Download Lesson 4 – Circuits and Resistance

Lesson 4 – Circuits and Resistance
© Lawrence B. Rees 2007. You may make a single copy of this document for personal use without written permission.
4.0 Introduction
The earliest experiments with electricity dealt more with static charge than with electrical
currents. By the middle of the 18th Century, scientists began to think of electricity in terms of
current flow. The first models described electricity as the simultaneous flow of two different
kinds of fluids. Benjamin Franklin first proposed what is called the “single fluid model” of
electricity. He suggested that positive objects have an excess of electrical fluid and negative
objects have a deficit. Electrical fluid, he postulated, is attracted to normal matter but repels other
electrical fluid. Conductors were thought to be materials that are “porous” to electrical fluid so
the fluid would flow easily through them. Insulators, however, were more solid and prevented
fluid flow. Franklin then explained current as the excess fluid of a positive object flowing
through a conductor to fill the fluid-poor negative object.
Think About It
Franklin’s idea of electrical fluid correctly modeled the repulsion of two positively charged
objects and the attraction of positive and negative charges. It, however, could not explain the
repulsion of two negative charges. Explain these statements.
There were two problems with Franklin’s model. First, Franklin failed to recognize that
normal matter is composed of components that have electrical charge. This means that when
there is a deficit of the electric fluid (electrons), there is a surplus of fixed charges (protons) in
the remaining matter. Second, Franklin had no way of knowing whether the electric fluid was
positive or negative. His guess that the electric fluid was positive had a 50/50 chance of being
right. But, as luck would have it, his guess was wrong. Ever since, we have had to deal with the
fact that if electrons moving right in a wire, they produce a current that goes left in the wire.
Generally, we will simply ignore the fact that the “electric fluid” is negative, and speak of
positive charges flowing in wires, as if Franklin were right after all.
4.1 Electric Fields in Wires
A wire is just a long, slender conductor. You know that by using wires, you are able to
transfer electricity from one point to another. But you have probably not thought about just what
happens when you turn on a switch and let electrical current flow through a wire in a circuit.
To make an electrical circuit, we first have to have a source of power. We usually use one
of two power sources: a battery or a generator. The power in our wall outlets is produced by
generators at power plants, often located hundreds of miles away from us. Since generators are a
little more complicated than batteries, we’ll consider them later in the course.
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For our purposes, batteries are basically black boxes that provide a constant voltage. Pick up
a battery and look at it. Your battery will have two terminals, a positive terminal and a negative
terminal. (See Fig 4.1.)
+
Figure 4.1. A typical dry cell battery
Usually positive terminals are labeled with a + sign. Sometimes positive terminals are colored
red. Negative terminals may be labeled with a – and may be colored black. A car battery, for
example, has a + by the positive terminal and the wires attached to it are red. The negative
terminal has a – by it and the wires attached to it are black. If you look on the battery label you
should be able to find something that indicates the battery’s voltage. The typical batteries you
use in radios, CD players, and similar small electronic devices are 1.5 V batteries. By connecting
1.5 V batteries together, we can produces larger voltages. For example, if you take a 9 V battery
apart, you would find six small 1.5 V batteries inside. (Fig. 4.2)
Photograph of a “dissected” 9 V battery
(Not available)
Fig. 4.2. The inside of a 9 V battery
So what does the voltage of a battery mean? Remember, that voltage is closely related to
potential energy. In order to get potential energy in joules, we need to multiply the voltage by a
charge. In an electrical circuit, the charges that move around the circuit are electrons. Thus, if
your battery is a 1.5 V battery, the difference in potential energy as it moves along a wire from
the negative terminal to the positive terminal is
∆U wire = −e × 1.5 V = −1.5eV = −2.40 × 10 −19 J
2
We could have called your battery a 2.4 × 10 −19 J battery, but that’s a bit awkward. Instead, we
call it a 1.5 V battery, and mean the same thing. Potential energy decreases because the electron
moves in the direction of the force, like a ball falling toward the surface of the earth. On the
other hand, as the electron moves from the positive terminal to the negative terminal inside the
battery, the voltage increases by 1.5 V, so:
∆U battery = +e × 1.5 V = +1.5eV = +2.40 × 10 −19 J
Let’s take a small wire and attach one end to the positive terminal of your battery and
leave the other end dangling in mid air, as in Fig. 4.3. What happens now? The reason the
positive terminal of a battery has positive voltage is because it has a deficit of electrons. For an
explanation of why it has a positive voltage, we have to turn to chemistry. Some atoms just want
electrons more than others. The difference between electron energies in different types of atoms
can be exploited to make a battery. So when you touch your wire to the positive terminal of the
battery, it sucks some electrons out of the wire. Within a short period of time, charges in the wire
rearrange themselves – just like they do on any other conductor – to cancel out the electric field
inside the wire. This leaves a net positive charge on the outside surface of the wire. When the
wire becomes sufficiently charged, the battery cannot pull out any more electrons, and all the
electrons in the battery and wire stop moving. (Actually, the electrons do continue to move and
to move at rather high speeds, but they move in random directions, so there is no net flow of
electrons in the wire. See Sect. 4.9 for details.)
+
++++++++++++
Figure 4.3. A wire attached to the positive terminal of the battery.
Similarly, if we were to attach a wire to the negative terminal of the battery, the excess electrons
in the negative terminal would go into the wire until the negative charge on the wire would
become large enough that no more electrons would move onto the wire.
Now, we take a piece of wire and connect it to both terminals of the battery, as in Fig.
4.4. The negative terminal pushes electrons onto one end of the wire and the positive terminal
sucks electrons off the other end of the wire. On the inside of the wire the net charge is zero, as
all the excess charge moves to the surface of the wire.
Let’s think what happens as we move along the wire from the positive terminal of the
battery toward the negative terminal. Near the positive terminal, there is a large positive surface
charge on the wire. This surface charge gradually gets smaller as we move away from the
battery. In the middle of the wire, the surface charge goes to zero. As we approach the negative
terminal of the battery, the surface charge becomes increasingly negative. Electrons within the
wire near the positive terminal of the battery feel attraction both ways from the positive surface
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charge; however, the charge near the terminal is larger, so they feel a net force toward the
positive terminal. Toward the middle of the wire, electrons within the wire feel a bit of positive
charge on one side of them pulling them toward the positive terminal and a bit of negative charge
on the other side pushing them toward the positive terminal. Electrons near the negative terminal
feel a large negative charge near the terminal pushing them away from the negative terminal. So
electrons all along the wire experience a force pushing them toward the positive terminal. If
electrons are moving within the wire, there must be an electric field inside the wire that causes
electrons to continue to move.
– – –
–
–
+
+
+
+
+
–
–– – – –
+
+
++ + + +
Figure 4.4 A wire attached to both terminals of the battery.
Since the wire is a conductor, you may be wondering why it is that an electric field can
remain in the wire. In earlier lessons, we emphasized the fact that the electric field inside static
conductors is zero. The word “static” is important. If there is current inside a conductor, we
know the electric field is not zero, or electrons would soon stop moving along the wire because
they would lose their energy in collisions with atoms in the wire. We’ll consider these collisions
in detail in Sect. 4.2.
When electrons continue to flow around a circuit, we say that there is an electric current.
Current is defined as the total amount of charge that passes a particular point in the circuit per
unit time. We use the symbol I (or sometimes i) for current. The SI unit of current is the ampere
(A), or amp, as it is often called.
1 ampere =
1 coulomb
second
Thus, if 300 mC of charge pass through the battery each second, then the current is 300 mA.
Note that charge is neither created nor destroyed in the circuit. Hence, if the current is constant in
time, the current must be the same at every point around the circuit. In practice, even when the
current is changing in time, the current is very nearly the same at every point along the wire. In
more complicated circuits, there may be junctions where the current can split into two different
wires. In this case, we have to modify our result a little:
At any given instant, the current at any point between two junctions in a circuit is the same.
Note that the battery must provide the energy to pull electrons off from the end of the
wire that is positive, force them toward the negative terminal of the battery, and then push them
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onto the end of the wire that is already negative. As you might suspect, if we use a battery with a
higher voltage, it is more effective at this process, producing a larger current in the wire. We now
want to see how current and voltage are related in our circuit.
Things to remember:
• Current is the rate charge flows through a circuit.
∆q
∆t
• Electrons flow in the opposite direction to current flow.
• The voltage of an electron in a circuit is related to its potential energy by the equation
U = qV.
• A battery is a device that increases the potential energy of electrons.
I=
4.2
Resistance, Ohm’s Law, and Power in Circuits
If you actually connect a wire from one terminal of a battery to the other, you’ll discover
that the battery and the wire both start getting hot. As we mentioned in our discussion above,
electrons going around the circuit collide with the atoms in the wire and in the battery. What
happens to the electrons in a wire is similar to what happens to a basketball dropped from an
airplane. At first the basketball accelerates downward. As it does, the basketball collides with air
molecules, transferring energy to them. The air molecules cause a drag force on the basketball.
The faster the basketball goes, the greater the drag force becomes until eventually the drag force
and the gravitational force cancel each other out. At this point, the basketball reaches terminal
velocity. Then, all the work done by gravity – or all the gravitational potential energy that is lost
– is transferred to the air in the form of kinetic energy and heat. Similarly, electrons in the wire
soon reach a terminal velocity and all the energy provided by the battery is lost to heat in the
circuit. If you’re making a toaster, that’s a good thing. In a motor or a computer, you’d rather not
lose energy in this way. But like it or not, collisions limit the amount of current flowing in the
circuit.
We can calculate how much energy is lost to heat every second if we know the current in
the wire and the voltage of the battery. From the voltage, we can determine how much energy
each electron is given by the battery, and hence how much energy is lost to heat. From the
current, we know how many electrons pass through the battery each second. Hence
P = energy given each electron ×
(4.1 Power Equation)
1
charge
1
×
= (e × V ) × I ×
second charge of 1 electron
e
P = IV
where
P is the power dissipated by the resistor or the power provided by a battery in watts (W).
I is the current passing through the resistor or battery in amperes (A).
V is the voltage across the resistor or battery in volts (V).
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This very important result tells us that power in a circuit is just the current multiplied by the
voltage. If we use SI units, power is in units of watts (W). Note that since the power dissipated by
the resistor is the same as the power provided by the battery, we can use the same relationship
for either batteries or resistors.
Not surprisingly, it turns out that different wires produce different amounts of heat. The
amount of heat produced depends on the length of the wire, the diameter of the wire, and the
material from which it is made. A convenient way to parameterize this is to define the resistance
of a wire, R, as the ratio of the voltage across the wire to the current in the wire. This is a
quantity that can be measured easily using electrical meters. That is:
R≡
(4.2 Definition of resistance)
V
I
In SI units, resistance is measured in units of ohms (S).
1Ω =
1 volt
1 ampere
In terms of resistance, power then can be written in two equivalent ways:
P = I 2R =
(4.3)
V2
R
Measurements of resistance show that for many materials the value of the resistance is
very nearly constant over a large range of currents and voltages. If resistance is constant, we say
the material is “Ohmic” and satisfies Ohm’s law. Ohm’s law is just the definition of resistance
with R taken to be a constant, but it usually written in a slightly different form:
(4.4 Ohm’s law)
V = IR
V is the voltage drop across a resistor in V.
I is the current flowing through the resistor in A.
R is the resistance of the resistor in S.
Things to remember:
• The potential energy of electrons decreases when the electrons lose energy in collision
with atoms in matter. Matter is said to have resistance.
• Resistance is defined by the relation V = IR. When R is constant, this is termed “Ohm’s
law.”
• The power dissipated in a resistor is P = IV.
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4.3 Resistivity and Resistors
Resistivity
If I’m building a house and using 12 AWG (American Wire Gauge) copper wire with a
diameter of 2.05 mm, I would like to know how much heat I’m going to be producing if I have a
20 A current going through the circuit. To calculate the power loss in the wire, I need to know
the resistance of the wire. We’d like to calculate the resistance.
We earlier proposed that the resistance of the wire should depend on the material from
which it is made and on the length and diameter of the wire. To predict from first principles how
resistance depends on material requires us to know the quantum mechanics of solids, so we’ll
just assume that we can make experimental measurements to determine this dependence. But we
can predict how the resistance should depend on the size and shape of an object.
For simplicity, let us take a block of material in the shape of a rectangular prism (that is,
shaped like a box). When a current I passes through this material, the power loss is P = I 2R.
Now let us take two such blocks and place them end to end to make one block that is twice as
long as the original block. We’ll adjust the voltage applied to the materials to that the original
current I is still passing through each of the blocks. The power loss in the blocks will be twice
the original power loss. That is:
Pnew = 2 Pold
2 I 2 Rold = I 2 Rnew
⇒ Rnew = 2 Rold
On the other hand, we could put the same two blocks together side by side. If each block still has
current I passing through it, the total current passing through the new block is 2I. Also the power
dissipated in the two blocks is twice the power originally dissipated in one block alone. Hence:
Pnew = 2 Pold
2
2
I new
Rnew = (2 I old ) Rnew = 2 I old
Rold
2
⇒ Rnew =
1
Rold
2
7
Rnew = 2 Rold
I
Rnew =
I
1
Rold
2
Figure 4.5. The resistance of two identical blocks arranged in two different orientations.
In summary, if we double the length of a resistor, we double its resistance. If we double
the cross-sectional area of a resistor, we halve its resistance. We can generalize this to an
arbitrary block of constant cross-sectional area:
R=ρ
(4.5 Resistivity)
ρ
ℓ
A
R is the resistance of the block in S.
is the resistivity of the material in units of units of m
is the length of a resistor in m.
A is the cross-sectional area of the material in m2.
The resistivity is just a constant that depends on the material. The resistivities of a few common
conductors are listed in Table 4.1.
Table 4.1 Resistance properties of a few selected materials
Material
Resistivity at 20°C
( m)
Temperature coefficient of
resistivity (°C–1,T0 = 20°C)
Carbon (C)
3.5 × 10–5
–0.5 × 10–3
Aluminum (Al)
2.82 × 10–8
3.9 × 10–3
Copper (Cu)
1.7 × 10–8
3.9 × 10–3
Silver (Ag)
1.59 × 10–8
3.8 × 10–3
Gold (Au)
2.44 × 10–8
3.4 × 10–3
Another quantity that is often defined is conductivity. The conductivity of a material is
defined as the inverse of its resistivity:
8
σ=
(4.6 Conductivity)
1
ρ
F is the conductivity of a material in (S m)—1 .
ρ
is the resistivity of the material in m
In practice, both conductivity and resistivity are used, so you should be familiar with their
definitions.
Example 4.1 Resistance and power in 12 AWG wire
Now we can actually calculate the resistance of the 12 AWG wire we discussed at the
beginning of this section. Of course, we need to know the length of the wire to calculate its
resistance, so let’s just calculate the resistance of a one-meter length. The diameter of the wire is
2.05 mm, so the radius is half that.
R=ρ
ℓ
1m
= 1.7 × 10 −8 Ωm
= 3.24 × 10 −3 Ω
A
π (0.001025m) 2
The power dissipated in one meter of this wire when 20 A of current passes through it is:
P = I 2 R = (20 A) 2 × 3.24 × 10 −3 Ω = 1.30W
Most elements of a circuit have resistance. Wires, batteries, light bulbs, etc., all have
resistance. However, sometimes we actually make circuit elements for the express purpose of
providing resistance in a circuit. It might seem strange to add circuit elements that dissipate
power; however, resistors can control how much current flows in different parts of a circuit and
how much voltage is available for circuit elements. So resistors are really quite important. Most
resistors are made of graphite cylinders. Precision resistors are often made by winding long
pieces of thin wire around an insulating core.
Temperature Dependence of Resistance
It is generally more difficult for an electron to pass through a resistor if the atoms in the
resistor are vibrating rapidly than if they are vibrating slowly. For this reason, the resistance of
most materials increases with temperature. However, the resistance of some materials actually
decreases with temperature. One such material is graphite, which is used to make most resistors.
Graphite resistors are sometimes a problem in circuits because as soon as they start overheating,
their resistance drops, the current passing through them increases, more power loss occurs, and
the resistor soon burns up.
Whether resistance increases or decreases with temperature, resistance often follows a
simple linear relationship over a fairly wide range of temperatures. Let us assume that we
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measure the resistance of a resistor to be R0 at a temperature T0 . Then we can write resistance as
a function of temperature as:
(4.7 Temperature dependence of R)
R(T ) = R0 [1 + α (T − T0 )]
R(T) is the resistance as a function of temperature in S.
" is the temperature coefficient of resistivity and has units of S /EC.
Some values of " are listed in Table 4–1.
T is the temperature in either K or EC (depending on which units you use).
R0 is the resistance at temperature T0.
Things to remember:
• The resistance of a resistor is given in terms of its resistivity as R = ρ
ℓ
A
• Conductivity is the reciprocal of resistivity.
• The temperature dependence of resistance is given by R = R0 [1 + α (T − T0 )]
4.4 Resistance Circuits
We are now ready to analyze circuits with batteries, resistors, and wires. In circuits that
contain wires and resistors, the resistance of the wires is generally so small that we ignore it. If
there is no resistance in a wire, then Ohm’s law tells us that there is no voltage drop across the
wire, either.
Assume that the resistance of wires is zero and that the voltage drop across wires is also zero
in circuits.
As we mentioned, circuit theory was all founded on the notion that the charge carriers are
positive. Thus we think of positive charges leaving the positive terminal of a battery when, in
reality, negative charges flow into the positive terminal. In most cases, it doesn’t make any real
difference, but since it’s hard to fight convention, we will usually pretend charge carriers are
positive.
Current flows from positive to negative (outside the battery) in a circuit, whereas
electrons flow from negative to positive.
First, consider a wire attached to the positive terminal of a battery, as illustrated in Fig.
4.3. Since the wire is not connected to anything, the wire soon gains an excess of positive charge
and no more charge flows from the battery. (Remember, we think of positive charge flowing out
10
the positive terminal!) Now let’s connect the loose wire to a huge conductor. In fact, the biggest
conductor we can think of is the earth itself. When we connect wires to the earth, we say we
connect the wires to “ground” or that we “ground” the wires. Will a steady current now flow out
of the battery? The answer is no, because as positive charge leaves the battery, the inside of the
battery becomes negative and provides an attractive force for positive charge. We say that the
circuit is “open.” Current will only momentarily flow in any open circuit.
+
Figure 4.6. An open circuit.
Now, instead of attaching the wire to ground, we attach the wire from the positive
terminal to the negative terminal of the battery. The negative terminal pulls in positive charge, so
there is no longer a net negative charge in the battery, and charge quickly flows through the wire.
Now the circuit is closed, and current flows.
Note that we could accomplish the same thing by attaching wires from both the positive
and the negative terminal to ground, as shown in Fig. 4.7.The ground then takes the place of one
wire segment. Just as we usually say that the gravitational potential energy of the earth is zero at
the earth’s surface, we usually consider the potential energy of the electric charges to be zero at
ground. But remember that in both cases, this choice is completely arbitrary. If we have no
ground in our circuit, which is the usual case in our drawings at least, we often choose the
negative terminal of the battery to be the zero point of potential energy. Many circuit boards
have a ground wire that runs around the outside of the circuit board, just to make it easy to make
connections to zero volts when needed. In a car, the entire chassis is connected to the negative
terminal of the battery, so the chassis becomes the ground for all the car’s circuits.
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+
+
Figure 4.7. Two closed circuits
Now let’s consider a single resistor connected to a battery. The easiest way to depict circuits
is to draw them in a circuit diagram or schematic. In such a diagram, circuit elements are
represented with standard symbols. Figure 4.8 shows this circuit both pictorially and schematically.
R
+
V
Figure 4.8 A simple resistance circuit
An equivalent circuit can be made by breaking the lower wire joining the resistor to the
battery and connecting both loose ends to ground. This circuit is illustrated in Fig. 4.9.
+
Figure 4.9 A circuit equivalent to that of Fig. 4.8.
12
In order to understand how circuits work, it is often useful to draw a voltage diagram of
the circuit. This is just a graph of the voltage in the circuit drawn alongside the circuit elements.
Equivalently, we can think of a voltage diagram as a representation of the potential energy of the
(positive) charge carriers as they pass through the circuit. In the circuit above, the charge carriers
gain energy as they pass through the battery, and lose energy as they pass through the resistor.
Let the battery be a standard 1.5 V dry cell. Since we usually take the voltage at ground to be 0
V, our voltage diagram for the simple circuit would be as shown below:
+
Voltage
1.5 V
0V
Figure 4.10. A voltage diagram for the circuit of Fig. 4.9.
As we mentioned above, we may safely ignore any voltage drop across the wire because
the resistance of the wire is very small compared to that of the resistor. Hence, the voltage across
the resistor is just the voltage across the battery, 1.5 V. This, of course, is independent of the
value of the resistor. What is affected by the resistance, however, is the amount of current that
flows through the circuit. Ohm’s law tells us that V = IR, so if the resistance is larger, the current
is smaller through the circuit.
Example 4.2.
In the circuit shown in Fig. 4.10, the resistor has a value of 6.00 . Find the current in the
circuit, the power dissipated by the resistor, and the power provided by the battery. How much
energy does the battery have to provide in one hour?
V 1.50 V
=
= 0.25 A
R 6.00 Ω
P = IV = 0.25 A × 1.50 V = 0.375W
I=
E = P∆ t = 0.375W × 3600 s = 1350 J
Since P = IV for both the battery and the resistor, the power provided by the battery is the same
as the power dissipated by the resistor. Conservation of energy, of course, requires this to be true.
13
On the label of a rechargeable AA battery, it states that it has about 2500 mAh of energy.
How many joules is this?
A mAh is a milliamp hour. Thus
2500mAh = 2.5 Ah × 3600
s
= 9000 J
hr
Things to remember:
• Current can flow into (or out of) a “ground” in unlimited amounts without changing the
grounds potential energy. We say the earth is at a potential of 0 V.
• The power dissipated by resistors in a circuit is the same as the power provided by the
batteries.
• In a resistor, voltage drops in the direction of current flow.
4.5 Resistors in Series and Parallel
Now let us see what happens when we have more complicated arrangements of resistors. Of
course, there are many different ways we can attach multiple resistors to a battery. In practice,
however, there are two ways of joining resistors that are much more important than other ways.
These ways are called “series” and “parallel.”
Two resistors joined in series have the same current passing through both resistors. That
is, there can be no junctions between the resistors. If there were, it would allow part of the
current passing through one resistor to be diverted before it passes through the second resistor.
Examples of resistors connected in series are illustrated in Fig. 4.11.
Test to Determine if Two Resistors are Connected in Series
Look at the wire connecting the two resistors. Is there
anything at all (a circuit element or a junction) along this
wire?
yes
no
The resistors
are connected n
series.
The resistors
are NOT
connected n
series.
14
(a) Resistors in series.
(b) Resistors NOT in series.
Figure 4.11 Examples of resistors connected in series.
Two resistors in parallel, on the other hand, have the same voltage across both of them.
This means that there can be no voltage changes (no batteries or resistors) between
corresponding ends of two resistors if they are in parallel. Examples of resistors connected in
parallel are illustrated in Fig. 4.12.
15
Test to Determine if Two Resistors are Connected in Parallel
Look at the wire connecting one end of the first resistor to
one end of the second resistor. Is there a circuit element (a
junction is OK and usually there) along this wire?
no
yes
The resistors are NOT connected n parallel.
Look at the wire connecting the other end of the first
resistor to the other end of the second resistor. Is there a
circuit element along this wire?
no
yes
The resistors
are NOT
connected in
parallel.
The resistors
are connected
in parallel.
A
B
+
(a) Resistors in parallel.
(b) Resistors NOT in parallel.
Figure 4.12 Examples of resistors connected in parallel.
16
Note that combinations of resistors in a circuit are still simply resistances. They can do
nothing more in the circuit than impede the flow of current. For that reason, we could put a
single resistor between points A and B in Fig. 4.12 that would have the same effect as the
parallel combination of the two resistors. Such a resistor is called an “equivalent resistor.” One
of the nice things about series and parallel combinations of resistance is that we can easily
replace a circuit (on paper, at least) with a simpler circuit containing equivalent resistors. Before
we see how we can do this in the general case, however, let us first find some rules for series and
parallel combinations of resistors.
First, let’s draw a voltage diagram for a circuit containing two resistors in series. As seen in Fig.
4.13, the voltage across both resistors combined must sum to the voltage provided by the battery
in the circuit. Any current passing through one resistor must pass simultaneously through the
second resistor (and through the battery as well in this particular circuit).
+
Voltage
1.5 V
0V
Figure 4.13. Voltage diagram for two resistors in series.
We know that two resistors in series have the same current passing through them. When
we make an equivalent resistor out of two series resistors, the equivalent resistor will also have
the same current passing through it. The voltages of the separate resistors sum to equal the
voltage of the equivalent resistor. Applying Ohm’s law, we have:
V = V1 + V2 , I =
IR = IR1 + IR2
⇒ R = R1 + R2
17
V V1 V2
=
=
R R1 R2
We see that the equivalent resistance is just the sum of the individual resistances. Note that we
already derived a special case of this relationship in Sect. 4.3 when we calculated the resistance
of two blocks placed end to end.
Resistors in Series
V2, I, R2
V1, I, R1
I = I1 = I 2
V = V1 + V2
R = R1 + R2
V, I, R
Next, let us draw a voltage diagram for two resistors in parallel. In this case, there are two
different paths through the resistors to ground. I’ve drawn the voltage of the upper path with a
dotted line and the voltage of the lower path with a dashed line. The voltage drop across each
path, however, is the same, since the resistors are in parallel. This time, however, the current
from the battery splits into two paths, with more current flowing through the smaller of the two
resistors.
+
Voltage
1.5 V
0V
Figure 4.14. Voltage diagram for two resistors in parallel.
18
We know that two resistors in parallel have the same voltage across them. The equivalent
resistor will also have the same voltage across it. The currents of the separate resistors sum to
equal the current of the equivalent resistor. This gives:
I = I 1 + I 2 , V = IR = I 1 R1 = I 2 R2
V V
V
=
+
R R1 R2
⇒
1
1
1
=
+
R R1 R2
Thus, in parallel, the reciprocals of resistance add.
Resistors in Parallel
V, I1, R1
I = I1 + I 2
V =V1 =V 2
V, I, R
1
1
1
=
+
R R1 R 2
V, I2, R2
Things to remember:
• In series and parallel circuits, the current, voltage, and resistance of equivalent resistors is given
by:
I = I1 = I 2
I = I1 + I 2
V = V1 + V2
V = V1 = V2
R = R1 + R2
1
1
1
=
+
R R1 R2
4.6 Series-Parallel Reduction
There are two types of problems that we frequently encounter with resistor networks. In
the first type of problem, we are given a network of resistors and batteries and are asked to find
the voltages and currents of each circuit element. From this, we can also calculate the power
provided by each battery and the power dissipated by each resistor. In the second type of
problem, we are given a network of resistors and asked to find the equivalent resistance. Really
the two types of problems are very similar. In fact, it is often easiest to solve problems of the
second type by assuming that the resistor network is connected to a battery of arbitrary but
known voltage. As we will see in the next section, resistors need not be connected in series or
parallel combinations; however, when they are, the method of series-parallel reduction can be
19
employed to solve the problem rather easily. The algorithm is simple: first, find all the series and
parallel combinations of resistors and replace these combinations with equivalent resistors. Once
this is done, new series and parallel combinations will appear. Replace these with equivalent
resistors, and so on, until you can simplify the network no more. Probably the easiest way to
explain the procedure is to work through a typical example.
____________________________________________
Example 4.3. Find all the voltages and currents in the
illustrated circuit.
The first thing we do is scan the circuit for series and
parallel combinations. The 7 and the 5 resistor are in
series and the 4 and the 6 resistor are in parallel. (There
are no other resistors in series or parallel. You should take a
moment to convince yourself of that.) First we find the
equivalent resistances for these two combinations.
+
12 V
7
4
1
6
12
1
1
1
5
=
+
=
, RA = Ω
R A 4Ω 6Ω 12Ω
5
5
RB = 7Ω + 5Ω = 12Ω
+
The circuit then reduces to:
12 V
12/5
1
Now, we have the two right resistors in parallel, and can combine them:
1
5
1
6
=
+
=
RC 12Ω 12Ω 12Ω
+
12 V
RC = 2Ω
1
The circuit now is:
20
12
2
+
These clearly reduce to one 3 resistor. The voltage across
this single resistor is 12V, so the current by Ohm’s law is
4A.
12 V
3
Now we can work our way back, one step at a time, to find the currents and voltages
across each resistor. We only need Ohm’s law and the result that the current through resistors in
series is the same as the current through the equivalent resistor, and the voltages across resistors
in parallel is the same as the voltage across the equivalent resistor. This gives:
12 V
+
4A
48 W
+
12 V
4A
48 W
12 V +
4A
48 W
1
4A
4V
16 W
3
12 V
4A
48 W
2
4A
8V
32 W
1
4A
4V
16 W
12 V +
4A
48 W
12
8V
2/3 A
16/3 W
1
4A
4V
16 W
12/5
8V
10/3 A
80/3 W
4
8V
2A
16 W
5
2/3 A
10/3 V
20/9 W
21
7
2/3 A
14/3 V
28/9 W
6
8V
4/3 A
32/3 W
Note that the total power dissipated is
32
20
28
16 W + 16 W + W + W + W = 48W
3
9
9
the same as the power supplied by the battery.
Things to remember:
• Study this example carefully and be able to apply series-parallel reduction to circuits.
4.7 Real Batteries
The voltage of a battery is entirely determined by the chemical reactions in the battery. Even
if a battery is nearly dead, a measurement with a high quality voltmeter will always yield
essentially the same number. What a dead battery cannot do is maintain that voltage once a load
is applied. Let’s see why this is true.
+
V
R
Rb
Figure 4.15 A real battery in series with a load resistor.
To this point, we have treated a battery as if it were a pure source of voltage. In reality, a
battery also has internal resistance. That is, a true battery can be considered as a series
combination of an idealized battery and a resistor. The voltage across the physical battery is
smaller than the voltage across the idealized battery by an amount equal to the voltage across the
internal resistance in the battery. (We know that the voltage across the resistors drop as shown in
Fig. 4.16 because voltage always drops in the direction of the current.) Using Ohm’s law, we see
that the effective voltage of the battery is:
Vb = V − IRb
I=
V
V
=
Rtotal R + Rb
Vb = V −
R
V
Rb = V
R + Rb
R + Rb
22
Current
+
Voltage
V
Vb
V
0V
Figure 4.16 Energy diagram of the circuit in Fig. 4.15
If there is no load on the battery ( R
), then the effective voltage is V, the nominal
voltage of the battery. But when a load is applied, we see that the voltage drops. The larger the
internal resistance of the battery, the greater is the voltage drop. As a battery gets old, its internal
resistance increases, causing the voltage of the battery under load to drop.
∞
→
Things to remember:
• Batteries have internal resistance.
• The voltage of a battery is divided between the load and the internal resistance.
• As batteries age, their internal resistance increases.
4.8 Kirchoff’s Laws
Some circuits, even simple circuits, cannot be solved by using series-parallel reduction.
An example is shown in Fig. 4.17. In this case, we can resort to two fundamental ideas that are
known as Kirchoff’s circuit laws.
23
12 V
+
+
4
2
1
6V
Figure 4.17 A circuit that cannot be reduced using series-parallel reduction.
Kirchoff’s Laws
1. Current coming into a junction must leave the junction.
2. The net change in voltage around any closed loop in a circuit must equal zero.
The first law is just conservation of charge. Charge can neither be created nor destroyed,
and it can not build up at a junction either. Therefore, the same amount of current that goes into a
junction must also leave it.
The second law is just a statement of the fact that, at any time, there is a well defined
voltage (relative to ground) at each point in the circuit. Therefore, when we draw an energy
diagram around a complete loop, we end up at the same voltage at which we started.
Kirchoff’s laws are conceptually simple – perhaps even obvious, but sometimes their
application can be a little difficult. The following outline is a suggested approach to solving a
circuit problem with Kirchoff’s laws.
1. Mark each junction in the circuit (any place wires intersect) with a circle.
12 V
1
+
+
2
24
6V
4
2. Select a direction for each current to travel and indicate these with arrows on the diagram.
Every segment of wire starting and ending on a junction must have a different current. It is not
important if you choose a current in the wrong direction. If you choose a current direction wrong,
you will get a negative number when you solve for the currents.
+
+
12 V
6V
I2
I1
4
2
1
I3
3. Write down the junction equations. If there are N junctions in the circuit, there will be N–1
independent junction equations. In this circuit, there are two junctions, so we only write one
junction equation. (Note that the second junction gives exactly the same equation in this simple
case.) We choose to use the top junction.
Current In = Current Out
I1 + I 2 = I 3
4. Put a + sign and a – sign on each resistor as a reminder that the voltage drops in the direction
of the current flow. Also be sure + and – signs are on each battery.
12 V
+
+
−
−
−
6V
−
I1
1
+
−
−
I2
2
+
I3
4
+
5. Draw as many loops as necessary to obtain enough equations to solve the system. Be sure that
every circuit element is in at least one loop. In this case, we have three currents, so we need three
equations. We have only one junction equation, so we must choose two loops for loop equations.
25
12 V
+
+
−
−
−
6V
−
I1
+
1
−
−
I2
2
+
I3
4
+
6. Using Ohm’s law for resistors (write IR) and remembering that the sum of all voltages around
any loop must be zero, write the loop equations. Be careful to get each sign correct: if you go +
to – across a battery or resistor, the sign is negative, if you go – to +, the sign is positive. If all
units are SI units, you may drop them for convenience.
Left loop :
− 12 + 1I 1 − 2 I 2 + 6 = 0
Right loop :
− 6 + 2I 2 + 4I 3 = 0
We can now obtain the currents by solving the three equations in three unknowns by
whatever method you like. The result is:
I1 =
24
9
15
A, I 2 = − A, I 3 =
A
7
7
7
Note that the sign of I2 is negative. This means that current is actually flowing downward
through the 6V battery. The 12V battery is strong enough that it forces current backward through
the 6V battery.
As a final exercise, let’s draw voltage diagrams for each loop, starting at the upper
junction. We will call the voltage at this junction 12 V so that the negative terminal of the left
battery will be 0V. For these diagrams, we’ll reverse the + and – signs on the 2 resistor in view
of the fact that the current is actually reversed from our original guess.
26
Left loop:
12V
1
+
−
2
+
+ −
−
6V
12 V
6V
24/7 V
0V
Right loop:
6V
+
2
−
4
+
− −
12 V
6V
24/7 V
0V
27
+
−
+
If we want, we could also have chosen a loop around the outside of circuit. The energy diagram
for this circuit looks like:
12V
+
−
1
4
+ −
−
+
12 V
6V
24/7 V
0V
Kirchoff’s laws can be used to solve for currents in any system of resistors and batteries.
Additionally, the laws still apply to cases where capacitors and other circuit elements are in a
circuit as well.
Things to remember:
• Current coming into a junction must leave the junction.
• The net change in voltage around any closed loop in a circuit must equal zero.
• If there are N junctions, there are N –1 independent junction equations.
• The remaining equations are loop equations. There must be at least one loop covering
every circuit element.
• You need to be able to set up a series of equations that can be solved for the currents,
but you do not need to solve those equations, except in very simple cases.
4.9 Drift Speed and Electron Velocity
If we want to get a rough idea of the speed of electrons, we can look at Bohr’s simple model
of a hydrogen atom where an electron orbits a proton in a circular orbit. Let us assume the
electron is at the Bohr radius, rb = 5.292 × 10 −11 m , and that it obeys Newtonian mechanics. Since
the Coulomb force is the real force that holds the electron in a circular orbit, we may equate it to
the general expression for the centripetal force:
28
Fcentripetal
mv 2 ke 2
=
= 2 = FCoulomb
rb
rb
8.99 × 10 9
k
−19
= 1.602 × 10 ×
m / s = 2.19 × 10 6 m / s
v=e
−31
−11
mrb
9.11 × 10 × 5.292 × 10
Although we should take this with a grain of salt, it is clear that electrons in atoms move quite
rapidly. There should not be surprising because electrons are so light that they are very easily
accelerated to great speeds.
In a metal, the conduction electrons are not really attached to individual atoms; however,
the thermal energy of the crystal lattice keeps them bouncing around. If you’ve had Physics 123,
you may recognize that the average kinetic energy of an electron in the metal lattice is, by the
equipartition theorem, equal to 32 kT where k is Boltzmann’s constant and T is the temperature in
K. We can then get a rough estimate of an electron’s velocity from this:
3kT
3 × 1.38 × 10 −23 × 300
=
m / s = 1.17 × 10 5 m / s
− 31
m
9.11 × 10
This is still quite a respectable speed.
1
2
mv 2 = 32 kT , v =
However, it turns out that an electron can’t travel very far without interacting with
another charged particle, usually another electron. So, even when there is no voltage on a wire,
an electron is moving around with a large average speed, but zero average velocity. That is, it
bounces around a lot, but it isn’t going anywhere. When we apply a voltage to a wire, however,
the electrons are put into motion along the wire with an average velocity that no longer equals
zero. This velocity is called the “drift speed” of electrons in the wire. For simplicity, we usually
think of all the electrons moving down the wire with an identical velocity equal to this drift
speed.
It turns out that with a little ingenuity, we can estimate the drift speed of electrons in a
wire. Let us take the case of a 20 A of current flowing in 12 AWG copper wire, the same values
we used in Example 4.1 above.
First, We calculate the total number of valence electrons in a one-meter length of this
wire.
29
Mass of 1 m of wire in g = V of 1 m of wire in cm 3 × density in
= π r 2 Lρ = π (0.1025cm ) × 100cm × 8.92
2
g
cm 3
g
= 294 g
cm 3
Moles of 1 m of wire = mass of 1 m of wire in g ÷ atomic wight in
=
g
mole
29.4 g
= 0.463 moles
63.546 g / mole
Cu atoms in 1 m of wire = moles in 1 m of wire × Avagadro' s number in
atoms
mole
atoms
= 2.79 × 10 23 atoms
mole
Free electrons in 1 m of wire = Cu atoms in 1 m of wire× valence electrons per atom
= 0.463 moles × 6.023 × 10 23
= 2.79 × 10 23 atoms × 1 = 2.79 × 10 23 electrons
Then we calculate how many electrons go past a point on the wire each second:
number of electrons
coulombs
1
= current in
×
second
coulombs / electron
second
1
= 20 A ×
= 1.25 × 10 20 electrons / second
−19
1.602 × 10 C
If we divide the number of electrons per second moving past a point in the wire by the
number of electrons per meter of wire, we get the length of wire in which that number of
electrons is found:
d=
1.25 × 10 20
= 4.47 × 10 −4 m .
23
2.79 × 10
In other words, the drift speed of the electrons is about a half millimeter per second!
Think About It
If electrons move very slowly down a wire, why does it take such a short time for a
light to turn on after we flip a light switch?
The velocity of electrical signals in wires is measured to be about 2×108 m/s, or
roughly 2/3 the speed of light in vacuum.
Things to remember:
• The drift speed of electrons is very slow (~10–4 m/s).
30