Download FUNDAMENTAL ELECTRICAL CONCEPTS

ELECTRONICS LEVEL 3
D.C. PRINCIPLES
Unit 20430 V1 L3 C7
1
2
INDEX
Introduction
Electrical Concepts
4
5
Charge
6
Voltage
7
Work Done
7
Power
7
Class Task
9
Worked Example
16
Student Examples
17
EMF and Volt Drop
20
Ohm’s Law Review Task
21
Workshop Ohm’s Law
22
Kirchoff’s Laws
25
Student Examples
28
Workshop Series Parallel Circuits
32
Tutor Sig _____________
Workshop potentiometer
37
Tutor Sig _____________
Supplementary Questions
39
Tutor Sig _____________
Resistivity
41
Class Task
43
Temperature Coefficient
48
Class Task
51
Student Examples
54
Workshop Temperature Probe
57
Crossword
59
Strain Gauge
60
Voltage Divider
62
Student Examples
64
Workshop Voltage Divider
67
Review Questions
70
Pair Definitions
83
Glossary
84
Assessment plan
86
Tutor Sig _____________
Tutor Sig _____________
Tutor Sig _____________
Workbook completed : Tutor Signature __________________ Date ________
3
Introduction
For many of you, the majority of this unit will be a revision of knowledge and skills
gained at Level 2. For others there may be many new terms and concepts.
Your first task is to skim through the booklet and identify any words or phrases that
you have not seen before or think that you don’t already have a good understanding
of. Try to find about 10 words. These should be written in the table below.
We will bring together all the words produced by the class (I may add some) and
keep them on display during the lessons. By the end of the unit you should all have a
good understanding of these key words.
The second column of the table should be completed at the end of the unit.
Key Words or Phrases
Word or Phrase
Explanation (to be completed at the end
of the unit)
4
FUNDAMENTAL ELECTRICAL CONCEPTS
Electric Charge
Electric charge is the build up and movement of negatively charged electrons. All
materials contain electrons but the electrons in some materials are more easily
moved than in others.
Conductors
Conductors, such as copper, will have 1 electron in the outer
shell of its atom, known as the Valence Level. This electron is
easily moved from this shell by the application of voltage and it
will easily move within the material. It is then known as a free
electron and if enough electrons are forced (generally by
voltage) to move in a particular direction it is known as current.
Current is therefore the movement of free electrons.
Insulators
Insulators have more than 5 electrons in their outer shell and
these are held tightly to the atom. Applying voltage to an
insulator will not move theses electrons and current will not
flow. However if sufficient energy (in the form of voltage or
heat) is applied the insulator can be made to break down and
the electrons will flow. A piece of glass which is a good insulator
can be made to conduct electricity by the application of heat.
The electrons are given energy from the heat and break away
from the atom. We see this as molten glass. At this point the
glass becomes a good conductor and current will flow through it.
Semi-Conductors
Semi-conductors (such as silicon shown opposite) have a
valence level of 4 and are neither good conductors nor
insulators. However their atomic structure can be changed by
the addition of another material (called doping) to make them
either conductors or insulators. We will look at this process
when we look at semi-conductors in more detail later in the
course.
5
CURRENT FLOW
As we know current flow is the movement of negatively charged electrons in a
material. The amount of electrons flowing determines the current value. The amount
of charged electrons moving is measured in coulombs (C) and is given the symbol Q.
One coulomb of charge is the combined charge of 6240000000000000000 electrons.
That’s 6.24 x 1018 electrons. If that ‘chunk’ of electrons passes a point in 1 second
then that is equal to 1A of current. Two amps of current would be twice as many
electrons passing a given point in one second. From this definition we get the
formula:
Where
I = current measured in amps
Q = charge measured in coulombs
t = time measured in seconds
In other words, the amount of current flowing in a circuit is equal to the total charge
of electrons divided by the time taken to pass a point in the circuit.
Example 1
An electric charge of 12C flows pass a point in 3 seconds. Determine the amount of
current flowing.
I=Q/t
I = 12 / 3
I = 4A
Example 2
How long will it take for a charge of 6C to pass a point in a circuit if the current value
is measured as 12A
From the formula I = Q / t we can transpose to give t = Q / I
t=Q/I
t = 6 / 12
t = 0.5 seconds or 500mS
6
Voltage
Voltage, or potential difference (PD) is an electromotive force (emf) that causes
current to flow in a circuit. In order for current to flow the emf (V) must use energy.
When energy is used, work (measured in joules) must have been done. We can say
that if an emf of one volt moves a charge of one coulomb, one joule of work would
have been done. We can show this as the formula
Where:
W = work done in joules (J)
Q = charge in coulombs (C)
V = emf in volts (V)
In words, the amount of work, measured in joules, taking place in a circuit is
calculated by the charge being moved and the voltage being applied to move it.
Example
An emf of 20V moves 4C of charge in an electric circuit. How much work is being
done?
W=QxV
W = 4 x 20
W = 80J
Power
The energy used up is expended in the form of power, measured in Watts (W).
Power is the rate at which energy is used and is calculated by the formula
P=W/t
In words, power is energy divided by time
Example
What is the power of a circuit that uses 30J in 6 seconds
P=W/t
P = 30 / 6 = 5W
7
POWER
We have the formulas
If we substitute the first two formulas in the third we will get P = I x V (Watts). This is
the more familiar and more useful power formula. The power developed in an
electrical circuit is the product of current and voltage and is measured in Watts.
See if you can derive the formula P = I V from the three formulas above.
We know from Ohm’s Law (Level 2 electrical) that I= V/R and V = IR and so we can
substitute these formulas into P = I V to give us two other Power formulas
P = I2R (Power is equal to current squared times resistance)
P = V2/R (Power is equal to the voltage squared times resistance)
Example
A voltage of 12V causes a current of 3A to flow in a 4Ω circuit. Calculate the power
P = I x V = 3 x 12 = 36 Watts
P = I2R = 3 x 3 x 4 = 36 Watts
P = V2/R = 12 x 12 / 4 = 36 Watts
8
Class Task
We have looked at 3 main formulae:
I=Q/t
W=QxV
P=W/t
You will need to be able to use these formulas in their current form as well as
transpose for any of the other terms.
You will also need to be able to explain the formula in words and state all the units
used.
To enhance your understanding of the formula and to provide you with a revision
sheet your task is to produce an A3 explanation of each formula. You will need to
provide the following information on each of the three sheets:

The original formula using symbols

The original formula explained in words

The units of all the terms

An example of the original formula being used

The transposition of the formula for each of the other two terms

An example of each of these transposed formula being used

A pictorial representation of the formula. You need to make this something
that will help you remember the formula.
The unit states that these formulas must be remembered and so they will NOT be
provided in the assessment.
The following pages should be used to produce the review sheets. You should then
use your revision sheets to complete the Student Example Questions
9
10
11
12
13
14
15
Worked Example
A 6Ω load is connected across an 18V load for 20 seconds.
Determine:



The charge (Q) on the circuit (Q = I x t)
How much work (W) has been done (W = Q x V)
The power (P) of the circuit (P = W / t)
Calculation of Charge (Q)
In order to calculate the charge we must first calculate the current flowing through
the load
I=V/R
I = 18 / 6
I = 3A
We now have current (3A) and time (20s)
Q=Ixt
Q = 3 x 20
Q = 60 Coulombs
Calculation of Work (W)
Work = Q x V
Work = 60 x 18
Work = 1080 joules
Work = 1.08kj
Calculation of Power
P=W/t
P = 1080 / 20
P = 54W
We can check this with P = V2 / R
P = 182 / 6 = 54W
(Also true for P = I x V and P = I2R)
16
Student Example 1
A 10Ω load is connected across a 120V supply for 5 seconds.
Determine:



The charge (Q) on the circuit (Q = I x t)
How much work (W) has been done (W = Q x V)
The power (P) of the circuit (P = W / t)
17
Student Example 2
A 0.05KΩ load is connected across a 150mV supply for 4.2mS.
Determine:



The charge (Q) on the circuit (Q = I x t)
How much work (W) has been done (W = Q x V)
The power (P) of the circuit (P = W / t)
18
Student Example 3
20V is applied to a circuit for 10 seconds. It is known that 1000j of work was done in
that time.
Determine:



The charge of the circuit
The current flowing
The resistance of the load
19
EMF and VOLT DROP
We know that in order for current to flow in a circuit we need an energy source. This
applied energy in the form of voltage is known as the electro-motive-force (emf).
This can come in the form of a battery or generator for example.
As current flows, it causes volt drops across components within the circuit. This
voltage is known as a volt drop and not an emf. An emf is the supplied voltage. A volt
drop is a consequence of an applied emf within a circuit.
I
EMF
Volt Drop
CONVENTIAL AND ACTUAL CURRENT FLOW
Before scientists were able to study the actual movement of electrons within circuits
it was assumed that current flowed from the positive connection of an emf through
the circuit to the negative.
When it was finally discovered that in actual fact electrons (which are negative)
move towards the positive of the emf it was too late to change all the thinking and
theory related to electrical science.
Scientists therefore decided to leave the idea of current flow from positive to
negative. This is known as CONVENTIONAL current. We now know that ACTUAL
current flow is the movement of electrons from negative to positive. Conventional
current flow is actually the movement of the holes left behind by the electrons.
It can be summarised as:

Conventional current flows from positive to negative

Actual current flow is from negative to positive

Electron flow is from negative to positive

Hole flow is from positive to negative
20
OHM’S LAW REVIEW TASK
This task will review series and parallel resistive circuits. Your tutor will split the class
into three groups.
Instructions
You will be given a resistive pack by your tutor.
Determine the maximum number of series / parallel circuits possible using the
resistors given.
Work as a team to calculate the resistance of each of the resistor combinations
Use the space below for working out but the combinations (diagrams) and the
resistance value for each must be transferred to A3 paper for display. It doesn’t have
to be ‘neat’ but it must be readable.
The group with the most correct combinations in 15 minutes will win the prize
21
WORKSHOP PRACTICAL
OHM’S LAW LAB
Aim
To prove Ohm’s law
Theory
Ohm’s Law states that the current through a resistor is
equal to the voltage across it, divided by the resistance
value.
Equipment Adjustable DC power Supply
Two Digital Multimeters
220Ω Resistor
Procedure
Task 1
Construct the following circuit
A
Adjustable
DC Power
Supply
V
22
220Ω
Task 2
Complete the table below using the following procedure:
1.
2.
3.
4.
5.
Adjust the power supply voltage to the required voltage across the resistor
Record the current value flowing through the resistor for each voltage step
Calculate the current using ohm’s law for each voltage step
Calculate the actual error and the percentage error for each voltage step
Calculate the power used in the resistor for each voltage step
Actual Error = calculated current – actual measured current
Percentage error = actual error / measured current x 100%
Results Table
Resistor
Voltage
Measured
Current
Calculated
Current
Actual Error
0V
1V
2V
4V
6V
8V
10V
12V
14V
16V
18V
20V
23
Percentage
error
Calculated
power
Task 3
Plot a graph showing the relationship between voltage and current
using the actual measured values from the table.
I
V
Conclusion
Do your results prove that Ohm’s Law is true?
Were there any large errors in your readings?
What have you learnt from completing this lab?
Are there any other conclusions you can make from this lab?
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
24
KIRCHOFF’S LAWS
Kirchoff is accredited with the discovery of 2 laws. His first law relates to current
within a circuit at a particular point. This point is also referred to as a node or
junction.
His law says that the sum of all the currents at the junction is equal to zero.
I1
I4
I2
Current at
this point is
always zero
I3
If we look at the circuit we can see that 2 currents are flowing into the circuit and
two are flowing away. If the directions of the currents are taken into account the
sum of the currents will always be zero.
For example if i1 = 5A and i4 = 3A the total current flowing into the junction would be
8A.
If i3 = 6A and i2 = 2A then current flowing out of the junction would be 8A.
The direction of the currents are opposite so the total current at the junction would
be 8 – 8 = 0A
If any of the currents flowing into the junction increase there will be a corresponding
increase in the current flowing away from the junction.
25
His law can also be expressed as the current flowing into a junction must be equal to
the current flowing out of the circuit. This is relevant when we look at parallel
circuits.
5A
8A
2A
1A
The 8A flowing into the circuit split up into 3 currents determined by the resistance
of each leg.
The sum of the three currents is equal to the current flowing into the circuit. Taking
into account the directions of the currents the sum is equal to zero.
26
KIRCHOFF’S SECOND LAW
Kirchoff’s second law relates to the volt drops within a circuit. He says that the sum
of the volt drops in a closed circuit is equal to the emf supplied.
Circuit Diagram
In the diagram below V1 + V2 + V3 would be equal to the emf supplied by the battery
V4.
V1
V2
V3
V4
V5
V5 is in parallel with V3 and so would have the same volt drop as V3
Kirchoff’s second law shows that the emf supplied to a circuit is ‘lost’ as volt drops
throughout the circuit.
For instance if V4 were equal to 12V this 12V would be dropped across the 3
resistances.
The voltage remaining after the third resistor would be equal to zero.
When you carry out series circuit calculations you should always check that the sum
of the voltages within the circuit is equal to the supply voltage.
27
Example 1
Three resistors of 6Ω, 12Ω and 18Ω are connected in series across a 24V supply.
Complete the circuit diagram below with this information and determine:




total resistance
total current flow
voltage drop across each resistor
power developed by each resistor.
Show how the circuit demonstrates Kirchoff’s Second Law.
28
Example 2
Three resistors of 10Ω, 15Ω and 20Ω are connected in series across a 32V supply.
Draw a circuit diagram using this information and determine:




total resistance
total current flow
voltage drop across each resistor
power developed by each resistor.
Show how the circuit demonstrates Kirchoff’s Second Law.
29
Example 3
Three resistors of 6Ω, 12Ω and 18Ω are connected in parallel across a 24V supply.
Draw a circuit diagram using this information and determine:




total resistance
current flow through each resistor
voltage drop across each resistor
power developed by each resistor.
Show how the circuit demonstrates Kirchoff’s First Law.
30
Example 4
Two resistors of 6Ω and 12Ω are connected in parallel. They are then connected in
series with an 18Ω resistor. Using this information draw the circuit diagram and
determine:




total resistance
current flow through each resistor
voltage drop across each resistor
power developed by each resistor.
Show how the circuit demonstrates both of Kirchoff’s Laws.
31
WORKSHOP PRACTICAL
SERIES / PARALLEL LAB
Aims
To prove Ohm’s law.
To prove the rules of calculating total resistance of series
and parallel circuits.
To prove Kirchoff’s Laws
Equipment Adjustable DC power Supply
5 resistors of different values between 800Ω and 15kΩ
Two Digital Multimeters
220Ω Resistor
Procedure
Determine the values of the colour coding of the resistor and complete the table
below
1st Colour
R
Colour
Digit
2nd Colour
Colour
3rd Colour
Digit
Colour
Exponent
4th Colour
Colour
Tolerance
1
2
3
4
5
Transfer the actual value to the table below and measure the resistance using an
ohmmeter.
R
Resistance
(from colour coding
above)
Measured
Resistance
Actual Error
% error
(difference between
colour coding and
measured)
(Actual
error/resistance from
colour coding x 100%)
1
2
3
4
5
Do the actual errors fall within the stated tolerances ______________
32
Stated
percentage
Tolerance
Connection
You will need to produce a series/parallel circuit using your chosen resistors.
Three of your resistors should be connected in parallel. There should be a resistor in
series before the parallel branch and a resistor after it. Draw the circuit diagram
below stating the values of resistors.
Circuit Diagram
Calculations
Calculate the total resistance of the circuit and the likely currents and voltages within
the circuit. Discuss your answers with your tutor as you made need to make
adjustments to the resistor positions to achieve adequate/safe readings.
33
Resistance Readings
Build your circuit but DO NOT CONNECT TO THE SUPPLY
Measure and record the resistance of your parallel branch ___________________
What was the calculated value of the parallel branch?
___________________
Is the measured resistance within 5% of the calculated
___________________
Measure and record the resistance of your circuit
___________________
What was the calculated resistance of your circuit
___________________
Is the measured resistance within 5% of the calculated
___________________
Voltage Readings
Connect your circuit to a 10V DC Supply. Use a voltmeter to measure the voltage
across each part of the circuit
Measured Voltage across R1
____________________
Calculated Voltage across R1
____________________
What is the percentage error?
____________________
Measured Voltage across parallel branch
____________________
Calculated Voltage across parallel branch
____________________
What is the percentage error?
____________________
Measured Voltage across R5
____________________
Calculated Voltage across R5
____________________
What is the percentage error?
____________________
What is the value of the addition of the 3 voltages? ________________________
Measure the voltage across the complete circuit __________________________
Does this prove Kirchoff’s Voltage Law __________________________________
34
Current Readings
Remember an ammeter used to measure current MUST be connected in SERIES
with the resistor being measured.
Connect an ammeter in series with R1 and measure the total current flowing
Measured Value
_____________________
Calculated Value
_____________________
What is the percentage error? ______________
Connect an ammeter in series with one of the three parallel resistors and measure
the current flowing through it
Measured Value
_____________________
Calculated Value
_____________________
What is the percentage error? ______________
Follow the same procedure for the remaining two parallel resistors
Measured Value
_____________________
Calculated Value
_____________________
What is the percentage error? ______________
Measured Value
_____________________
Calculated Value
_____________________
What is the percentage error? ______________
What is the value of the addition of all three parallel currents ______________
Does this prove Kirchoff’s Law? ______________________________________
On the next page produce a large circuit diagram of your circuit showing the values
at all points. Show how your results confirm Kirchoff’s Laws
35
36
WORKSHOP PRACTICAL
Potentiometer
Aims
To measure the resistance of a potentiometer
To observe the change in resistance of a potentiometer
To observe the change in voltage output of a potentiometer
Equipment
Potentiometer
20V DC Power Supply
Voltmeter
Task 1
Measure the resistance of the potentiometer supplied
1
Maximum resistance ________________
3
2
Measure the resistance for the following (approximate) positions.
Pot Position
Resistance Between 1 & 3
Resistance between 2 & 3
0% (fully turned left)
25%
50%
75%
100% (fully turned right)
What can you say about the results obtained from this test.
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
37
Task 2
Build the following circuit using your potentiometer and record the output voltage
for the (approximate) positions indicated.
10kΩ
20V
Supply
V
Pot Position
Output Voltage
0% (fully turned left)
25%
50%
75%
100% (fully turned right)
38
Supplementry Questions
Explain the following terms relating to resistors
Tolerance
___________________________________________________________
___________________________________________________________
preferred values
___________________________________________________________
___________________________________________________________
stability
___________________________________________________________
___________________________________________________________
power rating,
___________________________________________________________
___________________________________________________________
power dissipation
___________________________________________________________
___________________________________________________________
voltage rating
___________________________________________________________
___________________________________________________________
current rating
___________________________________________________________
_________________________________________________________
39
Describe the following non-linear resistors in terms of operating characteristics,
symbols, connections, and applications.
1. Negative temperature co-efficient (NTC) thermistor
2. Light dependent resistor (LDR)
40
RESISTIVITY
Resistivity is a measure of the resistive qualities of a material. It is an indication of
how well it is able to allow current to pass through it.
Different materials will have different values of
resistance for the same length and cross sectional area
because of their different resistivity values.
A piece of copper that is 1m long with an area of 1m2
will have a much lower resistance than a piece of iron
of the same dimensions because it’s resistivity value is
much less.
The value of resistivity can be used to calculate the resistance of a piece of wire of
known dimensions.
Resistivity (ρ) is equal to the value of resistance multiplied by the area divided by the
length of the material
ρ = Resistance x Area / length and its units are Ωm
The formula can be transposed to give us a formula for resistance.
This means that the resistance of a material can be found by multiply the resistivity
by the length and dividing by the area.
The table below shows the fixed values of resistivity for common electrical materials.
As you can see silver has the lowest value of resistivity and it therefore has the
lowest resistance for any given length and cross sectional area.
Material
Resistivity (Ωm) at 20oC
Silver
Copper
Gold
Aluminium
Tungsten
Nickel
Iron
Tin
Platinum
Lead
Manganin
1.59 x 10-8
1.72 x 10-8
2.44×10-8
2.82×10-8
5.60×10-8
6.99×10-8
1.0×10-7
1.09×10-7
1.06×10-7
2.2×10-7
4.82×10-7
41
When calculating resistance the units for length and cross sectional area must be the
same and as resistivity is given in Ωm we will uses metres for length and metres
squared for area.
Example
Find the resistance of a 100m of copper wire if its area is 2.5mm2
We will need to look at the table on the previous page to find the resistivity of
copper. The value given is 1.72 x 10-8 Ωm
The length is in metres (100m) but the area is in mm2. We therefore need to convert
mm2 into m2.
There are a thousand mm in one metre but as we are converting mm2 into m2 the
conversion is 10002 = 1000000
There are a million mm2 in one square metre. Therefore
2.5mm2 = 2.5 x 10-6 m2
R = 1.72 10-8 x 100 / 2.5 x 10-6 = 0.688Ω
The formula can be transposed to find any of the unknowns. Create a formula for
each of the variables (ρ , L and A) below by transposition.
42
Class Task
It is important that you are confident explaining and using the formula
You will need to be able to use this formula in its current form as well as transpose
for any of the other terms.
You will also need to be able to explain the formula in words and state all the units
used.
To enhance your understanding of the formula and to provide you with a revision
sheet your task is to produce an A3 explanation of the formula. You will need to
provide the following information on each of the three sheets:

The original formula using symbols

The original formula explained in words

The units of all the terms

An example of the original formula being used

The transposition of the formula for each of the other terms

An example of each of these transposed formula being used

A pictorial representation of the formula. You need to make this something
that will help you remember the formula.
The unit states that this formula must be remembered and so it will NOT be provided
in the assessment.
The following page should be used to produce the review sheet. You should then use
your revision sheet to complete the Student Example Questions
43
44
45
Student Example 1
Determine the resistance of a 500m length of 1.5mm2 copper wire.
Calculate the resistance if the length was doubled?
Calculate the resistance if the cross sectional area was doubled?
What do these results tell you?
46
Student Example 2
What length of copper wire of 4mm2 would give a resistance of 2.5Ω
How much aluminium wire would be needed to give the same resistance?
47
TEMPERATURE COEFFICIENT OF RESISTANCE
For most materials increasing its temperature will increase its resistance. This means
that it has a positive temperature coefficient.
A common exception to this rule is carbon which is used in motor brushes. It has a
negative temperature coefficient and its resistance goes down as its temperature
increases. You will see that it has a negative number in the table below.
Material
Temperature Coefficient (K-1)
Silver
0.0038
Copper
0.0039
Gold
0.0034
Aluminium
0.0039
Tungsten
0.0045
Iron
0.005
Tin
0.0045
Platinum
0.00392
Lead
0.0039
Manganin
0.000002
Constantan
0.00001
Mercury
0.0009
Nichrome
0.0004
Carbon
-0.0005
Germanium
-0.048
Silicon
-0.075
You will also notice that the two semi-conductors (Germanium and Silicon) also have
a negative temperature coefficient. This means that their resistance will decrease as
temperature increases.
48
Calculating Resistance Change
The new resistance for a change in temperature can be found using the following
formula
R = R0 (1 + α (T - T0) )
Where
R0 = initial resistance
α = temperature coefficient
T - T0 = change in temperature
Example
A copper wire has a resistance of 2Ω at a temperature of 20 degrees Celsius.
Calculate it’s new resistance if the temperature is increased to 400C.
We can see from the table that copper has a coefficient of 0.0039
The temperature change is 40 – 20 = 20
R = R0 (1 + α (T - T0) )
R = 2 (1 + 0.0039 (20))
R = 2 (1 + 0.078 )
R = 2 (1.078)
R = 2.156Ω
Another way of calculating it is to calculate the change in resistance
This is R x α x change in temperature
In this example it would be 2 x 0.0039 x 20 = 0.156Ω
This just then needs to be added to the original resistance (2 + 0.156 = 2.156) to give
us the same answer. It’s up to you which way you calculate it.
49
Student Example 1
Determine the new resistance of a piece of copper wire if its original resistance at 20
degrees was 3Ω and it is heated to 50 degrees.
What temperature would the same piece of wire need to be to measure 3.5Ω?
50
Class Task
It is important that you are confident explaining and using the formula
R = R0 (1 + α (T - T0) )
You will need to be able to use this formula in it’s current form as well as transpose
for any of the other terms.
You will also need to be able to explain the formula in words and state all the units
used.
To enhance your understanding of the formula and to provide you with a revision
sheet your task is to produce an A3 explanation of the formula. You will need to
provide the following information on each of the three sheets:

The original formula using symbols

The original formula explained in words

The units of all the terms

An example of the original formula being used

The transposition of the formula for each of the other terms

An example of each of these transposed formula being used

A pictorial representation of the formula. You need to make this something
that will help you remember the formula.
The unit states that this formula must be remembered and so it will NOT be provided
in the assessment.
The following page should be used to produce the review sheet. You should then use
your revision sheet to complete the Student Example Questions
51
52
53
Student Example 1
Determine the new resistance of a piece of copper wire if its original resistance at 20
degrees was 3Ω and it is heated to 50 degrees.
What temperature would the same piece of wire need to be to measure 3.5Ω?
54
Student Example 2
Determine the new resistance of a piece of tungsten wire if its original resistance at
20 degrees was 1.5Ω and it is heated to 75 degrees.
What temperature would the same piece of wire need to be to measure 2Ω?
55
Student Example 3
A conductor has a resistance of 10Ω at 20oC. It is heated to 50oC and its resistance
increases to 11.14Ω.
Determine the material of the conductor.
b) A material has a resistance of 1.1MΩ at at 20oC. It is heated to 25oC and its
resistance decreases to 836kΩ.
Determine the material of the conductor.
56
WORKSHOP PRACTICAL
Temperature Probe
Aims
To show the relationship between resistance and temperature.
To calculate the temperature coefficient of the probe.
Equipment
Jug
Temperature probe
Multi-meter capable of measuring temperature
Ohmmeter with probe
Procedure




Fill a jug with cold water
Place both probes in the water
Note the temperature and resistance
Switch on the jug and take readings of temperature and resistance at 5 0 steps
Temperature
Resistance Ω
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
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Plot a graph of temperature against resistance
R
t
What is the resistance at 20oC _____________________________
What is the change in temp from 20oC to the maximum temperature _____________
What is the change in resistance from 20oC to the maximum temperature _________
Using these figures calculate the temperature coefficient of the probe material
58
Crossword
59
Strain Gauge
Read through the text on the opposite page about strain gauges.
Identify and highlight key words and phrases.
Use your highlighted key words and phrases to produce a summary statement
explaining the strain gauge. Remember the SEED method (Statement, Explanation,
Example, Diagram). In order to produce the diagram you will need to visualise what
the gauge might look like from the information in the text.
Summary Statement
Diagram
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Strain Gauges
A strain gauge is a device used to measure the strain of an
object. Invented by Edward E. Simmons and Arthur C. Ruge in
1938, the most common type of strain gauge consists of an
insulating flexible backing which supports a metallic foil
pattern. The gauge is attached to the object by a suitable
adhesive. As the object is deformed, the foil is deformed,
causing its electrical resistance to change. This resistance
change is related to the strain by the quantity known as
the gauge factor.
A strain gauge takes advantage of the physical property
of electrical conductance's dependence on not merely
the electrical conductivity of a conductor, which is a property of
its material, but also the conductor's geometry. When
an electrical conductor is stretched within the limits of
its elasticity such that it does not break or permanently deform,
it will become narrower and longer, changes that increase its
electrical resistance end-to-end.
Conversely, when a conductor is compressed such that it does
not buckle, it will broaden and shorten, changes that decrease its
electrical resistance end-to-end. From the measured electrical
resistance of the strain gauge, the amount of applied stress may
be determined.
A typical strain gauge arranges a long, thin conductive strip in a
zig-zag pattern of parallel lines such that a small amount of
stress in the direction of the orientation of the parallel lines
results in a multiplicatively larger strain over the effective
length of the conductor—and hence a multiplicatively larger
change in resistance—than would be observed with a single
straight-line conductive wire.
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Voltage Dividers.
Resistors can be used to produce a simple voltage divider. It is a circuit that takes the
voltage from a supply and distributes it to other circuits at a lower value. For
instance, a car lighting system might only need 6 volts. A voltage divider would be
used to take the 12V from the battery and split it into 2 x 6V supplies.
12V Supply
R1
V1
6V Supply
R2
V2
6V Supply
Vs
The voltage will only split evenly if the resistances are the same. As long as both R1
and R2 above are the same resistance the voltage will be split evenly.
To vary the voltage output of each half, the resistances must be changed. The
voltage will be determined by the ratio of the resistors.
The voltage across R1 will be
It will be its resistance divided by the sum of the two resistors multiplied by the
supply voltage. For instance if R1 = 50Ω and R2 = 100Ω, the voltage across R1 would
be
50Ω / (50Ω + 100Ω) x 12V
50 / (150) x 120 = 4V
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The voltage across R2 would therefore be 12 – 4 = 8V
(from Kirchoff’s Law)
This can be checked using the formula
R2 / (R1 + R2) x 12
100 / (50 + 100) x 12 = 8V
Worked Example
A voltage divider is produced using 2 resistors of 1kΩ and 3.3kΩ. It is connected
across a 10V power supply. Determine the voltages produced at the outputs
1kΩ
Supply 1
3.3kΩ
Supply 2
10V Supply
Supply 1 = 1 x 103 / ( 1 x 103 + 3.3 x 103 ) x 10
Supply 1 = 1000 / (4300) x 10
Supply 1 = 2.33V (Remember to express your answer to 2 decimal places)
Supply 2 = 10V – 2.33V = 7.67V
This can be checked by
Supply 2 = 3300 / 4300 x 10 = 7.67V
63
Student Example 1
A voltage divider is produced using 2 resistors of 18kΩ and 39kΩ. It is connected
across a 24V power supply.
Draw the circuit diagram
Determine the voltages produced at the outputs.
64
Student Example 2
A voltage divider is produced using 2 resistors of 18Ω and 47Ω. It is connected across
a 12V power supply.
Draw the circuit diagram
Determine the voltages produced at the outputs.
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Student Example 3
A voltage divider connected across a 24V power supply has one resistor (R2) with a
value of 1kΩ. It is required to have outputs of 10V and 14V.
Draw the circuit diagram
Determine the value of resistor R1 required to produce the outputs.
(Hint : you will need to transpose the formula to make R1 the subject)
66
WORKSHOP PRACTICAL
Voltage Divider
Aims
To demonstrate how a voltage divider can be designed to produce a
required output
To calculate the value of resistance to produce a given output
Equipment
Various Resistors (one of which must be a 1kΩ)
30V DC Power Supply
Voltmeter
Task 1
Connect 2 resistors (one of which must be the 1kΩ) in series across the 30V supply.
The second resistor should be chosen by you but must be greater than 500Ω. Draw
the circuit diagram below.
Calculate the voltage you would expect across the 1kΩ resistor
Measure and record the voltage across the 1kΩ ___________________
Calculate the voltage you would expect across the 2nd resistor
Measure and record the voltage across the 2nd resistor _______________
Are your calculations and measured values within 5% _________________________
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Task 2
You are required to produce a 12V output from a voltage divider fed from a 30V
power supply. One of the resistors must be a 1kΩ.
You should calculate the value of the second resistor to produce 12V
Produce a circuit diagram of the voltage divider
Connect the voltage divider to the 30V supply and measure the output voltages
Voltage across R1 ______________________
Voltage across R2 ______________________
Is the voltage output within 5% of 12V ____________________________________
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Task 3
Calculate the expected load current if a 10kΩ load was connected across the 12V
output
Connect a 10kΩ resistor across the 12V output and measure the new voltage across
the load
Voltage across Load _____________________________
Use the clamp meter to measure the current flowing through the load ___________
Calculate the power used by the load _____________________________________
Conclusion
What have you learnt from carrying out this practical task?
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
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Introduction
The more examples you carry out the easier the assessment will be. This follows the
law that the more effort you put in the more you will get out. Unless you have a
photographic memory there is no substitute for hard work before the assessment.
Make sure you put in the work BEFORE the assessment!
1. State the formula for electric current using the symbols for charge and time
2. Explain the above formula in words
3. What are the units of each part of the formula
4. State the formula for voltage in a circuit using the symbols for energy and
charge
5. Explain this voltage formula using words
6. What are the units for each part of this formula
7. State the 4 formulas for Power
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8. Explain the formula for power that uses W and t
9. What is the difference between emf and volt drop
10. Explain the difference between conventional and actual current flow
11. Explain Kirchoff’s current law
12. Explain Kirchoff’s voltage law
71
13. Five resistances of 2, 5, 7, 12, 15 ohms are connected in series across a 12V
supply. Determine the total resistance, current flowing, voltage across each
resistor, power developed by each resistor and the total power of the circuit.
Show how Kirchoff’s law can be used to explain the circuit
Circuit Diagram
72
13a.
The same 5 resistors are then connected in parallel. Determine the current,
voltage and power at all points of the circuit. Show how Kirchoff’s law can be used to
explain the circuit
Circuit diagram
73
14. 2 resistors of 10K and 0.56K are connected in series with a parallel circuit
containing three resistors of 12 x 103Ω, 0.007MΩ and 3KΩ.
Determine the current, voltage and power at all parts of the circuit. Show how
Kirchoff’s Laws can be used to explain the circuit
74
15. Define resistivity
16. State the formula for resistivity
17. State the units of all the components of the resistivity formula
18. Determine the resistance of a 50m length of 1.5mm2 copper wire.
19. What would the resistance be if the material in Q18 were changed to iron.
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20. What would be the effect of increasing the area to 2.5mm2 for both copper
and iron.
21. What would be the length of a copper wire of 2.5mm2 if the resistance was
2Ω
22. What length of tungsten wire would be needed to create a resistance of
100Ω if the cross sectional area was 0.2mm2
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23. Describe the operation of a strain gauge
24. State the formula for the coefficient of resistance
25. State the units of all parts of the formula for coefficient of resistance
26. Determine the new resistance of a piece of copper wire if its original
resistance at 20 degrees was 3Ω and it is heated to 30 degrees.
77
27. What temperature would it need to be changed to, to raise its resistance to
3.7Ω
28. Carry out the same calculation but for a piece of carbon
78
Complete the following table
I
1
3A
2
400mA
3
5000μA
Q
5mS
5C
200mS
V
4
W
Q
20J
2C
5
100V
6
12V
0.0005KJ
P
W
7
20mC
25J
8
30W
9
60mW
t
t
3µS
20mS
100mJ
79
Material
R
A
l
10
Copper
1mm2
50m
11
Iron
4mm2
100m
12
Tungsten
2.5mm2
25m
13
Tungsten
1Ω
2.5mm2
14
Tin
2Ω
10mm2
15
Carbon
1Ω
100mm2
16
Copper
1Ω
50m
17
Iron
2Ω
50m
18
Tungsten
1Ω
50m
80
Original R
Material
Change in T
19
3Ω
Copper
20
20
5Ω
Platinum
30
21
2Ω
Tungsten
43
22
1.5Ω
Tin
10
23
2Ω
Carbon
100
24
3.7Ω
Copper
46.7
81
New R
Vs
R1
R2
25
12V
1KΩ
500Ω
26
24V
100Ω
25Ω
27
10mV
3.3kΩ
0.007V
28
250mV
2.2k
100mV
82
V1
Pair Definitions
Partner’s Guess
Definition
83
Term
GLOSSARY OF TERMS
Term
Definition
84
85
Assessment Plan
National Certificate Electronic Engineering Level 3
Student
____________________________________________
Assessment Title
____________________________________________
Unit
____________________________________________
Date of Assessment
___________________________________________
Location
___________________________________________
Type of Assessment ___________________________________________
Special Requirements ___________________________________________
Student Action Plan
Tutor Signature _______________________________
Date ________________
Student Signature _____________________________
Date ________________
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