Revision of Basic A.C. Theory 1
Revision of Basic AC Theory (Much of this material has come from Electrical & Electronic Principles & Technology by John Bird) Electricity is generated in power stations and distributed by vast networks of transmission lines. The electricity generated is in alternation current (a.c.) form since it is cheaper to produce and easier to distribute. In this first section, we take the opportunity to revise basic a.c. theory. A pure alternation current and voltage varies sinusoidally as shown below, however it should be noted that it is also possible to obtain a.c. with other graph shapes such as square of saw tooth. The peak to peak value changes f times per second where f is known as the frequency in cycles per second or hertz. 1 cycle/s = 1 Hz The arithmetic mean of the alternating voltage or current over one cycle is zero, however if the a.c. is applied to a heater it will continue heating over the positive and negative part of the cycle. We cannot use this mean value to calculate power using the familiar P = V2 or P = R I2 R, as we did with d.c. values since the result would be zero. Instead we use an effective value, ie a value which would produce the same heating effect as a d.c. value. This value is known as the root mean squared (r.m.s.) value. No proof is given here, but for a pure sinusoid the r.m.s. value is 0.707 of the peak value. Worked Example 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. P = V2 R therefore R = v2 P = 2402 2 10 3 = 28.8Ω Also P = I 2 R therefore I = P R = 2 103 28.8 = 8.33A r. m. s 2
Hence peak current 8.33 0.707 = 11.78A Equations of a Sinusoid In the diagram below a vector OA of length OB is rotating about O with an angular velocity ω rad/s. Such a rotating vector is also known as a phasor and we will see later that phasors are very useful in a.c. circuit analysis. You will notice that after a time t seconds the vector has moved through ωt radians. If the vertical of the rotating vector is projected onto a graph of y against ωt the result is a sine curve have the equation y = OB sin ωt. It is worth noting at this point that any quantity that varies sinusoidally can be represented as a phasor. You will notice in the diagram below that not all sinewaves start at 0 and when this is the case it becomes necessary to represent the function in the form y = sin(ωt ± ϕ) where ϕ is known as a phase difference. This is illustrated below together with the opportunity to illustrate the meaning of lead and lag. In (a) and (b) we see that y 2 leads y 1 by ϕ and in (c) and (d) y 4 lags y 3 by ϕ. General Points of Interest: For a general sinusoidal voltage given by v = V m sin(ωt ± ϕ) Vm is the amplitude or maximum value 2Vm is the peak to peak value ω is the angular velocity (or angular frequency) in rad/s T = 2π is the periodic time in seconds ω 3
f = ω is the frequency in Hz 2π Of course similar equations apply to sinusoidal current. Worked Example An alternating voltage is given by the equation v = 75 sin(200πt 0.25) volts. Determine (a) the amplitude, (b) the peak to peak value, (c) the r.m.s. value, (d) the periodic time, (e) the frequency, and (f) the phase angle in degrees, relative to 75sin 200ωt. (a) We have the general expression v = V m sin(ωt ± ϕ). Comparing with the above equation we see that the amplitude = Vm = 75v (b) Peak to peak value is 2 x 75 = 150v (c) R.M.S value is 0.707 x Vm = 0.707 x 75 = 53.02v (d) Periodic time T = 2π ω = 2π 200π = 1.0 10 2 = 10ms (e) Frequency f = ω 2π = 200π 2π = 100Hz (f) Phase angle ϕ = 0.25 rads = 0.25 180 π = 14.32 lagging 75sin 200ωt Combination of Waveforms It sometimes becomes necessary to add or subtract waveforms and this can be done a number of ways. 1) Plotting the periodic functions graphically 2) Resolution of phasor by drawing or calculation Method 2 (Drawing ) by Example Two alternating voltages are given by v 1 = 50 sin ωt volts and v 2 = 100 sin (ωt π 6 ) volts. Draw the phasors and determine by measurement the expression represented by v 1 + v 2. It is normal to select one of the phasors to be a reference phasor and to draw this at time t = 0. v1 = 50v 0 a 30 o (= π 6) v2 = 100v lagging by 30 o b Two ways of solving this are shown below 4
v1 = 50v 0 a vr v2 = 100v b Alternatively draw a parallelogram as shown v1 = 50v 0 a vr v2 = 100v b Method 2 Calculate The calculation method is either achieved by the use of the cosine rule or by using complex numbers. In these notes, only the complex method will be considered. Brief recap of Complex Numbers Complex numbers are of the form a + jb and this is called Rectangular Form. Such a complex number may be represented on an Argand Diagram. Imaginary or j-axis b m ϕ a (a + jb) Real axis From the argand diagram, we see that (a + jb) = m cos ϕ + mj sin ϕ = m(cos ϕ + j sin ϕ Where m is the modulus and is equal to a 2 + b 2 And ϕ is the argument and equal to tan 1 b a An alternative form is Polar Form which is given by m(cos ϕ = j sin ϕ) = m ϕ 5
Addition and Subtraction of Complex Numbers If z 1 = (2 + j3) and z 2 = (4 J2) Then z 1 + z 2 = (2 + j3) + (4 j2) = 2 + j3 + 4 j2 = 6 + j Similarly, z 1 z 2 = (2 + j3) (4 j2) = 2 + j3 4 + j2 = 2 + j5 Going back to the question in hand v 1 = 50 sin ωt and v 2 = 100 sin (ωt π 6 ) Expressing v 1 + v 2 in polar form v 1 + v 2 = 50 0 + 100 - π/6 = 50 0 + 100-30 Now to convert polar to rectangular a = m cos ϕ and b = m sin ϕ v 1 + v 2 = (50 + j0) + (86.6 j50) = 50 + 86.6 j50 = 136.6 j50 Converting back to polar form m = 136.6 2 + 50 2 = 145.46 1 50 φ = tan = 136.6 20.1o = 20.1 π = 0.35 rads 180 So, in conclusion v 1 + v 2 = 145.46 sin(ωt 0.35) volts Tutorial Problems 1) The instantaneous value of voltage in an a.c. circuit at any time t seconds is given by v = 100 sin(50πt 0.523) volts. Determine: (a) The peak to peak voltage, the frequency, the periodic time and the phase angle. (b) The voltage when t = 0 (c) The voltage when t = 8ms (d) The times in the first cycle when the voltage is 60v (e) The first time when the voltage is a maximum. (200v, 25Hz, 0.04s, 29.97 o lagging, -49.95v, 66.96v, 7.426ms, 19.23ms,13.33ms) 2) The instantaneous values for two alternating currents are given by i 1 = 20 sin ωt amperes and i 2 = 10 sin(ωt + π 3) amperes. Determine the sinusoidal expression for (i 1 + i 2 ) by calculation. (26.46sin(ωt + 0.333) amperes) 6
3) Two alternating voltages are given by v 1 = 120 sin ωt volts and v 2 = 200 sin(ωt π 4) volts. Obtain the sinusoidal expression for v 1 v 2 by calculation. (143sin(ωt + 1.73) volts) 7
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