Unit 39: Further Mathematics Unit Workbook 1...Pearson BTEC Levels 5 Higher Nationals in Engineering (RQF) Unit 39: Further Mathematics Unit Workbook 1 in a series of 4 for this unit

Unit WorkBook 1 – Level 5 ENG – U39 Further Mathematics – LO1 Applications of Number Theory

© 2018 UniCourse Ltd. All Rights Reserved.

Page 1 of 27

Pearson BTEC Levels 5 Higher Nationals in Engineering (RQF)

Unit 39: Further Mathematics

Unit Workbook 1 in a series of 4 for this unit

Learning Outcome 1

Applications of Number Theory

Sample



Page 4 of 27

1 Number theory:

1.1 Number Types and Systems The basic types of number we deal with regularly in engineering are:

▪ Natural – Everyday positive whole counting numbers (i.e. 0, 1, 2, 3, 4, 5, …)

▪ Integer – Similar to Naturals, but negatives are allowed (i.e. …-3, -2, -1, 0, 1, 2, 3, …)

▪ Rational – Can be expressed as a fraction of two integers (0 not allowed in denominator, i.e. −9 5⁄ )

▪ Irrational - Can NOT be expressed as a fraction of two integers (i.e. √2 )

▪ Real – ANY quantity along a continuous line (i.e. -16.76449)

▪ Complex – Have real and imaginary components. Read more on these later

Common number systems encountered in engineering are:

▪ Binary – Base 2

▪ Octal – Base 8

▪ Denary – Base 10 (what you’re most used to)

▪ Duodecimal (also known as Dozenal) – Base 12

▪ Hexadecimal – Base 16

To work efficiently in modern engineering, you need to be able to not only appreciate each of these

number systems but also convert between them. You can always check your answers by using the

Windows Calculator (click on View then Programmer). Using the calculator is fine, but Assignment 1 asks

that you demonstrate knowledge of the mechanics of these conversions. Let’s look at each system first and

then work out how to go about the conversions in detail.

Binary System (Base 2)

The Binary system only uses two digits, 0 and 1 (hence the term Binary). Let’s look at a template for the

Binary system…

𝟐𝟕 𝟐𝟔 𝟐𝟓 𝟐𝟒 𝟐𝟑 𝟐𝟐 𝟐𝟏 𝟐𝟎 How many

128’s How many

64’s How many

32’s How many

16’s How many

8’s How many

4’s How many

2’s How many

1’s

Octal System (Base 8)

The Octal system uses eight digits, 0 through to 7 (hence the term Octal). Let’s look at a template for the

Octal system…

Sample



Page 5 of 27

𝟖𝟕 𝟖𝟔 𝟖𝟓 𝟖𝟒 𝟖𝟑 𝟖𝟐 𝟖𝟏 𝟖𝟎 How many 2,097,152’s

How many 262,144’s

How many 32,768’s

How many 4,096’s

How many 512’s

How many 64’s

How many 8’s

How many 1’s

Denary System (Base 10)

The Denary system uses ten digits, 0 through to 9 (hence the term Denary). Let’s look at a template for the

Denary system (which you got used to as a child) …

𝟏𝟎𝟕 𝟏𝟎𝟔 𝟏𝟎𝟓 𝟏𝟎𝟒 𝟏𝟎𝟑 𝟏𝟎𝟐 𝟏𝟎𝟏 𝟏𝟎𝟎 How many 10 millions

How many millions


How many 10,000’s

How many 1,000’s

How many 100’s

How many 10’s

How many 1’s

Duodecimal System (Base 12)

The Duodecimal system uses twelve digits, 0 through to 9, then ‘T’ for ten and ‘E’ for eleven (hence the

alternative term Dozenal, meaning dozen). Let’s look at a template for the Duodecimal system …

𝟏𝟐𝟕 𝟏𝟐𝟔 𝟏𝟐𝟓 𝟏𝟐𝟒 𝟏𝟐𝟑 𝟏𝟐𝟐 𝟏𝟐𝟏 𝟏𝟐𝟎 How many

35,831,808’s How many 2,985,984’s


How many 20,736’s

How many 1,728’s

How many 144’s

How many 12’s

How many 1’s

Hexadecimal System (Base 16)

The Hexadecimal system uses sixteen digits, 0 through to 9 plus A through to F (hence the term

Hexadecimal). To employ 16 digits, we have used the letters A to F to represent the numbers 10 to through

to 15 respectively. Let’s look at a template for the Hexadecimal system…

𝟏𝟔𝟕 𝟏𝟔𝟔 𝟏𝟔𝟓 𝟏𝟔𝟒 𝟏𝟔𝟑 𝟏𝟔𝟐 𝟏𝟔𝟏 𝟏𝟔𝟎 How many

etc. How many

etc. How many

etc. How many

65,536’s How many

4,096’s How many

256’s How many

16’s How many

1’s

Representing Numbers in Each System

If we wanted to represent the everyday Denary (base 10) number 12 in Binary format then we would have

to work from left to right in the Binary table…

How many 128’s in 12? Answer: 0 so put 0 in that cell




Sample



Page 6 of 27

How many 8’s in 12? Answer: 1 so put 1 in that cell and subtract the 8 from the 12, leaving 4

How many 4’s in 4? Answer: 1 so put 1 in that cell and subtract the 4 from the 4, leaving 0

There is nothing left over here, so we’ve finished. We may now write 1210 = 000011002.

Some examples of starting with a Denary number and representing it in the Binary, Octal and Hexadecimal

formats now follow…

Binary System (Base 2)

Denary Examples 𝟐𝟕 𝟐𝟔 𝟐𝟓 𝟐𝟒 𝟐𝟑 𝟐𝟐 𝟐𝟏 𝟐𝟎 12 0 0 0 0 1 1 0 0

17 0 0 0 1 0 0 0 1

57 0 0 1 1 1 0 0 1

129 1 0 0 0 0 0 0 1

Octal System (Base 8)

Denary Examples 𝟖𝟕 𝟖𝟔 𝟖𝟓 𝟖𝟒 𝟖𝟑 𝟖𝟐 𝟖𝟏 𝟖𝟎 7 0 0 0 0 0 0 0 7

17 0 0 0 0 0 0 2 1

855 0 0 0 0 1 5 2 7

1,067,003 0 4 0 4 3 7 7 3

Hexadecimal System (Base 16)

Denary Examples 𝟏𝟔𝟕 𝟏𝟔𝟔 𝟏𝟔𝟓 𝟏𝟔𝟒 𝟏𝟔𝟑 𝟏𝟔𝟐 𝟏𝟔𝟏 𝟏𝟔𝟎 15 0 0 0 0 0 0 0 F

33 0 0 0 0 0 0 2 1

99,999 0 0 0 1 8 6 9 F

200,000,000 0 B E B C 2 0 0

Conversions between bases:

Denary to Binary, Denary to Octal and Denary to Hexadecimal conversions are all performed in the manner

described in the above tables.

Sample



Page 7 of 27

Binary to Denary, Octal to Denary and Hexadecimal to Denary conversions are all performed in the reverse

manner to those conversions already stated.

That leaves us with six types of conversion to deal with:

▪ Binary to Octal

▪ Octal to Binary

▪ Binary to Hexadecimal

▪ Hexadecimal to Binary

▪ Octal to Hexadecimal

▪ Hexadecimal to Octal

Binary to Octal Conversion

The quick way to do this is to group the Binary bits into groups of three and convert each group individually

into Octal. A quick example…

101100010111 can be grouped as 101 100 010 111. The first group on the left is 101 which we know to be

integer 5. Doing the same to the second group gives 4, then 2 for the third group and 7 for the last group.

We can then say that 1011000101112 = 54278. Easy enough!

Octal to Binary Conversion

That’s quite east too. Just turn each of the Octal digits in your number into their 3-digit binary equivalents,

reversing the previous process. A quick example…

Octal 3627: convert the 3 to 011, convert the 6 to 110, convert the 2 to 010, convert the 7 to 111. This

gives us 36278 = 0111100101112.

Binary to Hexadecimal Conversion

Just group your Binary bits into groups of four. A quick example…

101100010111 can be grouped as 1011 0001 0111. We now look at each group of four bits and convert

that group to hex:

10112 = 𝐵16, 00012 = 116, 01112 = 716

So, 1011000101112 = 𝐵1716

Hexadecimal to Binary Conversion

Again, this is really easy. Just reverse the previous process.

Octal to Hexadecimal Conversion

Just convert your Octal Number into Binary bits – groups of 4 from the right-hand side. Then convert again

to hex. Easy.

Sample



Page 8 of 27

Hexadecimal to Octal Conversion

Just convert your Hex Number into Binary bits – groups of 3 from the right-hand side. Then convert again

to Octal. Also, easy.

Please Note: Duodecimal is a very rarely used number system, and is not covered in the conversions above

for this reason. However, you may like to work out how to make such conversions by adapting the tables

above.

Hexadecimal Addition

To add hexadecimal numbers together we need to think of the hex number system (0 to F) and start at the

first digit to be added to. We then count to the right. Once we move right and pass F we restart (loop) to

the left at 0 until we have counted the dumber of digits for the number we wish to add.

Let’s clarify this process by first of all listing the digits in the hex system …

0 1 2 3 4 5 6 7 8 9 A B C D E F

Now assume we wish to add the hex numbers 8 and A. We start at 8 and then move right by A places (i.e.

10 actual places). Once we have moved right by 7 places we reach the end, so need to go back to the start

and move right by the remaining 3 places, which will take us to 2. Since we have looped once we note

down that 1 loop and then append where we ended up, i.e. that 2. The answer in hex is therefore 12 (one

lot of 16 plus 2 is 18, of course, in decimal).

Here is a useful table for hex addition …

+ 0 1 2 3 4 5 6 7 8 9 A B C D E F

0 0 1 2 3 4 5 6 7 8 9 A B C D E F

1 1 2 3 4 5 6 7 8 9 A B C D E F 10

2 2 3 4 5 6 7 8 9 A B C D E F 10 11

3 3 4 5 6 7 8 9 A B C D E F 10 11 12

4 4 5 6 7 8 9 A B C D E F 10 11 12 13

5 5 6 7 8 9 A B C D E F 10 11 12 13 14

6 6 7 8 9 A B C D E F 10 11 12 13 14 15

7 7 8 9 A B C D E F 10 11 12 13 14 15 16

8 8 9 A B C D E F 10 11 12 13 14 15 16 17

9 9 A B C D E F 10 11 12 13 14 15 16 17 18

A A B C D E F 10 11 12 13 14 15 16 17 18 19

B B C D E F 10 11 12 13 14 15 16 17 18 19 1A

C C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B

D D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C

E E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D

F F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E

Sample



Page 9 of 27

To use the table, locate the first digit to be added in the orange row. Then, locate the second digit to be

added in the green column. Where your selections meet is the answer to the addition. For example, to add

together hex 9 with hex A we locate the 9 in the orange row. Then, locate the A in the green column.

Where these two cells meet is the answer i.e. hex 13 in this case. Here is an example use of the table for

that particular problem …

+ 0 1 2 3 4 5 6 7 8 9 A B C D E F

0 0 1 2 3 4 5 6 7 8 9 A B C D E F

1 1 2 3 4 5 6 7 8 9 A B C D E F 10

2 2 3 4 5 6 7 8 9 A B C D E F 10 11

3 3 4 5 6 7 8 9 A B C D E F 10 11 12

4 4 5 6 7 8 9 A B C D E F 10 11 12 13

5 5 6 7 8 9 A B C D E F 10 11 12 13 14

6 6 7 8 9 A B C D E F 10 11 12 13 14 15

7 7 8 9 A B C D E F 10 11 12 13 14 15 16

8 8 9 A B C D E F 10 11 12 13 14 15 16 17

9 9 A B C D E F 10 11 12 13 14 15 16 17 18

A A B C D E F 10 11 12 13 14 15 16 17 18 19

B B C D E F 10 11 12 13 14 15 16 17 18 19 1A

C C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B

D D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C

E E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D

F F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E

Sample



Page 10 of 27

Hexadecimal Multiplication

Let’s now take a look at a useful hex multiplication table …

x 1 2 3 4 5 6 7 8 9 A B C D E F 10

1 1 2 3 4 5 6 7 8 9 A B C D E F 10

2 2 4 6 8 A C E 10 12 14 16 18 1A 1C 1E 20

3 3 6 9 C F 12 15 18 1B 1E 21 24 27 2A 2D 30

4 4 8 C 10 14 18 1C 20 24 28 2C 30 34 38 3C 40

5 5 A F 14 19 1E 23 28 2D 32 37 3C 41 46 4B 50

6 6 C 12 18 1E 24 2A 30 36 3C 42 48 4E 54 5A 60

7 7 E 15 1C 23 2A 31 38 3F 46 4D 54 5B 62 69 70

8 8 10 18 20 28 30 38 40 48 50 58 60 68 70 78 80

9 9 12 1B 24 2D 36 3F 48 51 5A 63 6C 75 7E 87 90

A A 14 1E 28 32 3C 46 50 5A 64 6E 78 82 8C 96 A0

B B 16 21 2C 37 42 4D 58 63 6E 79 84 8F 9A A5 B0

C C 18 24 30 3C 48 54 60 6C 78 84 90 9C A8 B4 C0

D D 1A 27 34 41 4E 5B 68 75 82 8F 9C A9 B6 C3 D0

E E 1C 2A 38 46 54 62 70 7E 8C 9A A8 B6 C4 D2 E0

F F 1E 2D 3C 4B 5A 69 78 87 96 A5 B4 C3 D2 E1 F0

10 10 20 30 40 50 60 70 80 90 A0 B0 C0 D0 E0 F0 100

Suppose we wished to multiply hex 7 with hex C (in the decimal world that would be the same as

multiplying 7 by 12, which is 84, of course – 84 in hex is 54. We simply perform a similar process as we did

for hex addition, finding the cell corresponding to our chosen row and column. Here is our current problem

highlighted …

x 1 2 3 4 5 6 7 8 9 A B C D E F 10

1 1 2 3 4 5 6 7 8 9 A B C D E F 10

2 2 4 6 8 A C E 10 12 14 16 18 1A 1C 1E 20

3 3 6 9 C F 12 15 18 1B 1E 21 24 27 2A 2D 30

4 4 8 C 10 14 18 1C 20 24 28 2C 30 34 38 3C 40

5 5 A F 14 19 1E 23 28 2D 32 37 3C 41 46 4B 50

6 6 C 12 18 1E 24 2A 30 36 3C 42 48 4E 54 5A 60

7 7 E 15 1C 23 2A 31 38 3F 46 4D 54 5B 62 69 70

8 8 10 18 20 28 30 38 40 48 50 58 60 68 70 78 80

9 9 12 1B 24 2D 36 3F 48 51 5A 63 6C 75 7E 87 90

A A 14 1E 28 32 3C 46 50 5A 64 6E 78 82 8C 96 A0

B B 16 21 2C 37 42 4D 58 63 6E 79 84 8F 9A A5 B0

C C 18 24 30 3C 48 54 60 6C 78 84 90 9C A8 B4 C0

D D 1A 27 34 41 4E 5B 68 75 82 8F 9C A9 B6 C3 D0

E E 1C 2A 38 46 54 62 70 7E 8C 9A A8 B6 C4 D2 E0

F F 1E 2D 3C 4B 5A 69 78 87 96 A5 B4 C3 D2 E1 F0

10 10 20 30 40 50 60 70 80 90 A0 B0 C0 D0 E0 F0 100

Sample



Page 11 of 27

Octal Addition

Here is the useful table for Octal addition …

+ 0 1 2 3 4 5 6 7

0 0 1 2 3 4 5 6 7

1 1 2 3 4 5 6 7 10

2 2 3 4 5 6 7 10 11

3 3 4 5 6 7 10 11 12

4 4 5 6 7 10 11 12 13

5 5 6 7 10 11 12 13 14

6 6 7 10 11 12 13 14 15

7 7 10 11 12 13 14 15 16

Let’s suppose we wish to add together octal 4 with octal 7. Again, locate these in the columns and rows

from the table, then find the corresponding intersection cell. This will result in locating the cell containing

octal 13, as below …

+ 0 1 2 3 4 5 6 7

0 0 1 2 3 4 5 6 7

1 1 2 3 4 5 6 7 10

2 2 3 4 5 6 7 10 11

3 3 4 5 6 7 10 11 12

4 4 5 6 7 10 11 12 13

5 5 6 7 10 11 12 13 14

6 6 7 10 11 12 13 14 15

7 7 10 11 12 13 14 15 16

Octal Multiplication

Here is the useful table for Octal multiplication …

x 1 2 3 4 5 6 7

1 1 2 3 4 5 6 7

2 2 4 6 10 12 14 16

3 3 6 11 14 17 22 25

4 4 10 14 20 24 30 34

5 5 12 17 24 31 36 43

6 6 14 22 30 36 44 52

7 7 16 25 34 43 52 61

Suppose we wished to multiply octal 5 with octal 7 (in the decimal world that would be the same as

multiplying 5 by 7, which is 35, of course – 35 in octal is 43 (i.e. 4 lots of 8 plus 3). We simply perform a

Sample



Page 12 of 27

similar process as we did for octal addition, finding the cell corresponding to our chosen row and column.

Here is our current problem highlighted …

x 1 2 3 4 5 6 7

1 1 2 3 4 5 6 7

2 2 4 6 10 12 14 16

3 3 6 11 14 17 22 25

4 4 10 14 20 24 30 34

5 5 12 17 24 31 36 43

6 6 14 22 30 36 44 52

7 7 16 25 34 43 52 61

Sample



Page 13 of 27

1.2 Introduction to Complex Numbers When we have circuits containing just resistors then life is so easy in terms of circuit analysis. Most useful

circuits also contain capacitors and inductors (usually coils and windings). The introduction of capacitors

and inductors into circuits causes ‘phase angles’ in our calculations. The study of these phase angles is

made much easier by introducing complex numbers.

From your level 3 studies you will have come across Inductive Reactance (XL) and Capacitive Reactance (XC).

These terms are used to quantify the amount of ‘opposition’ caused by capacitors and inductors to changes

in current or voltage. The term ‘reactance’ is brought about because capacitors cannot be fully charged or

discharged in zero time, and inductors cannot be fully energised or de-energised in zero time. A good

analogy for capacitors is the amount of water in a bathtub. It is impossible to fill a bathtub in zero time,

and it’s also impossible to empty a bathtub in zero time. The amount of reactance from capacitors and

inductors is a function of their manufactured properties and the frequency of operation. Let’s review the

equations for these reactances…

𝑋𝐿 = 2𝜋𝑓𝐿 [Ω]

𝑋𝐶 =1

2𝜋𝑓𝐶 [Ω]

where;

𝑋𝐿 = 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 (𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝑂ℎ𝑚𝑠, Ω)

𝑋𝐶 = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 (𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝑂ℎ𝑚𝑠, Ω)

𝑓 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝐻𝑒𝑟𝑡𝑧, 𝐻𝑧)

𝐿 = 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 (𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝐻𝑒𝑛𝑟𝑖𝑒𝑠, H)

𝐶 = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 (𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝐹𝑎𝑟𝑎𝑑𝑠, 𝐹)

Consider the RLC circuit below…

We can draw a phasor diagram for this circuit, as follows…

Sample



Page 14 of 27

The black arrow represents resistance. The current through a resistor is always in phase with the voltage

across it. We place resistance on the horizontal axis.

The red arrow represents inductive reactance. We see that this leads the resistance by 90 degrees

(𝜋 2 𝑟𝑎𝑑𝑠.⁄ ). We name this axis the ‘+j axis’. Mathematicians tend to designate this the ‘+i’ (imaginary) axis.

Engineers do not use i since it clashes with the current symbol, so we use ‘j’ instead.

The blue arrow represents capacitive reactance. We see that this lags the resistance by 90 degrees

(𝜋 2 𝑟𝑎𝑑𝑠.⁄ ). We designate this the ‘-j’ axis.

The dashed lines represent a graphical method of finding the resultant of these phasors, drawn in green.

We term this resultant the impedance of the circuit and mark it with ‘r’ for resultant. This resultant

impedance makes an angle with the horizontal axis, marked with 𝜙.

The resultant impedance is given the symbol Z for calculation purposes. We see that the green resultant

has both horizontal and vertical components. The horizontal contribution is known as the real component

and the vertical contribution is known as the imaginary component.

We may use Pythagoras’ theorem to denote impedance as follows…

𝑍2 = 𝑅2 + 𝑋2

∴ 𝑍 = √𝑅2 + 𝑋2 [Ω]

In complex number notation we represent Z as…

𝒁 = 𝑹 + 𝒋(𝑿𝑳 − 𝑿𝑪) [𝛀]

Sample



Page 21 of 27

1.5 Three-Phase Circuit Problems using j Notation

We are now armed with the mathematical tools needed to analyse three-phase circuits. Consider the

unbalanced 4-wire star system below…

We notice that phase voltage 1 is entirely across phase impedance 1. We may therefore say…

𝑖1 =𝑉1

𝑍1=

230∠0𝑜

115∠100= 2∠−10𝑜 [𝐴]

The situation is similar for the other two phases…

𝑖2 =𝑉2

𝑍2=

230∠120𝑜

5∠−600= 46∠180𝑜 [𝐴]

𝑖3 =𝑉3

𝑍3=

230∠240𝑜

10∠400= 23∠200𝑜 [𝐴]

The sum of these three phase currents is equal to the neutral current (𝐼𝑁) in this star configuration.

Unfortunately, we cannot add Polar quantities (division and multiplication are ok though, as we’ve seen) so

we need to convert each of them into 𝑎 + 𝑗𝑏 form. Some quick conversions on the calculator yield…

𝑖1 =𝑉1

𝑍1=

230∠0𝑜

115∠100= 2∠−10𝑜 ≡ 1.97 − 𝑗0.35 [𝐴]

𝑖2 =𝑉2

𝑍2=

230∠120𝑜

5∠−600= 46∠180𝑜 ≡ −46 + 𝑗0 [𝐴]

𝑖3 =𝑉3

𝑍3=

230∠240𝑜

10∠400= 23∠200𝑜 ≡ −21.61 − 𝑗7.87 [𝐴]

Adding the complex numbers gives…

Sample



Page 22 of 27

𝒊𝑵 = (𝟏. 𝟗𝟕 − 𝟒𝟔 − 𝟐𝟏. 𝟔𝟏) + 𝒋(−𝟎. 𝟑𝟓 + 𝟎 − 𝟕. 𝟖𝟕) = −𝟔𝟓. 𝟔𝟒 − 𝒋𝟖. 𝟐𝟐 ≡ 𝟔𝟔. 𝟏𝟓∠−𝟏𝟕𝟐. 𝟖𝟔𝒐 [𝑨]

Let us now consider the voltage between two separate phases. Look at the phasor diagram below…

The diagram on the left illustrates phase 1 in red (𝑉𝑃1). For the sake of analytical simplicity it is given a

magnitude of 1 volt, although this could be scaled up to any voltage you like. The solid blue phasor

represents phase 2, which is 120 degrees out of phase with phase 1. To work out the difference in voltage

between these two phases we must find invert phase 2, giving −𝑉𝑃2, shown dashed in blue. Our task is to

find the resultant of 𝑉𝑃1 and −𝑉𝑃2. This task is performed in the diagram on the right.

The green phasor represents the resultant line voltage, 𝑉𝐿. This is formed by drawing a parallelogram based

upon 𝑉𝑃1 and −𝑉𝑃2. To work out the magnitude of the line voltage in green we apply Pythagoras’

theorem…

|𝑉𝐿| = √[(1 + 0.5)2 + (√3

2)

2

]

|𝑽𝑳| = √[(𝟑

𝟐)

𝟐

+ (√𝟑

𝟐)

𝟐

] = √𝟗

𝟒+

𝟑

𝟒= √

𝟏𝟐

𝟒= √𝟑 𝒗𝒐𝒍𝒕𝒔

This then proves that the line voltage (from phase to phase) is √3 times the phase voltage. Therefore, if we

have a phase voltage of 230 V then the line voltage will be √3 × 230 = 398.37 𝑣𝑜𝑙𝑡𝑠. This figure tends to

Sample



Page 23 of 27

Worked Example 9

be rounded to 400 V in the UK since it is impossible to maintain an exact voltage on the distribution

system.

In the UK the consumer supply voltage is 230 V +10%/-6%. This means that the phase voltage can rise to

253V and fall to 216.2 V. If we look at the line voltage then it can be as high as 438.2 volts and as low as

374.5 volts. The figure of 438.2 volts for maximum line voltage tends to be rounded to 440 volts for normal

everyday use and signage.

Three identical coils, each of resistance 𝟐𝟎𝛀 and inductance 𝟐𝟎𝟎𝒎𝑯, are connected in a Delta

configuration to a 230 volt, 50Hz 3-phase supply. Determine the magnitude of each load current.

The first step here is to determine the load impedance on each phase…

𝑋𝐿 = 2𝜋𝑓𝐿 = 2𝜋 × 50 × 0.2 = 20𝜋 = 62.83Ω

∴ 𝑍𝐿 = 20 + 𝑗62.83 ≡ 65.94∠72.34𝑜 [Ω]

The circuit is shown below.

We already know that the line voltage magnitude is √3 times the phase voltage…

𝐿𝑖𝑛𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 = √3 × 230 = 398.37 𝑉

We also notice that each load impedance has a line voltage connected. We simply need to divide our

magnitudes for voltage and impedance to find the magnitude of each load current…

𝑴𝒂𝒈𝒏𝒊𝒕𝒖𝒅𝒆 𝒐𝒇 𝒆𝒂𝒄𝒉 𝒍𝒐𝒂𝒅 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 = 𝟑𝟗𝟖. 𝟑𝟕

𝟔𝟓. 𝟗𝟒= 𝟔. 𝟎𝟒 𝑨

Sample



Page 25 of 27

The total power is found in each case by adding the readings on the wattmeters. The two-wattmeter

measurement produces 1.14kW + 2.29kW = 3.43kW. The three-wattmeter method produces three lots of

1.14kW = 3.42kW. There is only a minor difference here, caused by our truncation whilst calculating the

inductance value.

Sample

Documents

Unit 39: Further Mathematics Unit Workbook 1...Pearson BTEC Levels 5 Higher Nationals in Engineering (RQF) Unit 39: Further Mathematics Unit Workbook 1 in a series of 4 for this unit