270 Higher Engineering Mathematics
Hence, by cosine and sine rules,
i R = i 1 + i 2 = 26.46 sin(ωt + 0.333) A
Now try the following exercise
Exercise 109 Resultant phasors by the sine
and cosine rules
1. Determine, using the cosine and sine rules, a
sinusoidal expression for:
y = 2 sin A + 4 cos A.
[4.5 sin(A + 63.5 ◦ )]
2. Given v 1 = 10 sin ωt volts and
v 2 =14 sin(ωt + π/3) volts use the cosine and
sine rules to determine sinusoidal expressions
for (a) v 1 + v 2 (b) v 1 − v 2 .
(a) 20.88 sin(ωt + 0.62) volts
(b) 12.50 sin(ωt − 1.33)volts
In Problems 3 to 5, express the given expressions
in the form A sin(ωt ± α) by using the cosine and
sine rules.
3. 12 sin ωt + 5 cos ωt
[13 sin(ωt + 0.395)]
4. 7 sin ωt + 5 sin
ωt +
π
4
[11.11 sin(ωt + 0.324)]
5. 6 sin ωt + 3 sin
ωt −
π
6
[8.73 sin(ωt − 0.173)]
25.5 Determining resultant phasors
by horizontal and vertical
components
If a right-angled triangle is constructed as shown in
Fig. 25.16, then 0a is called the horizontal component
of F and ab is called the vertical component of F.
From trigonometry (see Chapter 11),
cos θ =
0a
0b
from which,
0a = 0b cos θ = F cos θ
F
F sin
F cos
a
b
0
Figure 25.16
i.e. the horizontal component of F, H = F cos θ
and sin θ =
ab
0b
from which ab = 0b sin θ
= F sin θ
i.e. the vertical component of F, V = F sin θ
Determining resultant phasors by horizontal and vertical
components is demonstrated in the following worked
problems.
Problem 9. Two alternating voltages are given by
v 1 = 15 sin ωt volts and v 2 = 25 sin(ωt − π/6)
volts. Determine a sinusoidal expression for the
resultant v R = v 1 + v 2 by finding horizontal and
vertical components.
The relative positions of v 1 and v 2 at time t = 0 are
shown in Fig. 25.17(a) and the phasor diagram is shown
in Fig. 25.17(b).
The horizontal component of v R ,
H = 15 cos0 ◦ + 25 cos(−30 ◦ ) = 0a + ab = 36.65 V
The vertical component of v R ,
V = 15 sin 0 ◦ + 25 sin(−30 ◦ ) = bc = −12.50 V
Hence,
v R = 0c =
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 = 38.72sin(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.
Hence, by cosine and sine rules,
i R = i 1 + i 2 = 26.46 sin(ωt + 0.333) A
Now try the following exercise
Exercise 109 Resultant phasors by the sine
and cosine rules
1. Determine, using the cosine and sine rules, a
sinusoidal expression for:
y = 2 sin A + 4 cos A.
[4.5 sin(A + 63.5 ◦ )]
2. Given v 1 = 10 sin ωt volts and
v 2 =14 sin(ωt + π/3) volts use the cosine and
sine rules to determine sinusoidal expressions
for (a) v 1 + v 2 (b) v 1 − v 2 .
(a) 20.88 sin(ωt + 0.62) volts
(b) 12.50 sin(ωt − 1.33)volts
In Problems 3 to 5, express the given expressions
in the form A sin(ωt ± α) by using the cosine and
sine rules.
3. 12 sin ωt + 5 cos ωt
[13 sin(ωt + 0.395)]
4. 7 sin ωt + 5 sin
ωt +
π
4
[11.11 sin(ωt + 0.324)]
5. 6 sin ωt + 3 sin
ωt −
π
6
[8.73 sin(ωt − 0.173)]
25.5 Determining resultant phasors
by horizontal and vertical
components
If a right-angled triangle is constructed as shown in
Fig. 25.16, then 0a is called the horizontal component
of F and ab is called the vertical component of F.
From trigonometry (see Chapter 11),
cos θ =
0a
0b
from which,
0a = 0b cos θ = F cos θ
F
F sin
F cos
a
b
0
Figure 25.16
i.e. the horizontal component of F, H = F cos θ
and sin θ =
ab
0b
from which ab = 0b sin θ
= F sin θ
i.e. the vertical component of F, V = F sin θ
Determining resultant phasors by horizontal and vertical
components is demonstrated in the following worked
problems.
Problem 9. Two alternating voltages are given by
v 1 = 15 sin ωt volts and v 2 = 25 sin(ωt − π/6)
volts. Determine a sinusoidal expression for the
resultant v R = v 1 + v 2 by finding horizontal and
vertical components.
The relative positions of v 1 and v 2 at time t = 0 are
shown in Fig. 25.17(a) and the phasor diagram is shown
in Fig. 25.17(b).
The horizontal component of v R ,
H = 15 cos0 ◦ + 25 cos(−30 ◦ ) = 0a + ab = 36.65 V
The vertical component of v R ,
V = 15 sin 0 ◦ + 25 sin(−30 ◦ ) = bc = −12.50 V
Hence,
v R = 0c =
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 = 38.72sin(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.