Higher Engineering Mathematics, Sixth Edition

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.