12/22/12
Chapter 3: Electricity
Goals of Period 3
Section
Section
Section
Section
Section
3.1
3.1:
3.2:
3.3:
3.4:
3.5:
To define electric charge, voltage, and energy
To define electricity as the flow of charge
To explain the generation electricity
To discuss the transmission of electricity
To calculate the cost of using electricity
Electric Charge, Voltage, and Energy
Electric Charge
The world around us is full of electric charge. Evidence for the existence of
electric charge comes from the electric forces between charged objects.
Because
electric charges on two adjacent objects can cause the objects to move, we conclude
that a force must exist between the electric charges. An electric force can move
charged objects closer together (an attractive force) or move them apart (a repulsive
force). Since electric forces can move charged objects together or apart, we conclude
that two opposite types of electric charge must exist. We rarely notice electric charge
because most objects contain equal amounts of these two opposite types of electric
charge, which cancel one another.
A charged object results when a quantity of one type of charge is separated from
an equal quantity of the opposite type of charge. The sum of equal quantities of
opposite charge is zero. For this reason, the two types of electric charge are called
positive charge (+) and negative charge (–). Objects with equal numbers of positive
and negative charge have a total net charge of zero and are electrically neutral.
Objects with more positive than negative charge have a net positive charge, and objects
with more negative than positive charge have a net negative charge. Charge is
measured in units of coulombs (coul). We will use the variable Q to represent charge.
Electrical Potential Energy and Voltage
Chapter 2 defined work as the product of the force applied to an object times
the distance that force caused the object to move: W = F x D. If we ignore wasted
energy, the joules of energy required to do this work is equal to the joules of work
done. Work and energy are likewise xx in electric charge.
When the attractive electric force between positive and negative charges pulls
charge together, work must be done against this attractive force to separate the
charges. When separated charges are allowed to come back together, we can get work
back out. On the other hand, since two positive charges or two negative charges repel
one another, we must do work against this repulsive electric force to bring the charges
together. We can get work out as the two charges move apart.
24
12/22/12
Electrical potential energy is described in terms of charge, Q, and voltage, V,
rather than in terms of force times distance. The energy per charge is called the
voltage. Voltage is measured in units of volts (V), with 1 volt = 1 joule of energy per 1
coulomb of charge. Voltage is the ratio Epot/Q, that is, voltage is the potential energy
per unit of charge as shown in Equation 3.1.
(Equation 3.1)
E pot  Q V
E pot =
Q
=
V
=
where
the electrical potential energy (in joules)
the charge (in coulombs)
the voltage (in volts)
(Example 3.1)
Electric charge is placed on a metal surface. How much electrical potential
energy do 2 x 10–9 coulombs of electric charge have when maintained at a voltage of
2,000 volts?
Epot = Q V =
3.2
2 x 10–9 coul x 2,000 V = 4 x 10–6 joules
Electric Current
Moving Electric Charge Produces Electric Current
When separated electric charge flows through conducting material, the moving
charge constitutes an electric current in the conductor. Electric current usually
consists of flowing electrons, which can move freely in a conducting material. The
amount of electric current is the rate at which charge Q flows past a given point in the
conductor. Since charge is measured in coulombs, current I is measured in units of
coulombs per second. One coulomb per second is called one ampere (amp). The
greater the amount of charge that flows, the larger the electric current.
We can
express this definition of current with the equation:
Current
=
Amount of Charge moved
Elapsed Time
or
I
=
Q
t
current (in amperes)
Q
=
charge (in coulombs)
t
=
time elapsed (in seconds)
I 
with
25
(Equation 3.2)
12/22/12
(Example 3.2)
How much charge must flow to provide a current of 10 amps for 20 seconds?
Solve equation 3.2 for Q by multiplying both sides by t and canceling:
Q
t
t

I t

10 amps x 20 sec = 200 coul
Concept Check 3.1
a)
How much current is present if 10 coulombs of charge flow through a conductor
every 5 seconds?
______________________
b)
How long must a 5 amp current flow to provide 200 coul of charge?
_____________________
3.3
Generating Electricity
Moving Electric Charges and Magnetic Fields
When an electric charge, such as an electron, moves, the charge is surrounded
by a magnetic field. The magnetic field produced by an electric current can exert a
force and do work on nearby permanent magnets or on other moving electric charges.
Likewise, a nearby magnetic field can do work on a current-carrying wire.
But an additional effect also occurs – a changing magnetic field can produce a
current in a nearby conductor such as a piece of metal or a wire that is part of a closed
or complete circuit. The changing magnetic field can be produced by moving a magnet
or by moving the conductor. The process of generating a current in a conductor is
known as inducing a current.
In class, we induce a current by moving a magnet into and out of a coil of wire.
Moving the magnet into the coil induces a current in one direction in the wire. When
the magnet stops moving, the current stops flowing. Pulling the magnet out of the coil
produces a current flowing in the opposite direction. To produce this current, only the
relative motion of the coil and wire is important – the same current is induced whether
the coil moves or the magnet moves.
Electric Generating Plants
The principle that a magnet moving with respect to a wire induces a current in
the wire is used in power plants to generate electricity. Generators use magnets and
coils of wire to convert kinetic energy into electrical energy. Electric generating plants
convert the kinetic energy of rotating magnets into electrical energy by spinning large
magnets near coils of conducting wire. To rotate the magnets, generating plants use
26
12/22/12
kinetic energy from the sources described below to turn the blades of turbines.
Turbines are wheels with blades attached, similar in principle to waterwheels. A shaft
attached to the rotating turbine causes the magnets to spin.
The most common mechanism for turning
turbines is steam pressure. In steam generating plants,
water is heated in a closed container. As the water
changes to steam, the volume and pressure increase.
The steam exerts pressure on the turbine blades, turning
them. In many generating plants, water is heated by
burning coal, oil, natural gas, or other fuels. In nuclear
power plants, water is heated by the thermal energy that
results from reactions in radioactive fuels.
Fig. 3.1 Turbine
In hydroelectric plants, falling water rotates the
turbines. The gravitational potential energy of the water is converted into kinetic energy
of motion as the water falls and turns the turbines. In addition to the many
hydroelectric power plants built along rivers, a tidal powered generating plant along the
ocean shore can use tides to turn turbines. In tidal plants, water flows in during high
tide and is trapped behind gates. As the tide recedes, the trapped water is left at a
higher level than the surrounding ocean. The trapped water returning to the ocean
spills over the turbines, turning them.
Wind turbines use the kinetic energy of moving air molecules to spin large blades
that are attached to a turbine. Wind turbines are most effective when placed in a near
constant source of wind, such as along the shore of large bodies of water or on flat
plains.
3.4
Transmitting Electricity
Joule Heating
As electric current moves through wires, electrons collide with atoms of the wire
material and some electrical energy is converted into thermal energy. Conversion of
electrical energy into thermal energy in a resistor is known as joule heating. Unless the
purpose of transmitting current is to generate heat in a wire, such as in a toaster’s
filament, this thermal energy is wasted. The watts of joule heating power are the
product of the current squared times the resistance of the wire. Resistance is the
ability of a material to resist the flow of electric current. Resistance is measured in units
of ohms (  ).
(Equation 3.3)
P joule = I 2 R
with
P joule =
I
=
R
=
power (in watts)
current (in amperes)
resistance (in ohms)
27
12/22/12
Joule heating in power transmission lines represents both a waste of energy
resources and a loss of revenue for the power generating company. Even a voltage
drop of only 18 volts between the generating plant and the user represents a large
amount of power waste from joule heating. Example 3.3 calculates the amount of
wasted power.
(Example 3.3)
How much power would be wasted as joule heating of the transmission wires if
1.67 x 106 amps of current was transmitted with the very small resistance of 1.1 x 10-5
ohms?
Pjoule = I 2 R = (1.67 x 106 amps)2 x (1.1 x 10-5 ohms)
= (2.79 x 1012 amps2 ) x (1.1 x 10-5 ohms) = 3.1 x 107 watts
As shown in Example 3.3, transmitting large amounts of current even at very low
resistance would wastes large amounts of power as joule heating of the wires.
Fortunately, there is another alternative – power transmission at high voltages.
Transformers
One way to reduce the amount of current in transmission wires is to increase the
voltage of that current. Power is directly proportional to the product of current and
voltage as shown in Equation 3.4. A higher voltage results in a smaller current for a
given amount of power transmitted.
(Equation 3.4)
P = IV
with
P
I
V
=
=
=
power (in watts)
current (in amperes)
voltage (in volts)
Transformers make it possible to transmit electricity at a higher voltage by
changing the voltage of the electric current. The purpose of a transformer is to trade
high voltage for low current or low voltage for high current. From Equation 3.4, we see
that as the voltage is increased, the current must decrease to provide the same power.
Likewise, if the voltage is decreased, the current increases.
Step-up transformers increase voltage and decrease current for long distance
transmission. While high voltage reduces joule heating, it poses a safety hazard for
consumers. Step-down transformers are used to decrease voltages to safe levels at
the point of use of the electricity.
28
12/22/12
Step-down transformers reduce the voltage of transmitted power in stages to
progressively lower voltages. Power is transmitted over long distances at up to 700,000
volts. Over shorter distances or to small towns with low power requirements, power is
transmitted at 25,000 volts. Power lines along city streets typically operate at about
2,000 volts. Finally, a transformer about the size of a garbage can, often located on a
utility pole, reduces the voltage for household use. Households are normally supplied
240 volts, which is delivered in such a way as to provide 120 volts to regular outlets and
240 volts to outlets for devices that require more power, such as electric stoves, water
heaters, and clothes dryers.
(Example 3.5)
If a power company transmits electricity at 500,000 volts (5 x 105 V) rather than
120 volts, how much current would be needed to provide 200 megawatts of power (2 x
108 watts) to a city?
Solve P = I V for I by dividing both sides by V and canceling:
P
V

I
8
 2 x 10 watts
5 x 105 volts

400 amps
Just 400 amps of current could supply 200 megawatts of power when
transmitted at a high voltage, compared to the 2.79 x 1012 amps of current required to
transmit at 120 volts in Example 3.4. Transmitting at high voltage and low current
greatly reduces the power wasted as joule heating of the wires.
Transformers make it possible to transmit electricity at high voltages, then step
the electricity down to lower voltages that are safer for consumer use.
Concept Check 3.2
a)
A 400 amp current is transmitted through wires with a resistance of 0.8 ohms for
every mile of wire. How much power is wasted as joule heating of a
transmission wire 200 miles in length?
______________________
b)
If 200 megawatts (2 x 108 watts) is the total power input, what is the efficiency
of the transmission process?
_________________________
29
12/22/12
3.5
Cost of Electricity
Power Measurements and Requirements
Power companies provide electrical energy, or electricity, and charge for it in
units of kilowatt-hours (kWh). To measure the electrical energy supplied to
consumers, power companies use kilowatt-hour meters attached to residences and
businesses. The meter is a small electric motor with a rotating disc. Because the
rotation speed of the disc is proportional to the power being used at the time, electricity
can be measured by the number of rotations of the disc. The measurement of electricity
may be shown by a digital display or by an analog display with rotating pointers on
dials. The difference between the current month's dial reading and previous month's
reading measures the total energy provided, which is used to determine the electric bill.
To calculate the cost of electricity from an electric bill, we must know how many kilowatts
of electric power were used and for how many hours. Power is the energy transferred or work
done divided by the time required, as shown in Equation 3.5.
(Equation 3.5)
P = E = W
t
t
with
P
E
W
t
=
=
=
=
power (in watts)
energy transferred (in joules)
work done (in joules)
time elapsed (in seconds)
From Equation 3.5, you see that the faster energy is transferred or work is done, the
greater the power required.
(Example 3.6)
How many kilowatt-hours (kWh) of energy are required to operate a 1,000 watt
electric heater for 2 hours?
Find the energy used by an appliance by solving equation 3.5 for the energy
requirement, E.
P

E
t
or
E

Pt
E = 1,000 watts x 2 hours = 2,000 watt-hours.
Use a ratio to convert from watts to kilowatts:
2,000 watt-hours
x
1 kW =
1,000 watts
30
2 kilowatt-hours = 2 kWh
12/22/12
The total kilowatt-hours of energy used during a month is the sum of the power
requirement of each appliance multiplied by the hours that appliance operated.
(Example 3.7)
How many kilowatt-hours of energy are used when a 100 watt light bulb burns
for 7 hours, a 1,000 watt hair dryer operates for 15 minutes, and a 1,500 watt electric
heater operates for ½ hour?
Light bulb:
100 watts x
1 kilowatt
x 7 hours
1,000 watts
Dryer:
1,000 watts x
Heater:
1,500 watts x

0.70 kilowatt hours
1 kilowatt
1 hour
x 15 min x
 0.25 kilowatt hours
1,000 watts
60 min
1 kilowatt
1 hour
x 30 min x
1,000 watts
60 min

0.75 kilowatt hours
The total energy used is 0.70 kWh + 0.25 kWh + 0.75 kWh = 1.7 kilowatt-hours.
(Example 3.8)
If the power company charges $0.10 per kilowatt-hour of energy, how much
would you pay for the energy used by the light bulb, hair dryer, and heater in Example
3.7?
Energy Cost = number of kWh used x cost per kWh
1.7 kilowatt hours x
$ 0.10
1 kilowatt hour

$ 0.17
Concept Check 3.3
a)
How many kilowatt-hours are required to use a 600 watt oven for 45 minutes?
________________
b)
If electricity costs $0.10 per kilowatt-hour, how much would you pay to operate
the oven for 45 minutes?
_________________
31
12/22/12
Table 3.1: Metric and English Units and Their Symbols
Quantity
Symbol
Metric
Units
Metric Unit
English Unit
Abbreviation
English Unit
N
2
(1 N = 1 kg m/s )
pound
lb
foot-pound
ft-lb
Abbrev.
Force
F
newton
Energy
E
joule
Work
W
joule
J
foot-pound
ft-lb
Power
P
watt
W
(1 W = 1 J/s)
horsepower
hp
(1 hp = 550 ft-lb/s)
J
2
2
(1 J = 1 kg m /s )
32
12/22/12
Chapter 3 Summary
3.1
Electric charge Q is measured in coulombs. The voltage V is the energy per
charge. The electrical potential energy of charge Epot = Q V
3.2:
Moving electric charge is electric current I. Current is measured in units of
amperes or amps. Since current is the rate of flow of charge, I = Q/t
3.3:
Electric current is generated, or induced, when magnets and coils of wire move
relative to one another. In power generating plants, kinetic energy from steam,
flowing water, or wind turns turbines. The moving turbines spin magnets near
coils of wire.
Electric generating plants are often located far from heavily populated areas,
thus the electricity they generate must be transmitted long distances to its point
of use. To transmit electricity to consumers, the electricity is transmitted at high
voltage to minimize the power wasted as joule heating of the transmission wires.
3.4:
Joule heating (Pjoule = I2R) occurs when some electrical energy transmitted
through wires is wasted as thermal energy heating the wires.
Transformers trade current and voltage, while keeping the power nearly
constant. (P = I V). Step-up transformers are used to reduce the current and
increase the voltage for efficient long distance transmission of electricity.
Less current transmitted reduces power wasted as joule heating.
Near the point of electricity consumption, step-down transformers reduce the
voltage and increase the current for safer use of electricity.
3.5: Power is the rate at which work is done or energy is transferred.
P = W/t
or
P = E/t
Power is measured in joules/second, or watts, in the metric system and in footpounds/second, or horsepower, in the English system.
Electrical energy provided to homes is measured and billed to consumers in
units of kilowatt-hours. To find the cost of electricity for using an appliance:
1) Convert watts to kilowatts, by dividing watts by 1,000.
2) Find the total time of use in hours.
3) Multiply kilowatts times hours to find the total kWh.
4) Multiply kWh by the cost per kWh.
33
12/22/12
Solutions to Chapter 3 Concept Checks
3.1
a)
I=Q
t
= 10 coul. = 2 amps
5 sec
b)
I=Q
t
or t = Q =
200 coul. = 40 sec
5 amps
I
3.2
a)
Pjoule = I2R = (400 amps)2 x 0.8 ohms = 128,000 watts/mile
128,000 watts/mile x 200 miles = 25,600,000 watts = 2.56 x 107 watts
b)
2 x 108 watts = total power input
total power input = useful power out + wasted power out
useful power out = total power input – wasted power out
useful power out =2 x 108 – 2.56 x 107 = 2 x 108 – 0.256 x 108 = 1.74 x 108
(Note that 2.56 x 107 was converted to 0.256 x 108. Adding or subtracting
numbers in scientific notation requires each number to have the same power of
ten.)
Efficiency = useful power out = 1.74 x 108 watts = 0.87 = 87%
total power in
2 x 108 watts
3.3
a)
E = P t = 600 watts x 0.75 hours = 450 watt-hours.
Use a ratio to convert from watts to kilowatts:
450 watt-hours
b)
x
1 kW = 0.45 kilowatt-hours = 0.45 kWh
1,000 watts
0.45 kWh x $0.10 = $0.045
kWh
34