CHAPTER 40 ALTERNATING VOLTAGES AND CURRENTS
EXERCISE 187, Page 423
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
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EXERCISE 188, Page 426
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
420
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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
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© John Bird Published by Taylor and Francis
 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  v32  v 4 2 
(iii) R.m.s. value =  1
 =
4


(iv) Form factor =
(v) Peak factor =
 0.52  1.52  2.52  3.52  4.52 

 = 2.87 A
5


 20 2  20 2  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


422
© John Bird Published by Taylor and Francis
 32  9 2  152  212  242  242  242  242 

 = 19.56 A
8


r.m.s.
19.56

(iv) Form factor =
= 1.09
average
18
=
max imum value
24

= 1.23
r.m.s.
19.56
(v) Peak factor =
(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  v32  v 4 2 
(iii) R.m.s. value =  1
 =
4


(iv) Form factor =
(v) Peak factor =
 02  02  100 2  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 waveform is shown below.
1
12 103 s  300 V
area under curve 2

= 150 V
(a) Mean value =
length of base
12 103 s
 v 2  v 2 2  v32  v 4 2  .... 
(b) R.m.s. value =  1

6


=
 37.52  112.52  187.52  262.52  2252  752 

 = 170 V
6


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© John Bird Published by Taylor and Francis
4. 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
5. 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
6. A sinusoidal voltage has a maximum value of 120 V. Calculate its r.m.s. and average values.
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
7. 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,
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© John Bird Published by Taylor and Francis
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
EXERCISE 189, Page 426
Answers found from within the text of the chapter, pages 420 to 426.
EXERCISE 190, Page 427
1. (b)
2. (c) 3. (d) 4. (d) 5. (a) 6. (d)
7. (c) 8. (b) 9. (c)
10. (b)
425
© John Bird Published by Taylor and Francis