PHYSICS
TUTORIAL NOTES
ELECTRICAL
ENERGY IN THE HOME
SYLLABUS
TOPIC 8.3
These notes are meant as a
guide only and are designed to focus your thoughts on the dot points mentioned
in the syllabus. They give a very brief overview of the topic and should be
used in conjunction with your class notes, your textbooks and your research
from other areas such as the library and the Internet.
Notes compiled by:
CARESA EDUCATION SERVICES
“1. Society has become increasingly dependent on electricity over the last 200 years.” (syllabus)
“Students learn to; discuss how the main sources of domestic energy have changed over time.” (syllabus)
DOMESTIC POWER
Since fire was discovered
around 1½ million years ago wood
remained the major source of fuel until
the industrial revolution in the 18th century. Since then fossil
fuels in the form of coal and then oil and natural gas have mostly replaced
wood as the major source of fuel. These original power sources were augmented
to a certain extent by the domestication of animals and by mills powered by
running water or wind. Wind has also been used to power boats for the last few
thousand years.
While
the major source of power in the early days of the industrial revolution was
steam engines powered by burning wood or coal this began to change when Faraday
in
Early
humans were hunter-gatherers and led a nomadic lifestyle. As they began to grow
crops they built permanent dwellings and remained in the one area. Eventually
cottage industries grew up where such items as textiles or furniture were
produced. The industrial revolution produced factories where large amounts of
energy were used and the demand for human labour led to the growth of towns and
cities as the population moved to live near the factories. As electricity
became the major power source, the factories became more efficient and were
able to produce a wide range of affordable goods including household electrical
goods. This led to a huge demand for domestic electricity.
Power
stations use large alternators to produce electricity in the form of
alternating current. The voltage is raised to very high levels by means of
transformers and then carried through wires to the location where it is needed.
Here other transformers lower the voltage to the level at which it is used; 240
volts for household consumption in
Where
the cost of installing transmission lines is too expensive, remote locations
produce their own electricity. This can be done by portable generators that run
on petrol, oil or gas or by using methods such as solar cells or wind
generators.
While
nuclear fission power stations are capable of supplying large amounts of
electricity they pose a hazard to the environment with their radioactive waste
and the possibility of a meltdown. Controlled nuclear fusion does not have
these dangers and as soon as technology allows us to build fusion power stations
it appears to be the energy source of the future.
“Students learn to; assess some of the impacts of
changes in, and increased access to, sources of energy for a community.”
(syllabus)
The availability of sources
of energy enables the functioning of the modern societies of today. The most
significant impact would most likely be transportation. Raw materials are
collected from all over the world and transported to factories where they are
converted into goods to be sold all over the world. Without affordable sources
of energy we would revert to the local cottage industries of days gone by.
Overseas travel is now very
common due to plentiful supplies of energy. Contrast the speed and comfort of
today’s plane with that of a sailing ship of a couple of hundred years ago. The
distances that a large portion of the population commutes to work or school are
much greater than would be possible on horseback. Again the use of energy for
transportation has made this possible.
The centralisation of
industry has led to the growth of large cities which further rely on adequate
sources of energy to run their factories, transport their people and to power
their homes and offices.
When we think of electrical
energy in the home it is amazing to think of the number and variety of
electrical appliances that we use. While some are readily replaceable (e.g. an
electric stove can be replaced with a gas stove) others, such as the washing
machine, are not readily powered by alternate sources of energy. Most people
can make a list of around 20 different appliances in their home that rely on a
readily available supply of electrical energy.
“Students learn to; discuss some of the ways in
which electricity can be provided in remote locations.” (syllabus)
Electricity
can be provided in remote locations by using portable generators that are
powered by petrol or diesel fuel. Wind powered generators can also be used but
these are unreliable as the wind is not always blowing when electricity is
required. A battery storage system for the electricity can help overcome this
problem. A bank of solar cells can also provide electricity and again a battery
storage system will make the electricity available at night or on overcast
days. In days gone by the pedal wireless was used in remote locations of
“Students; identify data sources, gather, process and analyse secondary information about the differing views of Volta and Galvani about animal and chemical electricity and discuss whether their different views contributed to increased understanding of electricity” (syllabus)
Identify data sources:
Websites:
Learning Materials Production Centre:
http://www.lmpc.edu.au
Bob Emery notes:
Other Notes
‘Timesavers” by Greg Marsden
Open Training and Educational Network
Gather, process and analyse secondary information:
Luigi
Galvani was a lecturer in medicine at the
Alessandro
Volta was Professor of Experimental Physics at the
The debate continued until
the death of Galvani in 1798. In 1799
Discuss whether their different views…….
The different views led
“2. One of the main advantages of electricity is that it can be moved with comparative ease from one place to another through electric circuits” (syllabus)
“Students learn to; describe the behaviour of
electrostatic charges and the properties of the fields associated with them.”
(syllabus)
A charged particle has a region of influence around
it in which another charged particle will experience a force. This is known as
an electric field. There are two types of electric charge: positive and
negative. Like charges (two positive charges or two negative charges) will
repel each other while unlike charges (a positive and a negative charge) will
attract each other.
“Students learn to; define the unit of the electric
charge as the coulomb.” (syllabus)
The electric unit of charge is the coulomb.
One coulomb is defined as one ampere second i.e. the
amount of charge that passes a point when a current of one ampere flows for one
second.
q = It Where
q = charge (coulombs) I = current (amps) and t = time (seconds)
“Students learn to; define the electric field as a
field of force with a field strength equal to the force per unit charge at that
point: E = F / q” (syllabus)
An electric field is a region where an electric
charge will experience a force.
The electric field strength at a point is equal to
the force experienced by a unit positive charge (i.e. a charge of +1coulomb)
placed at that point.
Note that the force gives both the magnitude and
direction of the electric field.
E = F / q where E = electric field strength, F =
force and q = charge.
“Students learn to; define electric current as the
rate at which charge flows (coulombs/second or amperes) under the influence of
an electric field.” (syllabus)
Electric current is the rate of flow of charge.
I = q / t Where
q = charge (coulombs) I = current (amps) and t = time (seconds)
Charges need something to move them along the
conductor and this is an electric field.
The fundamental unit of electricity is the ampere
and the coulomb is defined in terms of the ampere. The ampere has an interesting
definition that is not on the syllabus:
The ampere is the unit of current intensity which
when maintained in two rectilinear (straight) conductors of infinite length and
negligible cross-section separated by one metre in vacuo (vacuum) produces
between the conductors a force of 2.0 x 10-7 S.I. units of force
(Newtons) per metre of length.
“Students learn to; identify that current can either
be direct with the net flow of charge carriers moving in one direction or
alternating with the charge carriers moving backwards and forwards
periodically.” (syllabus)
Direct Current, DC, involves the net movement of
charge carriers (electrons) in one direction along a conductor. The valency
electrons of the conductor are in constant random motion but in the presence of
an electric field the electrons experience a net movement in the opposite
direction to the electric field.
Note: electrons move in the opposite
direction to the direction of the electric field.
Alternating Current, AC, involves the oscillation of
electrons due to an electric field that is constantly changing direction. Most
generators produce alternating current by means of a coil rotating in a
magnetic field.
“Students learn to; describe electric potential
difference (voltage) between two points as the change in potential energy per
unit charge moving from one point to the other (joules/coulomb or volts)”
(syllabus)
The potential difference between two points is equal
to the work done, i.e. the change in potential energy, in moving a unit
positive charge (i.e. a charge of +1 coulomb) from one point to the other. The
potential difference is the work per unit charge and is expressed as joules per
coulomb. One joule per coulomb is known as one volt.
V = W / q where V = potential difference (volts), W
= work or energy (joules) and q = charge (coulombs)
“Students learn to; discuss how potential difference
changes at different points around a DC circuit.” (syllabus)
A DC circuit has a source of potential difference
such as a battery or power pack, and one or more components with a resistance
connected to the terminals. The
potential difference between points on the circuit is proportional to the
resistance between those two points.
“Students learn to; identify the difference between
conductors and insulators.” (syllabus)
A substance that allows charge carriers (electrons)
to readily flow along it is known as a conductor. A substance that does not
allow charges to readily flow along it is an insulator.
No substance is a perfect conductor and no substance
is a perfect insulator.
“Students learn to; define resistance as the ratio
of voltage to current for a particular conductor: R = V / I” (syllabus)
Resistance is anything that hinders the flow of
charge along a conductor and is measured in ohms (symbol W). For a constant potential difference a small
resistance will produce a large current and a large resistance will produce a
small current. This can be expresses quantitatively as R = V / I
where R = resistance (ohms W), V = potential difference
(volts V) and
I = current (amps A)
“Students learn to; describe qualitatively how each
of the following affects the movement of electricity through a conductor.
-
length
- cross sectional area
-
temperature
-
material” (syllabus)
The resistance of a conductor is directly
proportional to its length. The longer the wire the higher the resistance.
The resistance of a conductor is inversely
proportional to its area of cross section. A thin wire has a large resistance
while a thick wire has a low resistance.
The resistance of a conductor increases with
temperature. (When you get to semiconductors in year 12 you will find the
opposite to be true of semiconductors) The resistance of a light globe will
increase as it gets hotter.
The resistance of a conductor depends on the
material of which it is made. While all metals are conductors there is a
difference in their conductivity i.e. how well they conduct. Similarly there is
a difference in the efficiency of materials as insulators. There is a continuum
of resistance ranging from very low for conductors, medium for semiconductors
and very high for insulators.
“Students; present diagrammatic information to describe the electric field strength and direction
- between charged parallel plates
- about and between a positive and negative point charge” (syllabus)
Electric fields can be shown diagramatically as lines of force. These show the direction of the force on a positive charge and therefore the arrows point from positive to negative.
Lines of force between two charged parallel plates.
Lines of
force around a small positively charged particle.
Lines of force around a
small particle with a high positive charge.
Note that a strong electric
field has more lines of force than a weak one.
Lines of force around a
small negatively charged particle.
Lines of force between a
positive and a negative point charge.
“Students; solve problems and analyse information using: E = F / q” (syllabus)
Q.1.
A polystyrene ball carries a charge of 1.0mC and is subjected
to an electrostatic force of 2.0 x 10-4 N north. What is the
strength of the electric field?
A.1.
E = F / q E = 2.0 x 10-4 / 1.0
x 10-6 = 2.0 x 102
NC-1 north.
Q.2.
An alpha particle (charge = +3.2 x 10-19C) enters an electric field of
magnitude 3.0 x 10-6 NC-1 vertically down. What force
does it experience?
A.2.
E = F / q F = qE F = 3.2 x 10-19 x 3.0 x
10-6
= 9.6 x 10-25 N vertically down.
Q.3.
An electron (charge= –1.6 x 10-19C)
enters the electric field of Q.2. What force does it experience?
A.3.
E = F / q F = qE F = 1.6 x 10-19 x 3.0 x
10-6
= 4.8 x 10-25 N vertically up.
Q.4.A charged oil drop of mass 0.1 mg is held so
that is stationary between two charged plates.
(i)
What is the force due to gravity acting on the oil drop?
(ii)
What is the force due to the electric field acting on the oil drop?
(iii)
If the oil drop has a charge of –6.4 x 10-19 C, what is the
magnitude of the electric field strength between the plates?
(iv)
What is the direction of the electric field?
A.4. (i) F
= mg = 10-7 x 9.8 =
9.8 x 10-7 N down.
(ii)
F = mg = 10-7 x 9.8 =
9.8 x 10-7 N up
(iii)
E = F / q = 9.8 x 10-7 / 6.4
x 10-19 = 1.53 x 1012 NC-1
(iv)
Down
“Students; plan, choose equipment for and perform a first-hand investigation to gather data and use available evidence to show the relationship between voltage across and current in a DC circuit” (syllabus)
Plan: Look at what you are trying to show and then work
out a way to show it. From the syllabus you know you have to show the
relationship between voltage across and current in a DC circuit. You would have
to set up a DC circuit, measure the voltage and current and determine the
relationship between them. As you plan, you think about what equipment is
required to carry out your investigation.
Choose Equipment: Make an accurate list of all equipment
needed to carry out your planned investigation.
Perform a first-hand investigation: This is best done in the
laboratory. An example of a typical experiment is shown below.
Mary and John performed an experiment to show the
relationship between voltage across and current in a DC circuit. They set up
the circuit below containing a power pack, voltmeter, ammeter, rheostat, switch
and a resistor labelled 2.0W.
Circuit
Diagram
Mary set the power pack to setting “D”. Mary then
held the switch closed while John adjusted the rheostat until the reading on
the voltmeter was 1.0 volt. Mary then read the ammeter and then opened the
switch. Mary immediately recorded the result in a table in her science
workbook.
Again Mary held the switch closed while John
adjusted the rheostat until the reading on the voltmeter was 2.0 volts. Mary
then read the ammeter, opened the switch and immediately recorded her results.
The procedure was repeated for readings on the voltmeter of 3.0, 4.0 and 5.0
volts.
Mary then held the switch closed while John adjusted
the rheostat until the voltmeter again read 1.0 volt and Mary recorded the
ammeter reading. This was repeated for 2.0, 3.0, 4.0 and 5.0 volts. John and
Mary then swapped jobs and took two more sets of readings so that 4 current
readings were made for each voltage setting. A copy of the table of Mary and
John’s results is shown below.
Table of Results: Potential Difference and Current
|
Potential Difference (Volts) |
Current (Amps) |
||||
|
1st |
2nd |
3rd |
4th |
Average |
|
|
1.0 |
0.5 |
0.4 |
0.5 |
0.6 |
0.5 |
|
2.0 |
1.1 |
0.9 |
1.0 |
1.1 |
1.025 |
|
3.0 |
1.6 |
1.4 |
1.4 |
1.5 |
1.475 |
|
4.0 |
2.0 |
2.1 |
2.0 |
1.9 |
2.0 |
|
5.0 |
2.4 |
2.4 |
2.5 |
2.6 |
2.475 |
Mary and John then drew a graph of potential
difference (vertical axis) against current (horizontal axis).
Try it yourself, using Mary and John’s results, and
compare your graph to the one shown below.
Graph of
Potential Difference vs Current
See if you can answer the following questions about
Mary and John’s investigation.
Q.1. What is the independent variable in the above
experiment?
Q.2. What is the dependent variable in the above
experiment?
Q.3. Determine the gradient of the above graph.
Q.4. Why is the switch only closed when readings are
being taken and left open at other times.
Q.5. What safety precautions should be taken by John
and Mary in their experiment?
Q.6. Suggest a conclusion for Mary and John’s
experiment.
A.1. Potential Difference (Voltage) is the
independent variable.
A.2. Current is the dependent variable.
A.3. The gradient of the graph is 2.0 volts/amp.
A.4. The switch is closed for the minimum possible
time so that the components of the circuit do not heat up.
A.5. They should only use low voltages and currents,
they should ensure that the bench is dry and they should be careful not to
touch any hot wires.
A.6. The voltage is directly proportional to the
current.
“Students; solve problems and analyse information applying: R = V / I” (syllabus)
Q.1.
A torch globe operates on a potential difference of 3.0 volts and has a current
of 0.1Amp flowing through it. What is its resistance?
A.1.
R = V / I = 3.0 / 0.1 = 300 ohms.
Q.2.
A hair drier operates at a potential difference of 240 volts and has a
resistance of 960 ohms. How much current flows through it?
A.2.
R = V / I I = V / R = 240 /
960 = 0.25 Amps.
Q.3.
What is the potential difference between the ends of a 500 W resistor when a current of 3 A flows through it?
A.3. R = V / I V = IR = 3 x 500 = 1500 V.
“Students; plan, choose equipment for and perform a first-hand investigation to gather data and use available evidence to show the variations in potential difference between different points around a DC circuit” (syllabus)
Plan: Look at what you are trying to show and then work
out a way to show it. From the syllabus you know you have to show how potential
difference varies between different points in a DC circuit. You would have to
set up a DC circuit and measure the voltage between different points. As you
plan, you think about what equipment is required to carry out your
investigation.
Choose Equipment: Make an accurate list of all equipment
needed to carry out your planned investigation.
Perform a first-hand investigation: This is best done in the
laboratory. An example of a typical experiment is shown below.
Sarah and Bob carried out an experiment to
investigate how the potential difference varied between different points in a
DC circuit. They connected a power pack, an ammeter, a switch, 3 resistors of 1W, 2W and 5W respectively and 3
voltmeters into a circuit as shown below.
R1
= 1 ohm R2 = 2 ohms R3 = 5 ohms
Sarah closed the switch and read the ammeter while
Bob read the three voltmeters and recorded the results. Sarah opened the switch
as soon as the readings had been recorded. Sarah and Bob repeated the process
twice so that they had three sets of readings. A table showing their results is
shown below.
|
|
Ammeter (Amps) |
V1 Volts |
V2 Volts |
V3 Volts |
|
|
1.5 |
1.5 |
3.0 |
7.4 |
|
|
1.5 |
1.5 |
2.9 |
7.3 |
|
|
1.5 |
1.5 |
3.0 |
7.4 |
Q.1. Calculate the values (i) V1/R1, (ii) V2/R2 (iii)V3/R3
A.1. (i) 1.5 (ii) 1.5 (iii)
1.47
Sarah and Bob concluded that the ratio V/R was
constant and was equal to the current.
They also concluded that the voltage drop around the
circuit was proportional to resistance.
“Students; gather and process secondary information to identify materials that are commonly used as conductors to provide household electricity” (syllabus)
Websites:
Learning Materials Production Centre:
http://www.lmpc.edu.au
Bob Emery notes:
Other Notes
‘Timesavers” by Greg Marsden
Open Training and Educational Network
Gather and process secondary information
By far the most common conductor used to provide
household electricity is copper. It is relatively cheap and a better conductor
than all metals other than silver. Consequently copper is used in most
household wiring. Silver, instead of copper, is occasionally used in some high
quality electronic equipment due to its higher conductivity but it is not used
widely due to its high price. Aluminium is not as good a conductor as silver or
copper but it is used in the wires for overhead power line distribution because
of its light weight. The light weight allows the supporting structures to be
placed further apart and this reduces the overall cost.
“3. Series and parallel circuits serve different purposes in households” (syllabus)
“Students learn to; identify the difference between series and parallel circuits” (syllabus)
Resistors
are connected in series when there are no branches in the circuit. Any current
that flows through the first resistor must also flow through the second and
subsequent resistors since there are no branches in the circuit to allow it to
take alternate paths. Note that current is not used up (Current is a flow of
electrons so if it was used up, where would the electrons go to?) so the
current through all parts of an unbranched, i.e. series circuit, is the same.
Resistors in Series
Note
that in the second diagram the resistors lie parallel to each other but since
there are no branches in the circuit they are connected in series.
Resistors
are connected in parallel when there are branches in the circuit so that the
current can flow around the circuit via different pathways.
Resistors in Parallel
“Students learn to; compare parallel and series circuits in terms of voltage across components and current through them” (syllabus)
In a series circuit the current flowing
through all points is the same. Since the current through all resistors is the
same, if we apply Ohm’s Law, V = IR, we find that the potential difference is
directly proportional to resistance.
In
parallel circuits the potential difference across parallel components is the
same Remember that it is the potential difference between two points on a circuit
so this is the same regardless of which way the current travels between them.
Since the voltage is the same, by applying Ohm’s Law, I = V/R, we find that the
current is inversely proportional to resistance.
“Students learn to; identify uses of ammeters and voltmeters” (syllabus)
Ammeters are used to measure the current
through a point in a circuit.
Voltmeters
are used to measure the potential difference between two points in a circuit.
“Students learn to; explain why ammeters and voltmeters are connected differently in a circuit” (syllabus)
Ammeters
are connected in series with the component that we want to measure the current
through. Any current flowing through the ammeter also flows through the
component that is in series with it. Ammeters have a very low internal
resistance so that there is very little voltage drop across the ammeter and
nearly all the voltage drops across the resistor. Voltage drop is proportional
to resistance.
Voltmeters
are connected in parallel with the component that we want to measure the
voltage across. Voltmeters have a very high internal resistance so that when
the current splits up, nearly all of it will go through the component and very
little through the voltmeter. Remember that the current is inversely proportional
to resistance.
“Students learn to; explain why there are different circuits for lighting, heating and other appliances in a house” (syllabus)
All
households have at least two circuits, one for lights and another for power
points. If a house has a large number of lights or power points it may have two
or more circuits for each of these functions. In addition, many households have
separate circuits for such things as electric stoves, hot water systems or air
conditioners. These circuits carry different amounts of current and may have
wires of different thickness. The lighting circuit is normally designed to
carry around 8 amps while the power circuits carry around twice this amount.
All the lights or power points on one circuit are connected in parallel.
“Students; plan, choose equipment or resources for and perform first-hand investigations to gather data and use available evidence to compare measurements of current and voltage in series and parallel circuits in computer simulations or hands-on equipment” (syllabus)
Plan: Decide whether you are going to use a computer simulation. If you are
going to use a computer simulation find details of the website or make
arrangements to use a suitable program. If you are going to use hands-on
equipment, decide what circuit or circuits will best demonstrate relationships
involving current and voltage in both series and parallel circuits.
Choose equipment: If you are using a computer simulation you will
require a suitable program and a computer to run it.
The
following equipment is suggested for the hands-on approach using the circuits
below. Power pack, tapping key, 1 ohm resistor, 2 ohm resistor, 5 ohm resistor,
10 ohm resistor, 3 ammeters, 2 voltmeters, connecting leads.
Gather data and use available
evidence: This is best done in the laboratory.
An example of a typical experiment is shown below.
Shirley
and Greg wanted to test the variations in current and voltage around (i) a
series circuit and (ii) a parallel circuit.
They
first set up the series circuit below using a 1 ohm resistor as R1
and a 2 ohm resistor as R2. Shirley closed the tapping key and Greg
adjusted the settings on the power pack until a current of approximately 3 amps
flowed through the ammeters. Shirley then held the tapping key closed while
Greg read and recorded the three ammeters and the two voltmeters. Shirley then
released the tapping key. Greg then held the tapping key closed while Shirley
read and recorded the three ammeters and the two voltmeters. Their results are
shown in the table below.
Series Circuit
They
next set up the parallel circuit below using a 5 ohm resistor as R1
and a 10 ohm resistor as R2. Shirley closed the tapping key and Greg
adjusted the settings on the power pack until a current of approximately 3 amps
flowed through the ammeter A3. Shirley then held the tapping key
closed while Greg read and recorded the three ammeters and the two voltmeters.
Shirley then released the tapping key. Greg then held the tapping key closed
while Shirley read and recorded the three ammeters and the two voltmeters.
Their results are shown in the table below.
Parallel Circuit
Table of results: Series circuit
Powerpack Setting “E”
|
Meter |
|
|
V1 (Greg) |
3.0 Volts |
|
V1 (Shirley) |
3.1 Volts |
|
V1 (average) |
3.05 Volts |
|
V2 (Greg) |
6.1 Volts |
|
V2 (Shirley) |
6.1 Volts |
|
V2 (average) |
6.1 Volts |
|
A1 (Greg) |
3.15 Amps |
|
A1 (Shirley) |
3.1 Amps |
|
A1 (average) |
3.15 Amps |
|
A2 (Greg) |
3.0 Amps |
|
A2 (Shirley) |
3.1 Amps |
|
A2 (average) |
3.05 Amps |
|
A3 (Greg) |
3.1 Amps |
|
A3 (Shirley) |
3.05 Amps |
|
A3 (average) |
3.1 Amps |
Table of results: Parallel circuit
Powerpack Setting “E”
|
Meter |
|
|
V1 (Greg) |
9.15 Volts |
|
V1 (Shirley) |
9.1 Volts |
|
V1 (average) |
9.15 Volts |
|
V2 (Greg) |
9.15 Volts |
|
V2 (Shirley) |
9.15 Volts |
|
V2 (average) |
9.15 Volts |
|
A1 (Greg) |
1.8 Amps |
|
A1 (Shirley) |
1.85 Amps |
|
A1 (average) |
1.85Amps |
|
A2 (Greg) |
0.9 Amps |
|
A2 (Shirley) |
0.9 Amps |
|
A2 (average) |
0.9 Amps |
|
A3 (Greg) |
2.75 Amps |
|
A3 (Shirley) |
2.7 Amps |
|
A3 (average) |
2.75 Amps |
Shirley
and Greg noted the following about their results.
In
the series circuit, the readings on all three ammeters was the same, which
indicates that the current through all parts of a series (unbranched) circuit
is the same.
In
the series circuit, the potential difference across the 2 ohm resistor was twice
that across the 1 ohm resistor which suggests that the potential difference is
proportional to resistance.
In
the parallel circuit the voltage across both resistors was the same, and this
suggests that the potential difference across parallel components of a circuit
is the same.
In
the parallel circuit the current through the 5 ohm resistor was twice that
through the 10 ohm resistor. This suggests that the current through parallel
components of a circuit is inversely proportional to resistance.
“Students; plan, choose equipment or resources and perform a first-hand investigation to construct simple model household circuits using electrical components” (syllabus)
Plan: Decide what type of circuit you are going to construct. Draw a circuit
diagram. Remember that household electrical circuits are connected in parallel.
Choose equipment: Choose typical components for the circuit such as a
battery or power pack, switches, leads, light globes on stands, motors or
devices such as a fan that contains a motor, heating coils (immersed in water)
or resistors to represent any load in the circuit.
Perform a first-hand investigation: Hopefully you will plan your own circuit but a couple of suggestions are shown below. These circuits demonstrate how to connect components in parallel. Close the switches one at a time and both together to demonstrate that each component in a parallel circuit operates independently of other components connected in parallel with it. A variation to the circuit would be to include a short length of fine iron wire that would get red hot and melt as the current was increased and so demonstrate the operation of a fuse.
Circuit 1. Lighting
Circuit 2. Power
“4. The amount of power is
related to the rate at which energy is transformed” (syllabus)
“Students learn to; explain that power is the rate at which energy is transformed from one form to another” (syllabus)
Power
is the rate of doing work.
Power
= work/time
Since
work and energy are the same type of quantity and energy can not be created nor
destroyed then power is the rate at which energy is transformed from one type
to another e.g. From electrical to heat energy in a toaster.
“Students learn to; identify the relationship between power, potential difference and current” (syllabus)
Power
is equal to the product of the potential difference and the current. P = VI
where
P = power (watts W), V = potential difference (volts V) and I = current (amps
A)
“Students learn to; identify that the total amount of energy used depends on the length of time the current is flowing and can be calculated using: Energy = VIt” (syllabus)
Power
is the rate at which energy is transformed. Power = energy/time
It
follows then that Energy = power x time
Since P = VI then Energy = VIt
“Students learn to; explain why the kilowatt-hour is used to measure electrical energy consumption rather than the joule” (syllabus)
A
watt is a joule per second. Electrical appliances consume many watts so over a
period of time they would use a huge number of joules. Consequently a joule is
too small a quantity for convenient measurement of the amount of electrical
energy used. The kilowatt-hour has been adopted as the unit to measure
electrical energy consumption. The kilowatt-hour is the amount of electrical
energy used by a one kilowatt appliance operating for one hour i.e. 1000 joules
per second for 1 hour.
1
kilowatt hour = 3 600 000 joules.
“Students; perform a first-hand investigation, gather information and use available evidence to demonstrate the relationship between current, voltage and power for a model 6V to 12 V electric heating coil” (syllabus)
This
is best done in the laboratory. A description of a typical experiment is shown
below.
Rebecca
and Ben carried out an experiment to examine the relationship between current,
voltage and power for a model electric heating coil. They took about a metre of
resistance wire and wound it into a coil around a pencil. They left straight
lengths of about 10 cm of wire at each end. They used alligator clips to
connect the coil into a circuit with a power pack, rheostat, voltmeter and
ammeter as shown in the circuit below. They then used a measuring cylinder to
measure 100 mls of water into a styrofoam cup. They immersed the coil in the
water and measured the temperature with a thermometer.
They
set the power pack at setting “E”, turned on the power and adjusted the
rheostat until the current flowing through the circuit was exactly 3.0 amps.
They read the voltmeter and noted that the potential difference across the coil
was 2.6 volts. They kept the current constant at 3.0 amps by making minor
adjustments to the setting on the rheostat. They measured and recorded the
temperature of the water and the potential difference across the coil every minute
for ten minutes. A table of their results is shown below.
|
Time (minutes) |
Potential Difference (Volts) |
Temperature (oC) |
|
0 |
2.6 |
18.1 |
|
1 |
2.6 |
18.9 |
|
2 |
2.7 |
19.8 |
|
3 |
2.7 |
20.7 |
|
4 |
2.7 |
21.7 |
|
5 |
2.8 |
22.6 |
|
6 |
2.8 |
23.4 |
|
7 |
2.8 |
24.2 |
|
8 |
2.8 |
25.0 |
|
9 |
2.8 |
25.7 |
|
10 |
2.8 |
26.4 |
Table Showing Temperature and Voltage at 1- Minute Intervals
Q.1. Given
that it takes 4.2 joules of energy to raise the temperature of 1 gram of water
by 1oC and that 1 gram of water occupies 1 millilitre, calculate the
amount of energy absorbed by the water.
A.1. Temperature
rise = 26.4 – 18.1 = 8.3oC
4.2
joules per gram = 420 joules per 100 grams per oC
=
420 x 8.3 joules total energy = 3486 joules
Q.2. At
what rate in joules per second was energy being absorbed by the water?
A.2. 3486
joules per 10 minutes or 600 seconds = 3486/600 joules/second = 5.81 joules per
second.
Q.3. What
is another name for the units “joules per second”?
A.3.
Q.4. What
was the average potential difference (2 d.p.) across the coil over the 10
minute period.
A.4. Average
= (2.8 x 6 + 2.7 x 3 + 2.6 x 2)/11 =
2.74 volts
Q.5. Use
the formula P = IV to calculate the average power dissipated by the coil.
A.5. Average
voltage = 2.74V P = IV = 3 x 2.74 =
8.22W
Q.6. Why
do you think that the power dissipated by the coil was greater than the heat
absorbed by the water?
A.6. No
allowance was made for the heat absorbed by the styrofoam cup and by the metal
in the resistance wire. Heat was also lost to the surrounding air.
Q.7. Suggest
a reason why the voltage rose as the experiment progressed.
A.7. The
resistance of the coil rose as it heated up.
I = V/R Since
I is constant then as R rises so does V.
Q.8. What
is the independent variable in this experiment?
A.8. Time
Q.9. What
is the dependent variable in this experiment?
A.9. Temperature
Q.10. Suggest
a way that the accuracy of the experiment could be improved.
A.10. Place
a lid on the styrofoam cup and wrap it in insulating material to minimise heat
loss.
Determine
the heat absorbed by the coil and include this in the calculations.
“Students; solve problems and analyse information using: P = VI and Energy = VIt” (syllabus)
Q.1. (a)
What power is consumed by a light globe operating with a potential difference
of 12 volts across it if a current of 5 amps flows through it?
(b) How much energy would it use in
5 minutes?
A.1. (a) P = VI = 12 x 5 = 60 watts.
(b) Energy = VIt = 12 x 5 x 300 = 18
000 joules. (Note that t is in seconds)
Q.2. A heater dissipates 1000 watts and operates
from a 250 volt supply.
(a) How much current passes through it?
(b) How much energy would it use in 10 minutes?
A.2. (a) P = VI I
= P/V = 1000/250 = 4 amps.
(b) Energy = VIt = Pt = 1000 x 600 =
600 000 joules = 600 kJ.
“5. Electric currents also
produce magnetic fields and these fields are used in different devices in the
home” (syllabus)
“Students learn to; describe the behaviour of the magnetic poles of bar magnets when they are brought close together” (syllabus)
When
two bar magnets are brought close together the magnetic fields around their
poles will produce a force between the two magnets. If two like poles (north
pole and north pole OR south pole and south pole) are brought close together
they will repel each other while if two unlike poles (a north pole and a south
pole) are brought close together they will attract each other.
Like
poles repel: Unlike poles attract.
“Students learn to; define the direction of a magnetic field at a point as the direction of force on a very small north magnetic pole when placed at that point” (syllabus)
The
direction of a magnetic field at a point is the direction of force on a very
small north magnetic pole when placed at that point. However magnetic poles
don’t come as monopoles (one pole only) but come as pairs of north and south poles.
If you place a suspended magnet at the point then it will line up in the
direction of the magnetic field with the north pole of the magnet pointing in
the direction of the magnetic field. The direction of the magnetic field is the
direction that the north pole points. The practical method of determining the
direction of a magnetic field at a point is to place a compass at the point and
the direction of the magnetic field is the direction that the compass points.
“Students learn to; describe the magnetic field around pairs of magnetic poles” (syllabus)
The
magnetic field around pairs of magnetic poles is best described by describing
the lines of magnetic flux. Lines of flux are directed from the north pole to
the south pole and never cross. In a strong magnetic field they are close
together while in a weak magnetic field they are well apart.
The
simplest example of a magnetic field around a pair of magnetic poles is shown
by the bar magnet.
Two
examples of the magnetic field around pairs of magnets are shown below 
“Students learn to; describe the production of a magnetic field by an electric current in a straight current-carrying conductor and describe how the right hand grip rule can determine the direction of current and field lines” (syllabus)
Moving
electric charges produce a magnetic field. When these charges are moving
through a conductor the lines of magnetic flux form concentric circles around
the conductor. The direction of the magnetic field can be found by applying the
right hand grip rule. Grip the wire with the thumb pointing in the direction of
the flow of current. (i.e. conventional current, which is the direction of
positive charge flow). The fingers now wrap around the wire in the direction of
the magnetic field.
The
existence and direction of the magnetic field can be verified by holding a
compass at various positions around a straight, current-carrying conductor. The
existence of the magnetic field can also be verified by passing the conductor
through a sheet of cardboard that is perpendicular to the conductor and
sprinkling iron filings onto the cardboard, tapping gently and noting the
pattern formed by the iron filings.
“Students learn to; compare the nature and generation of magnetic fields by solenoids and a bar magnet” (syllabus)
A
solenoid is a coil of insulated wire of many turns. It is long compared to its
diameter. A current through the solenoid will produce a magnetic field and the
direction of the field can be determined by the solenoid rule. Grip the
solenoid with the fingers wrapped around the solenoid pointing in the direction
of the current. The thumb will point to the north pole of the solenoid. The
lines of magnetic flux are continuous and the direction of the magnetic field
is from north to south outside the solenoid but from south to north inside the
solenoid.
Magnetic Field Around a Solenoid
Since
magnetic fields are due to moving electric charges it seems puzzling why a
piece of iron but not other metals can become magnetised. In most metals the
atoms have a random alignment so that the electrons orbit the atom in random
directions and the minute magnetic fields due to each atom cancel out. In iron,
cobalt and nickel atoms tend to group together in domains where the atoms are
aligned so that their electrons spin in the same direction. In an unmagnetised piece of metal, the
domains are randomly orientated so they have no overall magnetic effect. In a
magnetised piece of iron the domains are aligned so that the electrons spin in
the same direction and produce a magnetic field. Consequently, when it is in a
magnetic field, an iron bar can be magnetised to produce a north pole at one
end and a south pole at the other. The lines of flux travel from the north pole
to the south pole. Cobalt and nickel can also be magnetised but not as readily
as iron.
Magnetic Field around a Single Bar Magnet
“Students; plan, choose equipment or resources for and perform a first-hand investigation to build an electromagnet” (syllabus)
Plan: You have to build an electromagnet. Find out what
you need and how it is assembled.
Choose Equipment: Make an accurate list of all equipment
needed to build your electromagnet.
Perform a first-hand investigation: You can do this in the
laboratory but you can also make an electromagnet at home.
A possible approach to making an electromagnet is
described below.
A soft iron core is the most suitable starting point
for an electromagnet since this will produce a temporary magnet that loses its
magnetic properties when the current is switched off. However most commercially
available iron products such as bolts or nails are hard steel and retain their
magnetic properties after the current has been switched off i.e. they become
permanent magnets. If no soft iron core is available then iron bolts and nails
can be softened to a certain extent by annealing them, i.e. heating them to red
heat and allowing them to cool slowly.
Procedure: Heat three long iron nails to red heat with a
bunsen burner and allow them to cool slowly. When cool, bind them together with
sticky tape. Wind about two metres of thin, insulated wire around the nails and
connect the ends of the wire to the terminals of a 6-volt lantern battery (This
is a large battery and can deliver a high current.) Test the effectiveness of
your electromagnet by seeing how many small nails or paperclips it can pick up.
Disconnect the wire from one terminal of the battery and again see how many
nails or paperclips can be picked up when no current is flowing.
“Students; perform a first-hand investigation to observe magnetic fields by mapping lines of force
- around a bar magnet
- surrounding a straight DC current-carrying conductor
- a solenoid
- present information using o and k to show the direction of a current and direction of a magnetic field ” (syllabus)
Bar Magnet: Place a bar magnet on a bench and cover it with a
sheet of thin, firm cardboard. Sprinkle iron filings over the cardboard and tap
the cardboard lightly a couple of times so that the iron filings align
themselves with the magnetic field. Take a photo of the iron filings showing
the lines of force.
An
alternative is to use diazo paper (a light sensitive paper). Place the diazo
paper over the bar magnet and sprinkle iron filings over the diazo paper. Leave
the paper exposed to sunlight for three minutes (the time depends on how sunny
it is). Return the iron filings to their container and develop and fix the
diazo paper by exposing it to ammonia. This is done by pouring some strong
ammonia solution in the bottom of a desiccator and placing the diazo paper
above the ammonia in the desiccator. An example of a magnetic field recorded
with diazo paper is shown below.
Magnetic Field around a Single Bar Magnet
Straight DC current-carrying
conductor: Set up a tall retort stand with a bosshead and clamp at the top and at the
bottom of the retort stand and with a large retort ring in the middle. Use a
small nail to make a small hole in the centre of a sheet of thin, firm
cardboard about 20 cm square. Thread a length of insulated wire through the
cardboard and place the cardboard on the retort ring. Wrap the wire around the
clamps at the top and bottom of the retort stand so that there is a vertical
length of wire running through the cardboard. Connect the ends of the wire to a
power pack. And sprinkle iron filings onto the cardboard. Turn on the power and
tap the cardboard lightly a couple of times. Take a photo of the iron filings
showing the lines of force.
Solenoid: Make a solenoid by winding a long length of
insulated wire around a cardboard tube. (The cardboard tube from the centre of
a toilet roll is ideal). Cut a sheet of firm cardboard so that it goes around
the solenoid with a tongue to insert into the solenoid at each end. Place the
solenoid between two books on a bench and place the cardboard on the books with
the tongues inserted into the solenoid. Connect the wire at the two ends of the
solenoid to the terminals of a power pack and switch it on. Sprinkle iron
filings over the cardboard and tap gently. Take a photo of the iron filings
showing the lines of force.
Note:
In the above demonstrations, the lines of force can be mapped by placing a
compass at various points in the magnetic field instead of using iron filings.
The compass has the advantage of showing the direction of the magnetic field as
well as being able to demonstrate weaker magnetic fields.
Direction of Magnetic Field:
Q.
Show the
direction of the magnetic field above and below a wire carrying a current to
the right in the plane of the paper.
A. k k k k k
k k k k k
I
o o o o o
o o o o o
Q. A wire carries a current vertically into the plane
of the page. Show the direction of the magnetic field above and below the wire.
A. By the
right-hand grip rule the direction is shown by the arrows below.
B
o
B
“Students; identify data sources, gather, process and analyse information to explain one application of magnetic fields in household appliances” (syllabus)
Applications
that can be studied include electric motors, switches and relays, electric
bells, magnetic audio and video tapes and magnetic computer discs.
The
application studied here will be the loudspeaker used in radios, televisions,
audio equipment and computers.
The loudspeaker consists of a powerful magnet with
light cylindrical coils, known as the voice coils, mounted over the central
pole piece without touching it. The output current from the audio device
fluctuates as it passes through the voice coil and causes a fluctuating
magnetic field in the voice coil. The interaction of the fluctuating magnetic
field of the voice coil with the fixed magnetic field of the pole piece
produces a changing force on the voice coil that causes it to vibrate. The rigid
paper cone is attached to the voice coils so the vibration is transmitted to
the paper cone. This causes the air around it to vibrate and this is detected
as sound.
“6. Safety devices are important in household circuits” (syllabus)
“Students learn to; discuss the dangers of an electric shock from both a 240 volt AC mains supply and various DC voltages, from appliances, on muscles of the body” (syllabus)
An
electric shock occurs when an electric current passes through the body. Most
schools use power packs with circuit breakers that prevent currents in excess
of 5 amps flowing through the circuit but a current of a fraction of an amp
flowing through the human body is sufficient to cause death. Thankfully the
human body is covered with skin that is a pretty good insulator but this
insulation breaks down if the skin is wet. The resistance of around 0.5 MW for dry skin falls to about 100W if the skin is
wet. Add this to the 100W internal resistance of the
human body and you can see that a wet person has a total resistance of only
around 200W.
While
The
severity of an electric shock depends on several factors, among them being
(i)
The amount of
current passing through the body.
(ii)
The length of
time that the current passes.
(iii)
The path of the
current through the body.
(iv)
Whether the
current is A.C. or D.C.
(v)
The general
fitness and well being of the victim.
The
muscles of the body operate in response to electrical impulses from nerve cells
that cause the muscles to contract. An electric shock will interrupt these
nerve impulses and can cause muscles to contract independently. For this reason
a person who grasps hold of a wire carrying a current can find that they are
unable to let go.
As
a general guide, currents of 100-200mA are most dangerous. These can cause
ventricular fibrillation, i.e. rapid irregular contractions of the heart
muscles, a situation where the heart flutters rapidly and doesn’t pump
properly. CPR cannot normally restart a fibrillating heart and death usually
results. A fibrillating heart can be restarted with a defibrillator, a device
designed to apply a controlled electric shock across the heart. These are
available in most NSW ambulances so their rapid response to electrocution cases
is vital. Currents above 200mA do not usually cause ventricular fibrillation
although they can cause the heart to stop beating. These victims will normally recover
with prompt CPR and efficient medical treatment.
The
longer a current flows through the human body the more dangerous it is. For
example, a current of 0.1A flowing for a tenth of a second would
normally have no ill-effects on the body but the same current flowing for ten
seconds or more would probably cause death.
A
current through the hand will cause the muscles of the hand to contract so that
the person is unable to let go of the wire. This will increase the time that
the current passes through the body with possible fatal consequences. A current
across the chest can cause paralysis of the chest muscles so that the victim is
unable to breathe. However a much more serious consequence of a current across
the chest is its disruption of the heart rhythm and the onset of ventricular
fibrillation. Current across the chest is most likely to occur when the current
enters through one hand and leaves through the other hand or a foot. This can
occur if a person experiencing a shock grabs something connected to earth such
as a tap or pipe or if a person experiencing a shock is standing on something
connected to earth such as a concrete floor.
A.C.
is much more dangerous than D.C. as its changing polarity is much more likely
to cause fibrillation. This is particularly true of the 50 Hz frequency used in
A
fit and robust person is more likely to survive an electric shock that someone
in poor health, particularly if the poor health is related to the heart.
However, just as there are unmeasurable individual differences that make some
people more susceptible to some diseases than others, so too some people are
more affected by an electric shock than others.
Time Effect of Shock on the Human Body
|
Electric Current (mA) |
Duration (ms) |
Effect |
|
50 |
10-200 200-4000 >4000 |
Usually no dangerous effect Fibrillation unlikely Up to 50% probability of fibrillation |
|
100 |
10-100 100-600 600-10 000 >10 000 |
Usually no dangerous effect Fibrillation unlikely Up to 50% probability of fibrillation More than 50% probability of fibrillation |
|
500 |
10-40 40-500 >500 |
Fibrillation unlikely Up to 50% probability of fibrillation More than 50% probability of fibrillation. |
Table
copied from O.T.E.N. notes:
Electrical
Energy in the Home Part 6: Electrical Safety
“Students learn to; describe the functions of circuit breakers, fuses, earthing, double insulation and other safety devices in the home” (syllabus)
Properly
constructed circuits have safety devices that cut off the current if it exceeds
a level that is deemed safe.
An
excess of current can
(i)
Cause expensive
components in a circuit to burn out.
(ii)
Cause wires in
inaccessable places to overheat and cause fires.
(iii)
Cause
appliances to become “live” and electrocute anyone who touches them.
The
most basic safety device in a circuit is a fuse. A fuse is just a piece of wire
in a circuit and will heat up and melt if too much current flows through it.
When the wire melts it breaks the circuit. While some older houses still have
fuses they have been superseded in household circuits by circuit breakers and
safety switches. Fuses have the disadvantage of being awkward to replace and
not cutting off the electricity supply as quickly as safety switches. Fuses are
still used in cars and in some appliances such as televisions.
Original
circuit breakers worked on one of the following principles. In the first type,
too much current would cause a bimetallic strip to heat and bend away from its
contact points. In the second type, too much current would increase the
strength of an electromagnet so that it pulled an iron bar away from its
contacts.
The
most efficient type of circuit breaker is earth leakage device known as a safety
switch. This detects a difference between the current flowing into the house
through the active line and the current leaving in the neutral line. If current
leaks to earth the safety switch will cut off the current in a few
milliseconds, which is less time than it takes to electrocute someone. Because
safety switches are safer than fuses or conventional circuit breakers they are
mandatory on all new buildings and on those undergoing renovations.
While
some double insulated appliances have only two pins in the plug that is
inserted into the power point, any appliance with a metal casing will still
have a three-pin plug. The vertical pin at the bottom is an earth connection.
Normally the earth wire does not carry any current. However, if there is a short
circuit in the appliance and the outer casing becomes “live” then the earth
wire will carry the excess current to earth and either blow the fuse or trip
the circuit breaker.
You
can recognise a double insulated appliance by its two-pin plug. The internal
components of a double insulated device are completely enclosed in plastic and
this prevents contact with any outside metal parts.