CHAPTER 4 ALTERNATING VOLTAGES AND CURRENTS Exercise 77, Page 28. Determine the periodic time for the following frequencies: (a) 2.5 Hz (b) 00 Hz (c) 40 khz (a) Periodic time, T = = 0.4 s f 2.5 (b) Periodic time, T = = 0.0 s or 0 ms f 00 (c) Periodic time, T = f 400 = 25 s 2. Calculate the frequency for the following periodic times: (a) 5 ms (b) 50 s (c) 0.2 s (a) Frequency, f = = 200 Hz or 0.2 khz T 5 0 (b) Frequency, f = 6 T 500 = 20 khz (c) Frequency, f = = 5 Hz T 0.2. An alternating current completes 4 cycles in 5 ms. What is its frequency? Time for one cycle, T = 5 ms =.25 ms 4 Hence, frequency, f = T.250 = 800 Hz John Bird Published by Taylor and Francis 54
Exercise 78, Page 22. An alternating current varies with time over half a cycle as follows: Current (A) 0 0.7 2.0 4.2 8.4 8.2 2.5.0 0.4 0.2 0 time (ms) 0 2 4 5 6 7 8 9 0 The negative half cycle is similar. Plot the curve and determine: (a) the frequency (b) the instantaneous values at.4 ms and 5.8 ms (c) its mean value, and (d) its r.m.s. value. The graph is shown plotted below. (a) Periodic time, T = 2 0 ms = 20 ms, hence, frequency, f = T 200 = 50 Hz (b) At.4 ms, current, i = 5.5 A and at 5.8 ms, i =. A (c) Mean value = area under curve length of base Using the mid-ordinate rule, area under curve = 0 0..4. 6.0 8.8 5.5.6 0.8 0. 0.2 = 0 28 28 0 John Bird Published by Taylor and Francis 55
Hence, mean value = 280 00 = 2.8 A (d) r.m.s. value = 0..4. 6.0 8.8 5.5.6 0.8 0. 0.2 0 2 2 2 2 2 2 2 2 2 2 = 58.68 0 =.98 A or 4.0 A, correct to 2 significant figures. 2. For the waveforms shown below, determine for each (i) the frequency (ii) the average value over half a cycle (iii) the r.m.s. value (iv) the form factor (v) the peak factor. (a) (b) (c) (d) (a) (i) T = 0 ms, hence, frequency, f = T 00 = 00 Hz (ii) Average value = 5 0 5 area under curve 2 length of base 50 = 2.50 A John Bird Published by Taylor and Francis 56
(iii) R.m.s. value = i i i i i 5 2 2 2 2 2 2 4 5 = 0.5.5 2.5.5 4.5 5 2 2 2 2 2 = 2.87 A (iv) Form factor = r.m.s. 2.87 =.5 average 2.50 (v) Peak factor = max imum value 5 =.74 r.m.s. 2.87 (b) (i) T = 4 ms, hence, frequency, f = = 250 Hz T 4 0 (ii) Average value = area under curve 20 2 = 20 V length of base 2 (iii) R.m.s. value = v v v v 4 2 2 2 2 2 4 = 20 20 20 20 4 2 2 2 2 = 20 V (iv) Form factor = r.m.s. 20 =.0 average 20 (v) Peak factor = max imum value 20 =.0 r.m.s. 20 (c) (i) T = 8 ms, hence, frequency, f = = 25 Hz T 8 0 (ii) Average value = 24 2 24 24 area under curve 2 2 72 = 8 A length of base 4 4 (iii) R.m.s. value = = 2 2 2 2 i i2 i i 4... 8 9 5 2 24 24 24 24 8 2 2 2 2 2 2 2 2 = 9.56 A John Bird Published by Taylor and Francis 57
(iv) Form factor = r.m.s. 9.56 =.09 average 8 (v) Peak factor = max imum value 24 =.2 r.m.s. 9.56 (d) (i) T = 4 ms, hence, frequency, f = = 250 Hz T 4 0 (ii) Average value = area under curve 0.500 = 25 V length of base 2 (iii) R.m.s. value = v v v v 4 2 2 2 2 2 4 = 0 0 00 0 4 2 2 2 2 = 50 V (iv) Form factor = r.m.s. 50 = 2.0 average 25 (v) Peak factor = max imum value 00 = 2.0 r.m.s. 50. An alternating voltage is triangular in shape, rising at a constant rate to a maximum of 00 V in 8 ms and then falling to zero at a constant rate in 4 ms. The negative half cycle is identical in shape to the positive half cycle. Calculate (a) the mean voltage over half a cycle, and (b) the r.m.s. voltage The voltage waveform is shown below. John Bird Published by Taylor and Francis 58
(a) Average value = area under curve 2 2 length of base 20 80 00 40 00 = 50 V (b) R.m.s. value = v v v v v v 6 2 2 2 2 2 2 2 4 5 6 = 7.5 2.5 87.5 262.5 225 75 6 2 2 2 2 2 2 = 70 V 4. An alternating e.m.f. varies with time over half a cycle as follows: E.m.f. (V) 0 45 80 55 25 20 20 95 0 time (ms) 0.5.0 4.5 6.0 7.5 9.0 0.5 2.0 The negative half cycle is identical in shape to the positive half cycle. Plot the waveform and determine (a) the periodic time and frequency (b) the instantaneous value of voltage at.75 ms (c) the times when the voltage is 25 V (d) the mean value, and (e) the r.m.s. value The waveform is shown plotted below. John Bird Published by Taylor and Francis 59
(a) Half the waveform is shown, hence periodic time, T = 2 2.0 ms = 24 ms Frequency, f = T 240 = 4.67 Hz (b) The instantaneous value of voltage at.75 ms = 5 V (c) The times when the voltage is 25 V = 4 ms and 0.0 ms (d) Mean value = area under curve length of base Using the mid-ordinate rule with 2 intervals, area under curve = 0 5 45 68 00 45 90 250 20 260 60 95 25 = 0 67.67.67 Hence, mean value = 20 = 9 V (e) R.m.s. value = 5 45 68 00 45 90 250 20 260 60 95 25 2 2 2 2 2 2 2 2 2 2 2 2 2 = 4749 2 = 69 V 5. Calculate the r.m.s. value of a sinusoidal curve of maximum value 00 V. R.m.s. value = 0.707 peak value = 0.707 00 = 22. V 6. Find the peak and mean values for a 200 V mains supply. 200 V is the r.m.s. value r.m.s. value = 0.707 peak value, from which, peak value = Mean value = 0.67 peak value = 0.67 282.9 = 80.2 V r.m.s. 200 = 282.9 V 0.707 0.707 7. Plot a sine wave of peak value 0.0 A. Show that the average value of the waveform is 6.7 A over half a cycle, and that the r.m.s. value is 7.07 A John Bird Published by Taylor and Francis 60
A sine wave of maximum value 0.0 A is shown below. Over half a cycle, mean value = area under curve length of base Using the mid-ordinate rule with 2 intervals, 6 area under curve =..8 6. 7.9 9.2 9.9 9.9 9.2 7.9 6..8. 2 = 76.4 = 20.0 Hence, mean value = 20.0 = 6.7 A R.m.s. value =..8 6. 7.9 9.2 9.9 9.9 9.2 7.9 6..8. 2 2 2 2 2 2 2 2 2 2 2 2 2 = 596.8 2 = 7.05 A With a larger scale and taking values to greater than decimal place, it may be shown that the r.m.s. value is 7.07 A 8. A sinusoidal voltage has a maximum value of 20 V. Calculate its r.m.s. and average values. John Bird Published by Taylor and Francis 6
R.m.s. value = 0.707 peak value = 0.707 20 = 84.8 V Average value = 0.67 peak value = 0.67 20 = 76.4 V 9. A sinusoidal current has a mean value of 5.0 A. Determine its maximum and r.m.s. values. Mean value = 0.67 maximum value, from which, maximum value = mean value 5.0 = 2.55 A 0.67 0.67 R.m.s. value = 0.707 maximum value = 0.707 2.55 = 6.65 A John Bird Published by Taylor and Francis 62
Exercise 79, Page 224. An alternating voltage is represented by v = 20 sin 57.t volts. Find (a) the maximum value (b) the frequency (c) the periodic time. (d) What is the angular velocity of the phasor representing this waveform? (a) Maximum value = 20 V (b) 57. = = 2f, from which, frequency, f = 57. 2 (c) Periodic time, T = = 0.04 s or 40 ms f 25 (d) Angular velocity = 57. rad/s = 25 Hz 2. Find the peak value, the r.m.s. value, the frequency, the periodic time and the phase angle (in degrees and minutes) of the following alternating quantities: (a) v = 90 sin 400t volts (b) i = 50 sin(00t + 0.0) amperes (c) e = 200 sin(628.4t 0.4) volts (a) Peak value = 90 V R.m.s. value = 0.707 peak value = 0.707 90 = 6.6 V 400 = = 2f, from which, frequency, f = 400 2 Periodic time, T = = 5 ms f 200 Phase angle = 0 (b) Peak value = 50 A R.m.s. value = 0.707 peak value = 0.707 50 = 5.5 A 00 = = 2f, from which, frequency, f = 00 2 = 200 Hz = 50 Hz John Bird Published by Taylor and Francis 6
Periodic time, T = = 0.02 s or 20 ms f 50 Phase angle = 0.0 radians = (c) Peak value = 200 V 80 0. = 7.9 leading R.m.s. value = 0.707 peak value = 0.707 200 = 4.4 V 628.4 = = 2f, from which, frequency, f = 628.4 2 Periodic time, T = = 0.0 s or 0 ms f 00 Phase angle = 0.4 radians = 80 0.4 = 2.49 lagging = 00 Hz. A sinusoidal current has a peak value of 0 A and a frequency of 60 Hz. At time t = 0, the current is zero. Express the instantaneous current i in the form i = I sin t. m i = 0sin2 (60)t If t = 0 when i = 0, thus 0 = 0 sin i.e. 0 = sin from which, = sin 0 0 Hence, i = 0 sin 20t A 4. An alternating voltage v has a periodic time of 20 ms and a maximum value of 200 V. When time t = 0, v = - 75 volts. Deduce a sinusoidal expression for v and sketch one cycle of the voltage showing important points. Frequency, f = T 200 = 50 Hz Hence, v = 200sin2 (50)t = 200sin 00t If t = 0 when v = - 75, thus - 75 = 200 sin from which, = 75 sin 200 = - 0.84 John Bird Published by Taylor and Francis 64
Hence, v = 200 sin(00t 0.84) volts 5. The voltage in an alternating current circuit at any time t seconds is given by v = 60 sin 40t volts. Find the first time when the voltage is (a) 20 V (b) - 0 V Voltage, v = 60 sin 40t volts (a) When v = 20 V, 20 = 60 sin 40t from which, Hence, time, t = 20 sin 40t 60 and 40t = 20 sin 60 = 0.98 0.98 40 8.496 0 s = 8.496 ms (b) When v = - 0 V, - 0 = 60 sin 40t from which, 0 sin 40t and 40t = 60 0 sin 60 Sine is negative in the rd and 4 th quadrants as shown in the diagram. 0 sin 60 = 0.526 rad and the first time this occurs is in the rd quadrant. Measuring from zero, the angle is π + 0.526 =.6652 rad Hence, time, t =.6652 0.096s = 9.6 ms 40 John Bird Published by Taylor and Francis 65
6. The instantaneous value of voltage in an a.c. circuit at an time t seconds is given by v = 00 sin(50t 0.52) V. Find: (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 = 8 ms (d) the times in the first cycle when the voltage is 60 V (e) the times in the first cycle when the voltage is 40 V, and (f) the first time when the voltage is a maximum. Sketch the curve for one cycle showing relevant points (a) Peak to peak voltage = 2 maximum value = 2 00 = 200 V 50 = = 2f, from which, frequency, f = 50 2 = 25 Hz Periodic time, T = = 0.04 s or 40 ms f 25 Phase angle = 0.52 rad lagging = 80 0.52 = 29.97 lagging or 2958 lagging (b) When t = 0, v = 00 sin[50(0) 0.52] = - 49.95 V (c) When t = 8 ms, v = 00 sin[50( 8 0 ) 0.52] = 00 sin 0.76 = 66.96 V (d) When v = 60 V, 60 = 00 sin[50t 0.52] from which, 60 00 = sin[50t 0.52] i.e. 50t 0.52 = sin 0.60 = 0.645 or - 0.645 (sine is positive in the st and 2nd quadrants, as shown) John Bird Published by Taylor and Francis 66
Hence, 50t = 0.645 + 0.52 and t = 0.645 0.52 50 = 7.426 ms or 50t = - 0.645 + 0.52 and t = (e) When v = -40 V, -40 = 00 sin[50t 0.52] 0.645 0.52 50 = 9.2 ms from which, 40 = sin[50t 0.52] 00 i.e. 50t 0.52 = sin ( 0.40) = + 0.45 or 2-0.45 (sine is negative in the rd and 4 th quadrants, as shown) Hence, 50t = + 0.45 + 0.52 and t = or 50t = 2-0.45 + 0.52 and t = 0.45 0.52 50 2 0.45 0.52 50 = 25.95 ms = 40.7 ms (f) The first time when the voltage is a maximum is when v = 00 V i.e. 00 = 00 sin[50t 0.52] i.e. = sin[50t 0.52] i.e. 50t 0.52 = from which, t = sin.5708.5708 0.52 =. ms 50 John Bird Published by Taylor and Francis 67
A sketch of v = 00 sin[50t 0.52] is shown below. John Bird Published by Taylor and Francis 68
Exercise 80, Page 227. The instantaneous values of two alternating voltages are given by v 5sin t and v2 8sin t 6. By plotting v and v 2 on the same axes, using the same scale, over one cycle, obtain expressions for (a) v + v 2 (b) v - v 2 John Bird Published by Taylor and Francis 69
(a) From the sketched graphs above, v v 2.6sin t 0.2 2 (b) From the sketched graphs above, v v 4.4sin t 2 2 2. Repeat Problem by calculation. (a) The relative positions of v and v 2 at time t = 0 are shown as phasors in diagram (i). (i) (ii) The phasor diagram is shown in diagram (ii). Using the cosine rule, 2 2 2 from which, ac = 2.58 ac 5 8 2 5 8 cos50 Using the sine rule, 8 2.58 sin sin50 from which, 8sin50 sin 0.7965 2.58 and sin 0.7965 8.54 or 0.24 radians Hence, v v 2.58sin t 0.24 2 (b) The relative positions of v and v 2 at time t = 0 are shown as phasors in diagram (iii). (iii) (iv) The phasor diagram is shown in diagram (iv). Using the cosine rule, 2 2 2 ac 5 8 2 5 8 cos0 from which, ac = 4.44 John Bird Published by Taylor and Francis 70
Using the sine rule, 8 4.44 sin sin 0 from which, 8sin 0 sin 0.90090 4.44 and sin 0.90090 64.28 or 80 64.28 5.72 From the phasor diagram, = 5.72 or 2.02 radians Hence, v v 4.44sin t 2.02 2. Construct a phasor diagram to represent i + i2 where i = 2 sin t and i2 = 5 sin(t + π/). By measurement, or by calculation, find a sinusoidal expression to represent i + i2 The phasor diagram is shown below. By drawing the diagram to scale and measuring, i R = 2.5 and ϕ = 4º or 0.59 rad By calculation, using the cosine rule, 2 2 2 i 2 5 2 2 5 cos20 R from which, i = 2.4 R Using the sine rule, 5 2.4 sin sin20 from which, 5sin20 sin 0.5544 2.4 and sin 0.5544.67 or 0.588rad Hence, i i 2.4sin t 0.588 2 John Bird Published by Taylor and Francis 7
4. Determine, either by plotting graphs and adding ordinates at intervals, or by calculation, the following periodic function in the form v V sin( t ) m 0sin t 4sin t 4 The following is determined by calculation. The relative positions of v and v 2 at time t = 0 are shown as phasors in diagram (i). (i) (ii) The phasor diagram is shown in diagram (ii). Using the cosine rule, from which, ac =.4 2 2 2 ac 0 4 2 0 4 cos5 Using the sine rule, 4.4 sin sin5 from which, 4sin5 sin 0.25.4 and sin 0.25 2.4 or 0.27 rad 4 Hence, 0sin t 4sin t.4sint 0.27 5. Determine, either by plotting graphs and adding ordinates at intervals, or by calculation, the following periodic function in the form v V sin( t ) m 80sin t 50sin t 6 The following is determined by calculation. The relative positions of v and v 2 at time t = 0 are shown as phasors in diagram (iii). John Bird Published by Taylor and Francis 72
(iii) (iv) The phasor diagram is shown in diagram (iv). Since abc is a right angled triangle, Pythagoras theorem is used. 2 2 ac 50 80 94.4 and 50 tan 2 80 Hence, in diagram (iv), = 60-2 = 28 or 0.489 rad. 6 Thus, 80sin t 50sin t 94.4sin t 0.489 6. Determine, either by plotting graphs and adding ordinates at intervals, or by calculation, the following periodic function in the form v V sin( t ) m 00sin t 70sin t The following is determined by calculation. The relative positions of v and v 2 at time t = 0 are shown as phasors in diagram (v). Since the waveform of maximum value 70 is being subtracted it phasor is reversed as shown. The phasor diagram is shown in diagram (vii). John Bird Published by Taylor and Francis 7
(v) (vi) Using the cosine rule, from which, ac = 88.88 2 2 2 ac 00 70 2 00 70 cos 60 Using the sine rule, 88.88 70 sin 60 sin from which, 70sin 60 sin 0.68206 88.88 and sin 0.68206 4 or 0.75rad Hence, 00sin t 70sin t 88.88sint 0.75 7. The voltage drops across two components when connected in series across an a.c. supply are v = 50 sin 4.2t and v 2 = 90 sin (4.2t - /5) volts respectively. Determine (a) the voltage of the supply, in trigonometric form, (b) the r.m.s. value of the supply voltage, and (c) the frequency of the supply. Cosine and sine rules or horizontal and vertical components could be used to solve this problem; however, an alternative is to use complex numbers, as shown below. (a) Supply voltage, v = v v2 = 50 sin 4.2t + 90 sin4.2t /5 = 500 90 6 = (50 + j0) + (72.8 j52.90) = 222.8 j52.90 John Bird Published by Taylor and Francis 74
= 229.6 2290.2 rad = 229 sin(4.2t 0.2) V (b) R.m.s value of supply = 0.707 229 = 6.9 V (c) = 4.2 = 2f from which, frequency, f = 4.2 2 = 50 Hz 8. If the supply to a circuit is 25 sin 628.t volts and the voltage drop across one of the components is 8 sin (628.t - 0.52) volts, calculate (a) the voltage drop across the remainder of the circuit, (b) the supply frequency, and (c) the periodic time of the supply. (a) Voltage, v2 v v 25sin 628.t 8sin(628.4t 0.52) = 250 8 0.52 rad using complex numbers = (25 + j0) (5.62 j8.944) = 9.79 + j8.944 = 2.96 0.76 rad = 2.96 sin(628.t + 0.762) V (b) = 628. = 2f from which, supply frequency, f = 628. 2 (c) Periodic time, T = = 0.0 s or 0 ms f 00 = 00 Hz John Bird Published by Taylor and Francis 75
9. The voltages across three components in a series circuit when connected across an a.c. supply are: v v 0sin 00t 6 volts, v2 40sin 00t volts and 4 50sin 00t volts. Calculate (a) the supply voltage, in sinusoidal form, (b) the frequency of the supply, (c) the periodic time, and (d) the r.m.s. value of the supply. (a) Supply voltage, v = v v2 v 0sin 00t 40sin 00t 50sin 00t 6 4 = 00 4045 5060 using complex numbers = 79.265 + j56.586 = 97.9 5.52 V or 97.9 0.620 V = 97.9sin 00 t 0.620 V (b) = 00 = 2f from which, supply frequency, f = 00 2 (c) Periodic time, T = = 0.0667 s or 6.67 ms f 50 (d) R.m.s value of supply = 0.707 97.9 = 68.85 V = 50 Hz John Bird Published by Taylor and Francis 76