EDEXCEL NATIONAL CERTIFICATE/DIPLOMA UNIT …

1 1 EDEXCEL NATIONAL CERTIFICATE/DIPLOMA unit 5 - electrical AND ELECTRONIC PRINCIPLES NQF LEVEL 3 OUTCOME 4 - ALTERNATING CURRENT 4 Understand single-phase alternating current (ac) theory Single phase AC circuit theory: waveform characteristics sinusoidal and non-sinusoidal waveforms, amplitude, period time, frequency, instantaneous, peak/peak-to-peak, root mean square ( ), average values, form factor; determination of values using phasor and algebraic representation of alternating quantities 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/peak-to-peak, and average values); circuits half and full wave rectifiers 2 1. REVISION OF BASIC THEORY SINUSOIDAL WAVE FORMS A pure alternating current or voltage varies with time sinusoidally as shown.

2 INSTANTANEOUS VALUES For a sinusoidal voltage and current the instantaneous value at any moment in time is given by:- v = V sin ( t) and i = I sin ( t) or v = V sin (2 f t) and i = I sin (2 f 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. 1 Hz = 1 cycle/second PERIODIC TIME 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 360o or 2 radian.

3 It follows that f cycles/second = 2 f radian/s and this is the angular frequency. = 2 f = 2 /T rad/s 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 x average value hence the average value is 2/ = If the peak value is something other than 1 then the average is x peak value FORM FACTOR This is defined as valueAverageValuePeak Factor Form Hence for a sinusoidal voltage or current the form factor is /2 = PHASE and DISPLACEMENT The sinusoidal graph is produced because we made = 0 when t = 0.

4 We could choose to make any value at t = 0. We would then write the equations as: x = A sin( + ) or v = V sin( + ) where is the phase angle. The plot shown has x = 0 at = 30o so it follows that (30 + ) = 0 and so = -30o. If we made = 90o 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 ( ) voltage so that V = Vdc + V sin( + ) 4 SELF ASSESSMENT EXERCISE No. 1 1 Mains electricity has a frequency of 50 Hz. What is the periodic time and angular frequency? ( 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 = s? ( V) 4. An alternating current is given by the equation I = 5 sin(600t). Determine the following.

5 I. the frequency ( Hz) ii. the periodic time ( ms) iii. the average value. ( 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, s, 4 rad/s, Hz, radian or ) 5 ROOT MEAN SQUARE VALUES ( ) 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 = I2R or = V2/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 we must use the average value of V2 or I2 and these are not zero.

6 The diagram shows how a plot of V2 or I2 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 i12, i22 ..in2. The mean value of the i2 is: ( i12+ i22+ 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 value. I( ) = ( i12+ i22+ in)/n It can be shown by the use of calculus that the value of a sinusoidal wave form is Vm/ 2. We use values with so that we may treat some calculations the same as for When you use a voltmeter or ammeter with , the values indicated are values. Vrms = Vm/ 2 = SELF ASSESSMENT EXERCISE No. 2 1. The periodic time of an ac voltage is s. Calculate the frequency. (500 Hz) 2. The value of mains electricity is 240 V. Determine the peak voltage (amplitude).

7 ( ) 3. An current varies between plus and minus 5 amps. Calculate the value. ( A) 4. An electric fire produces 2 kW of heat from a 240 V supply. Determine the current and the peak current. ( A and A) 5. An electric motor is supplied with 110 V at 60 Hz and produces 200W of power. Determine the periodic time, the current and the peak current. ( ms, A and A) 6 OTHER WAVE FORMS Cyclic variations may take many forms such as SQUARE, SAW TOOTH and TRIANGULAR as shown below. Square waveforms are really levels that suddenly change from plus to minus. The 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.

8 3 1. 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 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. ( , 2 and V) 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 values. The value of X depends on the frequency of the 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.

9 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. CAPACITIVE REACTANCE XC 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 current is directly proportional to the voltage V, the capacitance C and the frequency f. It follows that Irms = Constant x Vrms x f x C Vrms/Irms = 1/(constant x f C) The constant is 2 so Vrms/Irms = XC = 1/(2 f C) Note that when f = 0, XC is infinite and when f is very large XC tends to zero. This means that a pure capacitor presents a total barrier to but the impedance to gets less and less as the frequency goes up.

10 This makes it an ideal component for separating from If we put in a combined + signal as shown, we get out pure but with a reduced amplitude depending on the reactance. WORKED EXAMPLE No. 1 15 V applied across a capacitance of F. Calculate the current when the frequency is 20 Hz, 200 Hz and 2000 Hz SOLUTION 20 Hz A 169310 x x 20 x 21C f 21XC6-C 200 Hz A x x 200 x 21C f 21XC6-C 2000 Hz A x x 2000 x 21C f 21XC6-C 8 INDUCTIVE REACTANCE XL The back produced by a varying current is e = - L x rate of change of current. In order to overcome the back , 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 voltage needed to produce an current is directly proportional to the current, the inductance and the frequency so that Vrms = Irms (2 fL) Hence Vrms/Irms = XL = 2 f L Ohms Note that the reactance is zero when f = 0 and approaches infinity when f is very large.