Name…………………..
Class………….
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G482
Electrons, Photons and Waves
Module 2.2:
Resistance
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Module 2.2 Resistance
1.
Circuits
2.2.1 Circuit symbols
Candidates should be able to:
(a) recall and use appropriate circuit symbols as set out in SI Units, Signs,
Symbols and Abbreviations (ASE, 1981) and Signs, Symbols and Systematics
(ASE, 1995);
(b) interpret and draw circuit diagrams using these symbols.
As circuit symbols
2.
PD/Voltage and EMF
2.2.2 E.m.f. and p.d.
Candidates should be able to:
(a) define potential difference (p.d.);
(b) select and use the equation W = VQ;
(c) define the volt;
(d) describe how a voltmeter may be used to determine the p.d. across a
component;
(e) define electromotive force (e.m.f.) of a source such as a cell or a power
supply;
(f) describe the difference between e.m.f. and p.d. in terms of energy transfer.
Energy, Work Done,
Energy types,
transformations and
transfers, voltage, p.d.,
e.m.f, coulombs, charge,
current
3.
IV for a resistor
2.2.3 Resistance
Candidates should be able to:
(a) define resistance;
(b) select and use the equation for resistance R = V / I
(c) define the ohm;
(d) state and use Ohm’s law;
(e) describe the I–V characteristics of a resistor at constant temperature,
Resistance, resistor, pd,
current, ohm, ampere,
volt, temperature
4.
I/V characteristics for
lamps and light
emitting diodes
(f) describe an experiment to obtain the I–V characteristics of a resistor at
constant temperature, filament lamp and light-emitting diode (LED);
(g) describe the uses and benefits of using light-emitting diodes (LEDs).
Current, Voltage,
Resistance,
Temperature, filament,
LED, Diode
5.
Resistivity
2.2.4 Resistivity
Candidates should be able to:
(a) define resistivity of a material;
(b) select and use the equation R = ρL /A
Resistance, Ohms, PD,
Current, resistivity,
(rho), micrometer
6.
Effect of
temperature on
resistance
(c) describe how the resistivities of metals and semiconductors are affected by
temperature;
(d) describe how the resistance of a pure metal wire and of a negative
temperature coefficient (NTC) thermistor is affected by temperature.
Resistance, Resistivity,
semiconductor,
conductor, insulator,
thermistor
7.
Power
2.2.5 Power
Candidates should be able to:
(a) describe power as the rate of energy transfer;
(b) select and use power equations P = VI, P = I2 R and P = V2/ R
(c) explain how a fuse works as a safety device (HSW 6a);
(d) determine the correct fuse for an electrical device;.
Power, Rate, Energy,
Watt, PD, Current,
Resistance, Fuse
8.
Electrical energy
(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).
Power, Rate, Energy,
Watt, PD, Current,
Resistance.
9.
G482 Module 1:
2.2 Resistance
Test
Review and assess their knowledge and understanding
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Lesson 6 notes – circuits
Objectives
(a) recall and use appropriate circuit symbols as set out in SI Units, Signs, Symbols and
Abbreviations (ASE, 1981) and Signs, Symbols and Systematics (ASE, 1995);
(b) interpret and draw circuit diagrams using these symbols.
Wires and connections
Component
Circuit Symbol
Function of Component
Wire
To pass current very easily from
one part of a circuit to another.
Wires joined
A 'blob' should be drawn where
wires are connected (joined), but
it is sometimes omitted. Wires
connected at 'crossroads' should
be staggered slightly to form two
T-junctions, as shown on the
right.
Wires not joined
In complex diagrams it is often
necessary to draw wires crossing
even though they are not
connected. I prefer the 'hump'
symbol shown on the right
because the simple crossing on
the left may be misread as a join
where you have forgotten to add
a 'blob'!
Power Supplies
Component
Circuit Symbol
Cell
Battery
DC supply
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Function of Component
Supplies electrical energy.
The larger terminal (on the left) is
positive (+).
A single cell is often called a battery, but
strictly a battery is two or more cells
joined together.
Supplies electrical energy. A
battery is more than one cell.
The larger terminal (on the left) is
positive (+).
Supplies electrical energy.
DC = Direct Current, always
flowing in one direction.
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AC supply
Supplies electrical energy.
AC = Alternating Current,
continually changing direction.
Fuse
A safety device which will 'blow'
(melt) if the current flowing
through it exceeds a specified
value.
Transformer
Two coils of wire linked by an iron
core. Transformers are used to
step up (increase) and step down
(decrease) AC voltages. Energy is
transferred between the coils by
the magnetic field in the core.
There is no electrical connection
between the coils.
Earth
(Ground)
A connection to earth. For many
electronic circuits this is the 0V
(zero volts) of the power supply,
but for mains electricity and some
radio circuits it really means the
earth. It is also known as ground.
Output Devices: Lamps, Heater, Motor, etc.
Component
Circuit Symbol
Function of Component
Lamp (lighting)
A transducer which converts
electrical energy to light. This
symbol is used for a lamp
providing illumination, for
example a car headlamp or torch
bulb.
Lamp (indicator)
A transducer which converts
electrical energy to light. This
symbol is used for a lamp which
is an indicator, for example a
warning light on a car dashboard.
Heater
A transducer which converts
electrical energy to heat.
Motor
A transducer which converts
electrical energy to kinetic energy
(motion).
Bell
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A transducer which converts
electrical energy to sound.
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A transducer which converts
electrical energy to sound.
Buzzer
A coil of wire which creates a
magnetic field when current
passes through it. It may have an
iron core inside the coil. It can be
used as a transducer converting
electrical energy to mechanical
energy by pulling on something.
Inductor
(Coil, Solenoid)
Switches
Component
Circuit Symbol
Function of Component
Push Switch
(push-to-make)
A push switch allows current
to flow only when the button is
pressed. This is the switch
used to operate a doorbell.
Push-to-Break
Switch
This type of push switch is
normally closed (on), it is open
(off) only when the button is
pressed.
On-Off Switch
(SPST)
SPST = Single Pole, Single
Throw.
An on-off switch allows current
to flow only when it is in the
closed (on) position.
2-way Switch
(SPDT)
SPDT = Single Pole, Double
Throw.
A 2-way changeover switch
directs the flow of current to
one of two routes according to
its position. Some SPDT
switches have a central off
position and are described as
'on-off-on'.
Dual On-Off
Switch
(DPST)
DPST = Double Pole, Single
Throw.
A dual on-off switch which is
often used to switch mains
electricity because it can
isolate both the live and
neutral connections.
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DPDT = Double Pole, Double
Throw.
This switch can be wired up as
a reversing switch for a motor.
Some DPDT switches have a
central off position.
Reversing
Switch
(DPDT)
An electrically operated
switch, for example a 9V
battery circuit connected to the
coil can switch a 230V AC
mains circuit.
Relay
NO = Normally Open,
COM = Common, NC = Normally
Closed.
Resistors
Component
Circuit Symbol
Function of Component
Resistor
A resistor restricts the flow of
current, for example to limit the
current passing through an LED.
A resistor is used with a
capacitor in a timing circuit.
Variable Resistor
(Rheostat)
This type of variable resistor
with 2 contacts (a rheostat) is
usually used to control current.
Examples include: adjusting
lamp brightness, adjusting motor
speed, and adjusting the rate of
flow of charge into a capacitor in
a timing circuit.
Variable Resistor
(Potentiometer)
This type of variable resistor
with 3 contacts (a
potentiometer) is usually used to
control voltage. It can be used
like this as a transducer
converting position (angle of the
control spindle) to an electrical
signal.
Variable Resistor
(Preset)
This type of variable resistor (a
preset) is operated with a small
screwdriver or similar tool. It is
designed to be set when the
circuit is made and then left
without further adjustment.
Presets are cheaper than
normal variable resistors so they
are often used in projects to
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reduce the cost.
Capacitors
Component
Circuit Symbol
Function of Component
Capacitor
A capacitor stores electric
charge. A capacitor is used
with a resistor in a timing
circuit. It can also be used as a
filter, to block DC signals but
pass AC signals.
Capacitor,
polarised
A capacitor stores electric
charge. This type must be
connected the correct way
round. A capacitor is used with
a resistor in a timing circuit. It
can also be used as a filter, to
block DC signals but pass AC
signals.
Variable Capacitor
A variable capacitor is used in
a radio tuner.
Trimmer Capacitor
This type of variable capacitor
(a trimmer) is operated with a
small screwdriver or similar
tool. It is designed to be set
when the circuit is made and
then left without further
adjustment.
Diodes
Component
Circuit Symbol
Diode
LED
Light Emitting Diode
Function of Component
A device which only allows
current to flow in one direction.
A transducer which converts
electrical energy to light.
Zener Diode
A special diode which is used to
maintain a fixed voltage across
its terminals.
Photodiode
A light-sensitive diode.
Transistors
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Component
Circuit Symbol
Function of Component
Transistor NPN
A transistor amplifies current. It can be
used with other components to make an
amplifier or switching circuit.
Transistor PNP
A transistor amplifies current. It can be
used with other components to make an
amplifier or switching circuit.
Phototransistor
A light-sensitive transistor.
Audio and Radio Devices
Component
Circuit Symbol
Function of Component
Microphone
A transducer which converts sound to
electrical energy.
Earphone
A transducer which converts electrical
energy to sound.
Loudspeaker
A transducer which converts electrical
energy to sound.
Piezo Transducer
A transducer which converts electrical
energy to sound.
Amplifier
(general symbol)
Aerial
(Antenna)
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An amplifier circuit with one input.
Really it is a block diagram symbol
because it represents a circuit rather
than just one component.
A device which is designed to receive
or transmit radio signals. It is also
known as an antenna.
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Meters and Oscilloscope
Component
Circuit Symbol
Function of Component
A voltmeter is used to measure
voltage.
Voltmeter
The proper name for voltage is 'potential
difference', but most people prefer to say
voltage!
An ammeter is used to measure
current.
Ammeter
Galvanometer
A galvanometer is a very sensitive
meter which is used to measure
tiny currents, usually 1mA or less.
Ohmmeter
An ohmmeter is used to measure
resistance. Most multimeters have
an ohmmeter setting.
Oscilloscope
An oscilloscope is used to display
the shape of electrical signals and
it can be used to measure their
voltage and time period.
Sensors (input devices)
Component
Circuit Symbol
LDR
Thermistor
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Function of Component
A transducer which converts
brightness (light) to resistance (an
electrical property).
LDR = Light Dependent Resistor
A transducer which converts
temperature (heat) to resistance
(an electrical property).
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Lesson 7 notes – Potential difference and EMF
Objectives
(a) define potential difference (p.d.);
(b) select and use the equation W = VQ;
(c) define the volt;
(d) describe how a voltmeter may be used to determine the p.d. across a component;
(e) define electromotive force (e.m.f.) of a source such as a cell or a power
supply;
(f) describe the difference between e.m.f. and p.d. in terms of energy transfer.
Voltage
Voltage is defined as the amount of work done or the energy required (in
joules) in moving a unit of positive charge (1 coulomb) from a lower potential
to a higher potential. Voltage is also called potential difference (PD). When
you measure voltage you must have two points to compare, one of them
being the reference point.
1 volt = 1 joule/coulomb
V=ΔW/ΔQ
Voltage in an electrical system can be thought of as the same thing as
pressure in a water system; the Cell being the pump.
Two important pieces of terminology for you:
The voltage across the source of electrical energy is the E.M.F. (or
electromotive force)
The voltage across a component is the p.d. (potential difference)
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Lesson 7 questions – Potential difference and EMF
1)a)
Put a tick in the box for an alternative unit for Voltage
JC
Js-1
JC-1
(1)
b)
A 1.2kW water heater is switched on for 1500s. During this time, a charge of
7.5kC passes. Calculate the p.d. across the heater.
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
Total [4]
2)a)i) State the unit of electric charge
…………………………………………………………………………………………
(1)
ii)
Name an instrument that measures the p.d. across a component.
…………………………………………………………………………………………
(1)
b)
A lamp uses 36 Joules every second and draws a constant current of 3.0A over
a period of 600s from a battery. Calculate:
i)
the total amount of energy transferred to the lamp,
…………………………....……………………………………………………………
(2)
ii)
the charge passing through the lamp in one second,
…………………………....……………………………………………………………
……………………....…………………………………………………………………
…………………………....……………………………………………………………
(2)
iii)
the total charge passing through the lamp,
…………………………....……………………………………………………………
…………………………....……………………………………………………………
……………………………....………………………………………………………(3)
iv)
The total number of electrons passing through the lamp,
…………………………....……………………………………………………………
…………………………....……………………………………………………………
…………………………....………………………………………………………… (2)
v)
The potential difference across the lamp.
…………………………....……………………………………………………………
…………………………....……………………………………………………………
…………………………....………………………………………………………… (3)
Total [14]
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Lesson 8 notes – I/V characteristics of thin wires and
resistors.
Objectives
(a) define resistance;
(b) select and use the equation for resistance R = V / I
(c) define the ohm;
(d) state and use Ohm’s law;
(e) describe the I–V characteristics of a resistor at constant temperature,
Starter
Let’s imagine a simple circuit with a power pack, ammeter, resistor and bulb in
series. A voltmeter is placed across the bulb.
A
V
If we increase the resistance, the current will go down and the brightness of
the bulb will go down too. The voltage will stay the same.
The point to be made is simple: more resistance means less current for the
same voltage.
This illustrates that increased resistance reduces the flow of current around a
circuit and should leads us to the idea that it is sensible to measure resistance
in terms of ‘volts per amp’. The larger the resistance of the circuit the greater
the electrical ‘push’ needed to make a particular current flow (the more
resistance the more volts needed per amp).
This leads to the equation R = V/I.
This is the ratio of the pd across a component to the current flowing through it.
So Resistance = Volts per Amp
Writing that out: Resistance = Voltage / Current
And 1 Ohm is One Volt per Amp
1 Ω = 1 V A-1
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Kilo-ohms (k Ω) and mega-ohms (M Ω) are commonly used:
1 k Ω = 1000 Ω
1 M Ω = 1000 k Ω = 1 000 000 Ω
Worked example:
Calculating resistance
Calculate the resistance of a lamp when a pd of 10 V makes a current of 2 mA
flow through it. (This will give practice in handling powers of 10.)
R = V/I = 10 / 2 x 10-3 = 5 x 103 = 5000 W = 5 k Ω
Characteristics of Constantan wire
V
1.0
0.8
0.6
I
0.4
Draw a best-fit
straight line
through the points.
I
V
= average conductance
gradient =
0.2
0
–5
–4
–3
–2
–1
–0.2
In this quadrant
the terminals
were reversed.
1
2
3
4
5
potential difference/V
6
–0.4
–0.6
–0.8
A straight line shows that
the current through the
coils is directly proportional
to the applied p.d.
The wire is ohmic.
–1.0
A straight line graph through the origin shows that the current is proportional
to the potential difference. This result is known as Ohm's law, which applies to
metal or metal alloy wires as long as their temperature remains constant.
For any point on the graph the resistance R can be found by calculating V/I. If
the graph is a straight line then the resistance is constant – the same for
every value of current or potential difference. Under these conditions finding
1/gradient gives the average resistance of the wire. (Graphs using best-fit
lines are often the best way of averaging results.)
The resistance stays the same when the current is reversed).
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Ohm’s law
Historically, Ohm showed that the resistance of a metal under constant
physical conditions (particularly temperature) is constant. The experiment of
passing a varying current through a wire and measuring the voltage across it
demonstrates this by generating a straight line graph that passes through the
origin: if I is directly proportional to V (or the other way around) then Ohm’s
law is obeyed. Any conductor (metallic or otherwise) that behaves in this way
is described as an ‘ohmic conductor’.
Plenary
Wires are known as ohmic conductors because of the straight line through the
origin. Resistance is constant at constant temperature.
If temperature varies, the resistance will vary also. In metals there is a lattice
of ions with free electrons that conduct the electricity. If it is harder for the
electrons to move along the wire because the lattice is vibrating, the
resistance will increase. This is the case in a filament bulb. As the voltage
increases, the temperature rises in the wire and the ions in the lattice start to
vibrate making it harder for the electrons to move along the wire, therefore
increasing the resistance.
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Lesson 9 notes - Ohm’s law and I/V characteristics for
filament bulbs and light emitting diodes.
Objectives
(f) describe an experiment to obtain the I–V characteristics of a resistor at constant
temperature, filament lamp and light-emitting diode (LED);
(g) describe the uses and benefits of using light-emitting diodes (LEDs).
Filament Lamps
A filament lamp
(non-ohmic).
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As temperature changes, resistance changes. In metals, as temp
goes up, the atoms have more kinetic energy and get in the way of
flowing free electrons and so the resistance goes up.
Diode characteristics
Non-Ohmic conductor
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Plenary
Diodes allow electricity to flow in only one direction. The arrow of the circuit
symbol shows the direction in which the current can flow. Diodes are the
electrical version of a valve and early diodes were actually called valves.
Electricity uses up a little energy pushing its way through the diode, rather like
a person pushing through a door with a spring. This means that there is a
small voltage across a conducting diode, it is called the forward voltage drop
and is about 0.6V for all normal diodes which are made from silicon. The
forward voltage drop of a diode is almost constant whatever the current
passing through the diode so they have a very steep characteristic (currentvoltage graph).
When a reverse voltage is applied a perfect diode does not conduct,
resistance is infinite, but all real diodes leak a very tiny current of a few µA or
less. This can be ignored in most circuits because it will be very much smaller
than the current flowing in the forward direction. However, all diodes have a
maximum reverse voltage (usually 50V or more) and if this is exceeded the
diode will fail and pass a large current in the reverse direction, this is called
breakdown.
Thermistors at constant temperature behave like a wire and they are ohmic
conductors. But if the temperature increases their resistance will decrease.
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Lesson 9 questions – Wires, resistors, bulbs and
diodes
1) a) Define Resistance
…………………………………………………………………………………………
………………………………………………………………………………………(2)
b) Fig 1.1 shows the I/V characteristics of a filament lamp.
State how, and explain why the resistance of the filament lamp changes as the PD
across it changes
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………(2)
Total [6]
2) Fig 2.1 shows how the potential difference V varies with the resistance R of a
tungsten filament lamp.
fig 2.1
a)
Use fig 2.1 to calculate, for a p.d of 3.0 V the current in the lamp.
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
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b)i)
Suggest why the resistance of the lamp does not vary significantly over the
range 0 to 2.0 V.
……………………………………………………………………………………… (1)
ii)
The tungsten filament lamp is at room temperature when the p.d. across it is
zero. State the resistance of the lamp at room temperature.
……………………………………………………………………………………… (1)
Total [5]
3)a) State Ohm’s Law
…………………………………………………………………………………………
………………………………………………………………………………………
………………………………………………………………………………………(2)
b)
The I/V characteristic for a component is shown in fig 4.1 below:
fig 4.1
i)
Name the component
……………………………………………………………………………………… (1)
ii)
1 mark is available for written communication in this question.
Describe, referring to figure 4.1 how the resistance of the component depends on the
potential difference V across it. Show any calculations you make.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………(5)
+1 for quality of written communication
Total [9]
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Lesson 10 Notes –Resistivity
Objectives
You will:
(a) define resistivity of a material;
(b) select and use the equation R = ρL /A
Resistance
Resistance is a measure of the difficulty for a charge to get through a material.
Resistance is a property of a material that makes a moving charge dissipate energy.
It is the ratio: Resistance = Potential Difference/Current
R=V/I
Water Flow
With the same pressure in each tube we will describe the water flow in each
pipe.
A
D
B
D/2
Doubling diameter quadruples area (area = πr2).
So water flow quadruples – you can fit 4 times as much through at once!
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Resistivity
Variation of resistance with length and area.
Resistance R is proportional to length l
Resistance R is inversely proportional to area A
So,
R = constant x length / cross-sectional area
This constant is called Resistivity ρ
R = ρl/A
Units are Ωm (NOT Ωm-1)
Make sure you can rearrange the equation to get ρ = RA/l
Resistance question
Two wires, A and B are made of the same material. Wire A is twice as long as
B and has twice its diameter. Which wire has the greatest resistance?
Wire A:
R=ρ2L/(2D/2)2
R=2ρL/D2
Wire B:
R=ρL/(D/2)2
R=ρL/(D2/4)
R=4ρL/D2
Answer: Wire A has ½ the resistance of wire B.
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Lesson 10 questions –resistivity
1)a) Define electrical resistivity.
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
b)
Fig1.1 illustrates a metallic resistor constructed by depositing a thin layer of
metal on a plastic strip. This particular resistor has a resistance of 5.0Ω, length
1.2x10-2 m and width 2.0x10-3m.
fig1.1
i)
The resistivity of the metal is 4.3x10-6Ωm. Calculate the cross-sectional area A
of the resistor.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
ii)
What is the thickness t of the resistor?
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
Total [8]
2)
Fig2.1 shows a conducting paint cylindrical glass vessel.
vessel
h
conducting
paint
cross-sectional
base
Fig2.1
The volume of the paint is 1.2x10-5m3 and the vessel has a base area of 3.0x10-4m2.
i)
Show that the height h of the paint column is 4.0cm
…………………………………………………………………………………………
…………………………………………………………………………………………
…...………………………………………………………………………………… (1)
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ii)
Calculate the resistance of the paint column of height 4.0cm. The resistivity of
the paint is 6.9x10-2Ωm.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
(2)
c)
State and explain how your answer to (b)(ii) changes when the same volume
of paint is poured into a cylindrical glass vessel having a base of double the crosssectional area.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
(2)
Total [5]
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Lesson 11 Notes – Effects of temperature on Current.
Objectives
(c) describe how the resistivities of metals and semiconductors are affected by
temperature;
(d) describe how the resistance of a pure metal wire and of a negative temperature
coefficient (NTC) thermistor is affected by temperature.
Conventional current flow
Ohmic conductor
metal ion
+
free electron
Electron flow
In an ohmic conductor, free electrons carry the current when a potential
difference is applied across it. When the physical conditions stay constant,
current will be proportional to the p.d. and this will equal ratio V/I which is the
resistance R of the component.
Heated wire
vibration
(lamp filament)
metal ion
++
-free electron
Electron flow
When the potential difference increases, the current flow also increases to
start with as we’d expect, heating the wire. Meaning the ions in the lattice
vibrate more rapidly. This in turn leads to a greater resistance slowing the flow
of electrons so a smaller current flows.
©2011 science-spark.co.uk
RAB Plymstock School
Semiconductors
In semiconductors as the energy given increases, more free electrons are let
free and so the resistance goes down. In light dependent resistors, this
energy is light and in thermistors it is heat.
Thermistors
Thermistors are inexpensive, easily-obtainable temperature sensors. They
are easy to use and adaptable. Circuits with thermistors can have reasonable
outout voltages - not the millivolt outputs thermocouples have. Because of
these qualities, thermistors are widely used for simple temperature
measurements. They're not used for high temperatures, but in the
temperature ranges where they work they are widely used.
Thermistors are temperature sensitive resistors. All resistors vary with
temperature, but thermistors are constructed of semiconductor material with a
resistivity that is especially sensitive to temperature. However, unlike most
other resistive devices, the resistance of a thermistor decreases with
increasing temperature. That's due to the properties of the semiconductor
material that the thermistor is made from. For some, that may be
counterintuitive, but it is correct. Here is a graph of resistance as a function of
temperature for a typical thermistor. Notice how the resistance drops from
100 kW, to a very small value in a range around room temperature. Not only
is the resistance change in the opposite direction from what you expect, but
the magnitude of the percentage resistance change is substantial.
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 12 Notes – Electrical Power
Objectives
You will be able to recall and use V=P/I
You will be able to recall and use P=I2R and P=V2/R
You will be able to recall and use W=IVt
You will understand and use the kilowatt-hour (kWh) as a unit of energy.
Derivation of equation for power
We know that:
and
ΔW=VΔQ
ΔQ=IΔt
(1)
(2)
Therefore
ΔW=IVΔt
(3)
Now,
Power is the rate of doing work in a given time
So, in symbols:
P=ΔW/Δt
P=IVΔt/Δt
Substituting in equation (3)
Therefore:
(4)
P=IV
We also know that:
follow)
(5)
(6)
R=V/I
(7) (V=IR and I=V/R
If we substitute these into (6) we get:
P=I2R and P=V2/R
Power is measured in Watts.
1 watt is 1Joule per second.
This is quite a small amount when considering household electrical power
consumption and even kW would be an awkward unit to use by power
companies.
Instead, power companies use a unit for Energy called the kilowatt hour (kWh)
(1 kWh = 1 Unit).
They take the amount of kilowatts used and multiply this by the amount of
hours power was used and get a number of units.
Number of Units (kWh) = Power (kW) x time (hours)
©2011 science-spark.co.uk
RAB Plymstock School
They then price these units depending on different factors (for example the
price of oil etc.) and then can work out the cost for the household using this
equation:
Cost (p) = cost per unit (p/kWh) x number of units (kWh)
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 13 Notes – Electrical Energy
Objectives
(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)..
Derivation of equation for power
We know that:
and
ΔW=VΔQ
ΔQ=IΔt
(1)
(2)
Therefore
ΔW=IVΔt
(3)
Now,
Power is the rate of doing work in a given time
So, in symbols:
P=ΔW/Δt
P=IVΔt/Δt
Substituting in equation (3)
Therefore:
(4)
P=IV
We also know that:
follow)
(5)
(6)
R=V/I
(7) (V=IR and I=V/R
If we substitute these into (6) we get:
P=I2R and P=V2/R
Power is measured in Watts.
1 watt is 1Joule per second.
This is quite a small amount when considering household electrical power
consumption and even kW would be an awkward unit to use by power
companies.
Instead, power companies use a unit for Energy called the kilowatt hour (kWh)
(1 kWh = 1 Unit).
They take the amount of kilowatts used and multiply this by the amount of
hours power was used and get a number of units.
Number of Units (kWh) = Power (kW) x time (hours)
©2011 science-spark.co.uk
RAB Plymstock School
They then price these units depending on different factors (for example the
price of oil etc.) and then can work out the cost for the household using this
equation:
Cost (p) = cost per unit (p/kWh) x number of units (kWh)
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 13 Questions – Electrical Power
1)
A 1.2kW water heater is switched on for 1500s. During this time, a charge of
7.5kC passes.
a)
The electrical energy transformed by the heater in joules,
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
b)
the cost of using the heater given that the cost of 1kWh is 6.4p.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
Total [4]
2)
A convenient unit of energy is the kilowatt hour (kWh)
a)
Define the kilowatt hour.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
b)
A 120W filament lamp transforms 5.8kWh. Calculate the time in seconds for
which the lamp is operated.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
Total [3]
3)
A battery has an e.m.f of 9V and a negligible internal resistance. It is capable
of delivering a total charge of 1350C. Calculate:
a)
the maximum energy the battery could deliver,
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
b)
the power it would deliver to the components of a circuit if the current through
it was 0.5A,
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
c)
how long the battery would last for if it was to supply power at the rate
calculated in (b).
………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………(3)
Total [7]
©2011 science-spark.co.uk
RAB Plymstock School