Transcript Slide 1

P2.4.2 Current, Charge and
Power
P2 Physics
Mr D Powell
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
Mr Powell 2012
Index
P2.4.2 Current, Charge and Power
a) When an electrical charge flows through a resistor, the
resistor gets hot.
b) The rate at which energy is transferred by an appliance is
called the power: P = E/t
c) Power, potential difference and current are related by
the equation: P = VI
d) Energy transferred, potential difference and charge are
related by the equation: E = VQ (HT Only)
P is power in watts, W
E is energy in joules, J
t is time in seconds, s
I is current in amperes (amps), A
V is potential difference in volts, V
Q is charge in coulombs, C
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Index
P5.4 – Electrical Power & PD
a) When an electrical charge flows through a
resistor, the resistor gets hot.
b) The rate at which energy is transferred by
an appliance is called the power:
P = E/t
c) Power, potential difference and current
are related by the equation:
If the value for energy is less
that 1 J it is displayed in
millijoules (mJ), and when it
reaches 1000 J it is displayed in
kilojoules (kJ).
If the value for power is less that
1 W it is displayed in milliwatts
(mW).
If the value for current is less
that 1 A it is displayed in
milliamps (mA).
P = VI
P is power in watts, W
E is energy in joules, J
t is time in seconds, s
I is current in amperes (amps), A
V is potential difference in volts, V
Q is charge in coulombs, C
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Index
Power, what do we mean?
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Index
Mr Powell 2012
Index
Calculating Power
Millions of millions of electrons pass through the circuit of
an artificial heart every second. Each electron transfers a
small amount of energy to it from the battery. So the total
energy transferred to it each second is large enough to
enable the device to work.
For any electrical appliance:

the current through it is a measure of the number of
electrons passing through it each second (i.e. the
charge flow per second),

the potential difference across it is a measure of how
much energy each electron passing through it
transfers to it (i.e. the electrical energy transferred per
unit charge),

the power supplied to it is the energy transferred to it
each second. This is the electrical energy it transforms
every second.
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Index
Coulombs

Electrons are charged particles and each of them have a charge of 1.9 x 10-19 C.

It is a simple property which cannot be removed or changed. It is a useful to us
as charges make particles move i.e. opposites attract.

If we add a load of them together and think of them as a single “sphere of
charge” or ball we get a whole coulomb of charge and can think about defining
the ampere or amp 1A = 1C/s
C
ee- e-- e-- ee- - e- e- e-e e e- e e
e- e- ee-
1.9 x 10-19 C x 6.25 x 1018 electrons = 1C
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Power Delivery
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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Example
In this circuit the Voltmeter = 4V and the
Current = 1A
4V
Power = Voltage x Current
P
= VxI
P
= 4V x 1A
P
= 4J/C x 1C/s
P
= 4J/s
P
= 4W
1A
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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 – “Power
Modelling”
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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Index
Homework - Current Flowing?
We can work out the current flow
from the total power use;
Power = Voltage x Current
P = VI
I = P/V
I = 154W / 230V
I = 154J/s / 230J/C
I = 0.664 C/s
I = 0.7 C/s
I = 0.7 A
So if this is the
current flow
we can now
pick a fuse to
match i.e. 3A
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Fuses in use...
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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Simple Energy Transfers
When you use an electrical appliance, it transforms electrical
energy into other forms of energy. The power of the appliance,
in watts, is the energy it transforms, in joules, per second. The
more per second the higher the power.
Take a short
note of
this key
point...
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Summary Questions
power
current
p.d.
current
a)
i) 12V x 3A = 36W
ii) 230V x 2A = 460W
b)
i) 24V / 12A = 2A -> 3A fuse
ii) 800W/230V= 3.47A -> 5A fuse
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Power & Efficiency...
The table below shows the time taken by different electrical devices to
transfer a given amount of energy supplied. The useful energy
transferred in this time is also stated. Copy and Complete the table....
Device
Time
taken
(s)
Energy
supplied
(J)
Useful energy
transferred
(J)
useful
power (W)
Efficiency
Lamp
1000
100 000
20 000
20
0.2
Microwave
oven
120
96 000
48 000
400
0.5
Motor
300
18 182
6000
20
0.33
Computer
3000
900 000
99 000
33
0.11
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Index
P5.5 – Electrical Energy & Charge p172
FT: Revisit of Q = It
d) Energy transferred, potential difference
and charge are related by the equation:
E = VQ (HT Only)
P is power in watts, W
E is energy in joules, J
t is time in seconds, s
I is current in amperes (amps), A
V is potential difference in volts, V
Q is charge in coulombs, C
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Flow of Charge (FT)
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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Worked Examples..... FT
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Energy and Potential Difference (HT)
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
A*/B Questions – copy & complete
1.
A cordless drill operates using a 14.4 V battery pack. The battery is
rated at 2 amp hours which means that it can deliver a current of two
amps for a period of 1 hour. How much energy flows from the battery?
1 hour = 3600 s
Charge = current x time = 2 A x 3600 s = 7200 C
Energy = charge x voltage = 7200 C x 14.4 V = 103680 J
2.
A 12 volt heater takes a current of 3.6 A. It is left to heat up an
aluminium block for a period of 45 minutes. How much heat energy is
transferred to the aluminium block?
Time = 45 x 60 = 2700 s
Energy = VIt = 2700 x 12 x 3.6 = 116640J
3.
What current is consumed by a 60 W light bulb operating on the 230 V
mains?
I = P/V = 60 W / 230 V = 0.26 A
Mr Powell 2012
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Summary Questions
Mr Powell 2012
Index
P2.4.2 Current, Charge and Power
P2.4.2 Current, Charge and Power
P2.4.2 Current, Charge and Power
a) When an electrical charge flows
through a resistor, the resistor gets
hot.
a) When an electrical charge flows
through a resistor, the resistor gets
hot.
a) When an electrical charge flows
through a resistor, the resistor gets
hot.
b) The rate at which energy is
transferred by an appliance is called
the power:
b) The rate at which energy is
transferred by an appliance is called
the power:
b) The rate at which energy is
transferred by an appliance is called
the power:
P = E/t
P = E/t
P = E/t
c) Power, potential difference and
current are related by the equation:
c) Power, potential difference and
current are related by the equation:
c) Power, potential difference and
current are related by the equation:
P = VI
P = VI
P = VI
d) Energy transferred, potential
difference and charge are related
by the equation:
E = VQ (HT Only)
P is power in watts, W
E is energy in joules, J
t is time in seconds, s
I is current in amperes (amps), A
V is potential difference in volts, V
Q is charge in coulombs, C
d) Energy transferred, potential
difference and charge are related
by the equation:
E = VQ (HT Only)
P is power in watts, W
E is energy in joules, J
t is time in seconds, s
I is current in amperes (amps), A
V is potential difference in volts, V
Q is charge in coulombs, C
d) Energy transferred, potential
difference and charge are related
by the equation:
E = VQ (HT Only)
P is power in watts, W
E is energy in joules, J
t is time in seconds, s
I is current in amperes (amps), A
V is potential difference in volts, V
Q is charge in coulombs, C
Suggested ideas for practical work to
develop skills and understanding
include the following:

A lot of energy is wasted in filament bulbs
as heat.
measuring oscilloscope traces

demonstrating the action of fuse
wires

using fluctuations in light
intensity measurements from
filament bulbs to determine the
frequency of a.c.

Additional Guidance...
measuring the power of 12 V
appliances by measuring energy
transferred (using a joulemeter or
ammeter and voltmeter) in a set
time.
Less energy is wasted in power-saving
lamps such as Compact Fluorescent
Lamps (CFLs).
There is a choice when buying new
appliances in how efficiently they transfer
energy.
Should be able to calculate the current
through an appliance from its power and
the potential difference of the supply, and
from this determine the size of fuse
needed.