CHAPTER 14 ALTERNATING VOLTAGES AND CURRENTS
Exercise 77, Page 218
1. Determine the periodic time for the following frequencies: (a) 2.5 Hz (b) 100 Hz (c) 40 kHz
(a) Periodic time, T =
1
1

= 0.4 s
f 2.5
(b) Periodic time, T =
1
1

= 0.01 s or 10 ms
f 100
(c) Periodic time, T =
1
1

= 25 s
f 40  103
2. Calculate the frequency for the following periodic times: (a) 5 ms (b) 50 s (c) 0.2 s
(a) Frequency, f =
1
1

= 200 Hz or 0.2 kHz
T 5 103
(b) Frequency, f =
1
1

= 20 kHz
T 50  10 6
(c) Frequency, f =
1
1

= 5 Hz
T 0.2
3. An alternating current completes 4 cycles in 5 ms. What is its frequency?
Time for one cycle, T =
Hence, frequency, f =
5
ms = 1.25 ms
4
1
1

= 800 Hz
T 1.25 103
© John Bird Published by Taylor and Francis
154
Exercise 78, Page 221
1. 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 1.0 0.4 0.2 0
time (ms)
0
1
2
3
4
5
6
7
8
9
10
The negative half cycle is similar. Plot the curve and determine: (a) the frequency (b) the
instantaneous values at 3.4 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  10 ms = 20 ms, hence, frequency, f =
1
1

= 50 Hz
T 20  10 3
(b) At 3.4 ms, current, i = 5.5 A
and at 5.8 ms, i = 3.1 A
(c) Mean value =
area under curve
length of base
Using the mid-ordinate rule,
area under curve = 1103   0.3  1.4  3.1  6.0  8.8  5.5  1.6  0.8  0.3  0.2 
= 1103   28   28 103
© John Bird Published by Taylor and Francis
155
Hence, mean value =
(d) r.m.s. value =
=
28 103
= 2.8 A
10 103
 0.32  1.42  3.12  6.02  8.82  5.52  1.62  0.82  0.32  0.22 


10


158.68
= 3.98 A or 4.0 A, correct to 2 significant figures.
10
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 = 10 ms, hence, frequency, f =
1
1

= 100 Hz
T 10 103
1
5 103   5 
area under curve 2 

(ii) Average value =
= 2.50 A
length of base
5 103
© John Bird Published by Taylor and Francis
156
 i 2  i 2 2  i32  i 4 2  i5 2 
(iii) R.m.s. value =  1
 =
5


(iv) Form factor =
(v) Peak factor =
r.m.s.
2.87

= 1.15
average 2.50
max imum value
5

= 1.74
r.m.s.
2.87
(b) (i) T = 4 ms, hence, frequency, f =
(ii) Average value =
1
1

= 250 Hz
T 4  10 3
area under curve 20  2

= 20 V
length of base
2
 v 2  v 2 2  v3 2  v 4 2 
(iii) R.m.s. value =  1
 =
4


(iv) Form factor =
(v) Peak factor =
 0.52  1.52  2.52  3.52  4.5 2 

 = 2.87 A
5


 202  202  202  202 

 = 20 V
4


r.m.s.
20

= 1.0
average 20
max imum value 20

= 1.0
r.m.s.
20
(c) (i) T = 8 ms, hence, frequency, f =
1
1

= 125 Hz
T 8 103
1

1

1 24    2  24    1 24 

area under curve  2

2
  72 = 18 A
(ii) Average value =

length of base
4
4
 i12  i 2 2  i32  i 4 2  .... 
(iii) R.m.s. value = 

8


=
 32  92  152  212  242  242  24 2  24 2

8


 = 19.56 A

© John Bird Published by Taylor and Francis
157
(iv) Form factor =
r.m.s. 19.56
= 1.09

average
18
(v) Peak factor =
max imum value
24

= 1.23
r.m.s.
19.56
(d) (i) T = 4 ms, hence, frequency, f =
(ii) Average value =
1
1

= 250 Hz
T 4  10 3
area under curve 0.5 100

= 25 V
length of base
2
 v 2  v 2 2  v3 2  v 4 2 
(iii) R.m.s. value =  1
 =
4


(iv) Form factor =
(v) Peak factor =
 02  02  1002  02 

 = 50 V
4


r.m.s.
50

= 2.0
average 25
max imum value 100

= 2.0
r.m.s.
50
3. An alternating voltage is triangular in shape, rising at a constant rate to a maximum of 300 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
158
1
1
8 103   300    4 103   300 

area under curve 2
2

(a) Average value =
= 150 V
length of base
12 103
 v 2  v 2 2  v3 2  v 4 2  v5 2  v 6 2 
(b) R.m.s. value =  1

6


=
 37.52  112.52 187.5 2  262.5 2  225 2  75 2 

 = 170 V
6


4. An alternating e.m.f. varies with time over half a cycle as follows:
E.m.f. (V)
0
45
80
155
215
320
210
95
0
time (ms)
0
1.5
3.0
4.5
6.0
7.5
9.0
10.5
12.0
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 3.75 ms (c) the times when the
voltage is 125 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
159
(a) Half the waveform is shown, hence periodic time, T = 2 × 12.0 ms = 24 ms
Frequency, f =
1
1

= 41.67 Hz
T 24  10 3
(b) The instantaneous value of voltage at 3.75 ms = 115 V
(c) The times when the voltage is 125 V = 4 ms and 10.0 ms
(d) Mean value =
area under curve
length of base
Using the mid-ordinate rule with 12 intervals,
area under curve = 1103  15  45  68  100  145  190  250  320  260  160  95  25
= 110 3  1673   1.673
Hence, mean value =
(e) R.m.s. value =
=
1.673
= 139 V
12  103
 152  452  682  100 2 1452 190 2  250 2  320 2  260 2 160 2 95 2  25 2 


12


341749
= 169 V
12
5. Calculate the r.m.s. value of a sinusoidal curve of maximum value 300 V.
R.m.s. value = 0.707  peak value = 0.707  300 = 212.1 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 =
r.m.s.
200

= 282.9 V
0.707 0.707
Mean value = 0.637  peak value = 0.637  282.9 = 180.2 V
7. Plot a sine wave of peak value 10.0 A. Show that the average value of the waveform is 6.37 A over half a
cycle, and that the r.m.s. value is 7.07 A
© John Bird Published by Taylor and Francis
160
A sine wave of maximum value 10.0 A is shown below.
Over half a cycle, mean value =
area under curve
length of base
Using the mid-ordinate rule with 12 intervals,

area under curve =   1.3  3.8  6.1  7.9  9.2  9.9  9.9  9.2  7.9  6.1  3.8  1.3
6

=    76.4  = 20.0
 12 
Hence, mean value =
R.m.s. value =
=
20.0
= 6.37 A

 1.32  3.82  6.12  7.9 2  9.2 2  9.9 2  9.9 2  9.2 2  7.9 2  6.12 3.8 2 1.3 2 


12


596.8
= 7.05 A
12
With a larger scale and taking values to greater than 1 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 120 V. Calculate its r.m.s. and average values.
© John Bird Published by Taylor and Francis
161
R.m.s. value = 0.707  peak value = 0.707  120 = 84.8 V
Average value = 0.637  peak value = 0.637  120 = 76.4 V
9. A sinusoidal current has a mean value of 15.0 A. Determine its maximum and r.m.s. values.
Mean value = 0.637  maximum value,
from which, maximum value =
mean value 15.0

= 23.55 A
0.637
0.637
R.m.s. value = 0.707  maximum value = 0.707  23.55 = 16.65 A
© John Bird Published by Taylor and Francis
162
Exercise 79, Page 224
1. An alternating voltage is represented by v = 20 sin 157.1t 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) 157.1 =  = 2f, from which, frequency, f =
(c) Periodic time, T =
157.1
= 25 Hz
2
1 1

= 0.04 s or 40 ms
f 25
(d) Angular velocity = 157.1 rad/s
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 400t volts (b) i = 50 sin(100t + 0.30) amperes
(c) e = 200 sin(628.4t – 0.41) volts
(a) Peak value = 90 V
R.m.s. value = 0.707  peak value = 0.707  90 = 63.63 V
400 =  = 2f, from which, frequency, f =
Periodic time, T =
400
= 200 Hz
2
1
1

= 5 ms
f 200
Phase angle = 0
(b) Peak value = 50 A
R.m.s. value = 0.707  peak value = 0.707  50 = 35.35 A
100 =  = 2f, from which, frequency, f =
100 
= 50 Hz
2
© John Bird Published by Taylor and Francis
163
Periodic time, T =
1 1

= 0.02 s or 20 ms
f 50
Phase angle = 0.30 radians = 0.3
180
= 17.19 leading

(c) Peak value = 200 V
R.m.s. value = 0.707  peak value = 0.707  200 = 141.4 V
628.4 =  = 2f, from which, frequency, f =
Periodic time, T =
628.4
= 100 Hz
2
1
1

= 0.01 s or 10 ms
f 100
Phase angle = 0.41 radians = 0.41
180
= 23.49 lagging

3. A sinusoidal current has a peak value of 30 A and a frequency of 60 Hz. At time t = 0, the
current is zero. Express the instantaneous current i in the form i = Im sin t .
i = 30 sin  2(60)t  
If t = 0 when i = 0, thus
0 = 30 sin 
i.e. 0 = sin 
 = sin 1 0  0
from which,
Hence,
i = 30 sin 120t 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 =
1
1

= 50 Hz
T 20  10 3
Hence, v = 200 sin  2(50)t   = 200 sin 100t  
If t = 0 when v = - 75, thus
from which,
- 75 = 200 sin 
 75 
 = sin 1 
 = - 0.384
 200 
© John Bird Published by Taylor and Francis
164
v = 200 sin(100t – 0.384) volts
Hence,
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) - 30 V
Voltage, v = 60 sin 40t volts
(a) When v = 20 V, 20 = 60 sin 40t
from which,
Hence, time, t =
20
 20 
 sin 40t and 40t = sin 1   = 0.3398
60
 60 
0.3398
 8.496 10 3 s = 8.496 ms
40
(b) When v = - 30 V, - 30 = 60 sin 40t
from which,

30
 30 
 sin 40t and 40t = sin 1   
60
 60 
Sine is negative in the 3rd and 4th quadrants as shown in the diagram.
 30 
sin 1   = 0.5236 rad and the first time this occurs is in the 3rd quadrant. Measuring from zero, the
 60 
angle is π + 0.5236 = 3.6652 rad
Hence, time, t =
3.6652
 0.09163s = 91.63 ms
40
© John Bird Published by Taylor and Francis
165
6. The instantaneous value of voltage in an a.c. circuit at an time t seconds is given by
v = 100 sin(50t – 0.523) 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  100 = 200 V
50 =  = 2f, from which, frequency, f =
Periodic time, T =
50
= 25 Hz
2
1 1

= 0.04 s or 40 ms
f 25
Phase angle = 0.523 rad lagging = 0.523
180
= 29.97 lagging or 2958 lagging

(b) When t = 0, v = 100 sin[50(0) – 0.523] = - 49.95 V
(c) When t = 8 ms, v = 100 sin[50( 8 103 ) – 0.523]
= 100 sin 0.7336 = 66.96 V
(d) When v = 60 V,
from which,
i.e.
60 = 100 sin[50t – 0.523]
60
= sin[50t – 0.523]
100
50t – 0.523 = sin 1 0.60 = 0.6435 or  - 0.6435 (sine is positive in the 1st
and 2nd quadrants, as shown)
© John Bird Published by Taylor and Francis
166
Hence,
50t = 0.6435 + 0.523
or
50t =  - 0.6435 + 0.523 and t =
i.e.
  0.6435  0.523
= 19.23 ms
50
-40 = 100 sin[50t – 0.523]
(e) When v = -40 V,
from which,
0.6435  0.523
= 7.426 ms
50
and t =

40
= sin[50t – 0.523]
100
50t – 0.523 = sin 1 (0.40) =  + 0.4115 or 2 - 0.4115 (sine is negative in the
3rd and 4th quadrants, as shown)
Hence,
50t =  + 0.4115 + 0.523
and t =
  0.4115  0.523
= 25.95 ms
50
or
50t = 2 - 0.4115 + 0.523 and t =
2  0.4115  0.523
= 40.71 ms
50
(f) The first time when the voltage is a maximum is when v = 100 V
i.e.
i.e.
i.e.
from which,
100 = 100 sin[50t – 0.523]
1 = sin[50t – 0.523]
50t – 0.523 = sin 1 1  1.5708
t=
1.5708  0.523
= 13.33 ms
50
© John Bird Published by Taylor and Francis
167
A sketch of v = 100 sin[50t – 0.523] is shown below.
© John Bird Published by Taylor and Francis
168
Exercise 80, Page 227
1. The instantaneous values of two alternating voltages are given by v1  5sin t and


v 2  8sin  t   . By plotting v1 and v 2 on the same axes, using the same scale, over one
6

cycle, obtain expressions for (a) v1 + v 2 (b) v1 - v 2
© John Bird Published by Taylor and Francis
169
(a) From the sketched graphs above, v1  v2  12.6sin  t  0.32
(b) From the sketched graphs above, v1  v2  4.4sin  t  2
2. Repeat Problem 1 by calculation.
(a) The relative positions of v1 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,
 ac 
from which,
2
 52  82  2  5  8  cos150
ac = 12.58
Using the sine rule,
8
12.58

sin  sin150
from which,
sin  
8sin150
 0.317965
12.58
  sin 1 0.317965  18.54 or 0.324 radians
and
v1  v2  12.58sin  t  0.324
Hence,
(b) The relative positions of v1 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,
 ac 
from which,
2
 52  82  2  5  8  cos 30
ac = 4.44
© John Bird Published by Taylor and Francis
170
Using the sine rule,
8
4.44

sin  sin 30
from which,
sin  
8sin 30
 0.90090
4.44
  sin 1 0.90090  64.28 or 180  64.28  115.72
and
From the phasor diagram,  = 115.72 or 2.02 radians
v1  v2  4.44sin  t  2.02
Hence,
3. Construct a phasor diagram to represent i1 + i2 where i1 = 12 sin t and i2 = 15 sin(t + π/3). By
measurement, or by calculation, find a sinusoidal expression to represent i1 + i2
The phasor diagram is shown below.
By drawing the diagram to scale and measuring, i R = 23.5 and ϕ = 34º or 0.59 rad
By calculation, using the cosine rule,
iR 
Using the sine rule,
Hence,
 122  152  2 12 15  cos120
i R = 23.43
from which,
and
2
15
23.43

sin  sin120
from which,
sin  
15sin120
 0.55443
23.43
  sin 1 0.55443  33.67 or 0.588rad
i1  i 2  23.43sin  t  0.588 
© John Bird Published by Taylor and Francis
171
4. Determine, either by plotting graphs and adding ordinates at intervals, or by calculation, the
following periodic function in the form v  Vm sin(t  )


10sin t  4sin  t  
4

The following is determined by calculation.
The relative positions of v1 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,
 ac 
from which,
Hence,
 102  42  2 10  4  cos135
ac = 13.14
Using the sine rule,
and
2
4
13.14

sin  sin135
from which,
sin  
4sin135
 0.2153
13.14
  sin 1 0.2153  12.43 or 0.217 rad


10sin t  4sin  t    13.14sin  t  0.217 
4

5. Determine, either by plotting graphs and adding ordinates at intervals, or by calculation, the
following periodic function in the form v  Vm sin(t  )




80sin  t    50sin  t  
3
6


The following is determined by calculation.
The relative positions of v1 and v 2 at time t = 0 are shown as phasors in diagram (iii).
© John Bird Published by Taylor and Francis
172
(iii)
(iv)
The phasor diagram is shown in diagram (iv). Since abc is a right angled triangle, Pythagoras’
theorem is used.
ac  502  802  94.34
and
 50 
  tan 1    32
 80 
Hence, in diagram (iv),  = 60 - 32 = 28 or 0.489 rad.
Thus,




80sin  t    50sin  t    94.34sin  t  0.489 
3
6


6. Determine, either by plotting graphs and adding ordinates at intervals, or by calculation, the
following periodic function in the form v  Vm sin(t  )


100sin t  70sin  t  
3

The following is determined by calculation.
The relative positions of v1 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
173
(v)
(vi)
Using the cosine rule,
 ac 
from which,
Hence,
 1002  702  2 100  70  cos 60
ac = 88.88
Using the sine rule,
and
2
88.88
70

sin 60 sin 
from which,
sin  
70sin 60
 0.68206
88.88
  sin 1 0.68206  43 or 0.751rad


100sin t  70sin  t    88.88sin  t  0.751
3

7. The voltage drops across two components when connected in series across an a.c. supply are
v1 = 150 sin 314.2t and v2 = 90 sin (314.2t - /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 = v1  v 2 = 150 sin 314.2t + 90 sin  314.2t   / 5
= 1500  90 36
= (150 + j0) + (72.81 – j52.90)
= 222.81 – j52.90
© John Bird Published by Taylor and Francis
174
= 22913.36  229 0.233 rad
= 229 sin(314.2t – 0.233) V
(b) R.m.s value of supply = 0.707  229 = 161.9 V
(c)  = 314.2 = 2f
from which, frequency, f =
314.2
= 50 Hz
2
8. If the supply to a circuit is 25 sin 628.3t volts and the voltage drop across one of the components
is 18 sin (628.3t - 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  v1  25sin 628.3t  18sin(628.4t  0.52)
= 250  18  0.52 rad using complex numbers
= (25 + j0) – (15.621 – j8.944)
= 9.379 + j8.944
= 12.960.76 rad
= 12.96 sin(628.3t + 0.762) V
(b)  = 628.3 = 2f
from which, supply frequency, f =
(c) Periodic time, T =
628.3
= 100 Hz
2
1
1

= 0.01 s or 10 ms
f 100
© John Bird Published by Taylor and Francis
175
9. The voltages across three components in a series circuit when connected across an a.c. supply
are:


v1  30sin  300t   volts,
6



v 2  40sin  300t   volts and
4



v3  50sin  300t   volts.
3

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 = v1  v 2  v3  30sin  300t    40sin  300t    50sin  300t  
6
4
3



= 30 30  4045  5060 using complex numbers
= 79.265 + j56.586
= 97.3935.52 V
or
97.390.620 V
= 97.39sin  300t  0.620  V
(b)  = 300 = 2f
from which, supply frequency, f =
(c) Periodic time, T =
300
= 150 Hz
2
1
1

= 0.0667 s or 6.67 ms
f 150
(d) R.m.s value of supply = 0.707  97.39 = 68.85 V
© John Bird Published by Taylor and Francis
176