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The application of Appropriate Technology

# Part 8: A Prologue To AC Theory

Sections:

### 8.1 Resolution Of Forces

Before we look at AC theory we should take a look at how forces are resolved and how to formulate the equations for sine waves. Often in AC circuits the current and voltage are not in phase, when this happens it is not possible to simple add values up. The methods used are very similar to those used to add up forces that act in different direction and since forces are easier to visualise than voltages we will first find out how to add forces up.

Scalar Quantities

Scalar quantities have magnitude but no direction and some examples are mass, volume, density and energy etc. It makes no sence to say that something has a mass of 5kg to the left or an energy of 5J up. Such quantities can always be added together with a simple addition sum; for example a 5kg mass and a 3kg mass put together have a mass of 5 + 3 = 8kg. Note that speed is a scalar quantity having the units of metres per second and distance is also a scalar having units of metres.

Vector Quantities

Vector quantities have direction as well as magnitude. If we have a force of 5N we need to know in which direction the force is acting. Examples are force, acceleration and momentum. Such quantities can not be added together simple unless they act in one dimension. Note that velocity is a speed in a given direction but still has units of metres per second and displacement is a distance in a given direction but still has units of metres.

As we have just mentioned two vector quantities acting in a one dimensional plane (i.e. along a single line) can be added together with a simple addition sum. For example if we have a force of 5N from pulling left and a 10N force pulling from the right:

$\text{total force }=5+\left (-10 \right )=-5N$

Note that the minus sign indicates that the force is pulling to the right but we could easily say that a minus force is pulling left and get an answer of 5N; still indicating that the overall force acts to the right.

Now consider a piece of string attacked to a weight case (figure 8.1). If you pull the string straight up, all of the force (F) acts vertically and the weight is lifted (figure 8.1a). If you pull along the horizontal to the right, with the same force, all of the force acts horizontally and the weight is pulled along the table (figure 8.1b). Now, if you pull the string at an angle, somewhere between the vertical and horizontal, some of the force is lifting the weight and some of the force is pulling the weight along the table (figure 8.1c). As the angle from the horizontal (θ) decreases more force acts along the horizontal, until θ = 0° when all of the force acts along the horizontal and none of the force acts along the vertical.

Figure 8.1: Pulling a weight with a piece of string.

We can ‘resolve’ a force at an angle to give the vertical and horizontal ‘components’ (figure 8.2a). This is done by using simple trigonometry so that (since cos θ = adjacent/hypotenuse and sin θ = opposite/ hypotenuse):

The horizontal component:

$F_H=F\cos \theta$

The vertical component:

$F_V=F\sin \theta$

Figure 8.2: The resolving of forces.

If you want to add two forces together using this method you must:

• find the horizontal and vertical components of each force
• add the vertical components together
• add the horizontal components together
• calculate the force that results from the summed horizontal and vertical forces

To perform this last step, again simple trigonometry is used, as illustrated in figure 8.2. The magnitude of the resulting force (FR) is found using Pythagoras’ theorem and the direction of the force can by found by finding the tangent of θ (tan θ = opposite/adjacent). Thus:

The magnitude of the resultant force:

$F_R^2=F_V^2+F_H^2$

The direction of the resultant force:

$\tan \theta=\dfrac{F_V}{F_H}$

Example

Two force both pull on the same point, one is 5N acting at an angle of 30° to the horizontal and the other is 10N acting at 45° to the horizontal. Find the resulting force acting at the point.

Vertical component of the 5N force:

$F_{V1}=F\sin \theta=5\times \sin 30=2.5N$

Vertical component of the 10N force:

$F_{V2}=F\sin \theta=10\times\sin 45=7.07N$

The total vertical component of both forces:

$F_V=F_{V1}+F_{V2}=2.5+7.07=9.57N$

Horizontal component of the 5N force:

$F_{H1}=F\cos \theta=5\times\cos 30=4.33N$

Horizontal component of the 5N force:

$F_{H2}=F\cos \theta=10\times \cos 45=7.07N$

The total vertical component of both forces:

$F_H=F_{H1}+F_{H2}=4.3+7.07=11.37N$

Magnitude of the resultant force:

$F_R^2=F_V^2+F_H^2=9.57^2+11.37^2=217.07$
$F_R=14.73N$

Direction of the resultant force:

$\tan \theta=\dfrac {F_V}{F_H}=\dfrac {9.57}{11.37}=0.842$
$\theta=40.10^{\circ}$

Alternatively, once FH and FV have been found a graphical representation can be used to find FR. This method is illustrated in figure 8.3a, where FH and FV are drawn to scale so that FR and θ can be measured off directly. Figure 8.3b shown an alternative way to draw the vectors.

Figure 8.3: Graphical solutions to summing FH and FV.

### 8.2 The Parallelogram Rule

Another graphic method that can be used to find the sum of two vector is to use the Parallelogram rule. Using this method you have to know the magnitude of each vector as well as the angle (α) between them (figure 8.4).

Figure 8.4: The Parallelogram rule.

To use this method draw the two forces to scale and with the correct angle between them, then to find the resultant force complete the parallelogram and draw in the diagonal. The magnitude of the resultant force will be represented by the length of the diagonal and the direction of the resultant will be angle γ to F1 and angle θ to F2.

Example

Figure 8.5: Using the parallelogram rule.

From the previous example, there are 15° between the forces. Drawing the two forces to scale, with the correct angle between them, gives the result that the resultant force equals 14.7N at an angle of 10° to the 5N force (figure 8.5). Note that this method does not use horizontal or vertical references however, from the information given above we can calculate that the resultant is 30° + 10° = 40° to the horizontal. If you need to add three or more forces using this method, you must add two of the forces together and then add the the third to the resultant, a fourth can then be added to this new resultant etc. Using the resolving method you can resolve each force vertically and horizontally before adding the components together, then you can find the resultant of the totalled vertical and horizontal components.

### 8.3 Angles In A Circle

Consider a circle with a solid stick as a radius , rotating so that one end is held at the circle’s centre(OP in figure 8.6a). As the stick sweeps round, the angle from the horizontal position OA increases. This angle is always measured from the line OA anticlockwise around the circle. Now we split the circle into four quadrants and find the value for sin θ in each; note that the length r is always considered to be positive and the value y is obtained by measuring the position of P on the y-axis. From figure 8.6b we can see that (remembering sin θ = opposite/hypotenuse):

$1^{st}\text{quadrant:}\ \sin \theta_1=\dfrac {y_1}{r}=\text{ a positive value, since }x_1\text{ is positive}$
$\text{e.g.}\sin{45}^{\circ}=0.707$
$2^{nd}\text{ quadrant: }\sin \theta_2=\dfrac{y_2}{r}=\text{ a positive value, since }x_2\text{ is positive}$
$\text{e.g. }\sin {135}^{\circ}=0.707$
$3^{rd}\text{ quadrant: }\sin \theta_3=\dfrac{y_3}{r}=\text{ a negative value, since }x_3\text{ is negative}$
$\text{ e.g. }\sin {225}^{\circ}=-0.707$
$4^{th}\text{ quadrant: }\sin \theta_4=\dfrac{y_4}{r}=\text{ a negative value, since }x_4\text{ is negative}$
$\text{e.g. }\sin {315}^{\circ}=-0.707$

Figure 8.6: (a) a rotating stick in a circle; (b) angles in the four quadrants.

Often degrees are not use to measure angles within a circle and radians (rads) are preferred. Radians are derived so that 360° = 2π rads, therefore 180° = π. This means that 1 radian is equal to (360°/2π) 57.3° and:

$\text{radians }=\text{ dergrees}\times \dfrac{2 \pi}{360}$

### 8.4 Circular Motion

If we consider again the stick in figure 8.6a we can see that if it rotates at a steady speed the angular velocity (ω) is given by:

$\mathbf{\boldsymbol{\omega }=\dfrac{\boldsymbol{\theta}}{t}}$

where:
θ = the angle turned through (rads)
t = time taken (s)

Thus the angular velocity is the number of radians turned through in t seconds. The period (T) is the time taken for the stick to make one complete rotation, therefore:

$\mathbf{T =\dfrac {2\boldsymbol{\pi} }{\boldsymbol{\omega} }}$

The period is measured in second so that if we take its reciprocal we get the frequency ( f ), which is the number of rotations per second that the sick makes. Thus:

$\mathbf{\mathit{f}=\dfrac{\boldsymbol{\omega}}{2\boldsymbol{\pi}}}\quad \text { and }\quad \mathbf{\boldsymbol{\omega}=\mathit{f}2\boldsymbol{\pi}}$

Frequency has units of hertz (Hz), so that one hertz is one cycle per second. The angle θ (in radians) at any time (t) is therefore given by:

$\mathbf{\boldsymbol{\theta}=\boldsymbol{\omega}t=\mathit{f}2\boldsymbol{\pi}t}$

### 8.5 The Sine Wave

If we plot the y value of the end of our rotating sick from figure 8.6, against the angle θ we get a graph like the one shown in figure 8.6. The curve formed is known as a sine wave. At the maximum amplitude the displacement from the x-axis is equal to the length of the stick, or if we consider the circle swept by the tip of the stick the maximum displacement equals the radius (r). Similarly the minimum displacement equals –r. The maxima appear every 180° (π radians) with the first at 90°, the minima also appear every 180° starting at 270° and the whole cycle repeats every 360° (π radians).

The value on the y-axis is given by:

$y=\sin \theta$

In the first quadrant of the circle (figure 8.7) the value of sin θ = 0 when θ = 0° (try this on a calculator) and increases to 1 when θ = 90°, sin θ then decreases again through the second quadrant and equals 0 at θ = 180°. In the third quadrant sin θ becomes negative and reaches a minimum at -1 when θ = 270°, it them increases again through the fourth quadrant reaching 0 when θ = 0°.

Figure 8.7: A sine wave formed by a rotating stick.

If we re-plot the graph with the x-axis (figure 8.8) measuring time rather than θ we get a graph with exactly the same shape; θ changes with time because the stick rotates at a steady angular velocity of ω = θ/t. Hence, in the previous equation we can replace θ with θ = ωt:

$y=\sin \omega t$

Since:

$y=\sin \mathit{f}2\pi t$

So:

$y=\sin \mathit{f}2\pi t$

Figure 8.8: A sine wave plotted against time.

Figure 8.8 shows that, when plotted against time, the distance from peak to peak along the x-axis is equal to the period (T), which is the reciprocal of the frequency ( f ). Therefore, if we know the frequency at which the stick rotates we can plot the relevant since wave. However there is one more parameter that we must include in the equation and that is the amplitude (A). A sine wave plotted without an amplitude function will always peak at 1 on the y-axis. Adding the amplitude to the equation gives:

$\mathbf{y=A\boldsymbol{\sin} \mathit{f}2\boldsymbol{\pi}t}$

Note that if we want the equation in terms of our stick, A = r (figure 8.6), r being the radius of the circle and the length of the stick. Now, knowing the two parameters of frequency and amplitude we can plot the sine wave for any length of stick, rotating at any speed. Figure 8.9 shows the sine wave for two different amplitudes but the same frequency. In figure 8.9, A1 in greater than A2.

When we consider AC supplies (part 9) we will find that the rate at which the y value changes with equal changes in θ is very important. Figure 8.7 shows that when θ changes from 0° to 30° degrees a large change in y results however, a change in θ from 60° to 90° produces a much smaller change in y. Therefore, the rate of change in y with θ is greatest as the wave passes over the x-axis (i.e. when y = 0) and least as the wave reaches its maximum and minimum values. At the instant the wave is at the maximum and minimum value the rate of change of y with θ is zero (figure 8.8).

Figure 8.9: The effect of amplitude on a sine wave. A1 is greater than A2.

### 8.6 Waves In And Out Of Phase

The two waves in figure 8.9 both have the same frequency and are in phase; they both cross the x-axis at the same points. Two waves with the same frequency need not be in phase and figure 8.10 shows out of phase sine waves. The amount by which waves are out of phase is usually expresses in terms of a number of degrees or radians rather than time; that is if you imagine that the two sticks plotting the waves are fixed together at the point of rotation, the phase difference () will be the angle between them.

Figure 8.10: (a) The waves are 25° (25*2π/360 rads) out of phase; (b) The waves are 90° (π/2) out of phase; (c) The waves are 180° (π) out of phase.

Note that once two waves are over 180° out of phase the difference can be expressed as a negative angle, as can be seen from figure 8.7, 270° is the same as -90°; for example two waves that are 300° out of phase can be said to be (300 – 360) -60° out of phase. These observations show us that phase differences are normally expressed in between the values of -180° and 180°. Note that if two waves are 360° out of phase they are a whole cycle out, and therefore not out of phase at all.

Figure 8.10b shows two waves out of phase by 90° and figure 8.11 shows two waves out of phase by -90°. Although these two diagrams look similar they can be differentiated from each other in terms of leading an lagging. In figure 8.10b the solid wave is said to lead the dotted wave by 90°, or the dotted wave can be said to lag the solid wave by 90°; the solid line is leading because as it starts a new cycle the dotted line is still on the previous cycle. In figure 8.11 the dotted wave is said to lead the solid wave since as it starts a new cycle the solid wave is still on the previous cycle, or if you look at the start of the graph, when the solid wave starts the new cycle the dotted wave has already partially completed a cycle. When the terms lagging and leading are used is normally always quote being positive and the minus sign is not used; so that phase angle of 90° and -90° are both written as =90°.

Figure 8.11: Two waves 270° (3π/2 rads) or -90° out of phase.

When two waves are out of phase, the equation of the wave that starts a cycle at the beginning of the graph (the solid line in figure 8.11) will be given by:

$y_1=A_1\sin \theta$

So that when the second wave (the dashed line in figure 8.11) leads the first will have the equation:

$y_2=A_2\sin \left (\theta +\phi \right )$

If the second wave lags the first (as in figure 8.10a) the previous equation becomes:

$y=\left ( A_1+A_2 \right )\sin \theta$

If waves are in phase then the can be simply added together. This can be done by adding individual y values together or by adding the maximum amplitudes together and using the sine wave equation. Thus if we have two waves in phase, with maximum amplitudes of A1 and A2, the resultant wave can be found from:

$y=\left ( A_1+A_2\right )\sin \theta$

Two waves that are 180° out of phase can also be added together in this was as long as you make one of the maximum amplitudes a negative value. Note that two waves that are 180° out of phase and that have the same maximum amplitude will cancel each other out.

Figure 8.12: Adding together two waves that are in phase.

Example

A wave with a maximum amplitude of 100 is in phase with a wave with a maximum amplitude of 150. Find the resultant wave if these two waves are added together.

The resultant wave will have a maximum amplitude of 100 + 150 = 250. The two waves and the resultant wave are shown in figure 8.12.

When two waves are out of phase, although individual values can be simply added together, the maximum amplitudes can not. To find the resultant wave using the sine wave equation the maximum amplitudes must be summed using the parallelogram rule, this method will also tell you the phase relationship between the resultant wave and the two waves being summed.

Example

Wave 1 has a maximum amplitude of A1 = 100 and leads Wave 2, which has a maximum amplitude of A1 = 150, by 60°. Find the resultant wave if these two waves are added together.

Measuring from a diagram or using the separate sine wave equations we could work out each point individually and add them together, so that:

$\begin{array}{llll} \text{at }0^{\circ},&\quad y_1=0&\quad y_2=-130&\quad y_1+y_2=-130\\ \\ \text{at }30^{\circ},&\quad y_1=50&\quad y_2=-75&\quad y_1+y_2=-25\\ \\ \text{at }0^{\circ},&\quad y_1=0&\quad y_2=-130&\quad y_1+y_2=-130\\ \\ \text{at }60^{\circ},&\quad y_1=86.6&\quad y_2=0&\quad y_1+y_2=+86.6 \end{array}$

etc.

However, the maximum amplitude of the resultant wave can be found using the parallelogram rule: figure 8.13b shows that the maximum amplitude of the resultant wave is 217 and that this wave lags Wave 1 by 37.5. Since Wave 2 lags Wave 2 the phaser diagram is draw so that Wave 2 is 60° under Wave 1. Since the resultant wave is also below Wave 1 we can see that it also lags Wave 1. Figure 8.13a shows the three complete waves. The equations of the three waves are:

Wave 1:

$y_1=100\sin \theta$

Wave 2:

$y_2=150\sin\left ( \theta-60^{\circ} \right )$

Resultant:

$y=217\sin\left ( \theta-37.5^{\circ} \right )$

Figure 8.13: Adding together two waves that are out of phase. (a) The thin solid line i s Wave 1, the tin dotted line is Wave 2 and the thick solid line is the resultant; (b) the parallelogram rule applied to Wave 1 and Wave 2.