Contents

Current

Electric Current

Electric current is defined to be the rate at which charge flows. A large current, such as that used to start a truck engine, moves a large amount of charge in a small time, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge over a long period of time. In equation form, electric current $I I size 12{I } {}$ is defined to be

I = \frac{Δ Q}{Δ t}, I = \frac{Δ Q}{Δ t}, size 12{I =  {  {ΔQ}  over  {Δt} } ","} {}

20.1

where $Δ Q Δ Q size 12{ΔQ} {}$ is the amount of charge passing through a given area in time $Δ t Δ t size 12{Δt} {}$ . (As in previous chapters, initial time is often taken to be zero, in which case $Δ t = t Δ t = t size 12{Δt=t} {}$ .) (See Figure 20.2.) The SI unit for current is the ampere (A), named for the French physicist André-Marie Ampère (1775–1836). Since $I = Δ Q / Δ t I = Δ Q / Δ t size 12{I = ΔQ/Δt} {}$ , we see that an ampere is one coulomb per second:

1 A = 1 C/s 1 A = 1 C/s size 12{"1 A "=" 1 C/s"} {}

20.2

Not only are fuses and circuit breakers rated in amperes (or amps), so are many electrical appliances.

Charges are shown as small spheres moving through a section of a conducting wire. The direction of movement of charge is indicated by arrows along the length of the conductor toward the right. The cross-sectional area of the wire is labeled as A. The current is equal to the flow of charge.

Figure 20.2 The rate of flow of charge is current. An ampere is the flow of one coulomb through an area in one second.

Example 20.1

Calculating Currents: Current in a Truck Battery and a Handheld Calculator

(a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How long does it take 1.00 C of charge to flow through a handheld calculator if a 0.300-mA current is flowing?

Strategy

We can use the definition of current in the equation $I = Δ Q / Δ t I = Δ Q / Δ t size 12{I = ΔQ/Δt} {}$ to find the current in part (a), since charge and time are given. In part (b), we rearrange the definition of current and use the given values of charge and current to find the time required.

Solution for (a)

Entering the given values for charge and time into the definition of current gives

\begin{array}{l} I & = & \frac{Δ Q}{Δ t} = \frac{720 C}{4.00 s} = 180 C/s \\ = & 180 A. \end{array} \begin{array}{l} I & = & \frac{Δ Q}{Δ t} = \frac{720 C}{4.00 s} = 180 C/s \\ = 180 A. \end{array}

20.3

Discussion for (a)

This large value for current illustrates the fact that a large charge is moved in a small amount of time. The currents in these “starter motors” are fairly large because large frictional forces need to be overcome when setting something in motion.

Solution for (b)

Solving the relationship $I = Δ Q / Δ t I = Δ Q / Δ t size 12{I = ΔQ/Δt} {}$ for time $Δ t Δ t size 12{Δt} {}$ , and entering the known values for charge and current gives

\begin{array}{l} Δ t & = & \frac{Δ Q}{I} = \frac{1.00 C}{0.300 \times 10^{- 3} C/s} \\ = & 3.33 \times 10^{3} s. \end{array} \begin{array}{l} Δ t & = & \frac{Δ Q}{I} = \frac{1.00 C}{0.300 \times 10^{- 3}} \\ = 3.33 \times 10^{3} \end{array}

20.4

Discussion for (b)

This time is slightly less than an hour. The small current used by the hand-held calculator takes a much longer time to move a smaller charge than the large current of the truck starter. So why can we operate our calculators only seconds after turning them on? It’s because calculators require very little energy. Such small current and energy demands allow handheld calculators to operate from solar cells or to get many hours of use out of small batteries. Remember, calculators do not have moving parts in the same way that a truck engine has with cylinders and pistons, so the technology requires smaller currents.

Figure 20.3 shows a simple circuit and the standard schematic representation of a battery, conducting path, and load (a resistor). Schematics are very useful in visualizing the main features of a circuit. A single schematic can represent a wide variety of situations. The schematic in Figure 20.3 (b), for example, can represent anything from a truck battery connected to a headlight lighting the street in front of the truck to a small battery connected to a penlight lighting a keyhole in a door. Such schematics are useful because the analysis is the same for a wide variety of situations. We need to understand a few schematics to apply the concepts and analysis to many more situations.

Part a shows a bulb glowing when its terminals are connected to a battery through a wire. The voltage of the battery is labeled as V. The current through the bulb is represented as I, and the current direction is shown using arrows emerging from the positive terminal of the battery, passing through the bulb, and entering the negative terminal of the battery. Part b shows an electric circuit diagram with a resistance connected across the terminals of a battery of voltage V. The current is shown using arrows as emerging from the positive terminal of the battery, passing through the resistance, and entering the negative terminal of the battery.

Figure 20.3 (a) A simple electric circuit. A closed path for current to flow through is supplied by conducting wires connecting a load to the terminals of a battery. (b) In this schematic, the battery is represented by the two parallel red lines, conducting wires are shown as straight lines, and the zigzag represents the load. The schematic represents a wide variety of similar circuits.

Note that the direction of current flow in Figure 20.3 is from positive to negative. The direction of conventional current is the direction that positive charge would flow. Depending on the situation, positive charges, negative charges, or both may move. In metal wires, for example, current is carried by electrons—that is, negative charges move. In ionic solutions, such as salt water, both positive and negative charges move. This is also true in nerve cells. A Van de Graaff generator used for nuclear research can produce a current of pure positive charges, such as protons. Figure 20.4 illustrates the movement of charged particles that compose a current. The fact that conventional current is taken to be in the direction that positive charge would flow can be traced back to American politician and scientist Benjamin Franklin in the 1700s. He named the type of charge associated with electrons negative, long before they were known to carry current in so many situations. Franklin, in fact, was totally unaware of the small-scale structure of electricity.

It is important to realize that there is an electric field in conductors responsible for producing the current, as illustrated in Figure 20.4. Unlike static electricity, where a conductor in equilibrium cannot have an electric field in it, conductors carrying a current have an electric field and are not in static equilibrium. An electric field is needed to supply energy to move the charges.

Making Connections: Take-Home Investigation—Electric Current Illustration

Find a straw and little peas that can move freely in the straw. Place the straw flat on a table and fill the straw with peas. When you pop one pea in at one end, a different pea should pop out the other end. This demonstration is an analogy for an electric current. Identify what compares to the electrons and what compares to the supply of energy. What other analogies can you find for an electric current?

Note that the flow of peas is based on the peas physically bumping into each other; electrons flow due to mutually repulsive electrostatic forces.

In part a, positive charges move toward the right through a conducting wire. The direction of movement of charge is indicated by arrows along the length of the wire. The area of a cross section of the wire is labeled as A. The direction of the electric field E is toward the right, in the same direction as movement of positive charge. The current direction is also toward the right, shown by an arrow. In part b, negative charges move toward the left through a conducting wire. The direction of movement of charge is indicated by arrows along the length of the wire. The area of a cross section of the wire is labeled as A. The direction of the electric field E is toward the right, opposite the direction of movement of negative charge. The current direction is also toward the right, shown by an arrow.

Figure 20.4 Current

I I size 12{I } {}

is the rate at which charge moves through an area

A A

, such as the cross-section of a wire. Conventional current is defined to move in the direction of the electric field. (a) Positive charges move in the direction of the electric field and the same direction as conventional current. (b) Negative charges move in the direction opposite to the electric field. Conventional current is in the direction opposite to the movement of negative charge. The flow of electrons is sometimes referred to as electronic flow.

Example 20.2

Calculating the Number of Electrons that Move through a Calculator

If the 0.300-mA current through the calculator mentioned in the Example 20.1 example is carried by electrons, how many electrons per second pass through it?

Strategy

The current calculated in the previous example was defined for the flow of positive charge. For electrons, the magnitude is the same, but the sign is opposite, $I_{electrons} = - 0.300 \times 10^{-3} C/s I_{electrons} = - 0.300 \times 10^{-3} C/s$ .Since each electron $(e^{-}) (e^{-})$ has a charge of $–1 . 60 \times 10^{- 19} C –1 . 60 \times 10^{- 19}$ , we can convert the current in coulombs per second to electrons per second.

Solution

Starting with the definition of current, we have

I_{electrons} = \frac{Δ Q_{electrons}}{Δ t} = \frac{–0 . 300 \times 10^{- 3} C}{s} . I_{electrons} = \frac{Δ Q_{electrons}}{Δ t} = \frac{–0 . 300 \times 10^{- 3} C}{s} . size 12{I =  {  {ΔQ}  over  {Δt} }  =  {  {0 "." "300 " times " 10" rSup { size 8{ - 3} } " C"}  over  {"s"} } "."} {}

20.5

We divide this by the charge per electron, so that

\begin{array}{l} \frac{e^{–}}{s} & = & \frac{–0 . 300 \times 10^{- 3} C}{s} \times \frac{1 e^{–}}{–1 .60 \times 10^{- 19} C} \\ = & 1.88 \times 10^{15} \frac{e^{–}}{s} . \end{array} \begin{array}{l} \frac{e^{–}}{s} & = & \frac{–0 . 300 \times 10^{- 3}}{s} \times \frac{1}{–1 .60 \times 10^{- 19}} \\ = 1.88 \times 10^{15} \end{array}

20.6

Discussion

There are so many charged particles moving, even in small currents, that individual charges are not noticed, just as individual water molecules are not noticed in water flow. Even more amazing is that they do not always keep moving forward like soldiers in a parade. Rather they are like a crowd of people with movement in different directions but a general trend to move forward. There are lots of collisions with atoms in the metal wire and, of course, with other electrons.

Drift Velocity

Electrical signals are known to move very rapidly. Telephone conversations carried by currents in wires cover large distances without noticeable delays. Lights come on as soon as a switch is flicked. Most electrical signals carried by currents travel at speeds on the order of $10^{8} m/s 10^{8} size 12{"10" rSup { size 8{8} } `"m/s"} {}$ , a significant fraction of the speed of light. Interestingly, the individual charges that make up the current move much more slowly on average, typically drifting at speeds on the order of $10^{- 4} m/s 10^{- 4} size 12{"10" rSup { size 8{ - 4} } `"m/s"} {}$ . How do we reconcile these two speeds, and what does it tell us about standard conductors?

The high speed of electrical signals results from the fact that the force between charges acts rapidly at a distance. Thus, when a free charge is forced into a wire, as in Figure 20.5, the incoming charge pushes other charges ahead of it, which in turn push on charges farther down the line. The density of charge in a system cannot easily be increased, and so the signal is passed on rapidly. The resulting electrical shock wave moves through the system at nearly the speed of light. To be precise, this rapidly moving signal or shock wave is a rapidly propagating change in electric field.

Negatively charged electrons move through a conducting wire. Two electrons are shown entering the wire from one end, and two electrons are shown leaving the wire at the other end. The direction of movement of charge is indicated by arrows along the length of the wire toward the right. Some electrons are shown inside the wire.

Figure 20.5 When charged particles are forced into this volume of a conductor, an equal number are quickly forced to leave. The repulsion between like charges makes it difficult to increase the number of charges in a volume. Thus, as one charge enters, another leaves almost immediately, carrying the signal rapidly forward.

Good conductors have large numbers of free charges in them. In metals, the free charges are free electrons. Figure 20.6 shows how free electrons move through an ordinary conductor. The distance that an individual electron can move between collisions with atoms or other electrons is quite small. The electron paths thus appear nearly random, like the motion of atoms in a gas. But there is an electric field in the conductor that causes the electrons to drift in the direction shown (opposite to the field, since they are negative). The drift velocity $v_{d} v_{d} size 12{v rSub { size 8{d} } } {}$ is the average velocity of the free charges. Drift velocity is quite small, since there are so many free charges. If we have an estimate of the density of free electrons in a conductor, we can calculate the drift velocity for a given current. The larger the density, the lower the velocity required for a given current.

The diagram shows a section of a conducting wire. A free electron is shown in the wire, and the path of the electron is shown as zigzag arrows along the length of the wire. The path is shown beginning at one end of the wire and ending at the other end. The drift velocity, v sub d, is indicated by an arrow toward the right, opposite the direction of the electric field E and the current I.

Figure 20.6 Free electrons moving in a conductor make many collisions with other electrons and atoms. The path of one electron is shown. The average velocity of the free charges is called the drift velocity,

v_{d} v_{d} size 12{v rSub { size 8{d} } } {}

, and it is in the direction opposite to the electric field for electrons. The collisions normally transfer energy to the conductor, requiring a constant supply of energy to maintain a steady current.

Conduction of Electricity and Heat

Good electrical conductors are often good heat conductors, too. This is because large numbers of free electrons can carry electrical current and can transport thermal energy.

The free-electron collisions transfer energy to the atoms of the conductor. The electric field does work in moving the electrons through a distance, but that work does not increase the kinetic energy (nor speed, therefore) of the electrons. The work is transferred to the conductor’s atoms, possibly increasing temperature. Thus a continuous power input is required to keep a current flowing. An exception, of course, is found in superconductors, for reasons we shall explore in a later chapter. Superconductors can have a steady current without a continual supply of energy—a great energy savings. In contrast, the supply of energy can be useful, such as in a lightbulb filament. The supply of energy is necessary to increase the temperature of the tungsten filament, so that the filament glows.

Making Connections: Take-Home Investigation—Filament Observations

Find a lightbulb with a filament. Look carefully at the filament and describe its structure. To what points is the filament connected?

We can obtain an expression for the relationship between current and drift velocity by considering the number of free charges in a segment of wire, as illustrated in Figure 20.7. The number of free charges per unit volume is given the symbol $n n size 12{n} {}$ and depends on the material. The shaded segment has a volume $Ax Ax size 12{ ital "Ax"} {}$ , so that the number of free charges in it is $nAx nAx size 12{ ital "nAx"} {}$ . The charge $Δ Q Δ Q size 12{DQ} {}$ in this segment is thus $qnAx qnAx size 12{ ital "qnAx"} {}$ , where $q q size 12{q} {}$ is the amount of charge on each carrier. (Recall that for electrons, $q q size 12{q} {}$ is $- 1 . 60 \times 10^{- 19} C - 1 . 60 \times 10^{- 19} size 12{ - 1 "." "60" times "10" rSup { size 8{ - "19"} } "C"} {}$ .) Current is charge moved per unit time; thus, if all the original charges move out of this segment in time $Δ t Δ t size 12{Dt} {}$ , the current is

I = \frac{Δ Q}{Δ t} = \frac{qnAx}{Δ t} . I = \frac{Δ Q}{Δ t} = \frac{qnAx}{Δ t} .size 12{I =  {  {ΔQ}  over  {Δt} }  =  {  { ital "qnAx"}  over  {Δt} } "."} {}

20.7

Note that $x / Δ t x / Δ t size 12{x/Δt} {}$ is the magnitude of the drift velocity, $v_{d} v_{d}$ , since the charges move an average distance $x x size 12{x} {}$ in a time $Δ t Δ t size 12{Dt} {}$ . Rearranging terms gives

I = {nqAv}_{d}, I = {nqAv}_{d}, size 12{I=  ital "nqAv" rSub { size 8{"d"} } } {}

20.8

where $I I size 12{I } {}$ is the current through a wire of cross-sectional area $A A size 12{A} {}$ made of a material with a free charge density $n n size 12{n} {}$ . The carriers of the current each have charge $q q size 12{q} {}$ and move with a drift velocity of magnitude $v_{d} v_{d} size 12{v rSub { size 8{d} } } {}$ .

Charges are shown moving through a section of a conducting wire. The charges have a drift velocity v sub d along the length of the wire, shown by an arrow pointing to the right. The volume of a segment of the wire is equal to A times x, where x equals the product of the drift velocity, v sub d, and time t. A cross section of the wire is marked as A, and the length of the section is x.

Figure 20.7 All the charges in the shaded volume of this wire move out in a time

t t size 12{t} {}

, having a drift velocity of magnitude

v_{d} = x / t v_{d} = x / t size 12{v rSub { size 8{d} } =x/t} {}

. See text for further discussion.

Note that simple drift velocity is not the entire story. The speed of an electron is much greater than its drift velocity. In addition, not all of the electrons in a conductor can move freely, and those that do might move somewhat faster or slower than the drift velocity. So what do we mean by free electrons? Atoms in a metallic conductor are packed in the form of a lattice structure. Some electrons are far enough away from the atomic nuclei that they do not experience the attraction of the nuclei as much as the inner electrons do. These are the free electrons. They are not bound to a single atom but can instead move freely among the atoms in a “sea” of electrons. These free electrons respond by accelerating when an electric field is applied. Of course as they move they collide with the atoms in the lattice and other electrons, generating thermal energy, and the conductor gets warmer. In an insulator, the organization of the atoms and the structure do not allow for such free electrons.

Example 20.3

Calculating Drift Velocity in a Common Wire

Calculate the drift velocity of electrons in a 12-gauge copper wire (which has a diameter of 2.053 mm) carrying a 20.0-A current, given that there is one free electron per copper atom. (Household wiring often contains 12-gauge copper wire, and the maximum current allowed in such wire is usually 20 A.) The density of copper is $8 . 80 \times 10^{3} {kg/m}^{3} 8 . 80 \times 10^{3} size 12{8 "." "80" times "10" rSup { size 8{3} } `"kg/m" rSup { size 8{3} } } {}$ .

Strategy

We can calculate the drift velocity using the equation $I = {nqAv}_{d} I = {nqAv}_{d}$ . The current $I = 20.0 A I = 20.0 A$ is given, and $q = - 1.60 \times 10^{- 19} C q =$ is the charge of an electron. We can calculate the area of a cross-section of the wire using the formula $A = π r^{2}, A = π r^{2},$ where $r r$ is one-half the given diameter, 2.053 mm. We are given the density of copper, $8.80 \times 10^{3} {kg/m}^{3}, 8.80 \times 10^{3}$ and the periodic table shows that the atomic mass of copper is 63.54 g/mol. We can use these two quantities along with Avogadro’s number, $6.02 \times 10^{23} atoms/mol, 6.02 \times 10^{23}$ to determine $n, n,$ the number of free electrons per cubic meter.

Solution

First, calculate the density of free electrons in copper. There is one free electron per copper atom. Therefore, is the same as the number of copper atoms per $m^{3} m^{3}$ . We can now find $n n$ as follows:

\begin{array}{l} n & = & \frac{1 e^{-}}{atom} \times \frac{6 . 02 \times 10^{23} atoms}{mol} \times \frac{1 mol}{63 . 54 g} \times \frac{1000 g}{kg} \times \frac{8.80 \times 10^{3} kg}{{1 m}^{3}} \\ = & 8 . 342 \times 10^{28} e^{-} {/m}^{3} . \end{array} \begin{array}{l} n & = & \frac{1}{atom} \times \frac{6 . 02 \times 10^{23}}{mol} \times \frac{1 mol}{63 . 54 g} \times \frac{1000 g}{kg} \times \frac{8.80 \times 10^{3}}{{1 m}^{3}} \\ = 8 . 342 \times 10^{28} \end{array}

20.9

The cross-sectional area of the wire is

\begin{array}{l} A & = & π r^{2} \\ = & π {(\frac{2.053 \times 10^{−3} m}{2})}^{2} \\ = & 3.310 \times 10^{–6} m^{2} . \end{array} \begin{array}{l} A & = & π r^{2} \\ = π {(\frac{2.053 \times 10^{−3}}{2})}^{2} \\ = 3.310 \times 10^{–6} \end{array}

20.10

Rearranging $I = n q A v_{d} I = n q A v_{d}$ to isolate drift velocity gives

\begin{matrix} v_{d} = \frac{I}{nqA} \\ = \frac{20.0 A}{(8 . 342 \times 10^{28} {/m}^{3}) (–1 . 60 \times 10^{–19} C) (3 . 310 \times 10^{–6} m^{2})} \\ = –4 . 53 \times 10^{–4} m/s. \end{matrix} \begin{matrix} v_{d} = \frac{I}{nqA} \\ = \frac{20.0 A}{(8 . 342 \times 10^{28} {/m}^{3}) (–1 . 60 \times 10^{–19}} \\ = –4 . 53 \times 10^{–4} \end{matrix}

20.11

Discussion

The minus sign indicates that the negative charges are moving in the direction opposite to conventional current. The small value for drift velocity (on the order of $10^{- 4} m/s 10^{- 4} size 12{"10" rSup { size 8{ - 4} } `"m/s"} {}$ ) confirms that the signal moves on the order of $10^{12} 10^{12} size 12{"10" rSup { size 8{"12"} } } {}$ times faster (about $10^{8} m/s 10^{8} size 12{"10" rSup { size 8{8} } `"m/s"} {}$ ) than the charges that carry it.