EDEXCEL NATIONAL CERTIFICATE/DIPLOMA UNIT 5  ELECTRICAL AND ELECTRONIC PRINCIPLES NQF LEVEL 3 OUTCOME 4  ALTERNATING CURRENT


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1 EDEXCEL NATIONAL CERTIFICATE/DIPLOMA UNIT 5  ELECTRICAL AND ELECTRONIC PRINCIPLES NQF LEVEL 3 OUTCOME 4  ALTERNATING CURRENT 4 Understand singlephase alternating current (ac) theory Single phase AC circuit theory: waveform characteristics e.g. sinusoidal and nonsinusoidal waveforms, amplitude, period time, frequency, instantaneous, peak/peaktopeak, root mean square (r.m.s), average values, form factor; determination of values using phasor and algebraic representation of alternating quantities e.g. graphical and phasor addition of two sinusoidal voltages, reactance and impedance of pure R, L and C components ac circuit measurements: safe use of an oscilloscope eg setting, handling, health and safety; measurements (periodic time, frequency, amplitude, peak/peaktopeak, r.m.s and average values); circuits e.g. half and full wave rectifiers D.J.Dunn 1
2 1. REVISION OF BASIC A.C. THEORY 1.1 SINUSOIDAL WAVE FORMS A pure alternating current or voltage varies with time sinusoidally as shown. INSTANTANEOUS VALUES For a sinusoidal voltage and current the instantaneous value at any moment in time is given by: v = V sin (ωt) and i = I sin (ωt) or v = V sin (2πf t) and i = I sin (2πf t) Note that this assumes that t = 0 when v or i = 0 AMPLITUDE The maximum value of volts or current is called the peak volts or current and this is the amplitude of the wave form (V and I). The peak to peak value is double the amplitude as shown on the diagram. FREQUENCY The voltage or current changes from a maximum (plus) in one direction, through zero to a maximum (minus) in the other direction. This occurs at f times a second. f is the frequency in Hertz. PERIODIC TIME 1 Hz = 1 cycle/second The time it takes to complete 1 cycle is T seconds (the periodic time). It follows that T = 1/f ANGULAR FREQUENCY If we think of the voltage and current being generated by a machine that rotates one revolution per cycle, the 1 cycle corresponds to 360 o or 2π radian. It follows that f cycles/second = 2πf radian/s and this is the angular frequency. ω = 2πf = 2π/T rad/s D.J.Dunn 2
3 AVERAGE VALUES The average value of any true alternating current or voltage is zero since half the cycle is negative and half is positive. When an average value is stated it refers to the average over one half of the cycle. This may be determined from the area under the graph as illustrated below. For a sinusoidal waveform, the blue area is exactly 2 when the angle is in radians and the peak value is 1. The mean value is value that makes the blue rectangle contain the same area as the blue area of the half cycle. In other words the green area above the average is equal to the light blue area below the rectangle. The blue area must be equal to π x average value hence the average value is 2/π = If the peak value is something other than 1 then the average is x peak value FORM FACTOR Peak Value This is defined as Form Factor Average value Hence for a sinusoidal voltage or current the form factor is π/2 = PHASE and DISPLACEMENT The sinusoidal graph is produced because we made θ = 0 when t = 0. We could choose to make θ any value at t = 0. We would then write the equations as: x = A sin(θ + ) or v = V sin(θ + ) where is the phase angle. The plot shown has x = 0 at θ = 30 o so it follows that (30 + ) = 0 and so = 30 o. If we made = 90 o we would have a cosine plot. Often periodic functions are not based about a mean of zero. For example an alternating voltage might be added to a constant (d.c.) voltage so that V = V dc + V sin(θ + ) D.J.Dunn 3
4 SELF ASSESSMENT EXERCISE No. 1 1 Mains electricity has a frequency of 50 Hz. What is the periodic time and angular frequency? (0.02 s and 314 rad/s) 2. An alternating current has a periodic time of s. What is the frequency? (400 Hz) 3. A alternating voltage has a peak to peak amplitude of 300 V and frequency of 50 Hz. What is the amplitude and average value? (150 V and V) What is the voltage at t = 0.02 s? (16.4 V) 4. An alternating current is given by the equation I = 5 sin(600t). Determine the following. i. the frequency ( Hz) ii. the periodic time ( ms) iii. the average value. (3.183 A) 5. Determine the following from the graph shown. The amplitude. The offset displacement. The periodic time. The frequency. The angular frequency. The phase angle. (Answers 5, 2, 1.57 s, 4 rad/s, Hz, 0.2 radian or 11.5 o ) D.J.Dunn 4
5 1.2 ROOT MEAN SQUARE VALUES (R.M.S.) The mean value of an alternating voltage and current is zero. Since electric power is normally calculated with P = V I it would appear that the mean power should be zero. This clearly is not true because most electric fires use alternating current and they give out power in the form of heat. When you studied Ohms' Law, you learned that electric power may also be calculated with the formulae E.P. = I 2 R or E.P. = V 2 /R These formulae work with positive or negative values since a negative number is positive when squared and power is always positive. In the case of a.c. we must use the average value of V 2 or I 2 and these are not zero. The diagram shows how a plot of V 2 or I 2 is always positive. The mean value is indicated. The mean height may be obtained by placing many vertical ordinates on it as shown. Taking a graph of current with many ordinates i 1 2, i i n 2. The mean value of the i 2 is: ( i 1 2+ i 2 2+ i in )/n If we take the square root of this, we have a value of current that can be used in the power formula. This is the ROOT MEAN SQUARE or r.m.s. value. I(r.m.s.) = ( i 1 2+ i 2 2+ i in )/n It can be shown by the use of calculus that the r.m.s. value of a sinusoidal wave form is V m / 2. We use r.m.s. values with a.c. so that we may treat some calculations the same as for d.c. When you use a voltmeter or ammeter with a.c., the values indicated are r.m.s. values. V rms = V m / 2 = 0.707V m SELF ASSESSMENT EXERCISE No The periodic time of an ac voltage is s. Calculate the frequency. (500 Hz) 2. The r.m.s. value of mains electricity is 240 V. Determine the peak voltage (amplitude). (339.5V) 3. An a.c. current varies between plus and minus 5 amps. Calculate the r.m.s. value. (3.535 A) 4. An electric fire produces 2 kw of heat from a 240 V r.m.s. supply. Determine the r.m.s. current and the peak current. (8.33 A and A) 5. An electric motor is supplied with 110 V r.m.s. at 60 Hz and produces 200W of power. Determine the periodic time, the r.m.s. current and the peak current. (16.7 ms, 1.82 A and A) D.J.Dunn 5
6 1.3 OTHER WAVE FORMS Cyclic variations may take many forms such as SQUARE, SAW TOOTH and TRIANGULAR as shown below. Square waveforms are really d.c. levels that suddenly change from plus to minus. The r.m.s. value is the same as the peak value. They are typically used for digital signal transmission. Saw tooth waves are used for scanning a cathode ray tube. The electron beam moves across the screen at a constant rate and then flies back to the beginning. Triangular waves change at a constant rate first in one direction and then the other. SELF ASSESSMENT EXERCISE No Work out the average and form factor figures for a square wave. 2. A triangular voltage has a peak value of 15 V. Work out the average value, the form factor and the r.m.s. value. Note that shape of the triangle does not make a difference so you can assume a right angle triangle to make it easier. If you cannot do the maths try plotting and working out the areas by a graphical method. (7.5V, 2 and 10.6 V) D.J.Dunn 6
7 2. REACTANCE AND IMPEDANCE Capacitors and Inductors have a property called Reactance denoted with an X. On their own they may be used with a form of Ohm s Law such that V/I = X Both V and I are r.m.s. values. The value of X depends on the frequency of the a.c. and this is why they are called REACTIVE. It should be noted that a pure capacitor and inductor does not lose any energy. A resistor on the other hand, produces resistance by dissipating energy but the value of R does not change with frequency so a resistor is a PASSIVE component. When a circuit consists of Resistance, Capacitance and Inductance, the overall impedance is denoted with a Z. The units of R, X and Z are Ohms. 2.1 CAPACITIVE REACTANCE X C When an alternating voltage is applied to a capacitor, the capacitor charges and discharges with each cycle. This means that alternating current flows in the circuit but not across the dielectric. If the frequency of the voltage is increased the capacitor must charge and discharge more quickly so the current must increase with the frequency. The r.m.s. current is directly proportional to the r.m.s. voltage V, the capacitance C and the frequency f. It follows that I rms = Constant x V rms x f x C V rms /I rms = 1/(constant x f C) The constant is 2 so V rms /I rms = X C = 1/(2 f C) Note that when f = 0, X C is infinite and when f is very large X C tends to zero. This means that a pure capacitor presents a total barrier to d.c. but the impedance to a.c. gets less and less as the frequency goes up. This makes it an ideal component for separating d.c. from a.c. If we put in a combined a.c. + d.c. signal as shown, we get out pure a.c. but with a reduced amplitude depending on the reactance. WORKED EXAMPLE No V r.m.s. applied across a capacitance of 4.7 μf. Calculate the r.m.s. current when the frequency is 20 Hz, 200 Hz and 2000 Hz SOLUTION 20 Hz 1 1 V 15 XC 1693 Ω Irms A 6 2 π f C 2 π x 20 x 4.7 x 10 XC Hz 1 1 V 15 XC Ω Irms A 6 2 π f C 2 π x 200 x 4.7 x 10 XC Hz 1 1 V 15 XC Ω Irms A 6 2 π f C 2 π x 2000 x 4.7 x 10 X 1693 C D.J.Dunn 7
8 2.2 INDUCTIVE REACTANCE X L The back e.m.f. produced by a varying current is e =  L x rate of change of current. In order to overcome the back e.m.f., a forward voltage equal and opposite is required. Hence in order to produce alternating current, an alternating voltage is needed. It can be shown that the r.m.s. voltage needed to produce an r.m.s. current is directly proportional to the current, the inductance and the frequency so that V rms = I rms (2fL) Hence V rms /I rms = X L = 2 f L Ohms Note that the reactance is zero when f = 0 and approaches infinity when f is very large. This means that a pure inductor has no impedance to d.c. but the impedance to a.c. increases directly proportional to frequency. This is the opposite affect to that of a capacitor and an inductor may be used to reduce the a.c. component of a combined a.c. and d.c. signal as illustrated. WORKED EXAMPLE No V r.m.s. applied across an inductance of 4 μh. Calculate the r.m.s. current when the frequency is 200 Hz, 200 khz and 200 MHz SOLUTION 6 V Hz XC 2π fl 2π x 20 x 4x mω Irms ka 3 X x 10 6 V khz XC 2π fl 2π x 200 x 4x Ω Irms A X V MHz XC 2π fl 2π x 2000 x 4x Ω Irms 2.98 ma X 5027 C C C D.J.Dunn 8
9 SELF ASSESSMENT EXERCISE No Calculate the reactance of a capacitor with capacitance 60 F at a frequency of 50 Hz. (53 Ω) 2. The Voltage applied to the capacitor is 110 V r.m.s. Calculate the r.m.s. current. (2.073 A) 3. A capacitor is put in a circuit to limit the r.m.s. current to 2 ma when 10 V r.m.s. at 60 Hz is placed across it. What should the value of the capacitance be? (530 nf) 4. Calculate the r.m.s. current in an inductor of 60 mh when 110 V r.m.s. is applied at 60 Hz. (4.863 A) 5. An inductor passes 20 ma rms at 12 V r.m.s. and 1000 Hz. Calculate the inductance. (95 mh) 6. Calculate the inductance of a coil 25 mm diameter, 100 mm long with 30 turns. The core has a relative permeability of ( H) Calculate the energy stored when 10 A d.c. flow. (0.555 J) Calculate the reactance for ac with a frequency of 100 Hz. (6.97 ) Calculate the r.m.s. voltage needed to make 10 A r.m.s. flow. (69.7V rms) D.J.Dunn 9
10 3. PHASOR DIAGRAMS The way a sinusoidal voltage or current varies with time is given by the following equations. v = V sin (ωt) or v = V sin (2πf t) i = I sin (ωt) or i = I sin (2πf t) V and I is the amplitude and v and i are the instantaneous values at time t. f is the frequency in Hz. ω is the angular frequency in radian/s ω = 2πf. ωt = θ and this is an angle in radian. Consider a phasor representing a sinusoidal voltage. It is a vector of length V that can be drawn at an angle θ = ωt to represent the voltage at any instant in time t. If they were drawn in succession then they would be rotating anticlockwise (positive) at ω rad/s. Starting from the horizontal position after a time t it will have rotated an angle θ = ωt. The vertical component of the phasor is v = V sin (θ) This corresponds to the value of the sinusoidal graph at that angle. When θ = π/2 radian (90 o ) the peak value is V so V is the amplitude or peak value (not the r.m.s.value). You will find some animated phasor operations at the following addresses. Requires Microsoft excel to run ADDING AND SUBTRACTING Phasors can be added or subtracted from each other in the same way as vectors. Consider two voltage phasors with the same angular frequency. The diagram shows two such phasors V 1 and V 2 but V 2 lags V 1 by angle d. This is the PHASE DIFFERENCE BETWEEN THEM. We can express the vectors as: v 1 = V 1 sin(θ) and v 2 = V 2 sin(θ  d). D.J.Dunn 10
11 If we add the two vectors we do it as shown. It is normal to draw one of the phasors horizontal as the phase difference is the same at all angles. The easiest way to add vectors is to resolve them into horizontal and vertical components. v 2 has a vertical component V 2 sin d and a horizontal component V 2 cos d The vertical component of v 2 is V 2 sin d The vertical component of v is V 2 sin d The horizontal component of v is V 1 + V cos d WORKED EXAMPLE No. 3 If v 1 = 10 sin (θ) and v 2 = 7 sin (θ  30 o ) what is the resultant voltage? SOLUTION Let θ = 0 o when we add them. (V 1 in the horizontal position). The vertical component of v is v 2 sin d = 7 sin (30 o ) = 3.5 (down). The horizontal component of v is V 1 + V 2 cos d = sin 30 o = The resultant voltage is ( ) = The phase angle is = tan 1 (3.5/16.06) =12.3 o We can express the phasor as v = sin (θ 12.3 o ) D.J.Dunn 11
12 4. CIRCUITS CONTAINING RESISTIVE AND REACTIVE COMPONENTS A.C. AND RESISTANCE When a.c. is applied to a pure resistance R, Ohm s Law applies and since it is passive it is the same at all frequencies at all moments in time. The phasors for voltage and current must rotate together. They are said to be IN PHASE AC AND INDUCTANCE This requires a basic knowledge of differentiation. The voltage required to drive a current through an inductor is v = L x rate of change of current. L is the inductance in Henries. Suppose i = I sin t The rate of change of current is obtained by differentiating di/dt = I cos(t) It follows that v = I L cos(t) and the maximum value is V = I L If V and I are plotted together we see that that V is ¼ cycle displaced and it is said that the voltage leads the current by 90 o. The voltage phasor is 90 o anticlockwise of the current phasor. D.J.Dunn 12
13 4.3 AC WITH RESISTANCE AND INDUCTANCE Now consider a.c. applied to a resistor and inductor in series as shown. The current I flows through both so this is used as the reference. The voltage over the resistance is V R = I R and on the phasor diagram this must be in the same direction as the current. The voltage over the inductor is V L = I X L and this must lead the current by 90 o and also V R by 90 o. It is not true to say that V = V L + V R because they must be treated as phasors or vectors. The resultant voltage is V S and this is the hypotenuse of a rightangled triangle so V The angle is called the phase angle and is always measured from V R. It follows that = tan 1 (V L /V R ) 2 2 V R VL D.J.Dunn 13
14 4.4. AC AND CAPACITANCE Thgis requires knowledge of integration. When a.c. is applied across a capacitor, the voltage is given by the equation v C = q/c where q is the charge stored and C is the capacitance in Farads. Since q idt then v C idt C i varies sinusoidally so that i = I sin (t) idt I ω cosω t I ω Substitute and v C cosω t C The maximum value of v C is I/c so this will be the length of the phasor representing V C. If we plot V C and I we find that V C lags the current by ¼ cycle or 90 o. This is opposite to an inductor which leads by 90 o AC WITH RESISTANCE AND CAPACITANCE Now consider a resistor and capacitor in series as shown. The voltage over the resistance is I R and on the phasor diagram this must be in the same direction as the current. It follows that V C lags V R by 90 o. It is not true to say that V s = V C + V R because they must be treated as vectors. The resultant voltage is the hypotenuse of a rightangled triangle so V s V 2 R V The angle is called the phase angle and is always measured from V R. It follows that = tan 1 (V c /V R ) The only difference between this and the R L circuit is that V C lags V R and V L leads V R. This means that V L and V C are 180 o out of phase in a series circuit. 2 C D.J.Dunn 14
15 4.6. R L C IN SERIES The 3 voltages V R V L and V C are drawn as 3 phasors and the vector sum is found. It is convenient to draw V R horizontally but the vector diagram stays the same for all angles of rotation. Examining the small triangle, we see the vertical height is V L  V R and the horizontal length is V R. It follows that the resultant voltage is given by 2 2 V s VL Vc V and 1 V R L VC tan VR 4.7. REACTANCE AND IMPEDANCE REVISITED We know from previous studies that the relationship between current and voltage for any component is related as a ratio X = V/I. For a resistor this ratio is resistance R but for an inductor it is called inductive reactance X L and for a capacitor capacitive reactance X C. Inductive reactance increases with frequency and is given by X L = 2fL Capacitive reactance decreases with frequency and is given by X C = 1/ 2fC When current flows in a RLC circuit, the relationship between it and the resulting voltage is called the IMPEDANCE Z. Z = V/I where V and I are the resulting r.m.s. volts and current. Since reactance is V/I it follows that it is also a phasor. The phasor diagram for a series R L C circuit may be drawn as shown with R drawn horizontally to make it easier XL XC Z X L XC R and tan R D.J.Dunn 15
16 WORKED EXAMPLE No. 4 A resistor of value 470 is connected in series with a capacitor of 22 F and an inductor of 50 mh and a voltage is applied across it. A current of 100 ma (rms) is produced. Determine the impedance, the phase angle between the voltage and current and the applied voltage when the frequency is 50 Hz SOLUTION X L = 2πfL = 2π x 50 x 50 x 103 =15.71 Ω X C = 1/2πfC = 1/(2π x 50 x 22 x 106 ) =144.6 Ω Z X X R Ω L X C X R L C 1 o tan tan 15.3 V S = I Z = 0.1 x = 48.7 V rms SELF ASSESSMENT EXERCISE No A resistor of value 4 is connected in series with a capacitor of 47 F and an inductor of 20 μh and a voltage is applied across it. A current of 50 ma (r.m.s.) is produced. Determine the impedance, the phase angle between the voltage and current and the applied voltage when the frequency is 100 Hz. (34 Ω, o and 1.7 V) 2. A resistor of value 0.2 is connected in series with a capacitor of 4.7 F and an inductor of 5 mh and 0.5 V r.m.s. is applied across the ends. Determine the impedance, the phase angle between the voltage and current and the rms current when the frequency is 1000 Hz. (2.455 Ω, o and 204 ma) D.J.Dunn 16
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