PEGS Unit 2 Physics
Name:_________
Lighting flashes hit the desert near Tucson, Arizona in this time exposure photograph. Each lighting bolt is around one million volts and carries
enormous amounts of energy.
*Have a look at the LIGHTNING website.
1
Benjamin Franklin (1706-1790) was first to
use terms such as charge, battery, positive
and negative.
PEGS Electricity booklet
Tall structures are frequently
struck by lighting!
Introduction to Electricity
(i) Atomic structure: our understanding of the atom has changed dramatically over time.
(ii) Static electricity: it has been known since ancient times that some materials become charged when
rubbed. We now know that this is because outer-shell electrons are being rubbed on or off due to friction.
Before
After being rubbed together.
(iii) Attraction and repulsion: we know that forces (electrostatic) exist between charged objects.
The Van de Graaff generator is often used as a source of electrostatic charge.
*Refer to pg39 in your text and use notes and diagrams to explain how a Van de Graaff works.
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Just as we use units such as metres, kilograms, newtons, etc to measure physical quantities in kinematics, we
use electrical units and quantities during the study of electricity.
(iv) How do we measure electric charge?
A charged object has an excess or deficit of electrons. An electron has the smallest quantity of charge that can
exist in nature. Charge (Q or q) is measured in coulombs, C.
 Fundamental quantity of charge: 1 e- = 1.6×10-19 C
Example 1. A speck of dust has an excess of 5 electrons. What is its charge in coulombs?
(-8.0×10-19C)
Example 2. A balloon is rubbed and loses 20,000 electrons. What is its charge in coulombs?
(+3.2×10-15 C)
Each electron has 0.00000000000000000016 C of charge. This is a tiny amount! So if an object has a charge
of 1 coulomb, there must be an enormous number of electrons that have transferred. You should be able to
calculate how many!
 1 coulomb =
electrons
♥ Read worked example 2.1A on page 40 of your text.
How does the number of electrons on a fully charged Van de Graaff dome compare with the number on an
uncharged dome?
(v) Electrical potential energy
In mechanics, when a force acts on an object over a distance, work is done and a change in energy measured in
joules occurs. It is the same in electricity.
Consider a small negatively-charged particle B near to a large positively-charged object A.
B
A
There are forces acting on both objects, but we will only analyse B. We can say that B has electrical potential
energy because if it were free to move it would accelerate towards A and gain kinetic energy. The electric
force of attraction is doing work on B and causing it to lose potential energy and gain kinetic energy.
Q: How does electrical potential energy depend on:
the amount of charge involved?
the distance between the charges?
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(vi) Electrical potential at a point
This is defined as the amount of electrical energy (J) that one coulomb of charge would possess at that point.
Potential is measured in joules per coulomb. The usual way of saying this is volts, V.
For example, in the diagram above, if B had a charge of -1.0 C and 10 J of electrical potential energy, we
would say that its electrical potential is 10 joules per coulomb or 10 volts.
 1 volt = 1 joule per coulomb
1 V = 1 J/C
Example: If a charge of 0.10 C had 3.0 J of electrical potential energy, calculate its electrical potential
(in volts).
(30 V)
(vii) Potential difference (ΔV)
Consider two points A and B near to a positive central charge.
A
B
The electrical potentials at A and B are different. To compare the potential at these points, the convention is
that a positive sample charge is used.
The electrical potential at A is greater than the potential at B. Can you explain why?
If for example the electrical potential at A is 240 J/C (240 V) and the potential at B is 60 J/C (60V), the
electrical potential difference ΔV can be found: ΔV = 240 V – 60 V = 180 V
 By definition, positive charges will travel from high potential to low potential.
 Negative charges will move from low potential to high potential.
So as our charge moves from A to B, its potential difference is 180 V or 180 J/C. Each coulomb of charge will
lose 180 J of electrical potential energy and gain 180 J of kinetic energy.
(vii) Voltage and EMF
Cells and batteries provide an electrical potential difference across their terminals. So if a conducting path
was connected across these terminals, positive charges would tend to move from the high potential terminal to
the low potential terminal. This potential difference is known as the voltage of the cell. (An old term for
potential difference is EMF – electromotive force).
On a battery, the high potential terminal is marked with a + sign and the low potential with a – sign.
Convention states that positive charges will travel from the high potential (+)
to the low potential (-) terminal as they travel around the circuit.
(In reality, positive charges do not flow at all – the electrons do!)
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 Voltage (or EMF) of battery = energy (J) given to each coulomb of charge.
If a cell is marked 1.5 V, this means that each coulomb of
charge is given 1.5 J of electrical energy by the cell. This
energy is then carried by the charges and transformed into
another type of energy (heat, light, etc) depending on the
circuit
(ix) Electric current
When electric charges move between two points in a circuit, an electric current is said to flow. This will
happen if there is a potential difference between the points.
The direction of current flow is defined as the direction
in which positive charges would move. This is known as
conventional current, I. As previously discussed,
positive charges and so conventional current I will flow
around the circuit from the + to the – terminal of the battery.
NB: It is actually the electrons (outer shell) that travel around a circuit.
The positive charges are immobile and are locked in the nuclei
of the atoms. Draw this circuit diagram again, but showing the flow
of electrons (electron current) around the circuit.
 Current I is defined as the number of coulombs (C) of charge that pass a point in the circuit
each second.
I=
ch arg e (C ) Q
=
time (s)
t
Unit of current: ampere (A)
1 A = 1 C/s, so 1C = 1 A s
!
Example 1: Calculate the current in conductor if 10 C of charge flows past a point in 5.0 s.
(2.0 A)
Example 2: A current of 300 mA flows for 5 minutes and 20 seconds. How much charge has flowed?
(96 C)
*Refer to page 52 of your text, table 2.2 for some typical values of electric current.
♦ Homework: Text pg 58, 2.3 Q’s 1-8
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ELECTRIC CELLS AND BATTERIES
In 1790, Allesandro Volta developed the first way of reliably producing electricity. He invented a device
which transformed chemical energy into electrical energy. It was known as a wet cell.
An electric cell is a source of electrical potential energy.
A battery is a series of cells joined together.
TYPES OF CELLS
Primary cells use a chemical reaction and convert stored chemical energy into electrical energy in a process
that cannot be reversed.
Secondary or storage cells can be reused. Energy is put into the cell by applying a voltage causing a current
through in one direction during the charging process. Then the cell can discharge this energy as the current
flows in the opposite direction.
Charging reverses the chemical reaction and restores chemical energy to the cell. During use, the cell
discharges and loses energy as the chemical process runs down.
A solar cell (photovoltaic cell) converts light energy into electrical
energy. These are used in many situations.
A thermocouple transforms heat energy into electrical energy.
These are used for measuring very high temperatures where
thermometers are unsuitable.
Basis ideas about chemical cells and batteries
 All cells have positive and negative terminals called electrodes.
 When the positive and negative terminal is connected by a conducting path, the valence electrons in
the external circuit all flow towards the positive terminal of the cell.
 Conventional current I flows from the positive to the negative terminal.
 The two electrodes are made of different materials and are often metals. These are immersed in an
electrolyte – a substance that conducts electricity when in solution.
Internal circuit: Chemical reactions create a movement
of charge inside the cell.
Show this movement of + and – ions on the diagram.
External circuit: The potential difference created in the
internal circuit causes and electric current to flow in
the external circuit.
Show this current I on the diagram to the right.
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The first cell was a wet cell invented by Volta. It used copper electrodes and sulphuric acid as the electrolyte.
Wet cells that produce 12V are used in cars.
The cells that we use in our mobile phones, iPods, torches and calculators are dry cells. In these, the electrolyte
is in paste or solid form.
In a single cell, a chemical reaction provides the energy to separate the electrons from their parent nuclei.
Since each individual chemical reaction releases a particular amount of energy, each electron receives a
particular amount of electrical potential energy. Each coulomb of positive charge is therefore given the same
amount of potential energy.
 The energy that is supplied to each coulomb of charge is known as the voltage or EMF of the battery
and is measured in volts, V.
 1 volt = 1 joule per coulomb i.e. 1V = 1J/C
" energy
ch arg e
" energy = work W = q"V = "V # I # t
Potential difference "V =
The EMF or potential difference provided by a battery or cell will depend on the materials that are used for the
electrodes
! and electrolyte.
♦ Homework: Text pg 58: 2.3 Q’s 9-14.
Text pg 96: 3.3 Q’s 1-5.
Worksheet (PTO)
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TEST YOUR UNDERSTANDING
Question 1 Part of an electric circuit is shown below.
A
VA = 10.0 V
B
load
VB = 8.0 V
(a) How much electrical potential energy does a coulomb of charge have at: A?
B?
(b) What is the potential difference between A and B?
(c) Which way will conventional current I flow in this circuit? Show this on the diagram.
(d) When +1.0 C of charge moves from A to B, does it gain or lose potential energy? Calculate the magnitude
of this energy change.
(e) The +1.0 C of charge does not gain kinetic energy as it travels from A to B. Why not? What has happened
to the lost potential energy?
(f) If 0.75 C of charge moves between A and B, what is its change in potential energy (in joules)?
(g) If the current in the circuit is 10 A, how many coulombs pass a given point in 5.0 s?
(h) Calculate the total change in potential energy of the charges in the circuit during this 5.0 s.
Question 2
(a) Show the direction of the current flow I on the diagram.
(b) How much electrical potential energy does the battery give each
coulomb of charge?
(c) How much electrical potential energy does each coulomb of charge
lose in the external circuit?
(d) How much electrical potential energy does +5.0 C of charge lose in the external circuit?
(e) A current of 0.30 A flows around the circuit for 5.0 minutes. How much electrical energy (in joules) has
the battery provided in this time?
Answers: 1 (a) 10 J, 8.0 J (b) 2 V (c) A to B (d) loss of 2 J (e) Too many collision. Heat energy in load (f) -1.5
J (g) 50 C (h) – 100 J 2 (a) c’wise (b) 3 J (c) 3 J (d) 15 J (e) 270 J
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Taking measurements in electric circuits
A variety of meters can be used to measure characteristics of a circuit. These can be analogue (with a needle)
or digital (numerical display). The meters are designed so that they do not influence the circuit in any way.
Meters have a polarity (i.e. a + and – terminal) so it matters which way they are connected into a circuit.
Ammeters:
measure current flow (A, mA) in a circuit
are connected in series with the component being analysed
have very low resistance (≈ zero ohms) so as not to affect circuit
the + of the meter should be on the high potential side of the circuit
Question: In which of the circuits below is the ammeter incorrectly connected? (The dot on the meter is +)
A
Voltmeters:
B
C
measure potential difference (V) across a component
are connected in parallel with the device being analysed
have a very high resistance (≈ MΩ) so as not to draw any current
the + of the meter should be on the high potential side of the circuit
Problem: Draw a voltmeter on each circuit diagram (showing the + terminal) to measure the potential
difference across the:
(a) globe
(b) battery
(c) resistor
Problem: You connect 3 globes in series to a 9 V battery. You need to measure the current flowing through
the middle globe and the potential difference across the first globe. Draw a circuit diagram showing how this
circuit should be connected.
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Multimeters can measure a variety of quantities. These have a polarity and if used as an ammeter or
voltmeter, must be connected taking this into account. A typical digital multimeter is shown below.
1. Show where the control should be positioned to measure:
(a) DC current
(b) DC voltage
(c) resistance
2. It is on the 2000m voltage setting and the
meter reads 150. What is this voltage value?
NB: you must turn the dial to OFF after use!
 Two probes are used with a multimeter. The red probe (high potential or positive) connects into the
middle socket.
 The black probe (low potential, negative, common, earth or ground) connects into the bottom socket.
Show these coloured probes on the multimeter above.
Tips for wiring up a DC circuit
1. Make sure the power supply is turned to zero and that switches are in the off position.
2. Work from a circuit diagram.
3. Start at the high potential side or positive of the power supply or battery.
4. When you reach a meter, connect into the high potential (+) first. When using analogue meters, use the
largest (least sensitive) scale first. On ammeters (5A) or voltmeters (15 V)..
5. You should finish at the negative of the power supply or battery.
6. Connect voltmeters into the circuit last.
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Investigation of electrical c ircuits
In these 5 sections, use the multimeter to measure voltage and an analogue ammeter for current.
Make sure that you have 12 V globes and 6 V batteries. You should carefully observe the intensity of the
globes as you work through these circuits.
PART 1
Switch open
(off)
Switch closed
(on)
1. Use the ammeter to measure the current flowing
through the globe.
2. Position the voltmeter (multimeter) to measure the
potential difference across the battery.
3. With the voltmeter still connected, measure the current
flowing through the globe again.
4. Position the voltmeter to measure the potential
difference across the globe.
5. Use the voltmeter to measure the potential difference
across the ammeter.
6. Use the voltmeter to measure the potential difference
across the switch.
(a) Consider steps 1 and 3. Does putting a voltmeter into the circuit have any effect on the current flow? What
does this tell you about the resistance of a voltmeter?
(b) In step 5, is there a significant potential difference across the ammeter? What does this tell you about the
resistance of the ammeter?
(c) How easy is it for the charges to pass through a switch when it is closed?
(d) Compare the potential differences across the battery when the switch is open and closed. What does this
tell you about the energy output of the battery when a current is flowing? Can you think of a reason for this?
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PART 2: batteries in parallel
Note that the positive of one battery is connected to the positive of the other.
Battery 1
Battery 2
Switch open (off)
Switch closed (on)
1. Use the ammeter to measure the current flowing through
the globe.
2. Position the voltmeter so that it measures the potential
difference across battery 1.
3. Position the voltmeter so that it measures the potential
difference across battery 2.
4. Position the voltmeter so that it measures the potential
difference across the light globe.
5. Position the voltmeter so that it measures the potential
difference across the ammeter.
6. Compare the brightness of the globe to that in part one.
(a) Compare the current flowing through this globe to that in part 1. What do you notice? Is this what you
expected to happen
(b) Compare the potential difference across this globe to that in part 1. What do you notice? Is this what you
expected to happen?
(c) Given these observations, is there any advantage in using two batteries in parallel like this? Explain
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PART 3: batteries in series
Battery 1
Battery 2
Switch open (off)
Switch closed (on)
1. Use the ammeter to measure the current flowing through
the globe.
2. Position the voltmeter so that
difference across battery 1.
3. Position the voltmeter so that
difference across battery 2.
4. Position the voltmeter so that
difference across the light globe.
5. Position the voltmeter so that
difference across the ammeter.
it measures the potential
it measures the potential
it measures the potential
it measures the potential
6. Compare the brightness of the globe to that in part one.
(a) Compare the current flowing through this circuit to the current flowing through the globe in part 1. Is this
what you expected to happen?
(b) Compare the potential difference across the globe in this circuit to the globe in part 1. Is this what you
expected to happen?
(c) Explain using electrical terms why this globe is much brighter than the globe in part 1.
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PART 4: globe in series
Globe 1
Globe 2
Switch closed (on)
1. Use the ammeter to measure the current flowing from the battery.
2. Position the voltmeter so that it measures the potential difference
across globe 1.
3. Position the voltmeter so that it measures the potential difference
across globe 2.
4. Position the voltmeter so that it measures the potential difference
across the battery.
(a) Compare the brightness of globe 1 and globe 2. What does this tell you about the two globes?
(b) Compare the brightness to the globe in circuit 1. Is this what you expected?
(c) Compare the potential difference across globe 1 and globe 2. What does this tell you about the globes?
(d) Compare the potential differences across globes 1 and 2 with the potential difference across the battery? Is
this what you expected to find? Explain these observations in terms of energy per coulomb and the flow of
current in the circuits.
(e) By considering the current through and the potential difference across the globes, explain you earlier
observations about the intensity of these globes and the globe in part 1.
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PART 5: globes in parallel
Globe 1
Globe 2
Switch closed (on)
1. Use the ammeter to measure the current flowing from the battery.
2. Position the ammeter so that it measures the current flowing through globe 1.
3. Position the ammeter so that it measures the current flowing through globe 2.
4. Position the voltmeter so that it measures the potential difference across the
battery.
5. Position the voltmeter so that it measures the potential difference across globe
1.
6. Position the voltmeter so that it measures the potential difference across globe
2.
(a) Compare the intensity of the two globes. What does this tell you about your globes?
(b) Compare the globes intensities to the globe in part 1. Is this what you expected?
(c) Compare the potential difference across each globe. How does this compare with the potential difference
across the battery? Is this what you expected?
(d) Explain these observations in terms of the energy per coulomb and the flow of charge in this circuit.
(e) Compare the current flowing through globe 1 and globe 2 with the current flowing from the battery.
Explain this in terms of flow of charge in this circuit. Is this what you expected?
(f) Having now considered the current through and the potential difference across the globes, explain you
earlier observation about the intensity of these globes and the globe in part 1.
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ELECTRICAL RESISTANCE
A conductor is a material that electricity is able to travel through freely.
List some electrical conductors:
An insulator is a material that electricity does not pass through readily.
List some electrical insulators:
Question: Why is electricity able to pass easily through a conductor? What is different about insulators?
All materials conduct electricity to some degree. In metals and other conductors, even a small voltage will
cause electricity to flow. In insulators, a large voltage can result in a current flowing through a material. The
degree of difficulty for flow of charge through a material is called the electrical resistance. This is measured
in ohms, Ω
In a metal circuit wire with low resistance, the electrons through flow easily and so do not lose much energy
along the wire. In the filament of a light globe which has a high resistance, the electrons undergo many
collisions with atoms and other electrons. This causes the wire to heat up and glow. The electrical energy of
the charges has been transformed into heat and light.
Question: Label this circuit diagram. Given that the
switch is closed, show the parts of the circuit where the:
(a) electrical energy is produced.
(b) electrical energy is used.
(c) external circuit has low resistance.
(d) external circuit has high resistance.
The resistance R of a material depends on a number of factors; its length, thickness and the material itself.
Question: Which has higher resistance: 1 m of wire or 2 m of wire?
Question: Which has higher resistance: 1 m of thin wire or 1 m of thick wire?
Question: Which has higher resistance: 1 m of copper or 1 m of rubber (equal thickness)?
Resistance R = "l
A
where R = resistance of material (ohms Ω)
ρ =!resistivity of material (Ω m)
A = cross-sectional area (m2)
l = length (m)
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*Look at page 63 table 2.3 of your text.
 Which material is the best conductor? Why isn’t this metal used in many circuits?
 List a few semiconductors. What are these used for?
 Does temperature affect the resistance of a material? How?
The resistance of an electrical device is defined as the ratio of the potential difference across the device to the
current flow through it.
Resistance:
R=
V
I
Unit: ohms (Ω)
We can define 1 ohm as 1 volt/ampere i.e. 1 Ω = 1 V/A
!
Resistors are devices that are used in electric
and electronic circuits. They are used to control the size of the
current flowing in a circuit. A resistor is shown below.
♠ Work through Worked Example 2.4C on page 63 of your text.
Extension: resistor colour code!!
A resistor has bands: green, blue, orange. Its resistance is:
The resistor pictured has bands that are brown, black, red.
What is its resistance?
 Check out the Resistor Colour Codes (interactive) website
Label the conductors and insulators
on this photograph. Where does the
electricity flow?
In 1880 Thomas Edison was the first to realise that
the filaments in light globes should have a high
resistance. He originally used carbonised cotton
strands in an evacuated bulb.
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Ohm’s Law
If the potential difference V across an electrical component is increased, you would expect that the current I
flowing through the component would also increase. This is usually (but not always!) what happens.
A device for which a direct relationship exists between the potential difference V and current I is said to be
an ohmic device. When the potential difference doubles, the current flow also doubles. Resistors behave in
this way.
If a graph of I vs V was plotted for an ohmic device, it would look like the graph below.
Current (mA)
200
150
100
50
0
Ohm’s Law:
2
V = IR
4
6
where
8
potential difference(V)
V is potential difference (V)
I is current (A)
R is resistance (Ω)
Example: Use Ohm’s law and the above graph to calculate the resistance of the device when:
(a) 50 mA is flowing through it.
(b) 100 mA is flowing through it.
(c) 150 mA is flowing through it..
(d) 200 mA is flowing through it.
(e) What do you notice about the resistance of this device? Is it ohmic or non-ohmic?
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There are many electrical and electronic components that are not ohmic. These are known as non-linear or
non-ohmic devices. LEDs, diodes, transistors and light globes are all non-ohmic devices.
*Look at Figure 2.29 pg 60 of your text and then answer the following questions.
Which of the electronic devices is ohmic?
What do the I-V graphs tell you about the resistance of each component?
The resistance of a non-ohmic device does not remain constant. The ratio of V/I is not constant and
varies as the potential difference across the device changes.
Example: Consider the V-I graph below.
Voltage (V)
20
15
10
5
0
0.2
0.4
0.6
0.8
1.0
Current (A)
What is the resistance of this device when a current of:
(a) 0.1 A is flowing through it.
(b) 0.4 A flows through it.
(c) 1.0 A flows
(d) As the potential difference across this device increases, what happens to its resistance?
Questions
1. What is the potential difference across a 20 ohm resistor when a current of 4.0 A is flowing through it?
(80 V)
2. A current of 100 mA is flowing through a resistor when a potential difference of 25 V is applied
across it. What is the value of the resistance in: ohms? kΩ?
(250 Ω , 0.25 k Ω )
3. A banana has a resistance of 2.5 kΩ. A potential difference of 20 V is placed across the banana. What
is the size of the current flow through the banana (in mA)?
(8.0 mA)
♦ Homework: Text pg 65 Q’s 2.4 1-10
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Electrical energy and power
Electricity is used for transferring energy from one place to another. In Victoria, most of our energy is in the
form of brown coal in the La Trobe valley. This is burnt and the steam that results is used to rotate turbines in
generators. Electricity is created and carried to Melbourne along the transmission lines. Massive amounts of
carbon dioxide are released during this combustion process.
We use this electrical energy around the house for a wide variety of purposes. Discuss the energy
transformations that are involved when you use:
a light globe
a hair-dryer
a dishwasher
The amount of electrical energy used or created by the charges q can be found by using:
Δelectrical energy = ΔE = qV
where q = charge (C)
V = potential difference (V)
Δ E = work done (J)
This energy is released in the form of heat, light, kinetic, sound, etc. In most transformations, heat is created as
a waste of energy.
Since I = q/t
then change in electrical energy can be given by:
Work done W = Δ E = q V = ΔV × I × t
♠ Work through Worked example 2.5A on page 67 of your text.
During the Motion unit, we defined power as the rate at which work is done.
It was measured in joules per second or watts (W). The same applies with electrical power.
"energy VIt
=
= IV
"time
t
For ohmic conductors V = IR, so :
Power P =
Electrical power P = IV = I 2 R =
V2
R
Unit of power : watt(W )
*Look at table 2.4 on page 71 of your text for some typical power consumption values.
!
♠ Work through Worked example 2.5B on page 68.
• What determines the brightness of a globe?
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How power companies charge for electricity
Your family purchases its energy (electrical and/or gas) from a power company. They sell you 240 V (mains)
electricity. When you receive a bill from the power company, they will have looked at your electricity meter to
determine how much energy your house has used. The meter does not give a value in joules. It works in
kilowatt hours (kW h). This is also a measure of energy and is known as a “unit” of electricity.
 A 1 kW device running for 1.0 hour will use 1 kW h of energy
•
Show that 1 kW h is equal to 3.6 MJ of energy.
A typical power bill is shown on page 69 of your text.
♠ Now work through Worked example 2.5C on page 69 and Worked example 2.5D on page 70 of your text.
Question: The compact fluorescent globe uses five times less energy than the incandescent globe, even though
they are producing the same light intensity. Why?
POWER AND WORK PROBLEMS
Assume that there is no loss of energy in the circuit wires i.e. they have negligible resistance.
1. A current of 0.050 A flows from the 3.0 V battery.
(a) What is the power output of the battery?
(b) What is the power consumption of the globe?
2. A 2.5 ohm resistor has a current of 400 mA flowing through it. At what rate is energy being dissipated in the
resistor?
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A voltmeter was used to measure the potential difference across these resistors. There was found to be 2.0 V
across the 10 ohm resistor and 4.0 V across the 20 ohm resistor.
(a) What is the power usage of the 10 ohm resistor?
(b) At what rate is energy being used in the 20 ohm resistor?
(c) What is the power output of the battery?
4. An incandescent globe is marked ‘240V, 100W’.
(a) Does this globe run on batteries or mains electricity?
(b) What does the labelling tell you about how the globe runs?
(c) Calculate the current (in mA) flowing through the globe when it is operating to its specifications.
(d) Calculate the resistance of the globe.
(e) The globe is left on for 5.0 hours. How many kW h of energy does it use? How much does this cost if each
kilowatt hour costs 15c?
5. The same 240 V, 100 W globe is now used with a 60 V power supply.
(a) How will the brightness compare with that in Q4?
(b) Calculate the power dissipated in the globe now.
(c) How does this compare with the power dissipated in Q4?
Answers: 1 (a) 4.5 W (b) 4.5 W 2. 1.0 W 3 (a) 0.40 W (b) 0.80 W (c) 1.20 W 4 (a) mains (c) 420 mA
(d) 580 Ω 5 (a) dimmer (b) 6.3 W (c) lower by a factor of sixteen
♦ Homework: Text pg 75, 2.5 Q’s 1-5.
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Kirchoff’ s Laws
In any electric circuit, the following fundamental principles apply:
•
The total current entering a point is equal to the total current leaving that point i.e. current is never
lost in a circuit.
•
In any closed loop of a circuit, the algebraic sum of the potential differences is zero i.e. the
increase in potential difference provided by the battery is equal to the drop in potential differences
in the external loops.
Series circuits
Often electrical components are connected end to end. This is known as a series circuit. The charges give
some of their energy to each component in a series circuit. These are voltage-divider circuits.
•
Calculate the missing voltage values in these circuits:
Circuit A
Circuit B
In series circuits, the output voltage of the battery is equal to the sum of the potential differences across the
components in the circuit i.e. V = V1 +V2+V3 + ………….
In series circuits, there is only one path for the charges to take, so the current at all points is equal.
In circuit A, the current from the battery is 1A. Show the current in the other parts of circuit A.
In circuit B, the current from the battery is 50 mA. Show the current in other parts of circuit B.
In series circuits, adding more resistors makes it more difficult for charges to travel through the circuit and
so increases the resistance of the circuit as a whole. The equivalent resistance RE is equal to the sum of
the individual resistances:
Resistors in Series: RE = R1 + R2 +R3 + …
Example: For the series circuit shown, calculate:
(a) the effective resistance for the circuit.
(b) the current flowing from the battery.
(6 Ω )
(0.5A)
(c) the potential difference across each resistor.
(1V,2V)
(d) the power produced or used by each device.
(battery:1.5W, 2 ohm: 0.5W, 4 ohm: 1W)
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Parallel circuits
When components are connected in parallel with each other, it means that there is more than one path for
the current in the circuit. These are current-divider circuits.
R1
R2
R3
•
In this circuit, which resistor R1, R2 or R3 has the lowest resistance? How do you know this?
•
Which is the largest resistance resistor? Explain.
The charges lose an equal amount of energy through each parallel path, so the potential difference across
components in parallel is equal.
Having components in parallel makes it easier for the charges to flow around the circuit and so adding
resistors in parallel reduces the effective resistance of the circuit.
Re sistors in parallel
1
1
1
1
=
+
+
+ ...
R E R1 R 2 R 3
Example 1: Assume the switch is closed when answering these questions.
!
(a) Determine the effective resistance of the circuit.
(1.33 Ω )
(b) Redraw the equivalent circuit with the effective resistance.
(c) Calculate the current flowing from the power supply.
(3 A)
(d) Determine the potential difference across each resistor.
(4V across 2Ω and 4 Ω )
(e) Determine the current flowing through each resistor.
(4 Ω :1A, 2 Ω :2A)
(f) Calculate the power produced/consumed by each device.
(battery: 12W, 2 Ω :8W, 4 Ω :4W)
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Example 2: Assume the switch is closed.
(a) Calculate the effective resistance of the circuit.
(1.5 Ω )
(b) Draw the simplified circuit with just the effective resistance.
(c) Determine the current flowing from the battery.
(4A)
(d) Determine the potential difference across each resistor.
(1 Ω : 2V, 2 Ω : 4V, 3Ω : 6V)
(e) Calculate the current flowing through each resistor (in mA).
(1 Ω : 2000 mA, 2 Ω : 2000mA, 3 Ω : 2000mA)
♦ Homework: Series Circuits. Text pg 83, 3.1 Q’s 1-8.
Parallel Circuits. Text pg 88, 3.2 Q’s1,2,4-9.
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More series and parallel circuits
Question 1
(a) What is the equivalent resistance of this circuit?
(b) What is the current flowing from the battery (in mA)?
(c) What is the potential difference across:
(i) the 100 Ω resistor?
(ii) the 2 kΩ resistor?
Question 2
(a) Calculate the equivalent resistance.
(b) What is the potential difference across the 9 ohm
and 18 ohm resistors?
(c) Calculate the current flow through each resistor.
(d) What is the current flow from the battery?
Question 3
Two identical globes are rated “60W, 240V”. Assume the globes are ohmic.
(a) What is the resistance each of these globes?
(b) The globes are connected in series to a 12 V battery.
Draw a circuit diagram for this.
(c) Calculate the current flowing from the battery (in mA).
(d) What power (in mW) is used in each globe?
(e) What power (in mW) is provided by the battery?
(f) How much energy does the battery provide in 5.0 minutes?
(g) A third identical globe is added in series. How does this affect
the brightness of the other two globes? Explain.
Calculate the power (in mW) consumed by each.
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Question 4
The same three light globes are now connected in parallel to
the 12V battery. Draw a circuit diagram for this.
(a) Calculate the power dissipated in each globe (in mW)?
(b) What is the current flowing through each globe (in mA)?
(c) What is the total current flowing from the battery (in mA)?
(d) What is the power supplied by the battery (in mW)?
(e) Another identical globe is added in parallel. How does this affect the brightness of the other three
globes? Explain.
(f) What power does the battery provide in this circuit (in mW)?
Question 5
Lamp W is connected across a 9 V battery as shown in circuit 1 and glows with a certain intensity.
Y
Z
W
X
W
Circuit 1
Circuit 2
Three other identical lamps X, Y and Z are now added to the circuit as shown in circuit 2.
Use the answer key to answer these questions: KEY
B: brighter than W in circuit 1
S: same intensity as W in circuit 1
D: duller than W in circuit 1
U: the lamp is unlit
(a) Indicate the brightness of each lamp in circuit 2.
(b) Lamp W is unplugged in circuit 2. Indicate the intensities of lamps X, Y and Z now.
(c) Lamp W is replaced and lamp Y is now unplugged. Indicate the intensities of W, X and Z now.
Answers: 1 (a) 2.1 kΩ (b) 5.7 mA (c) 0.57 V, 11.4 V 2 (a) 6.0 Ω (b) 12V, 12V (c) 1.3 A, 0.7 A (d) 2.0 A 3 (a)
960 Ω (c) 6.3 mA (d) 38 mW (e) 75 mW (f) 23 J (g) duller, 17 mW 4 (a) 150 mW (b) 13 mA (c) 38 mA (d) 450
mW (e) no change (f) 600 mW 5 (a) W: S, X: S, Y: D, Z: D (b) X:D, Y:D, Z:S (c) W: S, X: S, Z: U
*Website: Series and parallel circuits (interactive)
♦ Homework: Worksheet- Electric circuits (PTO)
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Worksheet: Electric Circuits
1. Three resistors 10Ω, 15Ω and 25Ω are connected in series across a 200 V power supply.
(a) What is the equivalent resistance of this circuit?
(b) Calculate the current flowing through each resistor
(c) Determine the potential difference across each of the resistors.
(d) Calculate the power provided by the power supply.
(e) Calculate the power dissipated in each resistor.
(f) Is power conserved in this circuit?
(a) 50Ω (b) 4A (c) 40V, 60V, 100V (d) 800 W (e) 160W, 240W, 400W (f) yes
2. Three resistors 8Ω, 12Ω and 16 Ω are connected in parallel to each other and to a power supply. The
current flowing from the power supply is 19.5 A. Determine the current flowing through each of the resistors.
(9A, 6A, 4.5A)
3. (a) Calculate the equivalent resistance of this circuit. (25Ω)
(b) Determine the current flowing from the battery. (0.10 A)
(c) How much current is flowing through the 8Ω? (0.05A)
(d) How much current is flowing through the 1Ω? (O.05A)
(e) Determine the potential difference across each of
the resistors. (16Ω:1.6V, 8Ω :0.4V, 5Ω:0.5V, 1 Ω:0.05V, 3Ω:0.15V, 4Ω :0.2V)
4. (a) Determine the effective resistance of this compound circuit. (14Ω)
(b) Calculate the total current flowing? (0.30A)
(c) What is the voltage drop across the
parallel arrangement? (1.2V)
(d) Determine the current flowing through each resistor.
(0.3A, 0.12A, 0.12A, 0.06A)
5. (a) Calculate the current flowing from the
battery (in mA). (1000mA)
E
(b) Determine the current flowing through
one of the 3Ω resistors. (0.5A)
(c) Calculate the current in the 20Ω resistor.
(0.13A)
C
D
(d) What is the potential difference
between points: C and D?
D and E?
(16V, 6V)
(e) If the 20Ω resistor was removed from the circuit, what current would flow from the battery (in mA)?
(990mA)
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Pra ctic al in v es tig atio n: Anal y si s o f DC circ uit s
In this investigation, you will use Ohm’s law to determine the total resistance of series and parallel circuits.
This will then be compared with the theoretical value of total resistance obtained using the appropriate series
or parallel formula.
Apparatus: variable power supply, ammeter, multimeter (voltmeter), switch, resistors, leads
We will assume that the resistors being used are ohmic and so Ohm’s law (V = IR) holds for each.
Select three resistors:
R1 =
R2 =
R3 =
Part A: Series circuit
Connect your resistors in series as shown below. Set the power supply to approximately 10V.
Use the multimeter to measure the potential differences
across the power supply and each of the resistors.
Record the ammeter reading. This tells you the current
value from the power supply and through each resistor.
R1
Resistance given (Ω)
Potential difference
(V)
Current (A)
R2
R3
Resistance
calculated (Ω)
R1
R2
R3
1. Compare the given values of R1, R2 and R3 with the calculated values. Why might you expect their values
to be slightly different?
2. Now consider the circuit to be simplified with just the power supply and the equivalent or total resistance
RTOTAL. We can calculate the value of RTOTAL using Ohm’s law.
VTOTAL =
ITOTAL =
∴RTOTAL =
3. From theory, we know what the total resistance of a series circuit is: RTOTAL = R1 + R2 + R3 +………
Use this formula to calculate the theoretical value of RTOTAL for your circuit. Compare your two values of
RTOTAL. What could be the cause of any discrepancy?
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Part B: Parallel circuit
Now set the power supply to around 5V and connect the circuit shown below.
Use the multimeter to measure the
potential difference across the power
supply VTOTAL. This is also giving you the
potential difference across each resistor.
R1
R2
R3
Use the ammeter to measure the current
flowing from the power supply. This is
the total current ITOTAL.
1. We can consider the circuit to be simplified with just the power supply and the equivalent or total resistance
RTOTAL. We can calculate the value of RTOTAL using Ohm’s law.
VTOTAL =
ITOTAL =
∴RTOTAL =
2. By placing the ammeter adjacent to each of the resistors, the individual current values (I1, I2 I3) can be
measured. The potential difference values will be the same as for the power supply.
Record your results in the table:
Resistance given (Ω)
Potential difference (V)
Current (A)
R1
R2
R3
Kirchoff’s law and the theory for parallel circuits says that the current is conserved. Do your results agree with
this? Use calculations to support your answer.
3. Use the parallel formula to calculate the total resistance of your circuit. How well does this agree with the
value that you obtained in question 1 Discuss.
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Part C: Compound circuit
Set the power supply to around 6 V and connect the following circuit.
R1
R2
R3
1. Use theory to simplify this circuit by calculating the equivalent resistance. Show your working.
2. Show on the circuit diagram where an ammeter should be placed to measure the current from the power
supply ITOTAL . Also show the location of a voltmeter to measure VTOTAL.
3. Connect these meters into the circuit and record these values.
VTOTAL =
ITOTAL
Use Ohm’s law to calculate an experimental value for RTOTAL.
Compare the experimental and theoretical values:
Rtheoretical =
Rexperimental =
Conclusion: Do your experimental results for series, parallel and compound circuits support the theories
relating to these circuits that have been covered in class?
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EXTENDED PROBLEM: Crocodile Physics
(a) Use theory to simplify the circuit below and calculate the equivalent resistance RE.
(b) Calculate the current (in mA) that flows through each part of this circuit. Show these values on the
diagram.
(c) Calculate the power consumed by each resistor. Find the total power consumed. How does this compare
with the power produced by the battery?
(d) Use Crocodile Physics to set up this circuit. Check that all your answers are correct!!
PEGS Electricity booklet 1
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CHECK YOUR UNDERSTANDING!!
Complete these statements by including a < , = or > sign.
Question 1
(a) The current at A is
the current at C.
(b) The current at B is
the current at E.
(c) The current at E is
the current at G.
(d) The current at G is
the current at F.
(e) The current at B is
the current at F.
(f) The current at D is
the current at I.
(g) The current at A is
the current at L.
Question 2
In this question, you will be comparing the potential difference (voltage drop) across different parts of the
circuit.
(a) The voltage drop between B and C is
the voltage drop between J and K.
(b) The voltage drop between B and K is
the voltage drop between D and I.
(c) The voltage drop between E and F is
the voltage drop between G and H.
(d) The voltage drop between E and F is
the voltage drop between D and I.
(e) The voltage drop between J and K is
the voltage drop between D and I.
(f) The voltage drop between L and A is
the voltage drop between B and K.
Answers: 1 (a) = (b) > (c) > (d) < (e) = (f) = 2 (a) > (b) > (c) = (d) = (e) > (f) =
♦ Worksheets: What is current? What is voltage?
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IN T ER NA L R E S IS TA N C E
EMF ε -
+
r
•
Internal circuit
S
•
External circuit
When a current I flows from a battery of EMF ε and internal resistance r through an external circuit of R
and the resistance of the connecting wires is neglected then:
ε
ε
= Ir + IR
= Ir + V
then
VT = ε - Ir
where
VT = terminal voltage of the cell (V)
ε = ideal voltage (EMF) of the cell (V)
r = internal resistance of the cell (Ω )
I = current (A)
•
If the switch S is opened and no current is flowing, the potential difference across r will be zero
and the terminal voltage VT will equal of the equal the EMF ε of the cell. eg 12V battery will
have 12V across its terminals.
•
If the switch is closed and the current flows, there will now be a potential difference across the
internal resistance and the terminal voltage VT will be less than the EMF ε of the cell. e.g a 12V
battery may have 11.5V across its terminals. Energy is being lost due to heating effects inside the
battery.
•
What will happen when a larger current flows?
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Question
A battery has an EMF of 12V and an internal resistance of 0.1Ω and is connected to a resistor of 2.0Ω.
(a)
What is the energy per coulomb given to the external circuit ie what is the Terminal Voltage VT?
(11.4V)
(b)
What is the energy per coulomb VI consumed in the internal circuit (battery)?
(0.6V)
(c)
How are the above values for (a) and 1(b) affected if the resistor is replaced by a 1.0Ω resistor?
(11V; 1.1V)
(d)
How are the above values for (a) and (b) affected if the battery is replaced by another 12V battery
whose internal resistance is 0.20Ω?
(11V; 1.1V)
*Worksheet: The effects of Internal resistance.
♦ Homework: Text pg 96 3.3 Q’s 6, 7 8, and 9.
Chapter Review pg 102 Q’s1- 4, 9-2.
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