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OCR AA-Level Physics

9. Energy, Power and Resistance2019-2020

Name:________________

Physics Teacher:______________

Year 12

Checklist

Spec Ref.

Checklist Textbook Page

Reviewed Practiced Confident

4.2.1 a Can you recognise and draw circuit symbols? 139

4.2.2 a Can you define potential difference (p.d.) and the unit volt?

140

4.2.2 b Can you describe the electromotive force (e.m.f.) of a source such as a cell or a power supply?

141

4.2.2 c Can you explain the distinction between e.m.f. and p.d. in terms of energy transfer?

140 - 141

4.2.2 d Can you describe and calculate energy transfer using W = V Q and W = E Q?

140 – 141

4.2.2 e Can you describe and calculate energy transfer

using eV =

12 m v2 for electrons and other

charged particles?

143 - 144

4.2.3 a Can you define resistance and the unit ohm? 146

4.2.3 b Can you explain Ohm's law? 146

4.2.3 c i Can you describe the I–V characteristics of the resistor, filament lamp, thermistor, diode, and light-emitting diode (LED)?

149, 152, 159

4.2.3 c ii

Can you list and describe techniques and procedures used to investigate the electrical characteristics for a range of ohmic and non-ohmic components?

148

29. Energy, Power and Resistance

Spec Ref.

Checklist Textbook Page

Reviewed Practiced Confident

4.2.4 a i

Can you explain the resistivity of a material and

use the equation R= ρL

A ?

154

4.2.4 a ii

Can you list and describe techniques and procedures used to determine the resistivity of a metal?

155

4.2.4 b Can you explain how resistivity varies with temperature for metals and semiconductors?

155

4.2.4 cCan you explain how resistance varies with temperature for a negative temperature coefficient (NTC) thermistor?

157-158

4.2.5 a Can you use the equations P = V I, P = I2 R and

P=V 2

R ?

162-164

4.2.5 b Can you describe energy transfer using the equation W = V I t?

164

4.2.5 cCan you describe the kilowatt-hour (kW h) as a unit of energy and calculate the cost of energy?

165-166


Module 4: Electrons Waves and Photons 9. Energy, Power and Resistance In this topic you will gain a broader knowledge of circuits and some of the components that can be found in them. You will analyse the behaviour of components through experiments and graph analysis, using knowledge from 8. Charge and Current to explain this behaviour. This topic will also serve as an introduction to resistivity and the kilowatt-hour.

Lesson Content overview Specification reference:Learners should be able to demonstrate and apply their knowledge and understanding of:

1 Circuit Symbols, EMF and Potential Difference

4.2.1 Circuit symbols

(a) circuit symbols

(b) circuit diagrams using these symbols.

4.2.2 E.m.f. and p.d

(a) potential difference (p.d.); the unit volt

(b) electromotive force (e.m.f.) of a source such as a cell or a power supply

(c) distinction between e.m.f. and p.d. in terms of energy transfer

(d) energy transfer; W = VQ;W = εQ.

2 The Electron Gun 4.2.2 E.m.f. and p.d

(e) energy transfer eV=12mv2 for electrons and other charged particles.

3 Resistance 4.2.3 Resistance

(a) resistance; R=VI; the unit ohm

(b) Ohm’ slaw

4 Practical: Collecting data for I-V curves

4.2.3 Resistance

(c) (ii) techniques and procedures used to investigate the electrical characteristics for a range of ohmic and non-ohmic components.

5 Interpreting the I-V curves for filament lamps and resistors/wires

4.2.3 Resistance

(c) (i) I–V characteristics of resistor, filament lamp, thermistor, diode and light-emitting diode (LED)

6 Diodes 4.2.3 Resistance



Lesson Content overview Specification reference:Learners should be able to demonstrate and apply their knowledge and understanding of:

7 Resistivity 4.2.4 Resistivity

(a) (i) resistivity of a material; the equation R=ρLA

(a) (ii) techniques and procedures used to determine the resistivity of a metal.

(b) the variation of resistivity of metals and semiconductors with temperature

(c) negative temperature coefficient (NTC) thermistor; variation of resistance with temperature.

8/9 Practical: Finding the resistivity of a material

4.2.4 Resistivity

(a) (i) resistivity of a material; the equation R=ρLA

(a) (ii) techniques and procedures used to determine the resistivity of a metal.

10 The Thermistor 4.2.3 Resistance


4.2.4 Resistivity

(b) the variation of resistivity of metals and semiconductors with temperature

(c) negative temperature coefficient (NTC) thermistor; variation of resistance with temperature.

11 The LDR 4.2.3 Resistance

(d) light-dependent resistor (LDR); variation of resistance with light intensity.

12 Electrical Energy and Power

4.2.5 Power

(a) the equations P=VI ,P=I2 R∧P=V 2

R(b) energy transfer; W=VIt

13 Paying for Electricity

4.2.5 Power

(c) the kilowatt-hour (kW h) as a unit of energy; calculating the cost of energy.

Reminder: Notes in blue dashed boxes like this contain suggested follow-up/consolidation work. If these are not done in lessons, you should be working through these to ensure you have grasp content from lessons.


1. Circuit Symbols, EMF and Potential Difference

Drawing CircuitsTo make interpreting circuits easy and drawing the simple, we use a range of symbols to represent different components. These symbols are internationally recognised meaning that others can also interpret your diagrams.

Below is the range of symbols you are expected to know:

When drawing a circuit, the symbols are joined up with straight lines and junctions drawn at 90o to each other. If you struggle to draw straight lines free hand you should be using a ruler.

Potential Difference (p.d.)Potential difference is defined as:

‘The energy transferred from electrical energy to other forms (e.g. light, sound) per unit charge’.

It is measured in Volts [V] and has the base units JC-1. The volt is defined as the p.d. when 1J of energy is transferred per coulomb.

The equations for potential difference (p.d.) is:

V=WQ

where V is the p.d, Q is charge (measured in coulombs [C]) and W is energy transferred (measured in Joules [J]).

Electromotive Force (e.m.f)Electromotive force is the work done on charge carries as they pass through a source of power in a circuit. It is defined as:

‘The energy transferred from chemical energy to electrical energy per unit charge’.

Like p.d. it is measured in Volts [V] The equations for e.m.f is:

ε=WQ

where ε is the electromotive force.

Important: Electromotive force is not a force


Worked Examples

Equations Used:

I=∆Q∆ t

V=WQ

ε=WQ

1. Calculate the energy transferred in 3.0 hours by a electric heater that draws a current of 12A when the p.d. across it is 75V.

∆Q=I ∆ t ∆Q=12× (3×60×60 )∆Q=129600W=VQW=75×129600W=9720000W=9.7×106 J

2. What value of current is required for a cell with an electromotive force of 9.0V to transfer 20kJ in 1.0 minute?

∆Q=Wε

∆Q=20×103

9

∆Q=2.2×104C

I=∆Q∆ t

I=2.2×104

1×60

I=37 A

Worked Examples (in class)1. Calculate the energy transferred in 6.0 hours by an electric cooker that draws a current of 2kA when

the p.d. across it is 230V.


Convert time into seconds

Round to 2.s.f (as per questions values)


Convert energy to Joules

Convert time to seconds

2. What value of current is required for a cell with an electromotive force of 5.0V to transfer 18MJ in 1.0 hour?

Suggested Follow-up Work

Summary Questions on pages 139 and 142 (Blue Textbook) Test yourself to ensure you are confident with the circuit symbols.

2. The Electron Gun

An electron gun is an electrical devices used to produce a narrow beam of electrons. This beam can then be used to ionise atoms, in electron microscopes or even in oscilloscopes.

How it works

A small metal filament is heated by an electric current causing the electrons in the wire to gain kinetic energy. Some electrons will gain enough energy to escape from the surface of the metal in a process known as thermionic emission (the emission of electrons through the action of heat).

If we apply a high p.d. between the filament and an anode, the filament becomes a cathode. Putting this entire system into a vacuum allows the emitted electrons to accelerate away from the filament towards the anode gaining kinetic energy.


Inside the electron gun, the anode contains a small hole. This means that only electrons in line with this hole can pass through the anode, creating a beam of electrons. These electrons will have a specific kinetic energy depending on the accelerating p.d. of the electron gun.


How much energy do the electrons gain?

From the definition for p.d., the work done on a single electron while travelling from the cathode to the anode is:

W=Q×VW=e×V where e is the charge on each electron and V is the accelerating p.d.

Due to the law of conservations of energy:

work done onelectron=kinetic energy gained by electroneV=12mv2Note: we are assuming that the electron

has negligible kinetic energy as it leaves the filament.

You must learn this equation for your exam as it is not found in the data booklet.

From this expression we can see that the greater the accelerating p.d. in the electron gun, the more kinetic energy the electrons will gain and thus the faster they will move.

Worked Example

Calculate the velocity of an electron from an electron gun with an accelerating p.d. of 8.0kV.

v2=2eVm v=√ 2eVm v=√ 2×(1.6×10−19)×(8×103)

9.11×10−31 v=5.3×107ms−1


Summary Questions on page 144 (Blue Textbook) Practice Question 2 on page 167 (Blue Textbook)


Use your formula book to confirm values of e and m (elementary charge and mass of an electron)

Convert p.d to volts

Worked Examples (In Class)

1. Calculate the velocity of an electron from an electron gun with an accelerating p.d. of 5.0kV.

2. Calculate the magnitude of the accelerating p.d. that produces electrons with a velocity of 4.5 x 107 ms-1

.


Stretch Content: Particle Accelerators

A linear particle accelerator uses a series of electrodes (called drift tubes) to accelerate sub-atomic particles such as electrons.

The drift tubes alternate their polarity (positive to negative). They do so with a timing so that when the electron leaves one tube the polarity changes to ensure the electron is attracted to the next one.

Simplified circuit diagram of a particle accelerator

Every time the electron moves between tubes it gains kinetic energy equal to eV (V = maximum emf of the alternating source).

The more drift tube in the system the higher the velocity the electron can be accelerated to.

Linear particle accelerators are used in CERN as well as in hospitals to produce the necessary X-rays required for radiotherapy.

Particle accelerator at CERN

Some Questions to think about


3. Resistance

All components (not just resistors) have their own resistance as they all resist the flow of charge carries through them. The higher the resistance of a component the more energy it takes to push the charges (electrons) through that component.

Resistance (R) is defined as:

‘The ratio between the p.d. across the component and the current in the component.’

This gives the equation:

R=VI

where R is the resistance is Ohms [Ω], V is the p.d. across the component is Volts [V] and I is the current in the component in Amps [A].

The Ohm is defined as:

‘The resistance of a component when the p.d. of 1V is produced per Ampere of current.’

1Ω=1V A−1

Determining ResistanceTo determine the resistance of a component the following circuit should be used:

Note: The voltmeter is connected in parallel with the component whereas the ammeter is connected in series with the component.

Ohm’s Law Ohm’s law states that:

‘For a metallic conductor kept at room temperature, the current in the wire is directly proportional to the p.d. across it.’

In other others, if you double the p.d. across a piece of wire (while keeping it at a constant temperature) you will also double the current through that wire.


Resistance and TemperatureWhen current passes through a metallic conductor, the temperature of that metallic conductor increases. This causes the resistance of the metallic conductor to increase.

This happens because as the metallic conductor gets hotter, the positive ions in the conductor have more energy causing them to vibrate with greater amplitude around their fixed positions. As a result, there are more collisions between the charge carriers (electrons) and the ions, forcing the electrons to do more work as they move through the component. The electrons thus transfer more energy to the wire as they travel through it.

Worked Example

The current in a filament lamp is 20mA. It is operated for a time of 3.0 minutes. The charge flowing in the lamp transfers 1.2kJ of energy to the lamp. What is the resistance of the lamp?

∆Q=I ∆ t∆Q=(20×10−3)× (3×60 )∆Q=3.6CV=WQ V=1.2×10

3

3.6V=333.3…V R=V

IR= 333.3

20×10−3

R=16665R=16kΩ


Summary Questions on page 147 (Blue Textbook) Attempt the Resistance Calculations (Worksheet 1 – Pages 3435


Be aware of prefixes throughout

Convert time to seconds

Round your answer to 2 s.f. like those given in the question

Calculate charge 1 st

Use this to calculate p.d.

Finally calculate resistance

Worked Examples (In Class)1. The current in a resistor is 9mA when the p.d. across it is 5.0V. What is the resistance of the resistor?

2. The current in a heater is 8.0A. It is operated for a time of 2.0 hours. The charge flowing in the heater transfers 6.5MJ of energy to the lamp. What is the resistance of the heater?


4. Practical: Collecting Data for I-V Curves

In this experiment you will be determining the current – voltage characteristic of an electrical component .

Practical Aims

To set up the circuit correctly To obtain an appropriate set of data To plot the characteristic

Equipment (per group)

Power supply (max 12V) Rheostat/potentiometer Ammeter Voltmeter Leads Test component (resistor as shown in

circuit diagram, filament lamp, diode, LED)

Health and safety

Safe use of electrical circuits Work within the limits of voltage and current

given by your teacher Do not short the cells

Task

1. To measure the characteristic curve for a component you must change the voltage and measure the current, this

should include reversing the polarity of the supply to obtain readings for negative voltage.

2. Find the maximum and minimum values of voltage that give appropriate readings of current, and then select

the steps needed to give the required number of values.

3. Measure the current as the voltage is changed across the component.

4. Draw the current – voltage curves for your component on graph paper or excel.

5. Calculate the resistance of the component at multiple points.

6. Describe the characteristic of the component with relation to potential difference, current and resistance.

7. If there is time, complete a characteristic for both a resistor, a lightbulb (and a diode).


5. Interpreting the I-V curves for filament lamps and resistors/wires

The I-V characteristic for any component shows the relationship between the current in a component and the potential difference across it.

ResistorsFixed resistors are designed such that their resistance is constant (regardless of a change in temperature). See the I-V curve for a resistor below:

Looking at the graphs we can draw the following information:

For a resistor (and a metal wire at constant temperature) p.d. is directly proportional to current (V ∝ I)o The resistor therefore obeys Ohm’s Law [it is an ohmic conductor]o The resistance of a resistor is constant

A resistor behaves the same way regardless of the polarity.

Filament LampsSee below the I-V curve for a filament lamp.


The p.d across a filament lamp is not directly proportional to the current. o It does not obey Ohm’s Law [it is a non-ohmic component]o The resistance of a filament lamp is not constant

A filament lamp behaves the same way regardless of the polarity. By doing a few calculations at different points we can show that the resistance of a filament lamp increases

as the current increases. This is due to an increase in temperature. [see page 14]


Summary Questions on page 150 (Blue textbook) Attempt I-V Curve Questions (Worksheet 2 – Pages 3638


6. DiodesA diode is a component that only allows current to flow through it in one particular direction. That means that unlike the components we have studied so far, the diode is affected by polarity. Diodes should be connected such that the arrow in the circuit symbol or the silver stripe of the component are facing towards the negative terminal of the power supply.

I-V Characteristics Below is an I-V graph for a diode.

The p.d across a diode is not directly proportional to the current. o It does not obey Ohm’s Law [it is a non-ohmic component]o The resistance of a diode is not constant

A diode’s behaviour depends on the polarity. At A the resistance of the diode is extremely high (infinite practically). It will not conduct electricity when

reversed. At B, we can see that as the p.d. increases the resistances is gradually starting to decrease. [This is at around

0.7V for a silicon diode]. o We can this the threshold p.d.

Above the threshold p.d. (e.g. at C) the resistance of the diode decreases rapidly for small increases in the p.d. At C the diode has very little resistance.

Light-emitting DiodesLEDs are diodes which emit light when they are conducting electricity. Unlike a filament lamp, LEDs will only emit light of a single wavelength. This wavelength is determined by the threshold p.d. of the diode. LEDs are very efficient components and take very little energy to run. They can be used as indicators to let you know if current is running in a circuit.


Summary Questions on page 152 (Blue Textbook)


-+conventional current

7. ResistivityThe resistance of a wire can be affected by many things:

The temperature of the wire The material of the wire The length of wire (L) The cross-sectional area of the wire (A)

Resistance vs. ResistivityThe term resistance refers only to a specific component. Resistivity however is an electrical property of a material. For example you could have two copper components which would have the same resistivity (since they are both made of copper) but due to the make-up of the components they may have completely different resistances.

Factors Affecting ResistanceThe resistance R of a wire is directly proportional to its length L.

R∝L

If you double the length of a piece of wire, you double its resistance.

The resistance R of a wire is inversely proportional to its cross-sectional area A.

R∝ 1A

If you double the cross-sectional area of a piece of wire, you halve its resistance.

We can combine these relationships to give:

R∝ LA

ResistivityThe constant of proportionality for the above relation is the resistivity ρ of a particular material at a given temperature. If the temperature of a material increases, its resistivity will also increase. This yields the equation:

R=ρLA

Resistivity has the unit ‘ohm metre; [Ωm] and it is defined as:

‘Resistivity is the product of the resistance of a component made of the material and its cross-sectional area, divided by the length of the component.’

Note: This is simply a rearrangement of the above equation written in words!

Conductors, Semiconductors and InsulatorsThe resistivity of materials varies significantly. The table on the right gives some examples. Notice that good conductorshave resistivity of the order of 10-8Ωm, insulators have valuesof the order of 1016Ωm. Semiconductors have resistivitysomewhere in between these.


Key Experiment – Determining Resistivity

A typical way to determine the resistivity of a material is to investigate first how the length of a piece of wire (of the required material) varies with length. This can be done easily using the circuit to the right.

Following the collection of values of current and p.d. resistance can be calculated using the equation:

R=VI

A graph should then be plotted of Resistance (R) against length of wire (L).

Since R∝ L, we should get a straight line through the origin. By considering the general equation of a straight line:

y=mx+c

and by rearranging our equation for resistivity to give:

R= ρA×L

we can see that the gradient of the straight line is in fact equal to ρA . Therefore we can deduce that:

ρ=A× gradient


Summary Questions on page 156 (Blue Textbook) Practice Questions 3, 5, 8 and 9 (pages 167-169 of your blue textbook) Worksheet on Resistivity (Worksheet 3 – Pages 3940


Worked Examples (In Class)

1. Silver has a resistivity of 1.6 x 10-8 Ωm. Calculate the resistance of a wire made from silver that is 60cm long and has a radius of 4.0 x 10-4m.

2. A wire has a diameter of 1.0 mm and a resistivity of 5.0 10−6 Ω m.Calculate the length of wire that would have a resistance of 5.0 Ω.


8/9. Practical – Finding the Resistivity of a Material

BackgroundIn this experiment, you will measure the current through different lengths of a metal wire. You will then determine the resistivity of the metal wire.

The p.d. V across the wire is related to the length L of the wire by the expression

VI=ρL

A

where I, and A are constants for the experiment. I is the current in the wire, is the resistivity of the wire and A is the cross sectional area of the wire.

This expression may also be written as

V= ρIA×L

Equipment (per group)

switch S

1 m length of resistance wire

micrometer or Vernier calliper

2 crocodile clips

7 connecting leads

d.c. power supply or battery pack and rheostat connected as a potential divider

1 ammeter

Rheostat


metal wire

+

L

S

V

A

R

Health and safety

The metal wire may get hot

Record your planned procedure to minimise this hazard and get it authorised by your teacher before proceeding with the experiment.

Procedure

1. Connect the circuit as shown on page 29.

2. Adjust the length L of wire in the circuit so that it is 50.0 cm.

3. Close the switch S and adjust the power supply or potential divider so that the reading on the voltmeter is 3.0 V.

4. Note the reading on the ammeter. This must be kept constant throughout the experiment.

5. Record the reading on both the ammeter and voltmeter for a range of different lengths of the metal wire. (Remember, the ammeter reading should always be the same).

6. Record your data in a table.

7. Plot a graph of V against L

8. Determine the gradient of your graph.

9. By taking appropriate measurements, determine the diameter of the metal wire.

10. Calculate the cross-sectional area A of the metal wire.

11. Use your answers to steps 4, 5 and 10 to determine a value for the resistivity ρ.

Stretch/Opportunities for discussion - additional activities to expand on this activity can be considered.

Using the uncertainty in your measurements for the p.d. across the wire and the length of the wire add

range bars to your graph and draw a worst acceptable fit line.

Determine the uncertainty in your gradient.

Determine the uncertainty in the measurement of the diameter and hence in the measurement of the cross-

sectional area.

Give a final answer for the resistivity of the metal wire with a tolerance within which there is confidence that

the value lies.

Research a value for the resistivity of the metal wire you have been given. Quote your sources.

Calculate the percentage difference between your value for the resistivity and the researched value and

comment on the accuracy and reliability of your experiment.

Explain how the value for the resistivity would have been different if the p.d. had been kept constant and the

variation of current with length had been measured.


10. The Thermistor

The Thermistor is an electrical component that has a resistance which is affected by temperature. Unlike a wire however, as the thermistor gets hotter, its resistance decreases. We say that the thermistor has a negative temperature coefficient (NTC).

A thermistor is made of a material called a semiconductor. When the temperature of this semiconductor increases, the number density of the charge carriers in the semiconductor increases. This causes the resistance to therefore decrease. This decrease is often a dramatic one, which make the thermistor very useful in sensing circuits (e.g. inside incubators, thermostats, kettles, etc.).

Above is a graph showing the effect of temperature on the resistances of a resistor. Notices that the relationship is non-linear.

I-V CharacteristicsSee below the I-V curve for a NTC Thermistor.


The thermistor is a non-ohmic component. The thermistor acts the same way regardless of polarity. As the current through the thermistor increases, the resistance decreases.

o This happens because as the current in the thermistor increases, it heats up.o This increases the number density of the charge carriers, decreasing resistance.


Summary Questions on page 159 (blue textbook) Practice Question 6 (page 168 of your blue textbook)


Investigation – The Thermistor

Aim – To collect data in order to investigate the relationship between resistance and temperature in an NTC Thermistor.

Equipment

Thermistor Heater Water Ice Beaker Thermometer Multimeter (set to Ohmeter setting) Leads

Tasks

The below should be included in your working lab book while carrying out this investigation.

1. A brief risk assessment outlining the potential hazards of this investigation and the control measures you will

take to reduce their risk.

2. A description the procedure you will undertake to investigate the aim above.

3. A table of results

4. A sensible graph to display/analyse your results.

5. A summary of your findings.


11. The LDR

The Light Dependent Resistor (LDR) is an electrical component that has a resistance which is affected by the intensity of the light incident upon it.

A typical LDR is made from a semiconductor. This number density of the charge carriers in this semiconductor increases as the intensity of light incident on it increases, causing it’s resistance to decrease. This relationship is illustrated graphically below:

Notice that the relationship between light intensity and resistance is non-linear.

LDRs and Infrared Astronomy

Some LDRs are sensitive to Infrared radiation (a different electromagnetic wave). This means we can use them to detect very weak/dim infrared signals from space in ‘Infrared Telescopes’.

Since we are unable to detect infrared radiation with the naked eye, these telescopes allow us to see objects in space that we may have missed otherwise (e.g. those behind dense clouds of opaque gas).


Questions in Infrared Astronomy on page 161 (blue textbook) Summary Questions on page 161 (blue textbook) Practice Question 1 (page 167 of your blue textbook)


infrared

visible

Investigation – The LDR

Aim – To collect data in order to investigate the relationship between resistance and light intensity for a Light Dependent Resistor.

Equipment

LDR Light Source Ruler Luminosity Meter A narrow opaque tube (ideally black) Multimeter (set to Ohmeter setting) Leads

Tasks

The below should be included in your working lab book while carrying out this investigation.

1. A brief risk assessment outlining the potential hazards of this investigation and the control measures you will

take to reduce their risk.

2. A description the procedure you will undertake to investigate the aim above.

3. A discussion on the steps you will take to reduce the effects of outside light sources and why this is

important.

4. A table of results

5. A sensible graph to display/analyse your results.

[Why not try doing this using Excel?]

6. A summary of your findings.


12. Electrical Energy and Power

Electrical circuit are often used to transfer energy from one place to another (e.g. from a cell to a light bulb). Whenever there is current flowing in a component energy is being transferred to it (from the power source). If the circuit is broke (or a switch is open) there will be more current and therefor no energy is being transferred.

Electrical Power

Electrical Power is defined as:

‘The rate of energy transfer by an electrical component.’

It is dependent of the current in the component and the p.d. across that component and is measured in ‘Watts’ [W].

The equation for electrical power is:

P=V ×I

where current is measured in Amps [A] and p.d. is measured in Volts [V].

The equation is derived as shown below from the definition:

P=Wt W=QV P=QV

tP=Q

t×V P=I ×V

Additional Power Equations

The equation above can be combined with the equation V=I ×R to give two further equations for power.

Substituting in V=I ×R gives:

P=I ×R×I P=I 2R

Substituting in I=VR gives:

P=V × VR P=V 2

R


1. ‘Rate of energy transfer/work done’

2. Substitute in know formula for work done

3. Rearrange and substitute for Q=I ×t

Worked Examples

Equations Used:

P=V ×I P=I 2×R P=V 2

R

1. Calculate the current drawn by an 8.0W charger when the p.d. across the charge is 5.4V.

P=V ×I I= PVI=8.05.4 I=1.48148… .. I=1.5 A

2. The heating element on a toaster has a resistance of 25. When making toast it has a current of 13A running through it. What is the rate of energy transfer for the toaster?

P= I 2

R

P=132

25

P=6.76

P=6.8W


Rearrange your formulae



Note that is is the definition of ‘power’.

Worked Examples (in class)1. An alarm clock has a current of 18A through it when the p.d. across the clock is 5.6V. What is the

power rating of the alarm clock?

2. An appliance has 2kJ of energy transferred to it per second. If the resistance of the appliance is 120, what value of current must be flowing in the appliance when it is switched on?


Summary Questions on page 164 (Blue Textbook) Attempt the ‘Power and Circuits’ (Worksheet 4 – Pages 41 43)


13. Paying for Electricity

When you are paying your electricity bill, you are in fact paying for the total amount of energy transferred to you home. This will usually be measured using an electricity meter.

The energy that is transferred to each of the devices you use is dependent on:

1. The power rating of the device2. How long you run the device for.

Both these factors must thus be considered when calculating your energy bill.

From the definition, we can write power as:

power= energy transferredtime taken

So therefore, the energy transferred can be written as:

energy transferred=power× timetakenW=P×t

The SI unit for energy is the Joule, however this is a tiny amount of energy when compared to the total amount an average household uses in a month! To combat this, energy on your electricity statement is is measured in ‘kilowatt-hours’ [kWh].

The kilowatt-hour is defined as:

‘The energy transferred by a device with a power of 1kW operating for a time of 1 hour.’

1kWh=3600000 J=3.8MJ

We must therefore be aware that when calculating energy transferred we must watch out for units and use the correct of combination of units:

kWh: W in kilowatt-hours [kWh], P in kilowatts [kW], t in hours [h]

SI Units: W in joules [J], P in watts [W], t in seconds [s]

See below and example of an Electricity Statement. Notice that the company is charging 8.085p per kWh of energy used.

Worked Examples

Equations Used:


W=P×t

1. A 2400W washing machine is used for 35 minutes. Calculate the energy transferred in kWh.

W=P×tW=2.4× 3560W=1.4W=1.4kWh

2. Electricity 4U charges 9.7p per kWh to supply electricity to your home. Calculate how much it would cost to run a 160W fridge for 30 days if you left it on continuously?

W=P×t

W=0.16× (30×24 )

W=115.2kWh

total cost=115.2×0.097 total cost=11.1744total cost=£11.17


Convert minutes to hours (÷60¿

Round to 2.s.f (as per questions values)Convert time into hours and power into kW.

Note that the combination of units we will need will be P in kilowatts [kW] and t in hours [h]

Multiply the number of kWh used by the cost per kWh. [Note: I have converted the cost to pounds [£]

Worked Examples (in class)1. A 1200W washing machine is used for 5 minutes. Calculate the energy transferred in kWh.

2. Sparks ‘R’ Us charges 11.6p per kWh to supply electricity to your home. Calculate how much it would cost to run 55W lamp for 1 week if you left it on for 6 hours a day?


Summary Questions on page 166 (Blue Textbook) Past Paper Questions 7 on page 168 (Blue Textbook)


Worksheet 1 – Resistance Calculations [Score /30]

1. The current through a 12 V lamp when it is connected to a 12 V battery is 2.0 Aa. Calculate the resistance of the lamp at this current. [2]

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b. The current through a 12 Ω resistor in an electric circuit is 1.5 A. Calculate the potential difference across the resistor. [2]

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c. A 50 Ω resistor in an electric circuit has a potential difference across it of 20 V. Calculate the current through the resistor. [2]

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2. Complete the following table. [6]

Current (A) Potential difference (V) Resistance ()

a) 4.0 20

b) 3.0 15

c) 50 200

d) 0.50 12

e) 0.25 60

f) 6.0 30

3. Draw a circuit diagram to show how you would measure the current and the potential difference for a resistor in order to calculate its resistance. [4]

4. A resistor in a circuit has a current through it of 0.20 A when the potential difference across it is 15 V. Calculate the resistance of the resistor. [2]


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5. A student measured the potential difference across different lengths of a certain wire when a current of 0.5 A was passing through the wire. Her results are shown in the table below.

Length of wire (m) 1.00 0.80 0.60 0.40 0.20

Potential difference (V) 2.4 1.9 1.5 0.9 0.5

Resistance of wire ()

a. Calculate the resistance of each measured length of wire. [5]b. Plot a graph of the resistance of the wire against its length. [4]c. Use the graph to calculate the resistance of 5.0 m of the wire. [3]

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Worksheet 2 – I-V Characteristics

This first set of questions checks your ability to apply your understanding of the heating effect of a current to a range of scenarios.

Figure 1 shows the current–voltage (I–V) characteristics for a filament lamp.

Figure 1 Current–voltage characteristics for a filament lamp

1. State the value of the current in the lamp when it stops obeying Ohm’s law. [1]

2. Explain why the lamp stops obeying Ohm’s law as the current increases beyond this point. [1]

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3. Calculate the resistance of the lamp when the current is 0.2 A. [2]

A car is fitted with an automatic window defrosting device, consisting of several very thin metal filaments running through the front windscreen in parallel. These elements are activated remotely by the owner several minutes before returning to the car to ensure that the windows are defrosted.


4. When first activated, the system operates with a current of 1.20 A, and a potential difference of 12.0 V across it. Calculate the electrical power of the device. [2]

5. After the frost on the windscreen melts, the current into the device reduces

to 1.00 A. Explain this effect. [3]

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6. A chip in the windscreen breaks one of the filaments. Explain what effect,

if any, this will have on the current into the device. [2]

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Polystyrene foam can be cut and shaped using a hot metal wire. As the wire is moved through the foam, the polystyrene melts. Figure 2 shows how the temperature of a cutting wire varies with the current in it. The melting point of polystyrene is 220 °C.

Figure 2 Effect of current on the temperature of a cutting wire


7. What is the minimum current in the wire that will allow it to be used to cut polystyrene? [1]

8. The wire has a potential difference of 12 V and a current of 0.25 A in it during normal cutting. What is the temperature of the wire during normal operations? [1]

9. What is the resistance of the cutting wire during normal operations? [1]

10. What power is dissipated by the wire? [1]

11. Stretch: Whilst being used, the wire snaps and is replaced by a wire made from the same material and of the

same length but with twice the diameter of the original wire. The potential difference across the wire remains at

12 V. Discuss whether the wire will still be able to cut the polystyrene foam. [3]

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[Score /18]


Worksheet 3 – Resistivity Calculations

1. Calculate the resistivity of a wire of resistance 5.0 Ω, if the length of the wire is 2.0 m and the diameter of the wire 0.2 mm. [2]

2. Determine the length of a constantan wire, of resistivity 4.9 10−7 Ω m and diameter 0.1 mm, required to make a coil of resistance 20 Ω. . [2]

3. A. Calculate the resistance of 10 m of wire of diameter 0.36 mm and resistivity 44 10−8 Ω m. [2]

B. Deduce the resistance of another wire of the same material but of half the length and half the diameter. [2]


4. The potential difference across 5.0 m of nichrome wire is 3.2 V and the current flowing through it is 0.20 A. The resistivity of nichrome is 1.0 10−6 Ω m. Determine the diameter of the wire. [3]

5. STRETCH A generator produces 100 kW of power at 500 V. If the power lost in the 800 m long copper cables connected to the generator is 6% of the power produced, calculate the diameter of the cable.

Resistivity of copper = 1.6 10−8 Ω m [4]

[Score /15]


Worksheet 4 – Power and Circuits


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[Score /26]


3

Past Paper Questions


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Extra Notes/Working Space


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