online learning Preparing for the HNC Electrical Maths Components Page 1 of 15
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 Page 2 of 15
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 Page 3 of 15
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 10 9 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 = 3 10 12 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... 3 = 3 10 12 1012 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... Page 4 of 15
1,000 1 10 3 10,000 10 10 3 8,400,000 8.4 10 6 0.001 1 10 3 0.00015 150 10 6 5.000678 5.000678 (yyyyyy cccccccc mmmmmmmm tthaaaa oooooo cccccccccccccc) 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 10 3 and we also know that 1,000,000 is expressed as 1 10 6. Let s perform the multiplication in our new concise indexed form... 1,000 1,000 1 10 3 1 10 3 = 1 10 3 10 3 = 1 10 6 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... aa mm aa nn = aa mm+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... Here are some examples... Division of Indexed Quantities xx aa xx bb xx cc = xx aa+bb+cc 10 2 10 4 = 10 6 10 5 10 3 = 10 5+( 3) = 10 2 tt 2 tt 6 tt 3 = tt 2+6 3 = tt 5 pp 6 pp 2 pp 4 qq 5 qq 2 = pp 6 2+4 qq 5 2 = pp 8 qq 3 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... 100,000 100 = 1 105 1 10 2 = 105 = 1000 = 103 102 Page 5 of 15
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... 10 5 10 2 = 105 2 = 10 3 It looks like we have another rule then, this time for the division of indexed quantities... Here are some examples... aa mm = aamm nn aann 10 6 10 4 = 106 4 = 10 2 8 10 7 2 10 2 = 4 107 2 = 4 10 5 3.6 10 4 1.2 10 2 = 3 104 ( 2) = 3 10 6 12pp 4 3pp 3 = 4pp4 3 = 4pp 7 1000 1000 = 103 10 3 = 103 3 = 10 0 = 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... aa 2 aa 5 aa 3 aa 4 = aa2+( 5) aa 3+4 = aa 3 aa 7 = aa 3 7 = aa 10 cc 9 cc 4 dd 3 dd 6 = cc9+( 4) dd 3+6 = cc5 dd 3 = cc5 dd 3 In that last example we brought the dd 3 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 1000 2? You know that the answer is 1 million but how did we get that via indexes? Let s look at the problem another way... Page 6 of 15
1000 2 = (10 3 ) 2 = 10 3 10 3 = 10 6 What we have done is to multiply the 3 by the 2. This brings us to our third rule for indices... (aa mm ) nn = aa mm nn Some examples using this rule... (10 4 ) 2 = 10 4 2 = 10 8 (xx 2 ) 5 = xx 2 5 = xx 10 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... llllll(100) + llllll(1,000) = 2 + 3 = 5 = log(100 1,000) = llllll(100,000) = 5 Law 2: log(a) log(b) = log(a/b) Example... llllll(1,000,000) llllll(10,000) = 6 4 = 2 = llllll(1,000,000 10,000) = llllll(100) = 2 Law 3: log(a n ) = n log (A) Example... llllll(1000 2 ) = log(1,000,000) = 6 = 2 log(1000) = 2 3 = 6 Page 7 of 15
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... ii = VV1 1 ee RRRR LL RR 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... ii = VV bbbbbbbbbbbbbb RR ee tt RRRR Our laws of logs apply equally well to this new base of e. An example... llllll ee (4.556) 3 = 3 llllll ee (4.556) = 3 1.516 = 4.548 Page 8 of 15
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... Worked Example 1 yy = mmmm + 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... Page 9 of 15
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 = 3ii 6 Page 10 of 15
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. Worked Example 2 Solve the following pair of simultaneous linear equations... 2222 + 3333 = 2222 6666 2222 = 1111 The first thing we need to do is number the equations, so that we may refer to them easily in our developments... 2aa + 3bb = 23 [1] 6aa 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): 6aa 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... 6aa 2bb = 14 [2] 6aa 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]... 2aa + 3bb = 23 [1] 2aa + 3(5) = 23 2aa = 23 15 = 8 aa = 4 Page 11 of 15
We have now solved the simultaneous equations. We have aa = 44 and bb = 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 aa(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... aa(bb + cc) = aaaa + aaaa 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... (aa + bb)(cc + dd) = aaaa + aaaa + bbbb + bbbb 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... This is more efficiently written as... Again, these could have been letters... 2(8) + 2(9) 2(8 + 9) 2aa + 2bb = 2(aa + 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... 6aaaa + 12aa 2 bb 2 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... Page 12 of 15
6aaaa + 12aa 2 bb 2 = 6aaaa(1 + 2aaaa) If you re not sure then multiply it out and confirm the result. Here are some more examples of grouping... 32ff 3 gg + 16ff 2 gg 4 = 16ff 2 gg(2ff + gg 3 ) ff 3 gg 2 ff 4 gg 3 + 8 = ff 3 gg 2 (1 + ffff) + 8 ffff + ff 2 + ff 2 gg + ffgg 2 + 12 = ff(gg + ff + ffff + gg 2 ) + 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 = aaxx 2 + bbbb + 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 = xx 2 + 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... Page 13 of 15
(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 = 44. 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... yy = aaxx 2 + bbbb + cc xx = bb ± bb2 4aaaa 2aa This formula will always find you the roots. Let s check out the previous problem this way... yy = aaxx 2 + bbbb + cc yy = xx 2 + 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 4aaaa 2aa = 7 ± 72 4(1)(12) 2(1) = 7 ± 49 48 2 = 7 ± 1 2 = 3 oooo 4 Page 14 of 15
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! Page 15 of 15