Higher Engineering Mathematics, Sixth Edition

Methods of adding alternating waveforms 271

v 15 15 V

(a) (b)

v 25 25 V

/6 or 30 8 

0

vR

v 2

v 1

1508 308

ab

c

Figure 25.17

The horizontal component ofvR,
H=15cos0◦−25cos(− 30 ◦)=− 6 .65V

The vertical component ofvR,
V=15sin0◦−25sin(− 30 ◦)= 12 .50V

Hence,vR=

√

(− 6. 65 )^2 +( 12. 50 )^2

by Pythagoras’ theorem

=14.16 volts

tanφ=

V

H

=

12. 50

− 6. 65

=− 1. 8797

from which,φ=tan−^1 (− 1. 8797 )= 118. 01 ◦

or 2.06 radians.

Hence,

vR=v 1 −v 2 = 14 .16sin(ωt+ 2. 06 )V

The phasor diagram is shown in Fig. 25.18.

v 15 15 V

2 v 25 25 V

v 25 25 V



vR

308

308

Figure 25.18

Problem 11. Determine

20sinωt+10sin

(

ωt+

π

3

)

using horizontal and

vertical components.

From the phasors shown in Fig. 25.19:
Total horizontal component,
H=20cos0◦+10cos60◦= 25. 0

i 15 20 A

i 25 10 A

608

Figure 25.19

Total vertical component,

V=20sin0◦+10sin60◦= 8. 66

By Pythagoras, the resultant,iR=

√[

25. 02 + 8. 662

]

= 26 .46A

Phase angle,φ=tan−^1

(

8. 66

25. 0

)

= 19. 11 ◦

or0.333 rad

Hence, by using horizontal and vertical components,

20sinωt+10sin

(

ωt+

π

3

)

= 26 .46sin(ωt+ 0. (^333) )
Now try the following exercise
Exercise 110 Further problemson
resultant phasors by horizontal and vertical
components
In Problems 1 to 4, express the combination of
periodic functions in the formAsin(ωt±α)by
horizontal and vertical components:

1. 7sinωt+5sin

(

ωt+

π

4

)

[11.11sin(ωt+ 0. 324 )]

2. 6sinωt+3sin

(

ωt−

π

6

)

[8.73sin(ωt− 0. 173 )]

3. i=25sinωt−15sin

(

ωt+

π

3

)

[i= 21 .79sin(ωt− 0. 639 )]

4. x=9sin

(

ωt+

π

3

)

−7sin

(

ωt−

3 π

8

)

[x= 14 .38sin(ωt+ 1. 444 )]