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Level 3 NVQ Diploma in Electrotechnical Technology
Unit 309 Handout 1
Unit 309 (ELTK 08): Understanding the electrical principles
associated with the design, building, installation and
maintenance of electrical equipment and systems
Handout 1: Mathematical principles
Addition of fractions
Adding fractions is something that is required when calculating resistors in parallel, so it
is a skill required by engineers.
To add fractions, first find a common denominator. This is a number that the bottom of
the two fractions will both go into. The easiest way to find a common denominator is to
multiply the two numbers together.
Example:
1 2
+ Þ multiply the bottom will give a common denominator of 15
3 5
The next requirement is to find how many time each of the denominators go into
the 15 and multiply that by each of the top numbers (numerators) before adding
as follows:
2 1 ( 2 ´ 3) + (1´ 5) 11
+ =
=
5 3
5´ 3
15
Subtraction of fractions
The process is the same, find the common denominator, multiply but then take the
products away from each other.
Example:
2 1 ( 2 ´ 3) - (1´ 5) 1
- =
=
5 3
5´ 3
15
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SmartScreen
Level 3 NVQ Diploma in Electrotechnical Technology
Unit 309 Handout 1
Multiplication of fractions
To multiply fractions simply multiply the numerators and then the denominators. If
a common number goes into both the top and bottom, simplification can be
achieved.
Example:
2 1 2 ´1 2
´ =
=
5 3 5´ 3 15
Division of fractions
To divide fractions, simply turn one of the fractions upside down and then
multiply. Again, if a common number goes into both the top and the bottom,
simplification can be achieved by cancelation.
Example:
2 1 2 3 6 1
¸ = ´ = =1
5 3 5 1 5 5
Percentages
The simple approach way to remember percentages is ‘parts in 100’. If you need
to find 6% of a figure, divide the number by 100 to find 1/100th of it then multiply
by 6 to give you 6%. There are quick methods to find standard percentages, such
as 5% by finding 10% and then half of that gives 5%.
Example:
A lighting circuit has an allowed voltage drop of 3%. What is the actual voltage
drop?
230
´ 3 = 6.9
100
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SmartScreen
Level 3 NVQ Diploma in Electrotechnical Technology
Unit 309 Handout 1
Transposition of basic formulae
This is also known as ‘changing the subject of the formulae’.
I=V
R
In the example on the left, ‘I’ is the subject
of the formula and by inserting the values
for ‘V’ and ‘R’ the value of the subject ‘I’
can be calculated.
If we need to find ‘V’ for example, we must
transpose the formula to make ‘V’ the
subject.
There is one fundamental rule for
transposing a formula:
Whatever you do to one side of the formula you must do the same to the other
side
In other words:

add the same quantity to both sides of the formula

subtract the same quantity from both sides of the formula

multiply both sides of the formula by the same quantity

divide both sides of the formula by the same quantity

take ‘functions’ of both sides of the formula: for example, square both sides, find the
reciprocal of both sides.
EXAMPLE 1
I = V / R – make V the subject of the formula.
I
=
V
R
(I is currently the subject of the formula)
The question is ‘make V the subject of the formula’. This means that ‘V’ must be put on
its own on one side of the equals sign and the other terms must be on the other side.
To do this, first multiply both sides by R
Now cancel through:
I×R
=
I×R
=
I×R
=
V×R
R
V×R
R
V
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Level 3 NVQ Diploma in Electrotechnical Technology
Now reverse the formula:
𝐕
=
Unit 309 Handout 1
𝐈×𝐑
EXAMPLE 2
I = V / R – make R the subject of the formula.
I
=
V
R
(I is currently the subject of the formula)
The question is ‘make R the subject of the formula’. This means that ‘R’ must be put on
its own on one side of the equals sign and the other terms must be on the other side.
To do this, first multiply both sides by R
Now cancel through:
Now divide both sides by I:
Now cancel through :
I×R
=
I×R
=
I×R
I×R
I
I×R
I
=
𝐑
=
=
=
V×R
R
V×R
R
V
V
I
V
I
𝐕
𝐈
Triangles
Throughout electrical science there is a need to use maths related to triangles.
Applications include:

Power triangles

Impedance triangles

Light calculations
The main requirement is to find missing sides or angles on right hand triangles. This can
be done using Pythagoras’ theorem. This states that hypotenuse of a right angle triangle
can be found by taking the square root of the sum of the other sides squared.
𝐚 = √𝐛 𝟐 + 𝐜 𝟐
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SmartScreen
Level 3 NVQ Diploma in Electrotechnical Technology
Unit 309 Handout 1
a
b
Φ
c
Where:
a = hypotenuse
b = opposite
c = adjacent
If two sides are known, the third can be found by rearranging the Pythagoras formula.
𝐛 = √𝐚𝟐 − 𝐜 𝟐
𝐜 = √𝐚𝟐 − 𝐛
There are other ways to work out missing values if you know one side and one of the
angles is known other than the 90°.
SOHCAHTOA
This can be broken down into three parts:
𝐎𝐩𝐩𝐨𝐬𝐢𝐭𝐞
𝐬𝐢𝐧 𝜱 =
𝐡𝐲𝐩𝐨𝐧𝐞𝐧𝐮𝐬𝐞
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Level 3 NVQ Diploma in Electrotechnical Technology
𝐜𝐨𝐬 𝜱 =
Unit 309 Handout 1
𝐀𝐝𝐣𝐚𝐜𝐞𝐧𝐭
𝐇𝐲𝐩𝐨𝐭𝐞𝐧𝐮𝐬𝐞
𝐎𝐩𝐩𝐨𝐬𝐢𝐭𝐞
𝐭𝐚𝐧 𝜱 =
𝐀𝐝𝐣𝐚𝐜𝐞𝐧𝐭
Statistics
Average values are used in electrical science. There is a need to understand average
power for example and RMS (root mean square). This is needed so it is possible to work
out the heating effect of current in a circuit or equipment. The three types of averages
are:
Mean
Add the numbers up and divide by the amount of numbers in the sequence.
Number range:
2, 2 ,4 ,4 , 4, 6, 6
The mean is found as follows:
𝐦𝐞𝐚𝐧 =
𝟐+𝟐+𝟒+𝟒+𝟒+𝟔+𝟔
=𝟒
𝟕
Median
The median is the middle number when the number range are put in order from lowest to
highest as follows:
2, 2 ,4 , 4, 4, 6, 6
Counting from either end until the middle number is reached. If there were 2 numbers in
the middle, the mean is taken and that becomes the median.
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SmartScreen
Level 3 NVQ Diploma in Electrotechnical Technology
Unit 309 Handout 1
Mode
The most is sometimes referred to as the MOST common number in a number range.
For the same range used before the most common number is 4 also as it occurs 3 times.
Range
The range is defined as the largest number take away the smallest number when the
numbers are put in order from lowest to highest as follows:
2, 2 ,4 , 4, 4, 6, 6
6–2=4
The range is also 4 in this example.
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SmartScreen
Level 3 NVQ Diploma in Electrotechnical Technology
Unit 309 Handout 1
Useful Formulae
VOLTAGE
=
I.R
=
P/I
=
P.R
CURRENT
=
V/R
=
P/V
=
P/R
=
P/I2
=
V2/R
RESISTANCE
V/I
=
POWER
2
V /P
=
=
V.I
=
I2.R
In series:
R t = R1 + R 2 + R 3 etc
RESISTORS
1
1
1
1
=
+
+
R t R1 R 2 R 3
in parallel:
etc
KIRCHHOFF’S LAWS
Voltage:
VT = V1 + V2 + V3 + etc
in a Series circuit
Current:
IT = I1 + I2 + I3 + etc
in a Parallel circuit
RESISTIVITY
CHARGE (Quantity)
CAPACITORS
R=ρ×
RL
L
a
Q = It
or
R  1/a
Q = VC
in parallel:
Ct = C1 + C2 + C3
etc
in series:
1
1
1
1
= + +
Ct C1 C2 C3
etc

=
B.A
mmf
=
NxI
H
=
E
=
n×l
L
B × l × v (v = velocity)
E
=
ELECTROMAGNETISM
INDUCED EMF
FORCE ON A CONDUCTOR
F
=
BH
(I2 − I1)
t
B × I × l Newtons
−L ×
ENERGY
in a magnetic field:
½.L.I2 Joules
STORED
in a capacitor:
½.C.V2 Joules
EFFICIENCY
Efficiency
=
Output power
Input power
MECHANICS
Force
=
Mass x Newtons
Torque
=
Force x Distance
Work Done (WD)
=
Force x Distance
Energy
=
Joules
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Level 3 NVQ Diploma in Electrotechnical Technology
Unit 309 Handout 1
Joules
=
Watts x Seconds
Joules
=
Newton Metre
Z
=
√R2 + X 2
POWER FACTOR
p.f.
=
POWER FACTOR
p.f.
=
R
Z
P
VA
PYTHAGORAS
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