284 Basic Engineering Mathematics
Hence,
v R =
36.65 2 + (−12.50) 2
by Pythagoras’ theorem
= 38.72 volts
tan φ =
V
H
=
−12.50
36.65
= −0.3411
from which
φ = tan
−1
(−0.3411)
= −18.83
◦ or −0.329 radians.
Hence, v R = v 1 + v 2 = 38.72 sin(ωt − 0.329) V
Problem 10. For the voltages in Problem 9,
determine the resultant v R = v 1 − v 2 using
horizontal and vertical components
The horizontal component of v R ,
H = 15 cos0 ◦ − 25 cos(−30 ◦ ) = −6.65 V
The vertical component of v R ,
V = 15 sin0 ◦ − 25 sin(−30 ◦ ) = 12.50 V
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 Figure 30.18.
v 1 5 15 V
2v 2 5 25 V
v 2 5 25 V
v R
308
308
Figure 30.18
Problem 11. Determine
20 sin ωt + 10 sin
ωt +
π
3
using horizontal and vertical components
i 1 5 20 A
i 2 5 10 A
608
Figure 30.19
From the phasors shown in Figure 30.19,
Total horizontal component,
H = 20 cos0
◦
+ 10 cos 60
◦
= 25.0
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 sinωt + 10 sin
ωt +
π
3
= 26.46 sin(ωt + 0.333)
Now try the following Practice Exercise
Practice Exercise 121 Resultant phasors by
horizontal and vertical components (answers
on page 353)
In problems 1 to 5, 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
2. 6 sin ωt + 3 sin
ωt −
π
6
3. i = 25 sin ωt − 15 sin
ωt +
π
3
4. v = 8 sinωt − 5 sin
ωt −
π
4
5. x = 9 sin
ωt +
π
3
− 7 sin
ωt −
3π
8
6. The voltage drops across two components
when connected in series across an a.c.
supply are v 1 = 200 sin 314.2t and
v 2 = 120 sin(314.2t − π/5) volts
respectively. Determine
(a) the voltage of the supply (given by
v 1 + v 2 ) in the form A sin(ωt ± α).
Hence,
v R =
36.65 2 + (−12.50) 2
by Pythagoras’ theorem
= 38.72 volts
tan φ =
V
H
=
−12.50
36.65
= −0.3411
from which
φ = tan
−1
(−0.3411)
= −18.83
◦ or −0.329 radians.
Hence, v R = v 1 + v 2 = 38.72 sin(ωt − 0.329) V
Problem 10. For the voltages in Problem 9,
determine the resultant v R = v 1 − v 2 using
horizontal and vertical components
The horizontal component of v R ,
H = 15 cos0 ◦ − 25 cos(−30 ◦ ) = −6.65 V
The vertical component of v R ,
V = 15 sin0 ◦ − 25 sin(−30 ◦ ) = 12.50 V
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 Figure 30.18.
v 1 5 15 V
2v 2 5 25 V
v 2 5 25 V
v R
308
308
Figure 30.18
Problem 11. Determine
20 sin ωt + 10 sin
ωt +
π
3
using horizontal and vertical components
i 1 5 20 A
i 2 5 10 A
608
Figure 30.19
From the phasors shown in Figure 30.19,
Total horizontal component,
H = 20 cos0
◦
+ 10 cos 60
◦
= 25.0
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 sinωt + 10 sin
ωt +
π
3
= 26.46 sin(ωt + 0.333)
Now try the following Practice Exercise
Practice Exercise 121 Resultant phasors by
horizontal and vertical components (answers
on page 353)
In problems 1 to 5, 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
2. 6 sin ωt + 3 sin
ωt −
π
6
3. i = 25 sin ωt − 15 sin
ωt +
π
3
4. v = 8 sinωt − 5 sin
ωt −
π
4
5. x = 9 sin
ωt +
π
3
− 7 sin
ωt −
3π
8
6. The voltage drops across two components
when connected in series across an a.c.
supply are v 1 = 200 sin 314.2t and
v 2 = 120 sin(314.2t − π/5) volts
respectively. Determine
(a) the voltage of the supply (given by
v 1 + v 2 ) in the form A sin(ωt ± α).