Basic Engineering Mathematics, Fifth Edition

280 Basic Engineering Mathematics

2  angle t

19 

19 

i 1  20 sint

iR  20 sint10 sin (t )

90  180  270  360 

 30

 20

 10

10

20

26.5

30

3 

2



2



3



i 2  10 sin (t  3 )

Figure 30.4

3. Express 12sinωt+5cosωt in the form
   Asin(ωt±α)by drawing and measurement.

30.3 Determining resultant phasors by drawing

The resultant of two periodic functions may be found

from their relative positions when the time is zero.

For example, ify 1 =4sinωtandy 2 =3sin(ωt−π/ 3 )

then each may be represented asphasorsas shown in

Figure 30.5,y 1 being 4 units long and drawn horizon-

tallyandy 2 being3unitslong,laggingy 1 byπ/3radians

or60◦.Todeterminetheresultantofy 1 +y 2 ,y 1 isdrawn

horizontally as shown in Figure 30.6 andy 2 is joined to

the end ofy 1 at 60◦to the horizontal. The resultant

is given byyR. This is the same as the diagonal of a

parallelogram that is shown completed in Figure 30.7.

608 or /3 rads

y 154

y 253

Figure 30.5

ResultantyR, in Figures 30.6 and 30.7, may be deter-

mined by drawing the phasors and their directions to

scale and measuring using a ruler and protractor. In this

y 154

y 2

5

3

 608

yR

Figure 30.6

y 1  4

y 2  3



yR

Figure 30.7

example,yRis measured as 6 units long and angleφis

measured as 25◦.

25 ◦= 25 ×

π

180

radians= 0 .44 rad

Hence, summarizing, by drawing,

yR=y 1 +y 2 =4sinωt+3sin(ωt−π/ 3 )

=6sin(ωt− 0. 44 ).

If the resultant phasor,yR=y 1 −y 2 is required then

y 2 is still 3 units long but is drawn in the opposite

direction, as shown in Figure 30.8.

Problem 5. Two alternating currents are given by

i 1 =20sinωtamperes andi 2 =10sin

(

ωt+

π

3

)

amperes. Determinei 1 +i 2 by drawing phasors