ENGINEERING COUNCIL CERTIFICATE LEVEL ENGINEERING SCIENCE C103 TUTORIAL 18 ALTERNATING CURRENT

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1 ENGINEERING OUNIL ERTIFIATE LEVEL ENGINEERING SIENE 03 TUTORIAL 8 ALTERNATING URRENT On completion of this tutorial you should be able to do the following. Explain alternating current. Explain Root Mean Square (rms) values. Describe basic ac wave forms. Define Reactance and Impedance. Define and calculate apacitive Reactance. Define and calculate Inductive Reactance. Explain the use of apacitors and Inductors for filtering ac and dc components. D.J.Dunn

2 . REVISION OF BASI A.. THEORY Students already familiar with basic a.c. theory should go straight to the next section. A pure alternating current or voltage varies with time sinusoidally as shown. It is possible to obtain a.c. with other graph shapes such as square or saw tooth. 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. Hz cycle/second The time it takes to complete cycle is T seconds (the periodic time). It follows that T /f The maximum value of volts or current is called the peak volts or current and this is the amplitude of the wave form. The peak to peak value is double the amplitude as shown on the diagram. The arithmetic average of the volts and current over cycle is zero. However, if a.c. is applied to an electric fire for instance, electric energy is changed into heat on both the positive and negative halves of the cycle so the electric power is not zero. If direct current had been used, the electric power would be calculated from E.P. Volts x urrent. In the case of a.c., clearly we cannot use average volts and currents to do this calculation as the answer would always be zero. ROOT MEAN SQUARE VALES (RMS) 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 and has a mean height. D.J.Dunn 2

3 The mean height may be obtained by placing many vertical ordinates on it as shown. Taking a graph of current with many ordinates i 2, i i n 2. Mean value of i 2 ( i 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 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 of the peak value. We use r.m.s. values with a.c. so that we may treat 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. SELF ASSESSMENT EXERISE No.. The periodic time of an ac voltage is s. alculate bthe 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. alculate 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.79 A) 5. An electric motor is supplied with 0 V r.m.s. at 60 Hz and produces 200W of power. Determine the periodic time and the peak current. (6.7 ms and 2.57 A) D.J.Dunn 3

4 2. OMPLEX WAVEFORMS FUNDAMENTAL FREQUENY A sinusoidal voltage or current is described by the mathematical formula v V sin ωt or i I sin ωt ω is the angular frequency in radians/s and this is related to frequency f Hz by the formula ω 2πf The sinusoidal voltage formula is then v V sin(2πft) In this formula f is the fundamental frequency. HARMONIS A harmonic is a multiple of the fundamental frequency. 2f is the second harmonic. 3f is the third harmonic nf is the n th. harmonic. SYNTHESISING OMPLEX WAVES Waveforms with shapes that are not sinusoidal may be synthesised from one common sinusoidal waveform. The proof of this is not given here but the following is mathematically correct. This graph shows the result of adding the first and third harmonic with equal amplitudes. In reality the amplitude of the harmonic is likely to be less than the amplitude of the fundamental. This graph shows the affect of adding the third harmonic with /3 of the amplitude. GENERATION OF HARMONIS Harmonics are generated when a sinusoidal signal passes through a non-linear amplifier. An ideal amplifier increases a sinusoidal signal perfectly. D.J.Dunn 4

5 SQUARE WAVES Square waveforms are really d.c. levels that suddenly change from plus to minus. It can be shown that the following formula relates voltage and time. The formula is an infinite series. V V V v Vsin(ω t) + sin3(ω t) + sin5(ω t) + sin7(ω t) TRIANGULAR WAVES It can be shown that the following formula relates voltage and time. The formula is an infinite series. V V V v Vsin(ω t) + sin(3ω t π) + sin(5ω t) + sin(7ω t π) Note that in this series, a phase shift of π radians is added to each 3. REATANE AND IMPEDANE apacitors 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, apacitance and Inductance, the overall impedance is denoted with a Z. The units of R, X and Z are Ohms. 3. APAITIVE REATANE X 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 and the frequency f. Hence I rms constant x V x x frequency It follows that I rms onstant x V x f x V/I /(constant x f ) The constant is 2π so V/I X /(2 π f ) D.J.Dunn 5

6 Note that when f 0, X is infinite and when f is very large X 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. 5 V r.m.s. applied across a capacitance of 4.7 µf. alculate the r.m.s. current when the frequency is : 20 Hz, 200 Hz and 2000 Hz SOLUTION 20 Hz X 2 π f 2 π x 20 x 4.7 x Ω V Irms X Hz X -6 2 π f 2 π x 200 x 4.7 x Hz X 2 π f 2 π x 2000 x 4.7 x Ω V Irms X Ω V Irms X A A A SELF ASSESSMENT EXERISE No. 2. alculate the reactance of a capacitor with capacitance 60 µf at a frequency of 50 Hz. (53 Ω) 2. The Voltage applied to the capacitor is 0 V rms. alculate the rms current. (2.073 A) 3. A capacitor is put in a circuit to limit the r.m.s. current to 2 ma when 0 V r.m.s. at 60 Hz is placed across it. What should the value of the capacitance be? (530 nf) D.J.Dunn 6

7 3.2 INDUTIVE REATANE X L The back e.m.f. produced by a varying current is e - L di/dt and an equal and opposite e.m.f. or voltage is needed to make the current change in the circuit. In order to overcome the back emf, a forward voltage equal and opposite is required. Hence in order to produce alternating current, an alternating applied 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 I (2πfL) Hence V/I X L 2 π 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. 2 5 V r.m.s. applied across an inductance of 4 µh. alculate the r.m.s. current when the frequency is : 200 Hz, 200 khz and 200 MHz SOLUTION 6 V 5 20 Hz X 2π fl 2π x 20 x 4x mω Irms 3 X x 0 6 V khz X 2π fl 2π x 200 x 4x0 5.03Ω Irms A X V MHz X 2π fl 2π x 2000 x 4x Ω Irms 2.98 ma X ka D.J.Dunn 7

8 SELF ASSESSMENT EXERISE No. 3. alculate the rms current in an inductor of 60 mh when 0 V rms is applied at 60 Hz. (4.863 A) 2. An inductor passes 20 ma rms at 2 V rms and 000 Hz. alculate the inductance. (95 mh) 3. alculate the inductance of a coil 25 mm diameter, 00 mm long with 30 turns. The core has a relative permeability of (0.0 H) alculate the energy stored when 0 A d.c. flow. (0.555 J) alculate the reactance for ac with a frequency of 00 Hz. (6.97 Ω) alculate the rms voltage needed to make 0 A rms flow. (69.7V rms) 3.3 FILTERS You have seen that capacitors and inductors produce reactance that varies with frequency. If we use a suitable circuit containing both we can filter out unwanted components of mixed signals. The diagram shows a LOW and HIGH PASS FILTER. In order to understand such circuits thoroughly we need to move on to the next tutorial and study phasor diagrams. WORKED EXAMPLE No.3 alculate the ratio v o /v i in db for the circuit shown. SOLUTION x c /2πf /(2π x 50 x x 0-6 ) 383 Ω z R + x c 483 Ω i v i / x 0-6 v i v o i x c (239 x 0-6 v i ) v i v o /v i 0.76 In db v o /v i 20 log(0.76) db D.J.Dunn 8

9 WORKED EXAMPLE No.4 alculate the ratio v o /v i in db for the circuit shown. SOLUTION x c /2πf /(2π x 50 x x 0-6 ) 383 Ω for both capacitors. z 2 R + x c 483 Ω /z /z 2 + /x c z 808 Ω z t R + z 2808 Ω i v i / x 0-6 v i v v i ir v i (356. x 0-6 v i )000 v i ( 0.356) v i i v /x c v i / x 0-6 v i i 2 v /z v i / x 0-6 v i i 2 i i 356. x 0-6 v i x 0-6 v i 53.9 x 0-6 v i v o i 2 x c (53.9 x 0-6 v i ) v i or v o v i 2 R v i (53.9 x 0-6 v i ) v i v o /v i 0.49 In db v o /v i 20 log(0.49) -6.2dB SELF ASSESSMENT EXERISE No. 4 alculate the ratio v o /v i in db for the circuit shown. Answer -8.3 db D.J.Dunn 9