208 Higher Engineering Mathematics
(b) Area under waveform (b) for a half
cycle=( 1 × 1 )+( 3 × 2 )=7As.
Average value of waveform
=
area under curve
length of base
=
7As
3s
=2.33A
(c) A half cycle of the voltage waveform (c) is
completed in 4ms.
Area under curve=^12 {( 3 − 1 ) 10 −^3 }( 10 )
= 10 × 10 −^3 Vs
Average value of waveform
=
area under curve
length of base
=
10 × 10 −^3 Vs
4 × 10 −^3 s
=2.5V
Problem 6. Determine the mean value of current
over one complete cycle of the periodic waveforms
shown in Fig. 19.9.
(a)
0 481216202428
5
Current (mA)
t (ms)
(b)
0624 81012
2
Current (mA)
t (ms)
Figure 19.9
(a) One cycle of the trapezoidal waveform (a) is
completed in 10ms (i.e. the periodic time is
10ms).
Area under curve=area of trapezium
=^12 (sum of parallel sides) (perpendicular
distance between parallel sides)
=^12 {( 4 + 8 )× 10 −^3 }( 5 × 10 −^3 )
= 30 × 10 −^6 As
Mean value over one cycle
=
area under curve
length of base
=
30 × 10 −^6 As
10 × 10 −^3 s
=3mA
(b) One cycle of the sawtooth waveform (b) is com-
pleted in 5ms.
Area under curve=^12 ( 3 × 10 −^3 )( 2 )
= 3 × 10 −^3 As
Mean value over one cycle
=
area under curve
length of base
=
3 × 10 −^3 As
5 × 10 −^3 s
=0.6A
Problem 7. The power used in a manufacturing
process during a 6hour period is recorded at
intervals of 1hour as shown below.
Time(h) 0 1 2 3 4 5 6
Power(kW) 0 14 29 51 45 23 0
Plot a graph of power against time and, by using the
mid-ordinate rule, determine (a) the area under the
curve and (b) the average value of the power.
The graph of power/time is shown in Fig. 19.10.
(a) The time base is divided into 6 equal inter-
vals, each of width 1hour. Mid-ordinates are
erected (shown by broken lines in Fig. 19.10)
and measured. The values are shown in
Fig. 19.10.