27.6 electrical power
while the chemical potential energy in the battery decreases by the same amount.
(Recall from Eq. 25.3 that DU 5 q DV.) As the charge moves from c to d through the
resistor, however, the electric potential energy of the system decreases due to colli-
sions of electrons with atoms in the resistor. In this process, the electric potential
energy is transformed to internal energy corresponding to increased vibrational
motion of the atoms in the resistor. Because the resistance of the interconnect-
ing wires is neglected, no energy transformation occurs for paths bc and da. When
the charge returns to point a, the net result is that some of the chemical potential
energy in the battery has been delivered to the resistor and resides in the resistor as
internal energy E
associated with molecular vibration.
The resistor is normally in contact with air, so its increased temperature results
in a transfer of energy by heat Q into the air. In addition, the resistor emits thermal
, representing another means of escape for the energy. After some
time interval has passed, the resistor reaches a constant temperature. At this time,
the input of energy from the battery is balanced by the output of energy from the
resistor by heat and radiation, and the resistor is a nonisolated system in steady
state. Some electrical devices include heat sinks4 connected to parts of the circuit
to prevent these parts from reaching dangerously high temperatures. Heat sinks
are pieces of metal with many fins. Because the metal’s high thermal conductivity
provides a rapid transfer of energy by heat away from the hot component and the
large number of fins provides a large surface area in contact with the air, energy
can transfer by radiation and into the air by heat at a high rate.
Let’s now investigate the rate at which the electric potential energy of the system
decreases as the charge Q passes through the resistor:
where I is the current in the circuit. The system regains this potential energy when
the charge passes through the battery, at the expense of chemical energy in the bat-
tery. The rate at which the potential energy of the system decreases as the charge
passes through the resistor is equal to the rate at which the system gains inter-
nal energy in the resistor. Therefore, the power P, representing the rate at which
energy is delivered to the resistor, is
P 5 I DV
We derived this result by considering a battery delivering energy to a resistor. Equa-
tion 27.21, however, can be used to calculate the power delivered by a voltage source
to any device carrying a current I and having a potential difference DV between its
Using Equation 27.21 and DV 5 IR for a resistor, we can express the power deliv-
ered to the resistor in the alternative forms
When I is expressed in amperes, DV in volts, and R in ohms, the SI unit of power is
the watt, as it was in Chapter 8 in our discussion of mechanical power. The process
by which energy is transformed to internal energy in a conductor of resistance R is
often called joule heating;5 this transformation is also often referred to as an I2R loss.
4This usage is another misuse of the word heat that is ingrained in our common language.
5It is commonly called joule heating even though the process of heat does not occur when energy delivered to a resistor
appears as internal energy. It is another example of incorrect usage of the word heat that has become entrenched in
Pitfall Prevention 27.5
Charges Do Not Move all the Way
around a Circuit in a Short time
In terms of understanding the
energy transfer in a circuit, it is
useful to imagine a charge mov-
ing all the way around the circuit
even though it would take hours
to do so.
Pitfall Prevention 27.6
Misconceptions about Current
Several common misconceptions
are associated with current in a
circuit like that in Figure 27.11.
One is that current comes out
of one terminal of the battery
and is then “used up” as it passes
through the resistor, leaving
current in only one part of the
circuit. The current is actually
the same everywhere in the circuit.
A related misconception has the
current coming out of the resis-
tor being smaller than that going
in because some of the current
is “used up.” Yet another miscon-
ception has current coming out
of both terminals of the battery,
in opposite directions, and then
“clashing” in the resistor, deliver-
ing the energy in this manner.
That is not the case; charges flow
in the same rotational sense at all
points in the circuit.
Pitfall Prevention 27.7
Energy Is Not “Dissipated” In
some books, you may see Equation
27.22 described as the power “dissi-
pated in” a resistor, suggesting that
energy disappears. Instead, we say
energy is “delivered to” a resistor.
chapter 27 current and resistance
When transporting energy by electricity through power lines (Fig. 27.12), you
should not assume the lines have zero resistance. Real power lines do indeed have
resistance, and power is delivered to the resistance of these wires. Utility companies
seek to minimize the energy transformed to internal energy in the lines and maxi-
mize the energy delivered to the consumer. Because P 5 I DV, the same amount of
energy can be transported either at high currents and low potential differences or at
low currents and high potential differences. Utility companies choose to transport
energy at low currents and high potential differences primarily for economic rea-
sons. Copper wire is very expensive, so it is cheaper to use high-resistance wire (that
is, wire having a small cross-sectional area; see Eq. 27.10). Therefore, in the expres-
sion for the power delivered to a resistor, P 5 I2R, the resistance of the wire is fixed
at a relatively high value for economic considerations. The I2R loss can be reduced
by keeping the current I as low as possible, which means transferring the energy
at a high voltage. In some instances, power is transported at potential differences
as great as 765 kV. At the destination of the energy, the potential difference is usu-
ally reduced to 4 kV by a device called a transformer. Another transformer drops the
potential difference to 240 V for use in your home. Of course, each time the poten-
tial difference decreases, the current increases by the same factor and the power
remains the same. We shall discuss transformers in greater detail in Chapter33.
Q uick Quiz 27.5 For the two lightbulbs shown in Figure 27.13, rank the current
values at points a through f from greatest to least.
Example 27.4 Power in an Electric Heater
An electric heater is constructed by applying a potential difference of 120 V across a Nichrome wire that has a total
resistance of 8.00 V. Find the current carried by the wire and the power rating of the heater.
Conceptualize As discussed in Example 27.2, Nichrome wire has high resistivity and is often used for heating elements
in toasters, irons, and electric heaters. Therefore, we expect the power delivered to the wire to be relatively high.
Categorize We evaluate the power from Equation 27.22, so we categorize this example as a substitution problem.
Find the power rating using the expression P 5 I2R
from Equation 27.22:
51.803103 W5 5 1.80 kW
Use Equation 27.7 to find the current in the wire:
5 15.0 A
What if the heater were accidentally connected to a 240-V supply? (That is difficult to do because the
shape and orientation of the metal contacts in 240-V plugs are different from those in 120-V plugs.) How would that
affect the current carried by the heater and the power rating of the heater, assuming the resistance remains constant?
Answer If the applied potential difference were doubled, Equation 27.7 shows that the current would double. Accord-
ing to Equation 27.22, P 5 (DV)2/R, the power would be four times larger.
(Quick Quiz 27.5)
Two lightbulbs connected across
the same potential difference.
These power lines
transfer energy from the electric
company to homes and businesses.
The energy is transferred at a very
high voltage, possibly hundreds of
thousands of volts in some cases.
Even though it makes power lines
very dangerous, the high voltage
results in less loss of energy due to
resistance in the wires.
Example 27.5 Linking Electricity and Thermodynamics
An immersion heater must increase the temperature of 1.50 kg of water from 10.0°C to 50.0°C in 10.0 min while oper-
ating at 110 V.
(A) What is the required resistance of the heater?
Conceptualize An immersion heater is a resistor that is inserted into a container of water. As energy is delivered to the
immersion heater, raising its temperature, energy leaves the surface of the resistor by heat, going into the water. When
the immersion heater reaches a constant temperature, the rate of energy delivered to the resistance by electrical trans-
) is equal to the rate of energy delivered by heat (Q) to the water.
Categorize This example allows us to link our new understanding of power in electricity with our experience with
specific heat in thermodynamics (Chapter 20). The water is a nonisolated system. Its internal energy is rising because
of energy transferred into the water by heat from the resistor, so Equation 8.2 reduces to DE
5 Q. In our model, we
assume the energy that enters the water from the heater remains in the water.
Analyze To simplify the analysis, let’s ignore the initial period during which the temperature of the resistor increases
and also ignore any variation of resistance with temperature. Therefore, we imagine a constant rate of energy transfer
for the entire 10.0 min.
Substitute the values given in the statement of the
4 186 J/kg
5 28.9 V
Use Equation 20.4, Q 5 mc DT, to relate the energy
input by heat to the resulting temperature change
of the water and solve for the resistance:
Set the rate of energy delivered to the resistor equal
to the rate of energy Q entering the water by heat:
(B) Estimate the cost of heating the water.
Find the cost knowing that energy is purchased at
an estimated price of 11. per kilowatt-hour:
Cost 5 (0.069 8 kWh)($0.11/kWh) 5 $0.008 5 0.8.
Multiply the power by the time interval to find the
amount of energy transferred to the resistor:
5 69.8 Wh 5 0.069 8 kWh
Finalize The cost to heat the water is very low, less than one cent. In reality, the cost is higher because some energy
is transferred from the water into the surroundings by heat and electromagnetic radiation while its temperature is
increasing. If you have electrical devices in your home with power ratings on them, use this power rating and an
approximate time interval of use to estimate the cost for one use of the device.
The electric current I in a conductor is defined as
where dQ is the charge that passes through a cross sec-
tion of the conductor in a time interval dt. The SI unit
of current is the ampere (A), where 1 A 5 1 C/s.
chapter 27 current and resistance
The current density J
in a conductor is the cur-
rent per unit area:
For a uniform block
of material of cross-
sectional area A and
length ,, the resistance
over the length , is
where r is the resistivity
of the material.
The resistance R of a conductor is defined as
where DV is the potential difference across the conductor and I is the current it car-
ries. The SI unit of resistance is volts per ampere, which is defined to be 1 ohm (V);
that is, 1 V 5 1 V/A.
In a classical model of electrical conduction in metals, the electrons are treated as
molecules of a gas. In the absence of an electric field, the average velocity of the elec-
trons is zero. When an electric field is applied, the electrons move (on average) with
a drift velocity v
that is opposite the electric field. The drift velocity is given by
where q is the electron’s charge, m
is the mass of the electron, and t is the average
time interval between electron–atom collisions. According to this model, the resistiv-
ity of the metal is
where n is the number of free electrons per unit volume.
Concepts and Principles
The average current in a conductor
is related to the motion of the charge
carriers through the relationship
where n is the density of charge carri-
ers, q is the charge on each carrier, v
is the drift speed, and A is the cross-
sectional area of the conductor.
The resistivity of a conductor
varies approximately linearly with
temperature according to the
r 5 r
[1 1 a(T 2 T
is the resistivity at some
reference temperature T
is the temperature coefficient of
The current density in an ohmic conductor is proportional to the
electric field according to the expression
J 5 sE
The proportionality constant s is called the conductivity of the material
of which the conductor is made. The inverse of s is known as resistivity
r (that is, r 5 1/s). Equation 27.6 is known as Ohm’s law, and a mate-
rial is said to obey this law if the ratio of its current density to its applied
electric field is a constant that is independent of the applied field.
If a potential difference DV is maintained across a circuit element, the
power, or rate at which energy is supplied to the element, is
P 5 I DV
Because the potential difference across a resistor is given by DV 5 IR, we
can express the power delivered to a resistor as
The energy delivered to a resistor by electrical transmission T
the form of internal energy E
in the resistor.
2. Two wires A and B with circular cross sections are
made of the same metal and have equal lengths, but
the resistance of wire A is three times greater than that
of wire B. (i) What is the ratio of the cross-sectional
1. Car batteries are often rated in ampere-hours. Does
this information designate the amount of (a) current,
(b)power, (c) energy, (d) charge, or (e) potential the
battery can supply?
denotes answer available in Student Solutions Manual/Study Guide
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