Methods of adding alternating waveforms 271
v 1 5 15 V
(a)
(b)
v 2 5 25 V
␲/6 or 308
␾
0
v R
v 2
v 1
308
1508
a
b
c
Figure 25.17
The horizontal component of v R ,
H = 15 cos0 ◦ − 25 cos(−30 ◦ ) = −6.65V
The vertical component of v R ,
V = 15 sin0 ◦ − 25 sin(−30 ◦ ) = 12.50V
Hence, v R =
(−6.65) 2 + (12.50) 2
by Pythagoras’ theorem
= 14.16 volts
tan φ =
V
H
=
12.50
−6.65
= −1.8797
from which, φ = tan
−1
(−1.8797) = 118.01
◦
or 2.06 radians.
Hence,
v R = v 1 −v 2 = 14.16 sin(ωt + 2.06)V
The phasor diagram is shown in Fig. 25.18.
v 1 5 15 V
2v 2 5 25 V
v 2 5 25 V
␾
v R
308
308
Figure 25.18
Problem 11. Determine
20 sin ωt + 10 sin
ωt +
π
3
using horizontal and
vertical components.
From the phasors shown in Fig. 25.19:
Total horizontal component,
H = 20 cos0 ◦ + 10 cos60 ◦ = 25.0
i 1 5 20 A
i 2 5 10 A
608
Figure 25.19
Total vertical component,
V = 20 sin 0 ◦ + 10 sin 60 ◦ = 8.66
By Pythagoras, the resultant, i R =
25.0 2 + 8.66 2
= 26.46 A
Phase angle, φ = tan −1
8.66
25.0
= 19.11 ◦
or 0.333 rad
Hence, by using horizontal and vertical components,
20 sin ωt + 10 sin
ωt +
π
3
= 26.46 sin(ωt + 0.333)
Now try the following exercise
Exercise 110 Further problems on
resultant phasors by horizontal and vertical
components
In Problems 1 to 4, express the combination of
periodic functions in the form A sin(ωt ± α) by
horizontal and vertical components:
1. 7 sin ωt + 5 sin
ωt +
π
4
[11.11 sin(ωt + 0.324)]
2. 6 sin ωt + 3 sin
ωt −
π
6
[8.73 sin(ωt − 0.173)]
3. i = 25 sin ωt − 15 sin
ωt +
π
3
[i = 21.79 sin(ωt − 0.639)]
4. x = 9 sin
ωt +
π
3
−7 sin
ωt −
3π
8
[x = 14.38 sin(ωt + 1.444)]