Basic mathematics
for the trade.
Mathematics, Useful rules, hints and tables
Mathematics is a language, with defined grammar and syntax.
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Basic Rules of operation;
Rule 1
When carrying out calculations Work from left to right
Left → Right
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Rule 2
Follow the basic order of operation - BOMDAS
BBrackets
Work the calculations inside the brackets first. If there is more than one calculation inside the brackets, follow the order of operations (BOMDAS). There are 3 types of brackets used, each type indicating the order of operation.
In order of operation 1st ( ), 2nd[ square ] , 3rd { Braces}
OOf
Fraction of, Calculations containing, Index, Exponent, %, x², √¯¯
M Multiplication
Work multiplication and division calculations next, working L→R.
D Division
A Addition
Work addition and subtraction calculations next, working L→R.
S Subtraction
Fractions
Definition
A fraction is a way of expressing a quantity based on an amount or a whole that is divided into many equal parts.
A cake is divided into 4 equal parts = 4 equal parts of the whole.
The whole cake = 4/4 1 piece = 1/4
2 pieces = 2 x 1/4 or 2/4
Parts of a fraction Numerator > 2 < Number of parts out of the whole
Denominator > 4 < Number of equal parts in total
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Converting a fraction to a decimal
To express a fraction as a decimal, divide the numerator by the denominator.
Example1 → (1 ÷ 2 = ·5 ) ·5
2
1 = ·25 5 = ·625 3 = ·75
4 8 4
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Converting a fraction to a percentage
Definition
A percentage is a way of expressing a number as a fraction of 100.( parts per 100 ) Percentages are used to express a ratio of how large one quantity is relative to another quantity.
Example
If a student gets 42 marks out of a possible 50 marks in an assessment,
What percentage did the student achieve:
42 as a percentage of 50 , or as a ratio expressed as part of 100 % ,
42 → ( 40 = 0·8 ) → ( 0·8 x 100 = 80 % ) or 40 x 100 = 80 %
50 50 50
If a motor has an input of 120 Watts and an output of 84 watts,
what % efficiency does the motor achieve.
Out put = 84 84 x 100 =70 %
Input = 120 120
Scientific notation and Engineering notation
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We have seen that using Multiples and Submultiples is a simpler form of expressing very large or very small numbers and making calculator entries much easier.
We need to understand the difference between Scientific & Engineering notation. Both forms of notation express values in the power of 10 but there are fundamental differences between the two.______
Scientific notation
( sometimes referred to as standard notation )
A number is expressed in Scientific notation, when it is written with 1 unit to the left of the decimal point multiplied by the power of 10 with the index of the number of the multiplier. (either positive or negative index).
1 place to the↓left of the decimal point
Example 2.5 x10 3← Index( the number of multiples of 10 )
↑
To thePower of 10
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Examples
(1) 785 000 - expressed in scientific notation
← 5 places to the L ( + positive index)
∩∩∩∩∩
7 .85000. = 7.8 x 10 5
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(2) 390 000 000.
← 8 places to the L ( + index)
∩∩∩∩∩∩∩∩
390 000 000. = 3.9 x 10 8
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(3) 0.00025
→ 4 places to the R ( - negative index)
∩∩∩∩
0.0002.5 = 2.5 x 10 -4
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(4)0.0000000082
→ 9 places to the R (- index)
∩∩∩ ∩∩∩ ∩∩∩
0.000 000 008.2 = 8.2 x 10 -9
Engineering notation
In the Electrical industry, the fundamental and derived units that are used are based on the international system of measurement, called the SI units.(Système Internationale)
This system is based on metric units; therefore the multiplier for the power of 10 is always expressed to the power of 10 x threes or 1000.
Example
x103 (multiplied by 10 x10x10),
x106, (multiplied by 10 x10x10 x10x10x10),
x109, (multiplied by 10 x10x10 x10x10x10 x10x10x10)
Examples
Multiples
(1)2500 Watts
← 3 places to the L ( + positive index )
∩∩∩
2.500. Watts = 2.5x103 Watts
(2) 33 000 000 000 Volts
← 9 places to the L (+ index)
∩∩∩ ∩∩∩ ∩∩∩
33. 000 000 000. Volts = 33 x10 9 Volts
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Sub Multiples
(3)0.0145 Amps
→ 3 places to the R ( - negative index)
∩∩∩
0.014.5 Amps = 14.5 x 10 -3Amps
(4)0.000 000 000 368 Farads
→ 9places to the R ( - index)
∩∩∩ ∩∩∩ ∩∩∩
0.000 000 368. Farads = 368 x 10 -9 Farads
Multiples & Submultiples
In the Electrical industry, to simplify very large numbers and very small numbers, we use engineering notation.
For example; with regards to resistor values we need to be able to express either very small or very large values in a simpler form. For example
Resistors values may vary from 0.0001 Ω to 82 000 000 000. Ω
These can be very complex and difficult to work with, and enter into calculations.
To get around this problem we use multiplesand sub multiples
(Or multipliers) These multipliers are based on powers of 10 x threes.
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If we use the linear unit of the metre as an example, some linear terms you are familiar with are;
( The Kilometre – Km ) And ( The Millimetre - mm )
Kilo is another way of expressing 1000. – ( 1000 metres = 1km )
Kilo could be expressed as by 10 to the power of 3 ( x 103 )
or by the letter k for kilo.
milli is a way of expressing 1/1000th – ( 1/1000th of 1 metre = 1mm )
milli could be expressed as by 10 to the power of -3 ( x 10 -3 )
or by the letter m for milli.
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Examples
Multiples
5000. metres = by moving the decimal place 3 places to the left we are expressing 5000 as ( 5. by 3 times the power of 10) = 5 x 1000
← Decimal point moved 3 places to the left ( positive + index 3 )
∩∩∩
5000 = 5 .000. = 5 x 103 = 5 x 103 m. = 5 km
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Sub-multiples
0.008 metres = by moving the decimal place 3 places to the right we are expressing 0.008 as ( 8. by 3 times the power of – 10 ) = 8 x .001
Decimal point moved 3 places to the right ( negative - index 3 )
→ ∩∩∩
0.008 metres = 0.008. = 8 x 10-3 m. = 8 mm
Multiples and Sub multiples
12 9 6 3 - 3 - 6 - 9 - 12
10 10 1010 10 10 10 10
tera giga mega kilo milli micro nano pico
T G M k m μ n ρ
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12
101 000 000 000 000.teraT
9
10 1 000 000 000.gigaG
6
10 1 000 000.megaM
3
10 1 000.kilok
1.
-3
10.001millim
-6
10.000 001microμ
-9
10.000 000 001nanon
-12
10.000 000 000 001picoρ
Using your Calculator
We have seen that 2 milli Amps is 2/1000ths of 1 amp or ( 2 x 10-3 Amps)
Try entering 2x10-3 into your calculator using the following key strokes.
( Depending on the the type of calculator and the key displays )
2 3
Or 2 x 10-3
2 3
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Now try:
4.1 x108 + 8.7 x106
4.1 8 8.7 6 418700 000
418700 000 418.7 x106
↑
Engineering Notation Key
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Now try doing the following calculation
Find the power dissipated in a circuit with the following values.
A circuit has a current of 2.5 milli Amps and a Voltage of 36 kiloVolts.
using the formula ( P = I x V ) ------Power = 2.5 m Amps x 36 k volts
P = 2.5x 10-3 A x 36 x103 V
2.53 36 3 90
Answer 90 Watts
Exercises
BOMDAS & scientific and Engineering Notation
1) 2 + 3 x 6 ─ 5
2) 5 x 5 + ( 18 ÷ 3 ) x 2
3) 4 + 9 x 8 + 25 ÷ 5
4) 9² + 25 x 5
2
5)Complete the table below converting the units as indicated.
Express in Engineering notation / Express in normal form using decimals1267008 / 9.6 x 10 -3
0.001 / 6.2 x 106
2500 / 52.2 x 10 -6
10658 / 76.7 x 10 -3
0.1 / 13.2 x 10 -1
0.000 000 0026 / 65.75 x 10 3
6)Complete the table below converting the units as indicated.
6403 metres to / km / mm258.2 km to / m / Mm
35.063 M Watts to / kW / W
13688 Pascals to / k Pa / M Pa
227 Joules to / m J / kJ
2810 k Hertz to / M Hz / Hz
Transposition
Electrical calculations can seem complex, but are solved mathematically by the use of a formula. Formulas are equations or mathematical statements where S.I. unit values are blended in order to achieve a balanced defined outcome.
Let’s look at a basic formula like Ohms Law.
There is a fixed ratio between the component values of a circuit.
The applied voltage is equal to the circuit current multiplied by the circuit resistance.
V = I x R V = I x R Voltage = Current x Resistance
∆
If we are given 2 of the known values of the formula and wish to find an unknown value, we have to re-configure the equation to establish the unknown value. This is known as transposing the formula.
The transposition of the formula must be carried out while maintaining the balance of the equation thus ensuring that both sides of the equation are equal. To find an unknown value we must isolate that value on one side of the equation. To do this we must do to one side of the equation what we do to the other side.
Using V = I x R we wish to find the value for R
V = I x R( to isolate R we must remove I from this side )
V = I x R ( to cancel out I on this side we ÷ this side by I )
I I
we must do to the same to the other side so we must ÷ the other side by I
V = R therefore R = V
I I
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Example P = V2 find for value V
R
P x R = V2 x R( Isolate V2 )
R
√ P x R = √ V2 ( Bring V2 to its unit value V )
√ P x R = V therefore V =√P x R
Transposition Exercises
1. W = F dfind for F W = work
F = force
d= distance
2.F = m afind for m F = force
m = mass
a = acceleration
( 9.8 )
3. P = Wfind for W P = Power
t W = work
t = time
4. T = F rfind for r T = Torque
F = force
r = radius
5. P = 2 π n T find for T P = Power
60 T = Torque
n = Revs per min
6. V = I Rfind for R V = Voltage
I = Current
R = Resistance
7. V = I R find for I V = Voltage
I = Current
R = Resistance
8. P = V Ifind for V P = Power
V = Voltage
I = Current
9. P = I 2 Rfind for I P = Power
I = Current
R = Resistance
10.
η % = P.out x 100 find for Power in η % = % efficiency P. in P. out = Power output
P. in = Power input
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