2.2 Resistance

G482 Electricity, Waves & Photons

2.2.5 Power

KS5 OCR PHYSICS H158/H558Mr Powell 2012

Index

Mr Powell 2012Index

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 throughand voltage across the component;

Power = Voltage x CurrentP = V x I

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Analogy

Another way of thinking about it is saying that the current carries the energy;

4V

1A

C CC C

C C

C CC

1J

1J

1J

1J

C1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

As the Coulombs of Charge move they release their energy as heat and light (through the bulb)

C

1J

= 1 Coulomb of charge

= 1 Joule of energy

= 1 Second of time

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Analogy 2

If the voltage increases, more energy is delivered so the power increases;

5V

1A

C CC C

C C

C CC

1J

1J

1J

1J

C1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

Power = 5V x 1A = 5J/s = 5W

1J 1J 1J 1J

C

1J

= 1 Coulomb of charge

= 1 Joule of energy

= 1 Second of time

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Analogy 3

If the current increases, more energy is delivered so the power increases;

4V

2A

C CC C

C C

C CC

C1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

Power = 4V x 2A = 8J/s = 8W

C CC C

CC

C CC

C

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J 1J1J

1J

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).

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:

Q=It

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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.

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.

E = VQQ=ItE = VIt

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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;

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).

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)

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a) describe power as the rate of energy transfer;

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a) HSW - PC Energy Use Question

Energy = Power x Time

Energy = 154W x 1 hour

Energy = 154W x 3600s

E = 154J/s x 3600s

E = 554400J

E = 554.4kJ

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;

Current flow to the PC would then be…

P = VI or P/V = 0.67A

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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 RP = I V

P = I2 R

P = V2

R Energy per second transferred

to the component P:

In the steady state this equals the heat transfer to the surroundings

Resistance Heating...

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b) select and use power equations P = VI, P = I2R, P = V2/R

P = VIP = EtV = IRP = I2R

P = V2/RQ = ItE = VitE = VQ

TASK: Try and rearrange the equations and substitute so you can get from one to another!

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c) explain how a fuse works as a safety device (HSW 6a);

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Heater Circuit

3 Pin Plug / Mains

Inside 3 Pin Plug

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 thepower 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, Jm is mass in kilograms, kg is temperature change in degrees Celsius, °Cc is specific heat capacity in J / kg °C

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Diagram

Equipment two digital multimeters (V & A)

12V

Metal block

Heater

Thermometer AV

1 kg Metal blocks (Cu, Fe, Al)

thermometer, 0 – 100C

12 V heater

power supply, 0 – 12 V DC

electrical leads

stopwatch

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Using a Multi-meter

Volts or mA or

DC VoltsVarious Ranges

Resistance ranges in

DC mA various ranges

Normal Currents up to 10A

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Diagram

Homework - Design a new experiment for water?

two digital multimeters (V & A)

thermometer, 0 – 100Cpower supply, 0 – 12 V DC

electrical leads

stopwatch

?

?

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Results....A B C D E F G

Metal (1kg)

Start Temp / C

End Temp / C

Temp Rise / C

Current /A PD / V Time for

rise / sSHC

/ J /kg°C? 21.5 34 12.5 2.62 9.1 420 801AlFeCu

CkgJCkg

sAVCHS o

o/801

5.21341

42062.21.9..

Quoted....Aluminium (Al) is 902 J /kg°CIron (Fe) 450 J /kg°CCopper (Cu) is 385 J /kg°C

Analysis Thoughts?

1. Are there any flaws with your experiment where electrical or thermal energy are lost and don’t raise the temperature of the block?

2. What could you do to make the experiment more reliable?3. 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;

http://en.wikipedia.org/wiki/Kilowatt_hour

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.

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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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Answer….

Plenary Question….Two resistors, A and B, have different resistances but otherwise have identical physical properties. E is a cell of negligible internal resistance.

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)

iSlice

A B

E

A

B

E

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)

<

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P = VIP = EtV = IRP = I2R

P = V2/RQ = ItE = VitE = VQ

Can you connect the formulae...

Mr Powell 2012Index

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

• 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

Consolidation

• 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