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2.2 Resistance
G482 Electricity, Waves & Photons
2.2.5 Power
Mr Powell 2012
Index
2.2.5 Power (Prior Learning)
When we talk about Power what we mean
is “the amount of energy delivered per
second”
1 Joule / 1 Second = 1 Watt
It then makes sense that the Power used
by a component can be found from the
product of current through
and voltage across the component;
Power = Voltage x Current
P = VxI
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Analogy
Another way of thinking about it is saying that the current
carries the energy;
4V
1A
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
C
C
C
C
C
C
1J
C
C
C
C
C
1J
As the Coulombs of Charge move they release their
energy as heat and light (through the bulb)
= 1 Coulomb
of charge
= 1 Joule of
energy
= 1 Second of
Index
time
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Analogy 2
If the voltage increases, more energy is delivered so the power
increases;
5V
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
C
C
C
C
C
C
1A
1J
C
C
C
C
C
1J
Power = 5V x 1A = 5J/s = 5W
= 1 Coulomb
of charge
= 1 Joule of
energy
= 1 Second of
Index
time
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Analogy 3
If the current increases, more energy is delivered so the
power increases;
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
1J
1J
1J
1J
C
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
1J
C
1J
4V
2A
1J
C
C
1J
1J
Power = 4V x 2A = 8J/s = 8W
C
1J
= 1 Coulomb
of charge
= 1 Joule of
energy
= 1 Second of
time
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Flow of Charge
When an electrical appliance is on, electrons are
pushed through the appliance by the potential
difference
The potential difference causes a flow of charge
particles (free electrons). The rate of flow of charge is
the electric current through the appliance. 1A = 1C/s
The unit of charge, the coulomb (C).
Q=It
The charge passing along a wire or through a
component in a certain time depends on: the current,
and the time.
We can calculate the charge using the equation:
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Energy and Potential Difference
When a resistor is connected to a battery, electrons are made to
pass through the resistor by the battery.
Each electron repeatedly collides with the vibrating atoms of the
resistor, transferring energy to them.
E = VQ
Q=It
E = VIt
The atoms of the resistor therefore gain kinetic energy and vibrate
even more. The resistor becomes hotter.
When charge flows through a resistor, electrical energy is
transformed into heat energy.
The energy transformed in a certain time in a resistor depends on:
the amount of charge that passes through it, and the potential
difference across the resistor.
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Index
2.2.5 Power
Assessable learning outcomes
a)
describe power as the rate of energy transfer;
b) select and use power equations P = VI, P = I2R, P
= V2/R
c)
explain how a fuse works as a safety device (HSW
6a);
d) determine the correct fuse for an electrical
device;
A Joulemeter (V, I and t)
or a data-logger may be
used to determine
energy transfer.
A utilities statement can
be used to illustrate the
use of the kW h by
electricity companies.
(Homework)
e) select and use the equation W = IVt;
f)
define the kilowatt-hour (kW h) as a unit of
energy;
g)
calculate energy in kW h and the cost of this
energy when solving problems (HSW 6a).
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a) describe power as the rate of energy transfer;
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a) HSW - PC Energy Use Question
Here is an example of the energy
usage of a home computer. Each
internal component transfers a
variable amount of energy. If you left
this PC on for an hour it would
transfer;
Energy = Power x Time
Energy = 154W x 1 hour
Energy = 154W x 3600s
E = 154J/s x 3600s
E = 554400J
E = 554.4kJ
Current flow to the PC would then
be…
P = VI or P/V = 0.67A
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Resistance Heating...
Charge carriers transfer kinetic energy to positive ions through repeated collisions.
The pd across the material then provides an accelerating force to the charge carrier
which then collides with another positive ion.
V = I R
P = I V
P = V2
R
Energy per second transferred
to the component P:
In the steady state this equals
the heat transfer to the surroundings
P = I2 R
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b) select and use power equations P = VI, P = I2R, P = V2/R
TASK: Try and rearrange the equations and substitute so you can get from
one to another!
P = VI
P = Et
V = IR
2
P=I R
2
V /R
P=
Q = It
E = Vit
E = VQ
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c) explain how a fuse works as a safety device (HSW 6a);
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Heater Circuit
Inside 3 Pin Plug
3 Pin Plug
/ Mains
Appliance
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Facts Fuses
 Fuses are thin pieces of wire which resist a large current
 They melt when too much current flows
 Circuit is broken
 Have to be replaced
Facts Circuit Breakers
 Circuit breakers are designed to flip over when too large a current flows
 They break the circuit
 Rely on magnetic forces
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d) determine the correct fuse for an electrical device;
Domestic appliances are often fitted with a 3A, or
a 5A or a 13 A fuse.
If you don’t know which one to use for an
appliance, you can work it out from the
power rating of the appliance and its potential
difference (voltage).
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e) select and use the equation W = IVt; (SHC Practical)
The specific heat capacity of a substance is the amount of energy required to
change the temperature of one kilogram of the substance by one degree
Celsius.
E = mc
E is energy transferred in joules, J
m is mass in kilograms, kg
 is temperature change in degrees Celsius, °C
c is specific heat capacity in J / kg °C
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Equipment
two digital multimeters (V & A)
power supply, 0 – 12 V DC
thermometer, 0 – 100C
12 V heater
Diagram
Thermometer
stopwatch
12V
V
A
Heater
electrical leads
Metal block
1 kg Metal blocks
(Cu, Fe, Al)
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Using a Multi-meter
DC Volts
Various Ranges
DC mA various
ranges
Resistance
ranges in 
Normal Currents
up to 10A
Volts or mA or 
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Homework - Design a new experiment for water?
power supply, 0 – 12 V DC
?
thermometer, 0 – 100C
stopwatch
Diagram
?
two digital multimeters (V & A)
electrical leads
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Results....
A
B
C
Metal Start Temp End Temp Temp
(1kg)
/ C
/ C
Rise / C
?
21.5
34
12.5
Al
Fe
Cu
D
Current
/A
2.62
9.1V  2.62 A  420 s
o
S .H .C 

801
J
/
kg
C
o
1kg 34  21.5 C
E
F
Time for
PD / V
rise / s
9.1
420
G
SHC
/ J /kg°C
801
Quoted....
Aluminium (Al) is 902 J /kg°C
Iron (Fe) 450 J /kg°C
Copper (Cu) is 385 J /kg°C
Analysis Thoughts?
1.
2.
3.
Are there any flaws with your experiment where electrical or
thermal energy are lost and don’t raise the temperature of the
block?
What could you do to make the experiment more reliable?
Can you think of a similar experiment to find the SHC of water?
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f) define the kilowatt-hour (kW h) as a unit of energy;
The kilowatt hour, or kilowatt-hour, (symbol kW·h, kW h or kWh) is
a unit of energy equal to 1000 watt hours or 3.6MJ.
For constant power, energy in watt hours is the product
of power in watts and time in hours.
The kilowatt hour is most commonly known as a billing unit for
energy delivered to consumers by electric utilities.
The kilowatt-hour (symbolized kWh) is a unit of energy equivalent to
one kilowatt (1 kW) of power expended for one hour (1 h) of time.
Inversely, one watt is equal to 1 J/s.
One kilowatt hour is 3.6 megajoules, which is the amount of energy
converted if work is done at an average rate of one thousand watts
for one hour.
http://en.wikipedia.org/wiki/Kilowatt_hour
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When Maths starts to hurt!
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g) calculate energy in kW h and the cost of this energy
when solving problems (HSW 6a).



Poor Dr Frankenstein did not look at his electricity bill and check the cost of
each unit of electricity.
1 Unit = 1KW hour of electricity.
They are shown on the bill here..
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E
Plenary Question….
Two resistors, A and B, have different resistances but
otherwise have identical physical properties. E is a cell of
negligible internal resistance.
A
B
E
When the resistors are connected in the circuit shown in
figure 1, A reaches a higher temperature than B. When
connected in the circuit shown in figure 2, B reaches a
higher temperature than A. Explain these observations fully,
stating which resistance is greater. (6 marks)
A
B
Answer…. power determines heat produced (1)
in series, current is same (1)
 I2 RA must be > I2 RB (1)
 in parallel, p.d. is same (1)
iSlice
<
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Can you connect the formulae...
P = VI
P = Et
V = IR
2
P=I R
2
V /R
P=
Q = It
E = Vit
E = VQ
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Connection
•
•
•
Connect your learning to the
content of the lesson
Share the process by which the
learning will actually take place
Explore the outcomes of the
learning, emphasising why this will
be beneficial for the learner
Demonstration
• Use formative feedback – Assessment for
Learning
• Vary the groupings within the classroom
for the purpose of learning – individual;
pair; group/team; friendship; teacher
selected; single sex; mixed sex
• Offer different ways for the students to
demonstrate their understanding
• Allow the students to “show off” their
learning
Activation
Consolidation
• Construct problem-solving
challenges for the students
• Use a multi-sensory approach – VAK
• Promote a language of learning to
enable the students to talk about
their progress or obstacles to it
• Learning as an active process, so the
students aren’t passive receptors
• Structure active reflection on the lesson
content and the process of learning
• Seek transfer between “subjects”
• Review the learning from this lesson and
preview the learning for the next
• Promote ways in which the students will
remember
• A “news broadcast” approach to learning
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