investing billions of dollars in developing superconducting devices for cellular phone relay stations based on these materials.

There is currently no satisfactory theory of superconductivity in general, although superconductivity in metals is understood fairly well. Unfortunately I have yet to find a fundamental explanation of superconductivity in metals that works at the introductory level.

### Constant voltage throughout a conductor

The idea of a superconductor leads us to the question of how we should expect an object to behave if it is made of a very good conductor. Superconductors are an extreme case, but often a metal wire can be thought of as a perfect conductor, for example if the parts of the circuit other than the wire are made of much less conductive materials. What happens if $$R$$ equals zero in the equation $$R=\Delta V/I$$? The result of dividing two numbers can only be zero if the number on top equals zero. This tells us that if we pick any two points in a perfect conductor, the voltage difference between them must be zero. In other words, the entire conductor must be at the same voltage.

n / 1. The finger deposits charges on the solid, spherical, metal doorknob and is then withdrawn. 2. Almost instantaneously, the charges' mutual repulsion makes them redistribute themselves uniformly on the surface of the sphere. The only excess charge is on the surface; charges do exist in the atoms that form the interior of the sphere, but they are balanced. Charges on the interior feel zero total electrical force from the ones at the surface. Charges at the surface experience a net outward repulsion, but this is canceled out by the force that keep them from escaping into the air. 3. A voltmeter shows zero difference in voltage between any two points on the interior or surface of the sphere. If the voltage difference wasn't zero, then energy could be released by the flow of charge from one point to the other; this only happens before equilibrium is reached.

Constant voltage means that no work would be done on a charge as it moved from one point in the conductor to another. If zero work was done only along a certain path between two specific points, it might mean that positive work was done along part of the path and negative work along the rest, resulting in a cancellation. But there is no way that the work could come out to be zero for all possible paths unless the electrical force on a charge was in fact zero at every point. Suppose, for example, that you build up a static charge by scuffing your feet on a carpet, and then you deposit some of that charge onto a doorknob, which is a good conductor. How can all that charge be in the doorknob without creating any electrical force at any point inside it? The only possible answer is that the charge moves around until it has spread itself into just the right configuration so that the forces exerted by all the little bits of excess surface charge on any charged particle within the doorknob exactly cancel out.

We can explain this behavior if we assume that the charge placed on the doorknob eventually settles down into a stable equilibrium. Since the doorknob is a conductor, the charge is free to move through it. If it was free to move and any part of it did experience a nonzero total force from the rest of the charge, then it would move, and we would not have an equilibrium.

Excess charge placed on a conductor, once it reaches its equilibrium configuration, is entirely on the surface, not on the interior. This should be intuitively reasonable in figure n, for example, since the charges are all repelling each other. A proof is given in example 15 on p. 636.

Since wires are good conductors, constancy of voltage throughout a conductor provides a convenient freedom in hooking up a voltmeter to a circuit. In figure o, points B and C are on the same piece of conducting wire, so $$V_B=V_C$$. Measuring $$V_B-V_A$$ gives the same result as measuring $$V_C-V_A$$.

##### Example 8: The lightning rod
Suppose you have a pear-shaped conductor like the one in figure p/1. Since the pear is a conductor, there are free charges everywhere inside it. Panels 1 and 2 of the figure show a computer simulation with 100 identical electric charges. In 1, the charges are released at random positions inside the pear. Repulsion causes them all to fly outward onto the surface and then settle down into an orderly but nonuniform pattern.

We might not have been able to guess the pattern in advance, but we can verify that some of its features make sense. For example, charge A has more neighbors on the right than on the left, which would tend to make it accelerate off to the left. But when we look at the picture as a whole, it appears reasonable that this is prevented by the larger number of more distant charges on its left than on its right.

There also seems to be a pattern to the nonuniformity: the charges collect more densely in areas like B, where the surface is strongly curved, and less densely in flatter areas like C.

To understand the reason for this pattern, consider p/3. Two conducting spheres are connected by a conducting wire. Since the whole apparatus is conducting, it must all be at one voltage. As shown in problem 43 on p. 604, the density of charge is greater on the smaller sphere. This is an example of a more general fact observed in p/2, which is that the charge on a conductor packs itself more densely in areas that are more sharply curved.

Similar reasoning shows why Benjamin Franklin used a sharp tip when he invented the lightning rod. The charged stormclouds induce positive and negative charges to move to opposite ends of the rod. At the pointed upper end of the rod, the charge tends to concentrate at the point, and this charge attracts the lightning. The same effect can sometimes be seen when a scrap of aluminum foil is inadvertently put in a microwave oven. Modern experiments (Moore et al., Journal of Applied Meteorology 39 (1999) 593) show that although a sharp tip is best at starting a spark, a more moderate curve, like the right-hand tip of the pear in this example, is better at successfully sustaining the spark for long enough to connect a discharge to the clouds.

### Short circuits

So far we have been assuming a perfect conductor. What if it is a good conductor, but not a perfect one? Then we can solve for $$\Delta V=IR$$. An ordinary-sized current will make a very small result when we multiply it by the resistance of a good conductor such as a metal wire. The voltage throughout the wire will then be nearly constant. If, on the other hand, the current is extremely large, we can have a significant voltage difference. This is what happens in a short-circuit: a circuit in which a low-resistance pathway connects the two sides of a voltage source. Note that this is much more specific than the popular use of the term to indicate any electrical malfunction at all. If, for example, you short-circuit a 9-volt battery as shown in figure q, you will produce perhaps a thousand amperes of current, leading to a very large value of $$P=I\Delta V$$. The wire gets hot!

self-check:

What would happen to the battery in this kind of short circuit?

### Resistors

Inside any electronic gadget you will see quite a few little circuit elements like the one shown in the photo. These resistors are simply a cylinder of ohmic material with wires attached to the end.

At this stage, most students have a hard time understanding why resistors would be used inside a radio or a computer. We obviously want a lightbulb or an electric stove to have a circuit element that resists the flow of electricity and heats up, but heating is undesirable in radios and computers. Without going too far afield, let's use a mechanical analogy to get a general idea of why a resistor would be used in a radio.

The main parts of a radio receiver are an antenna, a tuner for selecting the frequency, and an amplifier to strengthen the signal sufficiently to drive a speaker. The tuner resonates at the selected frequency, just as in the examples of mechanical resonance discussed in chapter 18. The behavior of a mechanical resonator depends on three things: its inertia, its stiffness, and the amount of friction or damping. The first two parameters locate the peak of the resonance curve, while the damping determines the width of the resonance. In the radio tuner we have an electrically vibrating system that resonates at a particular frequency. Instead of a physical object moving back and forth, these vibrations consist of electrical currents that flow first in one direction and then in the other. In a mechanical system, damping means taking energy out of the vibration in the form of heat, and exactly the same idea applies to an electrical system: the resistor supplies the damping, and therefore controls the width of the resonance. If we set out to eliminate all resistance in the tuner circuit, by not building in a resistor and by somehow getting rid of all the inherent electrical resistance of the wires, we would have a useless radio. The tuner's resonance would be so narrow that we could never get close enough to the right frequency to bring in the station. The roles of inertia and stiffness are played by other circuit elements we have not discusses (a capacitor and a coil).

Many electrical devices are based on electrical resistance and Ohm's law, even if they do not have little components in them that look like the usual resistor. The following are some examples.

#### Lightbulb

There is nothing special about a lightbulb filament --- you can easily make a lightbulb by cutting a narrow waist into a metallic gum wrapper and connecting the wrapper across the terminals of a 9-volt battery. The trouble is that it will instantly burn out. Edison solved this technical challenge by encasing the filament in an evacuated bulb, which prevented burning, since burning requires oxygen.

#### Polygraph

The polygraph, or “lie detector,” is really just a set of meters for recording physical measures of the subject's psychological stress, such as sweating and quickened heartbeat. The real-time sweat measurement works on the principle that dry skin is a good insulator, but sweaty skin is a conductor. Of course a truthful subject may become nervous simply because of the situation, and a practiced liar may not even break a sweat. The method's practitioners claim that they can tell the difference, but you should think twice before allowing yourself to be polygraph tested. Most U.S. courts exclude all polygraph evidence, but some employers attempt to screen out dishonest employees by polygraph testing job applicants, an abuse that ranks with such pseudoscience as handwriting analysis.

#### Fuse

A fuse is a device inserted in a circuit tollbooth-style in the same manner as an ammeter. It is simply a piece of wire made of metals having a relatively low melting point. If too much current passes through the fuse, it melts, opening the circuit. The purpose is to make sure that the building's wires do not carry so much current that they themselves will get hot enough to start a fire. Most modern houses use circuit breakers instead of fuses, although fuses are still common in cars and small devices. A circuit breaker is a switch operated by a coiled-wire magnet, which opens the circuit when enough current flows. The advantage is that once you turn off some of the appliances that were sucking up too much current, you can immediately flip the switch closed. In the days of fuses, one might get caught without a replacement fuse, or even be tempted to stuff aluminum foil in as a replacement, defeating the safety feature.

#### Voltmeter

A voltmeter is nothing more than an ammeter with an additional high-value resistor through which the current is also forced to flow. Ohm's law relates the current through the resistor is related directly to the voltage difference across it, so the meter can be calibrated in units of volts based on the known value of the resistor. The voltmeter's two probes are touched to the two locations in a circuit between which we wish to measure the voltage difference, u/2. Note how cumbersome this type of drawing is, and how difficult it can be to tell what is connected to what. This is why electrical drawing are usually shown in schematic form. Figure u/3 is a schematic representation of figure u/2.

The setups for measuring current and voltage are different. When we are measuring current, we are finding “how much stuff goes through,” so we place the ammeter where all the current is forced to go through it. Voltage, however, is not “stuff that goes through,” it is a measure of electrical energy. If an ammeter is like the meter that measures your water use, a voltmeter is like a measuring stick that tells you how high a waterfall is, so that you can determine how much energy will be released by each kilogram of falling water. We do not want to force the water to go through the measuring stick! The arrangement in figure u/3 is a parallel circuit: one in there are “forks in the road” where some of the current will flow one way and some will flow the other. Figure u/4 is said to be wired in series: all the current will visit all the circuit elements one after the other. We will deal with series and parallel circuits in more detail in the following chapter.

If you inserted a voltmeter incorrectly, in series with the bulb and battery, its large internal resistance would cut the current down so low that the bulb would go out. You would have severely disturbed the behavior of the circuit by trying to measure something about it.

Incorrectly placing an ammeter in parallel is likely to be even more disconcerting. The ammeter has nothing but wire inside it to provide resistance, so given the choice, most of the current will flow through it rather than through the bulb. So much current will flow through the ammeter, in fact, that there is a danger of burning out the battery or the meter or both! For this reason, most ammeters have fuses or circuit breakers inside. Some models will trip their circuit breakers and make an audible alarm in this situation, while others will simply blow a fuse and stop working until you replace it.

##### Discussion Questions

In figure u/1, would it make any difference in the voltage measurement if we touched the voltmeter's probes to different points along the same segments of wire?

Explain why it would be incorrect to define resistance as the amount of charge the resistor allows to flow.

## 21.7 Applications of calculus (optional calculus-based section)

As discussed in example 1 on page 564, the definition of current as the rate of change of charge with respect to time must be reexpressed as a derivative in the case where the rate of change is not constant,

$\begin{equation*} I = \frac{dq}{dt} . \end{equation*}$

##### Example 9: Finding current given charge

$$\triangleright$$ A charged balloon falls to the ground, and its charge begins leaking off to the Earth. Suppose that the charge on the balloon is given by $$q=ae^{-bt}$$. Find the current as a function of time, and interpret the answer.

$$\triangleright$$ Taking the derivative, we have

\begin{align*} I &= \frac{d q}{d t} \\ &= - abe^{- bt} \end{align*}

An exponential function approaches zero as the exponent gets more and more negative. This means that both the charge and the current are decreasing in magnitude with time. It makes sense that the charge approaches zero, since the balloon is losing its charge. It also makes sense that the current is decreasing in magnitude, since charge cannot flow at the same rate forever without overshooting zero.

## 21.8 Series and parallel circuits

w / The two shaded areas shaped like the letter “E” are both regions of constant voltage.

Example 11.

Example 13: Uniting four resistors in parallel is equivalent to making a single resistor with the same length but four times the cross-sectional area. The result is to make a resistor with one quarter the resistance.

y / A voltmeter is really an ammeter with an internal resistor. When we measure the voltage difference across a resistor, 1, we are really constructing a parallel resistance circuit, 2.

z / 1. A battery drives current through two resistors in series. 2. There are three constant-voltage regions. 3. The three voltage differences are related. 4. If the meter crab-walks around the circuit without flipping over or crossing its legs, the resulting voltages have plus and minus signs that make them add up to zero.

Example 15.

Example 17. Doubling the length of a resistor is like putting two resistors in series. The resistance is doubled.

### Schematics

I see a chess position; Kasparov sees an interesting Ruy Lopez variation. To the uninitiated a schematic may look as unintelligible as Mayan hieroglyphs, but even a little bit of eye training can go a long way toward making its meaning leap off the page. A schematic is a stylized and simplified drawing of a circuit. The purpose is to eliminate as many irrelevant features as possible, so that the relevant ones are easier to pick out.

v / 1. Wrong: The shapes of the wires are irrelevant. 2. Wrong: Right angles should be used. 3. Wrong: A simple pattern is made to look unfamiliar and complicated. 4. Right.

An example of an irrelevant feature is the physical shape, length, and diameter of a wire. In nearly all circuits, it is a good approximation to assume that the wires are perfect conductors, so that any piece of wire uninterrupted by other components has constant voltage throughout it. Changing the length of the wire, for instance, does not change this fact. (Of course if we used miles and miles of wire, as in a telephone line, the wire's resistance would start to add up, and its length would start to matter.) The shapes of the wires are likewise irrelevant, so we draw them with standardized, stylized shapes made only of vertical and horizontal lines with right-angle bends in them. This has the effect of making similar circuits look more alike and helping us to recognize familiar patterns, just as words in a newspaper are easier to recognize than handwritten ones. Figure v shows some examples of these concepts.

The most important first step in learning to read schematics is to learn to recognize contiguous pieces of wire which must have constant voltage throughout. In figure w, for example, the two shaded E-shaped pieces of wire must each have constant voltage. This focuses our attention on two of the main unknowns we'd like to be able to predict: the voltage of the left-hand E and the voltage of the one on the right.

### Parallel resistances and the junction rule

One of the simplest examples to analyze is the parallel resistance circuit, of which figure w was an example. In general we may have unequal resistances $$R_1$$ and $$R_2$$, as in x/1. Since there are only two constant-voltage areas in the circuit, x/2, all three components have the same voltage difference across them. A battery normally succeeds in maintaining the voltage differences across itself for which it was designed, so the voltage drops $$\Delta V_1$$ and $$\Delta V_2$$ across the resistors must both equal the voltage of the battery:

$\begin{equation*} \Delta V_1=\Delta V_2=\Delta V_{battery} . \end{equation*}$

Each resistance thus feels the same voltage difference as if it was the only one in the circuit, and Ohm's law tells us that the amount of current flowing through each one is also the same as it would have been in a one-resistor circuit. This is why household electrical circuits are wired in parallel. We want every appliance to work the same, regardless of whether other appliances are plugged in or unplugged, turned on or switched off. (The electric company doesn't use batteries of course, but our analysis would be the same for any device that maintains a constant voltage.)

x / 1. Two resistors in parallel. 2. There are two constant-voltage areas. 3. The current that comes out of the battery splits between the two resistors, and later reunites. 4. The two resistors in parallel can be treated as a single resistor with a smaller resistance value.

Of course the electric company can tell when we turn on every light in the house. How do they know? The answer is that we draw more current. Each resistance draws a certain amount of current, and the amount that has to be supplied is the sum of the two individual currents. The current is like a river that splits in half, x/3, and then reunites. The total current is

$\begin{equation*} I_{total} = I_1 + I_2 . \end{equation*}$

This is an example of a general fact called the junction rule:

##### the junction rule

In any circuit that is not storing or releasing charge, conservation of charge implies that the total current flowing out of any junction must be the same as the total flowing in.

Coming back to the analysis of our circuit, we apply Ohm's law to each resistance, resulting in

\begin{align*} I_{total} &= \Delta V/R_1+ \Delta V/R_2 \\ &= \Delta V\left(\frac{1}{R_1}+\frac{1}{R_2}\right) . \end{align*}

As far as the electric company is concerned, your whole house is just one resistor with some resistance $$R$$, called the equivalent resistance. They would write Ohm's law as

$\begin{equation*} I_{total} = \Delta V/R , \end{equation*}$

from which we can determine the equivalent resistance by comparison with the previous expression:

\begin{align*} 1/R &= \frac{1}{R_1}+\frac{1}{R_2} \\ R &= \left(\frac{1}{R_1}+\frac{1}{R_2}\right)^{-1} \end{align*}

[equivalent resistance of two resistors in parallel]

Two resistors in parallel, x/4, are equivalent to a single resistor with a value given by the above equation.

##### Example 10: Two lamps on the same household circuit

$$\triangleright$$ You turn on two lamps that are on the same household circuit. Each one has a resistance of 1 ohm. What is the equivalent resistance, and how does the power dissipation compare with the case of a single lamp?

$$\triangleright$$ The equivalent resistance of the two lamps in parallel is

\begin{align*} R &= \left(\frac{1}{ R_1}+\frac{1}{ R_2}\right)^{-1} \\ &= \left(\frac{1}{1\ \Omega}+\frac{1}{1\ \Omega}\right)^{-1} \\ &= \left(1\ \Omega^{-1} + 1\ \Omega^{-1}\right)^{-1} \\ &= \left(2\ \Omega^{-1}\right)^{-1} \\ &= \text{0.5}\ \Omega \end{align*}

The voltage difference across the whole circuit is always the 110 V set by the electric company (it's alternating current, but that's irrelevant). The resistance of the whole circuit has been cut in half by turning on the second lamp, so a fixed amount of voltage will produce twice as much current. Twice the current flowing across the same voltage difference means twice as much power dissipation, which makes sense.

The cutting in half of the resistance surprises many students, since we are “adding more resistance” to the circuit by putting in the second lamp. Why does the equivalent resistance come out to be less than the resistance of a single lamp? This is a case where purely verbal reasoning can be misleading. A resistive circuit element, such as the filament of a lightbulb, is neither a perfect insulator nor a perfect conductor. Instead of analyzing this type of circuit in terms of “resistors,” i.e., partial insulators, we could have spoken of “conductors.” This example would then seem reasonable, since we “added more conductance,” but one would then have the incorrect expectation about the case of resistors in series, discussed in the following section.

Perhaps a more productive way of thinking about it is to use mechanical intuition. By analogy, your nostrils resist the flow of air through them, but having two nostrils makes it twice as easy to breathe.

##### Example 11: Three resistors in parallel
$$\triangleright$$ What happens if we have three or more resistors in parallel?

$$\triangleright$$ This is an important example, because the solution involves an important technique for understanding circuits: breaking them down into smaller parts and them simplifying those parts. In the circuit 21.8.2/1, with three resistors in parallel, we can think of two of the resistors as forming a single resistor, 21.8.2/2, with equivalent resistance

$\begin{equation*} R_{12} = \left(\frac{1}{ R_1}+\frac{1}{ R_2}\right)^{-1} . \end{equation*}$

We can then simplify the circuit as shown in 21.8.2/3, so that it contains only two resistances. The equivalent resistance of the whole circuit is then given by

$\begin{equation*} R_{123} = \left(\frac{1}{ R_{12}}+\frac{1}{ R_3}\right)^{-1} . \end{equation*}$

Substituting for $$R_{12}$$ and simplifying, we find the result

$\begin{equation*} R_{123} = \left(\frac{1}{ R_1}+\frac{1}{ R_2}+\frac{1}{ R_3}\right)^{-1} , \end{equation*}$

which you probably could have guessed. The interesting point here is the divide-and-conquer concept, not the mathematical result.

##### Example 12: An arbitrary number of identical resistors in parallel

$$\triangleright$$ What is the resistance of $$N$$ identical resistors in parallel?

$$\triangleright$$ Generalizing the results for two and three resistors, we have

$\begin{equation*} R_{N} = \left(\frac{1}{ R_1}+\frac{1}{ R_2}+...\right)^{-1} , \end{equation*}$

where “...” means that the sum includes all the resistors. If all the resistors are identical, this becomes

\begin{align*} R_{N} &= \left(\frac{ N}{ R}\right)^{-1} \\ &= \frac{ R}{ N} \end{align*}

##### Example 13: Dependence of resistance on cross-sectional area
We have alluded briefly to the fact that an object's electrical resistance depends on its size and shape, but now we are ready to begin making more mathematical statements about it. As suggested by figure 13, increasing a resistors's cross-sectional area is equivalent to adding more resistors in parallel, which will lead to an overall decrease in resistance. Any real resistor with straight, parallel sides can be sliced up into a large number of pieces, each with cross-sectional area of, say, 1 $$\mu \text{m}^2$$. The number, $$N$$, of such slices is proportional to the total cross-sectional area of the resistor, and by application of the result of the previous example we therefore find that the resistance of an object is inversely proportional to its cross-sectional area.

A fat pipe has less resistance than a skinny pipe.

An analogous relationship holds for water pipes, which is why high-flow trunk lines have to have large cross-sectional areas. To make lots of water (current) flow through a skinny pipe, we'd need an impractically large pressure (voltage) difference.

##### Example 14: Incorrect readings from a voltmeter

A voltmeter is really just an ammeter with an internal resistor, and we use a voltmeter in parallel with the thing that we're trying to measure the voltage difference across. This means that any time we measure the voltage drop across a resistor, we're essentially putting two resistors in parallel. The ammeter inside the voltmeter can be ignored for the purpose of analyzing how current flows in the circuit, since it is essentially just some coiled-up wire with a very low resistance.

Now if we are carrying out this measurement on a resistor that is part of a larger circuit, we have changed the behavior of the circuit through our act of measuring. It is as though we had modified the circuit by replacing the resistance $$R$$ with the smaller equivalent resistance of $$R$$ and $$R_v$$ in parallel. It is for this reason that voltmeters are built with the largest possible internal resistance. As a numerical example, if we use a voltmeter with an internal resistance of 1 $$M\Omega$$ to measure the voltage drop across a one-ohm resistor, the equivalent resistance is 0.999999 $$\Omega$$, which is not different enough to make any difference. But if we tried to use the same voltmeter to measure the voltage drop across a 2 $$M\Omega$$ resistor, we would be reducing the resistance of that part of the circuit by a factor of three, which would produce a drastic change in the behavior of the whole circuit.

This is the reason why you can't use a voltmeter to measure the voltage difference between two different points in mid-air, or between the ends of a piece of wood. This is by no means a stupid thing to want to do, since the world around us is not a constant-voltage environment, the most extreme example being when an electrical storm is brewing. But it will not work with an ordinary voltmeter because the resistance of the air or the wood is many gigaohms. The effect of waving a pair of voltmeter probes around in the air is that we provide a reuniting path for the positive and negative charges that have been separated --- through the voltmeter itself, which is a good conductor compared to the air. This reduces to zero the voltage difference we were trying to measure.

In general, a voltmeter that has been set up with an open circuit (or a very large resistance) between its probes is said to be “floating.” An old-fashioned analog voltmeter of the type described here will read zero when left floating, the same as when it was sitting on the shelf. A floating digital voltmeter usually shows an error message.

### Series resistances

The two basic circuit layouts are parallel and series, so a pair of resistors in series, aa/1, is another of the most basic circuits we can make. By conservation of charge, all the current that flows through one resistor must also flow through the other (as well as through the battery):

$\begin{equation*} I_1 = I_2 . \end{equation*}$

The only way the information about the two resistance values is going to be useful is if we can apply Ohm's law, which will relate the resistance of each resistor to the current flowing through it and the voltage difference across it. Figure aa/2 shows the three constant-voltage areas. Voltage differences are more physically significant than voltages, so we define symbols for the voltage differences across the two resistors in figure aa/3.

We have three constant-voltage areas, with symbols for the difference in voltage between every possible pair of them. These three voltage differences must be related to each other. It is as though I tell you that Fred is a foot taller than Ginger, Ginger is a foot taller than Sally, and Fred is two feet taller than Sally. The information is redundant, and you really only needed two of the three pieces of data to infer the third. In the case of our voltage differences, we have

$\begin{equation*} |\Delta V_1| + |\Delta V_2| = |\Delta V_{battery}| . \end{equation*}$

The absolute value signs are because of the ambiguity in how we define our voltage differences. If we reversed the two probes of the voltmeter, we would get a result with the opposite sign. Digital voltmeters will actually provide a minus sign on the screen if the wire connected to the “V” plug is lower in voltage than the one connected to the “COM” plug. Analog voltmeters pin the needle against a peg if you try to use them to measure negative voltages, so you have to fiddle to get the leads connected the right way, and then supply any necessary minus sign yourself.

Figure aa/4 shows a standard way of taking care of the ambiguity in signs. For each of the three voltage measurements around the loop, we keep the same probe (the darker one) on the clockwise side. It is as though the voltmeter was sidling around the circuit like a crab, without ever “crossing its legs.” With this convention, the relationship among the voltage drops becomes

$\begin{equation*} \Delta V_1 + \Delta V_2 = -\Delta V_{battery} , \end{equation*}$

or, in more symmetrical form,

$\begin{equation*} \Delta V_1 + \Delta V_2 + \Delta V_{battery} = 0 . \end{equation*}$

More generally, this is known as the loop rule for analyzing circuits:

##### the loop rule

Assuming the standard convention for plus and minus signs, the sum of the voltage drops around any closed loop in a circuit must be zero.

Looking for an exception to the loop rule would be like asking for a hike that would be downhill all the way and that would come back to its starting point!

For the circuit we set out to analyze, the equation

$\begin{equation*} \Delta V_1 + \Delta V_2 + \Delta V_{battery} = 0 \end{equation*}$

can now be rewritten by applying Ohm's law to each resistor:

$\begin{equation*} I_1R_1 + I_2R_2 + \Delta V_{battery} = 0 . \end{equation*}$

The currents are the same, so we can factor them out:

$\begin{equation*} I\left(R_1 + R_2\right) + \Delta V_{battery} = 0 , \end{equation*}$

and this is the same result we would have gotten if we had been analyzing a one-resistor circuit with resistance $$R_1+R_2$$. Thus the equivalent resistance of resistors in series equals the sum of their resistances.

##### Example 15: Two lightbulbs in series
$$\triangleright$$ If two identical lightbulbs are placed in series, how do their brightnesses compare with the brightness of a single bulb?

$$\triangleright$$ Taken as a whole, the pair of bulbs act like a doubled resistance, so they will draw half as much current from the wall. Each bulb will be dimmer than a single bulb would have been.

The total power dissipated by the circuit is $$I\Delta V$$. The voltage drop across the whole circuit is the same as before, but the current is halved, so the two-bulb circuit draws half as much total power as the one-bulb circuit. Each bulb draws one-quarter of the normal power.

Roughly speaking, we might expect this to result in one quarter the light being produced by each bulb, but in reality lightbulbs waste quite a high percentage of their power in the form of heat and wavelengths of light that are not visible (infrared and ultraviolet). Less light will be produced, but it's hard to predict exactly how much less, since the efficiency of the bulbs will be changed by operating them under different conditions.

##### Example 16: More than two equal resistances in series

By straightforward application of the divide-and-conquer technique discussed in the previous section, we find that the equivalent resistance of $$N$$ identical resistances $$R$$ in series will be $$NR$$.

##### Example 17: Dependence of resistance on length
In the previous section, we proved that resistance is inversely proportional to cross-sectional area. By equivalent reason about resistances in series, we find that resistance is proportional to length. Analogously, it is harder to blow through a long straw than through a short one.

Combining the results of examples 13 and 17, we find that the resistance of an object with straight, parallel sides is given by

$\begin{equation*} R = \text{(constant)} \cdot L/ A \end{equation*}$

The proportionality constant is called the resistivity, and it depends only on the substance of which the object is made. A resistivity measurement could be used, for instance, to help identify a sample of an unknown substance.

##### Example 18: Choice of high voltage for power lines
Thomas Edison got involved in a famous technological controversy over the voltage difference that should be used for electrical power lines. At this time, the public was unfamiliar with electricity, and easily scared by it. The president of the United States, for instance, refused to have electrical lighting in the White House when it first became commercially available because he considered it unsafe, preferring the known fire hazard of oil lamps to the mysterious dangers of electricity. Mainly as a way to overcome public fear, Edison believed that power should be transmitted using small voltages, and he publicized his opinion by giving demonstrations at which a dog was lured into position to be killed by a large voltage difference between two sheets of metal on the ground. (Edison's opponents also advocated alternating current rather than direct current, and AC is more dangerous than DC as well. As we will discuss later, AC can be easily stepped up and down to the desired voltage level using a device called a transformer.)

Now if we want to deliver a certain amount of power $$P_L$$ to a load such as an electric lightbulb, we are constrained only by the equation $$P_{L} = I\Delta V_L$$. We can deliver any amount of power we wish, even with a low voltage, if we are willing to use large currents. Modern electrical distribution networks, however, use dangerously high voltage differences of tens of thousands of volts. Why did Edison lose the debate?

It boils down to money. The electric company must deliver the amount of power $$P_L$$ desired by the customer through a transmission line whose resistance $$R_T$$ is fixed by economics and geography. The same current flows through both the load and the transmission line, dissipating power usefully in the former and wastefully in the latter. The efficiency of the system is

\begin{align*} \text{efficiency} &= \frac{\text{power paid for by the customer}} {\text{power paid for by the utility}} \\ &= \frac{ P_{L}}{ P_L+ P_{T}} \\ &= \frac{1}{1+ P_{T}/ P_L} \end{align*}

Putting ourselves in the shoes of the electric company, we wish to get rid of the variable $$P_T$$, since it is something we control only indirectly by our choice of $$\Delta V_T$$ and $$I$$. Substituting $$P_{T}= I\Delta V_T$$, we find

$\begin{equation*} \text{efficiency} = \frac{1} {1+\frac{ I \Delta V_T}{ P_L}} \end{equation*}$

We assume the transmission line (but not necessarily the load) is ohmic, so substituting $$\Delta V_T=IR_T$$ gives

$\begin{equation*} \text{efficiency} = \frac{1}{1+\frac{I^2R_T}{P_L}} \end{equation*}$

This quantity can clearly be maximized by making $$I$$ as small as possible, since we will then be dividing by the smallest possible quantity on the bottom of the fraction. A low-current circuit can only deliver significant amounts of power if it uses high voltages, which is why electrical transmission systems use dangerous high voltages.

Example 19.

##### Example 19: A complicated circuit
$$\triangleright$$ All seven resistors in the left-hand panel of figure ab are identical. Initially, the switch S is open as shown in the figure, and the current through resistor A is $$I_\text{o}$$. The switch is then closed. Find the current through resistor B, after the switch is closed, in terms of $$I_\text{o}$$.

$$\triangleright$$ The second panel shows the circuit redrawn for simplicity, in the initial condition with the switch open. When the switch is open, no current can flow through the central resistor, so we may as well ignore it. I've also redrawn the junctions, without changing what's connected to what. This is the kind of mental rearranging that you'll eventually learn to do automatically from experience with analyzing circuits. The redrawn version makes it easier to see what's happening with the current. Charge is conserved, so any charge that flows past point 1 in the circuit must also flow past points 2 and 3. This would have been harder to reason about by applying the junction rule to the original version, which appears to have nine separate junctions.

In the new version, it's also clear that the circuit has a great deal of symmetry. We could flip over each parallel pair of identical resistors without changing what's connected to what, so that makes it clear that the voltage drops and currents must be equal for the members of each pair. We can also prove this by using the loop rule. The loop rule says that the two voltage drops in loop 4 must be equal, and similarly for loops 5 and 6. Since the resistors obey Ohm's law, equal voltage drops across them also imply equal currents. That means that when the current at point 1 comes to the top junction, exactly half of it goes through each resistor. Then the current reunites at 2, splits between the next pair, and so on. We conclude that each of the six resistors in the circuit experiences the same voltage drop and the same current. Applying the loop rule to loop 7, we find that the sum of the three voltage drops across the three left-hand resistors equals the battery's voltage, $$V$$, so each resistor in the circuit experiences a voltage drop $$V/3$$. Letting $$R$$ stand for the resistance of one of the resistors, we find that the current through resistor B, which is the same as the currents through all the others, is given by $$I_\text{o}=V/3R$$.

We now pass to the case where the switch is closed, as shown in the third panel. The battery's voltage is the same as before, and each resistor's resistance is the same, so we can still use the same symbols $$V$$ and $$R$$ for them. It is no longer true, however, that each resistor feels a voltage drop $$V/3$$. The equivalent resistance of the whole circuit is $$R/2+R/3+R/2=4R/3$$, so the total current drawn from the battery is $$3V/4R$$. In the middle group of resistors, this current is split three ways, so the new current through B is $$(1/3)(3V/4R)=V/4R=3I_\text{o}/4$$.

Interpreting this result, we see that it comes from two effects that partially cancel. Closing the switch reduces the equivalent resistance of the circuit by giving charge another way to flow, and increases the amount of current drawn from the battery. Resistor B, however, only gets a 1/3 share of this greater current, not 1/2. The second effect turns out to be bigger than first, and therefore the current through resistor B is lessened over all.

##### Example 20: Getting killed by your ammeter

As with a voltmeter, an ammeter can give erroneous readings if it is used in such a way that it changes the behavior the circuit. An ammeter is used in series, so if it is used to measure the current through a resistor, the resistor's value will effectively be changed to $$R+ R_a$$, where $$R_a$$ is the resistance of the ammeter. Ammeters are designed with very low resistances in order to make it unlikely that $$R+ R_a$$ will be significantly different from $$R$$.

In fact, the real hazard is death, not a wrong reading! Virtually the only circuits whose resistances are significantly less than that of an ammeter are those designed to carry huge currents. An ammeter inserted in such a circuit can easily melt. When I was working at a laboratory funded by the Department of Energy, we got periodic bulletins from the DOE safety office about serious accidents at other sites, and they held a certain ghoulish fascination. One of these was about a DOE worker who was completely incinerated by the explosion created when he inserted an ordinary Radio Shack ammeter into a high-current circuit. Later estimates showed that the heat was probably so intense that the explosion was a ball of plasma --- a gas so hot that its atoms have been ionized.

##### Discussion Question

We have stated the loop rule in a symmetric form where a series of voltage drops adds up to zero. To do this, we had to define a standard way of connecting the voltmeter to the circuit so that the plus and minus signs would come out right. Suppose we wish to restate the junction rule in a similar symmetric way, so that instead of equating the current coming in to the current going out, it simply states that a certain sum of currents at a junction adds up to zero. What standard way of inserting the ammeter would we have to use to make this work?

## Vocabulary

charge — a numerical rating of how strongly an object participates in electrical forces

coulomb (C) — the unit of electrical charge

current — the rate at which charge crosses a certain boundary

ampere — the metric unit of current, one coulomb pe second; also “amp”

ammeter — a device for measuring electrical current

circuit — an electrical device in which charge can come back to its starting point and be recycled rather than getting stuck in a dead end

open circuit — a circuit that does not function because it has a gap in it

short circuit — a circuit that does not function because charge is given a low-resistance “shortcut” path that it can follow, instead of the path that makes it do something useful

voltage — electrical potential energy per unit charge that will be possessed by a charged particle at a certain point in space

volt — the metric unit of voltage, one joule per coulomb

voltmeter — a device for measuring voltage differences

ohmic — describes a substance in which the flow of current between two points is proportional to the voltage difference between them

resistance — the ratio of the voltage difference to the current in an object made of an ohmic substance

ohm — the metric unit of electrical resistance, one volt per ampere

## Notation

$$q$$ — charge

$$I$$ — current

$$A$$ — units of amperes

$$V$$ — voltage

V — units of volts

$$R$$ — resistance

$$\Omega$$ — units of ohms

## Other Notation

electric potential — rather than the more informal “voltage” used here; despite the misleading name, it is not the same as electric potential energy

\notationitem{eV}{a unit of energy, equal to $$e$$ multiplied by 1 volt; $$1.6\times10^{-19}$$ joules}

## Summary

{}

All the forces we encounter in everyday life boil down to two basic types: gravitational forces and electrical forces. A force such as friction or a “sticky force” arises from electrical forces between individual atoms.

Just as we use the word “mass” to describe how strongly an object participates in gravitational forces, we use the word “charge” for the intensity of its electrical forces. There are two types of charge. Two charges of the same type repel each other, but objects whose charges are different attract each other. Charge is measured in units of coulombs (C).

Mobile charged particle model: A great many phenomena are easily understood if we imagine matter as containing two types of charged particles, which are at least partially able to move around.

Positive and negative charge: Ordinary objects that have not been specially prepared have both types of charge spread evenly throughout them in equal amounts. The object will then tend not to exert electrical forces on any other object, since any attraction due to one type of charge will be balanced by an equal repulsion from the other. (We say “tend not to” because bringing the object near an object with unbalanced amounts of charge could cause its charges to separate from each other, and the force would no longer cancel due to the unequal distances.) It therefore makes sense to describe the two types of charge using positive and negative signs, so that an unprepared object will have zero total charge.

The Coulomb force law states that the magnitude of the electrical force between two charged particles is given by $$|\mathbf{F}| =k |q_1| |q_2| / r^2$$.

Conservation of charge: An even more fundamental reason for using positive and negative signs for charge is that with this definition the total charge of a closed system is a conserved quantity.

All electrical phenomena are alike in that that arise from the presence or motion of charge. Most practical electrical devices are based on the motion of charge around a complete circuit, so that the charge can be recycled and does not hit any dead ends. The most useful measure of the flow of charge is current, $$I=\Delta q/\Delta t$$.

An electrical device whose job is to transform energy from one form into another, e.g., a lightbulb, uses power at a rate which depends both on how rapidly charge is flowing through it and on how much work is done on each unit of charge. The latter quantity is known as the voltage difference between the point where the current enters the device and the point where the current leaves it. Since there is a type of potential energy associated with electrical forces, the amount of work they do is equal to the difference in potential energy between the two points, and we therefore define voltage differences directly in terms of potential energy, $$\Delta V=\Delta PE_{elec}/q$$. The rate of power dissipation is $$P=I\Delta V$$.

Many important electrical phenomena can only be explained if we understand the mechanisms of current flow at the atomic level. In metals, currents are carried by electrons, in liquids by ions. Gases are normally poor conductors unless their atoms are subjected to such intense electrical forces that the atoms become ionized.

Many substances, including all solids, respond to electrical forces in such a way that the flow of current between two points is proportional to the voltage difference between those points. Such a substance is called ohmic, and an object made out of an ohmic substance can be rated in terms of its resistance, $$R=\Delta$$V/I. An important corollary is that a perfect conductor, with $$R=0$$, must have constant voltage everywhere within it.

A schematic is a drawing of a circuit that standardizes and stylizes its features to make it easier to understand. Any circuit can be broken down into smaller parts. For instance, one big circuit may be understood as two small circuits in series, another as three circuits in parallel. When circuit elements are combined in parallel and in series, we have two basic rules to guide us in understanding how the parts function as a whole:

the junction rule: In any circuit that is not storing or releasing charge, conservation of charge implies that the total current flowing out of any junction must be the same as the total flowing in.

the loop rule: Assuming the standard convention for plus and minus signs, the sum of the voltage drops around any closed loop in a circuit must be zero.

The simplest application of these rules is to pairs of resistors combined in series or parallel. In such cases, the pair of resistors acts just like a single unit with a certain resistance value, called their equivalent resistance. Resistances in series add to produce a larger equivalent resistance,

$\begin{equation*} R_{series} = R_1+R_2 , \end{equation*}$

because the current has to fight its way through both resistances. Parallel resistors combine to produce an equivalent resistance that is smaller than either individual resistance,

$\begin{equation*} R_{parallel} = \left(\frac{1}{R_1}+\frac{1}{R_2}\right)^{-1} , \end{equation*}$

because the current has two different paths open to it.

An important example of resistances in parallel and series is the use of voltmeters and ammeters in resistive circuits. A voltmeter acts as a large resistance in parallel with the resistor across which the voltage drop is being measured. The fact that its resistance is not infinite means that it alters the circuit it is being used to investigate, producing a lower equivalent resistance. An ammeter acts as a small resistance in series with the circuit through which the current is to be determined. Its resistance is not quite zero, which leads to an increase in the resistance of the circuit being tested.

## Homework Problems

aa / Problem 1. Top: A realistic picture of a neuron. Bottom: A simplified diagram of one segment of the tail (axon).

ab / Problem 2.

ac / Problem 3.

ae / Problem 9.

af / Problem 11.

ag / Problem 16.

ah / A printed circuit board, like the kind referred to in problem 17.

ai / An LP record, problem 19.

aj / Problem 21.

ak / Problem 28.

al / Problem 29.

am / Problem 31.

an / Problems 36 and 37.

ao / Problem 38.

ap / Problem 39.

aq / Problem 40.

ar / Problem 44.

\begin{homeworkforcelabel}{neuron-stick}{1}{}{1}The figure shows a neuron, which is the type of cell your nerves are made of. Neurons serve to transmit sensory information to the brain, and commands from the brain to the muscles. All this data is transmitted electrically, but even when the cell is resting and not transmitting any information, there is a layer of negative electrical charge on the inside of the cell membrane, and a layer of positive charge just outside it. This charge is in the form of various ions dissolved in the interior and exterior fluids. Why would the negative charge remain plastered against the inside surface of the membrane, and likewise why doesn't the positive charge wander away from the outside surface? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{heline}{1}{}{2}A helium atom finds itself momentarily in this arrangement. Find the direction and magnitude of the force acting on the right-hand electron. The two protons in the nucleus are so close together ($$\sim1$$ fm) that you can consider them as being right on top of each other. As discussed in chapter 26, the charge of an electron is $$-e$$, and the charge of a proton $$e$$, where $$e=1.60\times10^{-19}\ \text{C}$$. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{hetriangle}{1}{}{3}The helium atom of problem 2 has some new experiences, goes through some life changes, and later on finds itself in the configuration shown here. What are the direction and magnitude of the force acting on the bottom electron? (Draw a sketch to make clear the definition you are using for the angle that gives direction.)(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{hands-repulsion}{1}{}{4}[See ch. 1 for help on how to do order-of-magnitude estimates.] Suppose you are holding your hands in front of you, 10 cm apart.
(a) Estimate the total number of electrons in each hand.
(b) Estimate the total repulsive force of all the electrons in one hand on all the electrons in the other.
(c) Why don't you feel your hands repelling each other?
(d) Estimate how much the charge of a proton could differ in magnitude from the charge of an electron without creating a noticeable force between your hands. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{leadbound}{1}{}{5}Suppose that a proton in a lead nucleus wanders out to the surface of the nucleus, and experiences a strong nuclear force of about 8 kN from the nearby neutrons and protons pulling it back in. Compare this numerically to the repulsive electrical force from the other protons, and verify that the net force is attractive. A lead nucleus is very nearly spherical, is about 6.5 fm in radius, and contains 82 protons, each with a charge of $$+e$$, where $$e=1.60\times10^{-19}\ \text{C}$$. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{muonic}{1}{}{6}The subatomic particles called muons behave exactly like electrons, except that a muon's mass is greater by a factor of 206.77. Muons are continually bombarding the Earth as part of the stream of particles from space known as cosmic rays. When a muon strikes an atom, it can displace one of its electrons. If the atom happens to be a hydrogen atom, then the muon takes up an orbit that is on the average 206.77 times closer to the proton than the orbit of the ejected electron. How many times greater is the electric force experienced by the muon than that previously felt by the electron?(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{earthmoonelec}{1}{}{7}The Earth and Moon are bound together by gravity. If, instead, the force of attraction were the result of each having a charge of the same magnitude but opposite in sign, find the quantity of charge that would have to be placed on each to produce the required force.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{nacl}{2}{}{8} (answer check available at lightandmatter.com) The figure shows one layer of the three-dimensional structure of a salt crystal. The atoms extend much farther off in all directions, but only a six-by-six square is shown here. The larger circles are the chlorine ions, which have charges of $$-e$$, where $$e=1.60\times10^{-19}\ \text{C}$$. The smaller circles are sodium ions, with charges of $$+e$$. The center-to-center distance between neighboring ions is about 0.3 nm. Real crystals are never perfect, and the crystal shown here has two defects: a missing atom at one location, and an extra lithium atom, shown as a grey circle, inserted in one of the small gaps. If the lithium atom has a charge of $$+e$$, what is the direction and magnitude of the total force on it? Assume there are no other defects nearby in the crystal besides the two shown here. [Hints: The force on the lithium ion is the vector sum of all the forces of all the quadrillions of sodium and chlorine atoms, which would obviously be too laborious to calculate. Nearly all of these forces, however, are canceled by a force from an ion on the opposite side of the lithium.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{croquet}{1}{1}{9} In the semifinals of an electrostatic croquet tournament, Jessica hits her positively charged ball, sending it across the playing field, rolling to the left along the $$x$$ axis. It is repelled by two other positive charges. These two equal charges are fixed on the $$y$$ axis at the locations shown in the figure. (a) Express the force on the ball in terms of the ball's position, $$x$$. (b) At what value of $$x$$ does the ball experience the greatest deceleration? Express you answer in terms of $$b$$. [Based on a problem by Halliday and Resnick.](answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{hail-an-electron-cab}{1}{}{10}(solution in the pdf version of the book) In a wire carrying a current of 1.0 pA, how long do you have to wait, on the average, for the next electron to pass a given point? Express your answer in units of microseconds. The charge of an electron is $$-e=-1.60\times10^{-19}\ \text{C}$$. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{neuroncurrent}{1}{}{11}Referring back to our old friend the neuron from problem 1 on page 597, let's now consider what happens when the nerve is stimulated to transmit information. When the blob at the top (the cell body) is stimulated, it causes Na$$^+$$ ions to rush into the top of the tail (axon). This electrical pulse will then travel down the axon, like a flame burning down from the end of a fuse, with the Na$$^+$$ ions at each point first going out and then coming back in. If $$10^{10}$$ Na$$^+$$ ions cross the cell membrane in 0.5 ms, what amount of current is created? The charge of a Na$$^+$$ ion is $$+e=1.60\times10^{-19}\ \text{C}$$. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{bulbresistance}{1}{}{12}If a typical light bulb draws about 900 mA from a 110-V household circuit, what is its resistance? (Don't worry about the fact that it's alternating current.) (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{jouleheating}{1}{}{13}A resistor has a voltage difference $$\Delta V$$ across it, causing a current $$I$$ to flow.
(a) Find an equation for the power it dissipates as heat in terms of the variables $$I$$ and $$R$$ only, eliminating $$\Delta V$$. (answer check available at lightandmatter.com)
(b) If an electrical line coming to your house is to carry a given amount of current, interpret your equation from part a to explain whether the wire's resistance should be small, or large. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{drier}{1}{}{14}(a) Express the power dissipated by a resistor in terms of $$R$$ and $$\Delta V$$ only, eliminating $$I$$. (answer check available at lightandmatter.com)
(b) Electrical receptacles in your home are mostly 110 V, but circuits for electric stoves, air conditioners, and washers and driers are usually 220 V. The two types of circuits have differently shaped receptacles. Suppose you rewire the plug of a drier so that it can be plugged in to a 110 V receptacle. The resistor that forms the heating element of the drier would normally draw 200 W. How much power does it actually draw now? (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{pointchargevoltage}{1}{}{15}Use the result of problem 42 on page 604 to find an equation for the voltage at a point in space at a distance $$r$$ from a point charge $$Q$$. (Take your $$V=0$$ distance to be anywhere you like.) (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{battery-and-bulb}{1}{}{16}You are given a battery, a flashlight bulb, and a single piece of wire. Draw at least two configurations of these items that would result in lighting up the bulb, and at least two that would not light it. (Don't draw schematics.) If you're not sure what's going on, borrow the materials from your instructor and try it. Note that the bulb has two electrical contacts: one is the threaded metal jacket, and the other is the tip (at the bottom in the figure). [Problem by Arnold Arons.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{measure-on-printed-circuit}{1}{}{17} (solution in the pdf version of the book) You have to do different things with a circuit to measure current than to measure a voltage difference. Which would be more practical for a printed circuit board, in which the wires are actually strips of metal inlaid on the surface of the board? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{teardrop}{1}{}{18}Problem 18 has been replaced with problem 43.\end{homeworkforcelabel}

\begin{homeworkforcelabel}{lpcurrent}{2}{}{19}(a) You take an LP record out of its sleeve, and it acquires a static charge of 1 nC. You play it at the normal speed of $$33\frac{1}{3}$$ r.p.m., and the charge moving in a circle creates an electric current. What is the current, in amperes? (answer check available at lightandmatter.com)
(b) Although the planetary model of the atom can be made to work with any value for the radius of the electrons' orbits, more advanced models that we will study later in this course predict definite radii. If the electron is imagined as circling around the proton at a speed of $$2.2\times10^6$$ m/s, in an orbit with a radius of 0.05 nm, what electric current is created? The charge of an electron is $$-e=-1.60\times10^{-19}\ \text{C}$$. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{equilateralcharges}{1}{}{20}Three charges, each of strength $$Q$$ ($$Q>0$$) form a fixed equilateral triangle with sides of length $$b$$. You throw a particle of mass $$m$$ and positive charge $$q$$ from far away, with an initial speed $$v$$. Your goal is to get the particle to go to the center of the triangle, your aim is perfect, and you are free to throw from any direction you like. What is the minimum possible value of $$v?$$ You will need the result of problem 42. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{vnacl}{2}{}{21} Referring back to problem 8 on page 598 about the sodium chloride crystal, suppose the lithium ion is going to jump from the gap it is occupying to one of the four closest neighboring gaps. Which one will it jump to, and if it starts from rest, how fast will it be going by the time it gets there? (It will keep on moving and accelerating after that, but that does not concern us.) You will need the result of problem 42. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{negative-resistance}{1}{}{22}We have referred to resistors dissipating heat, i.e., we have assumed that $$P=I\Delta V$$ is always greater than zero. Could $$I\Delta V$$ come out to be negative for a resistor? If so, could one make a refrigerator by hooking up a resistor in such a way that it absorbed heat instead of dissipating it? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{car}{1}{}{23}Today, even a big luxury car like a Cadillac can have an electrical system that is relatively low in power, since it doesn't need to do much more than run headlights, power windows, etc. In the near future, however, manufacturers plan to start making cars with electrical systems about five times more powerful. This will allow certain energy-wasting parts like the water pump to be run on electrical motors and turned off when they're not needed --- currently they're run directly on shafts from the motor, so they can't be shut off. It may even be possible to make an engine that can shut off at a stoplight and then turn back on again without cranking, since the valves can be electrically powered. Current cars' electrical systems have 12-volt batteries (with 14-volt chargers), but the new systems will have 36-volt batteries (with 42-volt chargers).
(a) Suppose the battery in a new car is used to run a device that requires the same amount of power as the corresponding device in the old car. Based on the sample figures above, how would the currents handled by the wires in one of the new cars compare with the currents in the old ones?
(b) The real purpose of the greater voltage is to handle devices that need more power. Can you guess why they decided to change to 36-volt batteries rather than increasing the power without increasing the voltage? (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{eel}{1}{}{24}(a) Many battery-operated devices take more than one battery. If you look closely in the battery compartment, you will see that the batteries are wired in series. Consider a flashlight circuit. What does the loop rule tell you about the effect of putting several batteries in series in this way?
(b) The cells of an electric eel's nervous system are not that different from ours --- each cell can develop a voltage difference across it of somewhere on the order of one volt. How, then, do you think an electric eel can create voltages of thousands of volts between different parts of its body? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{stove}{1}{}{25}The heating element of an electric stove is connected in series with a switch that opens and closes many times per second. When you turn the knob up for more power, the fraction of the time that the switch is closed increases. Suppose someone suggests a simpler alternative for controlling the power by putting the heating element in series with a variable resistor controlled by the knob. (With the knob turned all the way clockwise, the variable resistor's resistance is nearly zero, and when it's all the way counterclockwise, its resistance is essentially infinite.) (a) Draw schematics. (b) Why would the simpler design be undesirable? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{toaster}{1}{}{26}A 1.0 $$\Omega$$ toaster and a 2.0 $$\Omega$$ lamp are connected in parallel with the 110-V supply of your house. (Ignore the fact that the voltage is AC rather than DC.)
(a) Draw a schematic of the circuit.
(b) For each of the three components in the circuit, find the current passing through it and the voltage drop across it. (answer check available at lightandmatter.com)
(c) Suppose they were instead hooked up in series. Draw a schematic and calculate the same things.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{wiregauges}{1}{}{27}Wire is sold in a series of standard diameters, called “gauges.” The difference in diameter between one gauge and the next in the series is about 20%. How would the resistance of a given length of wire compare with the resistance of the same length of wire in the next gauge in the series? (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{two-switching-methods}{1}{}{28}The figure shows two possible ways of wiring a flashlight with a switch. Both will serve to turn the bulb on and off, although the switch functions in the opposite sense. Why is method 1 preferable? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{short-across-bulb}{1}{}{29}In the figure, the battery is 9 V.
(a) What are the voltage differences across each light bulb?(answer check available at lightandmatter.com)
(b) What current flows through each of the three components of the circuit? (answer check available at lightandmatter.com)
(c) If a new wire is added to connect points A and B, how will the appearances of the bulbs change? What will be the new voltages and currents?
(d) Suppose no wire is connected from A to B, but the two bulbs are switched. How will the results compare with the results from the original setup as drawn? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{circuit-puzzle}{1}{}{30}You have a circuit consisting of two unknown resistors in series, and a second circuit consisting of two unknown resistors in parallel.
(a) What, if anything, would you learn about the resistors in the series circuit by finding that the currents through them were equal?
(b) What if you found out the voltage differences across the resistors in the series circuit were equal?
(c) What would you learn about the resistors in the parallel circuit from knowing that the currents were equal?
(d) What if the voltages in the parallel circuit were equal? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{messy-wires}{1}{}{31}A student in a biology lab is given the following instructions: “Connect the cerebral eraser (C.E.) and the neural depolarizer (N.D.) in parallel with the power supply (P.S.). (Under no circumstances should you ever allow the cerebral eraser to come within 20 cm of your head.) Connect a voltmeter to measure the voltage across the cerebral eraser, and also insert an ammeter in the circuit so that you can make sure you don't put more than 100 mA through the neural depolarizer.” The diagrams show two lab groups' attempts to follow the instructions. (a) Translate diagram a into a standard-style schematic. What is correct and incorrect about this group's setup? (b) Do the same for diagram b. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{combineresistors}{1}{}{32}How many different resistance values can be created by combining three unequal resistors? (Don't count possibilities where not all the resistors are used.) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{extensioncord}{1}{1}{33}A person in a rural area who has no electricity runs an extremely long extension cord to a friend's house down the road so she can run an electric light. The cord is so long that its resistance, $$x$$, is not negligible. Show that the lamp's brightness is greatest if its resistance, $$y$$, is equal to $$x$$. Explain physically why the lamp is dim for values of $$y$$ that are too small or too large. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{combine-unequal-resistors}{1}{}{34}What resistance values can be created by combining a 1 $$\text{k}\Omega$$ resistor and a 10 $$\text{k}\Omega$$ resistor? (solution in the pdf version of the book) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{tetrahedron-resistors}{2}{}{35}Suppose six identical resistors, each with resistance $$R$$, are connected so that they form the edges of a tetrahedron (a pyramid with three sides in addition to the base, i.e., one less side than an Egyptian pyramid). What resistance value or values can be obtained by making connections onto any two points on this arrangement?(solution in the pdf version of the book) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{many-v-measurements}{1}{}{36} The figure shows a circuit containing five lightbulbs connected to a battery. Suppose you're going to connect one probe of a voltmeter to the circuit at the point marked with a dot. How many unique, nonzero voltage differences could you measure by connecting the other probe to other wires in the circuit? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{many-i-measurements}{1}{}{37}The lightbulbs in the figure are all identical. If you were inserting an ammeter at various places in the circuit, how many unique currents could you measure? If you know that the current measurement will give the same number in more than one place, only count that as one unique current. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{tricky-unlit-bulb}{2}{}{38}The bulbs are all identical. Which one doesn't light up? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{network}{2}{}{39}Each bulb has a resistance of one ohm. How much power is drawn from the one-volt battery? (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{junctions}{1}{}{40}The bulbs all have unequal resistances. Given the three currents shown in the figure, find the currents through bulbs A, B, C, and D. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{falling-charged-string}{1}{1}{41}A silk thread is uniformly charged by rubbing it with llama fur. The thread is then dangled vertically above a metal plate and released. As each part of the thread makes contact with the conducting plate, its charge is deposited onto the plate. Since the thread is accelerating due to gravity, the rate of charge deposition increases with time, and by time $$t$$ the cumulative amount of charge is $$q=ct^2$$, where $$c$$ is a constant. (a) Find the current flowing onto the plate.(answer check available at lightandmatter.com)
(b) Suppose that the charge is immediately carried away through a resistance $$R$$. Find the power dissipated as heat.(answer check available at lightandmatter.com)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{pe-of-point-charges}{1}{}{42} (a) Recall from example 9 on p. 333 that the gravitational energy of two gravitationally interacting spheres is given by $$PE=-Gm_1m_2/r$$, where $$r$$ is the center-to-center distance. Sketch a graph of $$PE$$ as a function of $$r$$, making sure that your graph behaves properly at small values of $$r$$, where you're dividing by a small number, and at large ones, where you're dividing by a large one. Check that your graph behaves properly when a rock is dropped from a larger $$r$$ to a smaller one; the rock should lose potential energy as it gains kinetic energy.
(b) Electrical forces are closely analogous to gravitational ones, since both depend on $$1/r^2$$. Since the forces are analogous, the potential energies should also behave analogously. Using this analogy, write down the expression for the electrical potential energy of two interacting charged particles. The main uncertainty here is the sign out in front. Like masses attract, but like charges repel. To figure out whether you have the right sign in your equation, sketch graphs in the case where both charges are positive, and also in the case where one is positive and one negative; make sure that in both cases, when the charges are released near one another, their motion causes them to lose PE while gaining KE.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{lightning-rod}{1}{}{43}In example 8 on p. 576, suppose that the larger sphere has radius $$a$$, the smaller one $$b$$. (a) Use the result of problem 15 show that the ratio of the charges on the two spheres is $$q_a/q_b=a/b$$. (b) Show that the density of charge (charge per unit area) is the other way around: the charge density on the smaller sphere is greater than that on the larger sphere in the ratio $$a/b$$. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{disguised-parallel}{1}{}{44}Find the current drawn from the battery. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{handson}{A}{Electrical measurements}{\onecolumn} 1. How many different currents could you measure in this circuit? Make a prediction, and then try it.

What do you notice? How does this make sense in terms of the roller coaster metaphor introduced in discussion question 21.5A on p. 571?

What is being used up in the resistor?

2. By connecting probes to these points, how many ways could you measure a voltage? How many of them would be different numbers? Make a prediction, and then do it.

What do you notice? Interpret this using the roller coaster metaphor, and color in parts of the circuit that represent constant voltages.

3. The resistors are unequal. How many different voltages and currents can you measure? Make a prediction, and then try it.

What do you notice? Interpret this using the roller coaster metaphor, and color in parts of the circuit that represent constant voltages. \end{handson}

\begin{handson}{B}{Voltage and current}{\twocolumn}

This exercise is based on one created by Virginia Roundy.

Apparatus:

DC power supply

1.5 volt batteries

lightbulbs and holders

wire

highlighting pens, 3 colors

When you first glance at this exercise, it may look scary and intimidating --- all those circuits! However, all those wild-looking circuits can be analyzed using the following four guides to thinking:

1. A circuit has to be complete , i.e., it must be possible for charge to get recycled as it goes around the circuit. If it's not complete, then charge will build up at a dead end. This built-up charge will repel any other charge that tries to get in, and everything will rapidly grind to a stop.

2. There is constant voltage everywhere along a piece of wire. To apply this rule during this lab, I suggest you use the colored highlighting pens to mark the circuit. For instance, if there's one whole piece of the circuit that's all at the same voltage, you could highlight it in yellow. A second piece of the circuit, at some other voltage, could be highlighted in blue.

3. Charge is conserved, so charge can't “get used up.”

4. You can draw a rollercoaster diagram, like the one shown below. On this kind of diagram, height corresponds to voltage --- that's why the wires are drawn as horizontal tracks.

A Bulb and a Switch

Look at circuit 1, and try to predict what will happen when the switch is open, and what will happen when it's closed. Write both your predictions in the table on the following page before you build the circuit. When you build the circuit, you don't need an actual switch like a light switch; just connect and disconnect the banana plugs. Use one of the 1.5 volt batteries as your voltage source.


 color meaning black 0 brown 1 red 2 orange 3 yellow 4 green 5 blue 6 violet 7 gray 8 white 9 silver ±10% gold ±5%
 color meaning black 0 brown 1 red 2 orange 3 yellow 4 green 5 blue 6 violet 7 gray 8 white 9 silver ±10% gold ±5%

} \newcommand{\circuittableonecol}{

 color meaning black 0 brown 1 red 2 orange 3 yellow 4 green 5 blue 6 violet 7 gray 8 white 9 silver ±10% gold ±5%

}

* \textsf{Circuit 1}

\circuittabletwocols{switch open}{switch closed}

Did it work the way you expected? If not, try to figure it out with the benefit of hindsight, and write your explanation in the table above.

* \textsf{Circuit 2 (Don't leave the switch closed for a long time!)}

\circuittabletwocols{switch open}{switch closed}

* \textsf{Circuit 3}

\circuittabletwocols{switch open}{switch closed}

* \textsf{Circuit 4}

\circuittabletwocols{switch open}{switch closed}

Two Bulbs

Instead of a battery, use the DC power supply, set to 2.4 volts, for circuits 5 and 6. Analyze this one both by highlighting and by drawing a rollercoaster diagram.

* \textsf{Circuit 5}

\circuittabletwocols{bulb a}{bulb b}

* \textsf{Circuit 6}

\circuittabletwocols{bulb a}{bulb b}

Two Batteries

Use batteries for circuits 7-9. Circuits 7 and 8 are both good candidates for rollercoaster diagrams.

* \textsf{Circuit 7}

\circuittableonecol

* \textsf{Circuit 8}

\circuittableonecol

A Final Challenge

* \textsf{Circuit 9}

\circuittabletwocols{bulb a}{bulb b}

\end{handson} \begin{handson}{C}{Reasoning about circuits}{\onecolumn} The questions in this exercise can all be solved using some combination of the following approaches:

a) There is constant voltage throughout any conductor.

b) Ohm's law can be applied to any part of a circuit.

c) Apply the loop rule.

d) Apply the junction rule.

In each case, discuss the question, decide what you think is the right answer, and then try the experiment.

If you've already done exercise 21B, skip number 1.

1. The series circuit is changed as shown.

Which reasoning is correct?

• Each bulb now has its sides connected to the two terminals of the battery, so each now has 2.4 V across it instead of 1.2 V. They get brighter.
• Just as in the original circuit, the current goes through one bulb, then the other. It's just that now the current goes in a figure-8 pattern. The bulbs glow the same as before.

2. A wire is added as shown to the original circuit.

What is wrong with the following reasoning?

The top right bulb will go out, because its two sides are now connected with wire, so there will be no voltage difference across it. The other three bulbs will not be affected.

3. A wire is added as shown to the original circuit.

What is wrong with the following reasoning?

The current flows out of the right side of the battery. When it hits the first junction, some of it will go left and some will keep going up The part that goes up lights the top right bulb. The part that turns left then follows the path of least resistance, going through the new wire instead of the bottom bulb. The top bulb stays lit, the bottom one goes out, and others stay the same.

4. What happens when one bulb is unscrewed, leaving an air gap?

5. This part is optional. You can do it if you finished early and would like an extra challenge.

Predict the voltage drop across each of the three bulbs in part 4, and also predict how the three currents will compare with one another. (You can't predict the currents in units of amperes, since you don't know the resistances of the bulbs.) Test your predictions. If your predictions are wrong, try to figure out what's going on.

\end{handson}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.