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:
- 7sinωt+5sin
(
ωt+
π
4
)
[11.11sin(ωt+ 0. 324 )]
- 6sinωt+3sin
(
ωt−
π
6
)
[8.73sin(ωt− 0. 173 )]
- i=25sinωt−15sin
(
ωt+
π
3
)
[i= 21 .79sin(ωt− 0. 639 )]
- x=9sin
(
ωt+
π
3
)
−7sin
(
ωt−
3 π
8
)
[x= 14 .38sin(ωt+ 1. 444 )]