Basic Engineering Mathematics, Fifth Edition

Methods of adding alternating waveforms 281

60 

60 



y 1  4

y y^2 ^3

R

y 2

Figure 30.8

The relative positions ofi 1 andi 2 at timet=0are

shown as phasors in Figure 30.9, where

π

3

rad= 60 ◦.

The phasor diagram in Figure 30.10 is drawn to scale
with a ruler and protractor.

i 15 20 A

i 25 10 A

608

Figure 30.9

i 25 10 A

i 1520 A

iR

^608

Figure 30.10

The resultantiRis shown and is measured as 26A and
angleφas 19◦or 0.33rad leadingi 1. Hence, by drawing
and measuring,

iR=i 1 +i 2 =26sin(ωt+ 0. 33 )A

Problem 6. For the currents in Problem 5,

determinei 1 −i 2 by drawing phasors

At timet=0, currenti 1 is drawn 20 units long hori-
zontally as shown by 0ain Figure 30.11. Currenti 2 is
shown, drawn 10 units long in a broken line and leading
by 60◦. The current−i 2 is drawn in the opposite direc-
tion to the broken line ofi 2 ,shownasabin Figure 30.11.
The resultantiRisgivenby0blagging by angleφ.

i 15 20 A

i 25 10 A

iR

a

b

0

2 i 2

608



Figure 30.11

By measurement,iR=17A andφ= 30 ◦or 0.52rad.

Hence, by drawing phasors,

iR=i 1 −i 2 =17sin(ωt− 0. 52 )A

Now try the following Practice Exercise

PracticeExercise 119 Determining

resultant phasors by drawing (answerson

page 352)

1. Determine a sinusoidal expression for
   2sinθ+4cosθby drawing phasors.


2. Ifv 1 =10sinωtvolts and
   v 2 =14sin(ωt+π/ 3 ) volts, determine by
   drawing phasors sinusoidal expressions for
   (a)v 1 +v 2 (b)v 1 −v 2


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

30.4 Determining resultant phasors

by the sine and cosine rules

As stated earlier, the resultant of two periodic func-

tions may be found from their relative positions when

the time is zero. For example, if y 1 =5sinωt and

y 2 =4sin(ωt−π/ 6 )then each may be represented by

phasors as shown in Figure 30.12,y 1 being 5 units

long and drawn horizontally andy 2 being 4 units long,

laggingy 1 byπ/6 radians or 30◦. To determine the

resultant ofy 1 +y 2 ,y 1 is drawn horizontally as shown

in Figure 30.13 andy 2 is joined to the end ofy 1 atπ/ 6

radians; i.e., 30◦to the horizontal. The resultant is given

byyR.

Using the cosine rule on triangle 0abof Figure 30.13

gives

y^2 R= 52 + 42 −[2( 5 )( 4 )cos150◦]

= 25 + 16 −(− 34. 641 )= 75. 641