Preparing for the HNC Electrical Maths Components for the HNC Electrical Maths... · Preparing for the HNC Electrical Maths Components author: Michael Lopez BEng(Hons) MSc PGCert

www.unicourse.org Preparing for the HNC Electrical Maths Components author: Michael Lopez BEng(Hons) MSc PGCert CertEd MIFL MIET FHEA ©UniCourse Ltd 2016

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online learning

Preparing for the HNC Electrical Maths

Components

http://www.unicourse.org/


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Contents INTRODUCTION .................................................................................................................................................. 3

1 Algebraic Methods .................................................................................................................................... 4

1.1 Indices and Logarithms ....................................................................................................................... 4

1.1.1 Indices ......................................................................................................................................... 4

1.1.2 Logarithms ................................................................................................................................... 7

1.1.3 Exponential Growth and Decay ................................................................................................... 8

1.2 Linear Equations and Straight Line Graphs ........................................................................................ 9

1.2.1 Linear Equations .......................................................................................................................... 9

1.2.2 Straight Line Graphs .................................................................................................................... 9

1.2.3 Linear Simultaneous Equations ................................................................................................. 11

1.3 Factorisation and Quadratics ........................................................................................................... 12

1.3.1 Multiplication by Bracketed Expressions .................................................................................. 12

1.3.2 Common Factors ....................................................................................................................... 12

1.3.3 Grouping .................................................................................................................................... 12

1.3.4 Quadratics and Roots of Equations ........................................................................................... 13

Summary .......................................................................................................................................................... 15



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INTRODUCTION This Workbook guides you through the learning outcomes related to:

Indices and logarithms: laws of indices, laws of logarithms e.g. common logarithms (base 10), natural logarithms (base e), exponential growth and decay

Linear equations and straight line graphs: linear equations; straight line graph (coordinates on a pair of labelled Cartesian axes, positive or negative gradient, intercept, plot of a straight line); experimental data e.g. Ohm’s law, pair of simultaneous linear equations in two unknowns

Factorisation and quadratics: multiply expressions in brackets by a number, symbol or by another expression in a bracket; by extraction of a common factor; by grouping; quadratic expressions; roots of an equation e.g. quadratic equations with real roots by factorisation, and by the use of formula



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1 Algebraic Methods

1.1 Indices and Logarithms 1.1.1 Indices In engineering we often use very large quantities, and very small quantities. Let’s look at a couple of examples...

A large power station may supply 4,000,000,000 Watts of power to the national grid. We normally term that as 4 GigaWatts (4GW). If we were manipulating that figure in mathematical calculations it would be very tiresome to have to write 4,000,000,000 each time. Wouldn’t it be better if we could represent that number in a more concise way? The number contains lots of zeros, nine of them, in fact. Maybe we could express the number 4 and multiply by a series of tens. Here’s a way to express the number by multiples of 10...

4,000,000,000 ≡ 4 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10

Wow, that looks even worse. However, when we notice that there are 9 multiples of 10 there then we could employ an index, as follows...

4,000,000,000 ≡ 4 × 109

The signal from a remote spacecraft is picked up from Earth. The signal is much attenuated and only registers 3 picoWatts (3pW) on the instrumentation. That’s a really small signal, and the number may be written in decimal form as...

0.000000000003

How are going to express that in index form? Well, we could express it as...

0.000000000003 =3

10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10=

31012

Since we try to avoid fractions we may look at that bottom part (the denominator) and bring it to the top (numerator) simply by changing the sign on the index...

31012

= 3 × 10−12

The best trick to remember with this number is to think of how many places do you need to move that decimal point to the right until it appears just after the 3? You need to move it 12 places, of course.

Let’s look at some further examples of employing this neat notation...



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1,000 ≡ 1 × 103

10,000 ≡ 10 × 103

8,400,000 ≡ 8.4 × 106

0.001 ≡ 1 × 10−3

0.00015 ≡ 150 × 10−6

5.000678 ≡ 5.000678 (𝑦𝑦𝑦𝑦𝑦𝑦 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑚𝑚𝑐𝑐𝑚𝑚𝑚𝑚 𝑐𝑐ℎ𝑐𝑐𝑐𝑐 𝑦𝑦𝑐𝑐𝑚𝑚 𝑐𝑐𝑦𝑦𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑚𝑚)

An extremely important index to know about is zero. When we raise a quantity to the power zero we always get 1 (there is an exception to this rule and that is when we raise zero to the power zero – something which you will never do so that we leave that one to the theorists). We shall see this proven shortly.

Multiplication of Indexed Quantities

If we multiply 1000 by 1000 we get a million. We know that 1,000 is expressed as 1 × 103 and we also know that 1,000,000 is expressed as 1 × 106. Let’s perform the multiplication in our new concise indexed form...

1,000 × 1,000 ≡ 1 × 103 × 1 × 103 = 1 × 103 × 103 = 1 × 106

What did we do with those 3’s to get the 6? We must have added them, right? That then brings us to our first rule of indices, for any general problem...

𝒂𝒂𝒎𝒎 × 𝒂𝒂𝒏𝒏 = 𝒂𝒂𝒎𝒎+𝒏𝒏

Here, ‘a’ is any base (could be 10 as before, or could be a quantity merely represented as a letter, like ‘v’ for voltage). The indexes m and n can be any number you like, including negative numbers. This rule can be extended further to incorporate the multiplication of three quantities...

𝑥𝑥𝑎𝑎 × 𝑥𝑥𝑏𝑏 × 𝑥𝑥𝑐𝑐 = 𝑥𝑥𝑎𝑎+𝑏𝑏+𝑐𝑐

Here are some examples...

102 × 104 = 106

105 × 10−3 = 105+(−3) = 102

𝑐𝑐2𝑐𝑐6𝑐𝑐−3 = 𝑐𝑐2+6−3 = 𝑐𝑐5

𝑝𝑝6𝑝𝑝−2𝑝𝑝4𝑞𝑞5𝑞𝑞−2 = 𝑝𝑝6−2+4𝑞𝑞5−2 = 𝑝𝑝8𝑞𝑞3

Division of Indexed Quantities

If you had 100,000 objects and divided them fairly amongst 100 people then each person would, of course, receive 1000 objects. Let’s have a look at this operation in index form...

100,000100

=1 × 105

1 × 102=

105

102= 1000 = 103



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We knew the answer was 1000, so what did we do with the 5 index and the 2 index to get the 3 index? We must have subtracted them. What we did was...

105

102= 105−2 = 103

It looks like we have another rule then, this time for the division of indexed quantities...

𝒂𝒂𝒎𝒎

𝒂𝒂𝒏𝒏= 𝒂𝒂𝒎𝒎−𝒏𝒏

Here are some examples...

106

104= 106−4 = 102

8 × 107

2 × 102= 4 × 107−2 = 4 × 105

3.6 × 104

1.2 × 10−2= 3 × 104−(−2) = 3 × 106

12𝑝𝑝4

3𝑝𝑝−3= 4𝑝𝑝4−−3 = 4𝑝𝑝7

10001000

=103

103= 103−3 = 100 = 1

That last one proves that raising a quantity to the power zero gives 1, as suggested earlier.

We may also combine our multiplication and division rules, as in the following examples...

𝑐𝑐2𝑐𝑐−5

𝑐𝑐3𝑐𝑐4=𝑐𝑐2+(−5)

𝑐𝑐3+4=𝑐𝑐−3

𝑐𝑐7= 𝑐𝑐−3−7 = 𝑐𝑐−10

𝑐𝑐9𝑐𝑐−4

𝑑𝑑−3𝑑𝑑6=𝑐𝑐9+(−4)

𝑑𝑑−3+6=𝑐𝑐5

𝑑𝑑3= 𝑐𝑐5𝑑𝑑−3

In that last example we brought the 𝑑𝑑3 in the denominator up to the numerator, simply by changing the sign on its index.

Raising Indexed Quantities to Powers

What if you saw something like 10002? You know that the answer is 1 million but how did we get that via indexes? Let’s look at the problem another way...



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10002 = (103)2 = 103 × 103 = 106

What we have done is to multiply the 3 by the 2. This brings us to our third rule for indices...

(𝒂𝒂𝒎𝒎)𝒏𝒏 = 𝒂𝒂𝒎𝒎×𝒏𝒏

Some examples using this rule...

(104)2 = 104×2 = 108

(𝑥𝑥2)5 = 𝑥𝑥2×5 = 𝑥𝑥10

1.1.2 Logarithms Many people shy away from logarithms at school. What are they for? In engineering we use logarithms to find answers to circuit problems involving capacitors and inductors (more on this in the next section). Now you know why they are needed let’s look at what they are and how we use them.

When we mention the term ‘logarithm’ what we are really saying is ‘what’s the logarithm of a particular number?’ The logarithm of a particular number is the index that needs to be applied to our base to get the number we start with. For example, if we should ask ‘what’s the logarithm to base 10 of 1000?’ we are wondering what power we need to raise 10 by to get 1000 – the answer is 3, as we already know from previous examples. We can therefore say that the log to base 10 of 1000 is 3.

In engineering we tend to work either in base 10 (used for decibels) and base ‘e’ (the universal constant, associated with natural growth and decay – as related to capacitors and inductors).

These days we tend to use calculators to work out logarithms. That is no bad thing but to be able to use and transpose many engineering expressions we need a knowledge of the rules of logarithms.

Law 1: log(A) + log(B) = log(AB)

Example...

𝑙𝑙𝑦𝑦𝑙𝑙(100) + 𝑙𝑙𝑦𝑦𝑙𝑙(1,000) = 2 + 3 = 5 = log(100 × 1,000) = 𝑙𝑙𝑦𝑦𝑙𝑙(100,000) = 5

Law 2: log(A) – log(B) = log(A/B)

Example...

𝑙𝑙𝑦𝑦𝑙𝑙(1,000,000) − 𝑙𝑙𝑦𝑦𝑙𝑙(10,000) = 6 − 4 = 2 = 𝑙𝑙𝑦𝑦𝑙𝑙(1,000,000 10,000⁄ ) = 𝑙𝑙𝑦𝑦𝑙𝑙(100) = 2

Law 3: log(An) = n log (A)

Example...

𝑙𝑙𝑦𝑦𝑙𝑙(10002) = log(1,000,000) = 6 = 2 log(1000) = 2 × 3 = 6



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1.1.3 Exponential Growth and Decay We have mentioned that exponential growth and decay are related, amongst other things, to the characteristics of capacitors and inductors. Here are some characteristic features of exponential growth...

The graph shows how the current in the LR circuit grows exponentially. The formula which relates the current to time, the inductor value, and the resistor value, is...

𝑐𝑐 =𝑉𝑉1𝑅𝑅�1 − 𝑚𝑚

−𝑅𝑅𝑅𝑅𝐿𝐿 �

Notice that base 10 is not used. The base is ‘e’, a natural constant equal to 2.71828 (to 5 decimal places). Components behave as nature/physics dictate they should. Base 10 is a human invention, convenient because we have 10 fingers/thumbs, which we use for counting.

We can also see exponential decay in circuits...

The current in this RC circuit experiences exponential decay. This is given by...

𝑐𝑐 =𝑉𝑉𝑏𝑏𝑎𝑎𝑅𝑅𝑅𝑅𝑏𝑏𝑏𝑏𝑏𝑏

𝑅𝑅𝑚𝑚−𝑅𝑅𝑅𝑅𝑅𝑅

Our laws of logs apply equally well to this new base of ‘e’. An example...

𝑙𝑙𝑦𝑦𝑙𝑙𝑏𝑏(4.556)−3 = −3 𝑙𝑙𝑦𝑦𝑙𝑙𝑏𝑏(4.556) = −3 × 1.516 = −4.548



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Worked Example 1

1.2 Linear Equations and Straight Line Graphs 1.2.1 Linear Equations A linear equation has an index of 1 on the independent variable. For example...

𝑦𝑦 = 𝑚𝑚𝑥𝑥 + 𝑐𝑐

Here, the independent variable is 𝑥𝑥 and 𝑚𝑚 indicates the slope of the graph’s line. The constant c indicates where the line crosses the y axis. (If you let 𝑥𝑥 = 0 then y will equal c and this will indicate that crossing point).

1.2.2 Straight Line Graphs Considering the information just discussed about linear equations, let’s look at a worked example involving such a straight line graph...

RMS measurements were taken for the current (i) through a resistor and the voltage (v) across it. The resistor experiences both AC and DC influences. The measurement data is as follows;

v [V] -6 -3 0 3 6 9 12 i [A] 0 1 2 3 4 5 6

a) Use the experimental data to plot a graph of voltage (v) on the vertical axis against current (i) on the

horizontal axis. b) Hence, deduce the equation of the line by determining its gradient and intercept with the voltage axis.

The first thing to do is to plot the graph from the table. This is most easily done by downloading and using the free Graph application.

Insert the data into a point series, as follows...


https://www.padowan.dk/


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Marker style and colour can be chosen as preferred. Then insert a curve of best fit using the icon at the top of the application. What results is shown below...

We see that the graph intercepts the vertical axis at v= -6 so our equation will have a value for c of -6. We then need to complete a triangle anywhere on the graph, as shown below with the aid of the blue dotted line...

To find the slope of the red line we measure the height of the triangle and divide by its width. This gives us 9/3 = 3, so m (slope) has a value of +3. We now have the equation of the straight line...

𝑣𝑣 = 3𝑐𝑐 − 6



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Worked Example 2

1.2.3 Linear Simultaneous Equations Such equations can be formed by analysing electrical circuits. Solution of the equations can yield unknown quantities, such as voltage or current. There are a number of ways to solve simultaneous equations. One method is by multiplication and substitution, which we shall now analyse.

Solve the following pair of simultaneous linear equations...

𝟐𝟐𝒂𝒂 + 𝟑𝟑𝟑𝟑 = 𝟐𝟐𝟑𝟑

𝟔𝟔𝒂𝒂 − 𝟐𝟐𝟑𝟑 = 𝟏𝟏𝟏𝟏

The first thing we need to do is number the equations, so that we may refer to them easily in our developments...

2𝑐𝑐 + 3𝑏𝑏 = 23 [1]

6𝑐𝑐 − 2𝑏𝑏 = 14 [2]

The aim at this point is to look at the two equations and think of ways in which we may eliminate one of the unknowns (‘a’ and ‘b’). What is apparent from the equations is that if we multiply equation [1] by ‘-3’ we will have ‘-6a’ in the new equation, which we may then add to equation [2] to eliminate the ‘a’ term. Maths is much more understandable when not expressed in sentences, so let’s try to do what we meant...

[1] X (-3): −6𝑐𝑐 − 9𝑏𝑏 = −69 [3]

The notation above is very useful. It says that when we take equation [1] and multiply it by ‘-3’ we will get what is now called equation [3].

The reason for that will quickly become apparent when we write equations [2] and [3] together...

6𝑐𝑐 − 2𝑏𝑏 = 14 [2]

−6𝑐𝑐 − 9𝑏𝑏 = −69 [3]

Notice that if we should ADD these two equations the ‘a’ will disappear. That will leave one unknown (‘b’), which we may easily find...

[2] + [3]: −11𝑏𝑏 = −55 [4]

∴ 𝑏𝑏 = 5

Since we now know the value of ‘b’ we can substitute this value into any previous equation we like to find the value of ‘a’. Let’s pick, say, equation [1]...

2𝑐𝑐 + 3𝑏𝑏 = 23 [1] ∴ 2𝑐𝑐 + 3(5) = 23

∴ 2𝑐𝑐 = 23 − 15 = 8 ∴ 𝑐𝑐 = 4



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We have now solved the simultaneous equations. We have 𝒂𝒂 = 𝟏𝟏 and 𝟑𝟑 = 𝟓𝟓.

1.3 Factorisation and Quadratics 1.3.1 Multiplication by Bracketed Expressions Let’s look at some obvious mathematical expressions and see if we can see how to multiply by bracketed terms...

15 = 7 + 8

2 × 15 = 30

∴ 2(15) = 2(7 + 8) = 14 + 16 = 30

The term 2(7 + 8) could easily have been a bunch of letters, like 𝑐𝑐(𝑏𝑏 + 𝑐𝑐) so we know we must multiply the letter on the left (‘a’) by EACH of the letters in the bracket. We then have...

𝑐𝑐(𝑏𝑏 + 𝑐𝑐) = 𝑐𝑐𝑏𝑏 + 𝑐𝑐𝑐𝑐

Let’s look at another example, which extends this concept a little further...

8 = 3 + 5

18 = 4 + 14

∴ (8)(18) = 144 = (3 + 5)(4 + 14) = (3)(4) + (3)(14) + (5)(4) + (5)(14) = 144

This problem could easily have been expressed in letters also...

(𝑐𝑐 + 𝑏𝑏)(𝑐𝑐 + 𝑑𝑑) = 𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑑𝑑 + 𝑏𝑏𝑐𝑐 + 𝑏𝑏𝑑𝑑

1.3.2 Common Factors Where we have common factors in an expression we tend to use efficient notation so that we are not writing out the same term multiple times. Consider...

2(8) + 2(9)

This is more efficiently written as...

2(8 + 9)

Again, these could have been letters...

2𝑐𝑐 + 2𝑏𝑏 = 2(𝑐𝑐 + 𝑏𝑏)

We have brought out a common factor of 2 and the expression is now more elegant.

1.3.3 Grouping Consider groups of letters, such as those below...

6𝑐𝑐𝑏𝑏 + 12𝑐𝑐2𝑏𝑏2

What’s common there, which we can group? Well, the 6 is a factor of the first and second term (two lots of six in twelve). Also, ab is a factor of both terms. We may now express this more elegantly as...



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6𝑐𝑐𝑏𝑏 + 12𝑐𝑐2𝑏𝑏2 = 6𝑐𝑐𝑏𝑏(1 + 2𝑐𝑐𝑏𝑏)

If you’re not sure then multiply it out and confirm the result.

Here are some more examples of grouping...

32𝑓𝑓3𝑙𝑙 + 16𝑓𝑓2𝑙𝑙4 = 16𝑓𝑓2𝑙𝑙(2𝑓𝑓 + 𝑙𝑙3)

−𝑓𝑓3𝑙𝑙2 − 𝑓𝑓4𝑙𝑙3 + 8 = −𝑓𝑓3𝑙𝑙2(1 + 𝑓𝑓𝑙𝑙) + 8

𝑓𝑓𝑙𝑙 + 𝑓𝑓2 + 𝑓𝑓2𝑙𝑙 + 𝑓𝑓𝑙𝑙2 + 12 = 𝑓𝑓(𝑙𝑙 + 𝑓𝑓 + 𝑓𝑓𝑙𝑙 + 𝑙𝑙2) + 12

1.3.4 Quadratics and Roots of Equations A quadratic equation is non-linear. What this means is that the highest index on the independent variable is 2. Such equations do not yield straight lines, they have bends. A quadratic equation has the following general form...

𝑦𝑦 = 𝑐𝑐𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐

You see that power of 2 there, that categorises this as a quadratic. We normally have two approaches to solving quadratic equations...

Factorise to find the roots Use the quadratic formula

Finding the Roots by Factorisation

We always look to do this first. If it’s not possible, or difficult, then we turn to the formula method, which you shall see shortly.

Let’s say we have the quadratic equation...

𝑦𝑦 = 𝑥𝑥2 + 7𝑥𝑥 + 12

We first look at the number on the RHS and ask ‘what factors of 12, when added together, will give that number in the middle (+7)?’ Let’s examine the factors of 12...

1 x 12 = 12 1 + 12 is not +7

-1 x -12 = 12 -1+ -12 is not +7

2 x 6 = 12 2+6 is not +7

-2 x -6 = 12 -2+ -6 is not +7

3 x 4 =12 3 + 4 IS +7

We therefore select +3 and +4 and place these into brackets, equating to zero, as follows...



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(𝑥𝑥 + 3)(𝑥𝑥 + 4) = 0

When the first bracket is zero then the expression above is true. For the first bracket to be zero the value of 𝑥𝑥 must be -3. We also look at the second bracket and do the same. In that case the value of 𝑥𝑥 needs to be -4. What we have done here is to find the roots of the given quadratic, which means we have found the points on the graph of the quadratic where the curve crosses the 𝑥𝑥 axis (i.e. where y is zero). Our proposed roots are 𝒙𝒙 = −𝟑𝟑 and 𝒙𝒙 = −𝟏𝟏. Let’s use the Graph application to check this is so...

Our calculations check out very well.

Formula Method for finding the Roots of a Quadratic

We have the general expression for a quadratic as follows...


The formula to find the roots is...

𝑥𝑥 =−𝑏𝑏 ± √𝑏𝑏2 − 4𝑐𝑐𝑐𝑐

2𝑐𝑐

This formula will always find you the roots. Let’s check out the previous problem this way...


𝑦𝑦 = 𝑥𝑥2 + 7𝑥𝑥 + 12

We see that a is 1, b is 7 and c is 12. Let’s use these figures in the formula...

𝑥𝑥 =−𝑏𝑏 ± √𝑏𝑏2 − 4𝑐𝑐𝑐𝑐

2𝑐𝑐=−7 ± �72 − 4(1)(12)

2(1)=−7 ± √49 − 48

2=−7 ± 1

2= −3 𝑦𝑦𝑜𝑜 − 4



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Summary Well done for getting this far. If you have understood this material, it is very likely that you are ready to take on the challenge of the Edexcel HNC course in Electrical and Electronic Engineering, delivered by UniCourse. Please then go to our website and complete the short online application form.

Our HNC course provides lots of workbooks, most of which are longer than this one, and useful tutorial videos which we have produced to ease your passage through the course.

If you feel that you need more lessons in Maths, at any level, to build your confidence, then we highly recommend the wonderful KhanAcademy website. We look forward to welcoming you to UniCourse!



https://www.khanacademy.org/math

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Preparing for the HNC Electrical Maths Components for the HNC Electrical Maths... · Preparing for the HNC Electrical Maths Components author: Michael Lopez BEng(Hons) MSc PGCert