Methods of adding alternating waveforms 269
y 1 5 5
␲/6 or 308
y 2 5 4
Figure 25.12
y 1 5 5
y 2 5 4
␾
0
y R
a
b
308
Figure 25.13
Using the cosine rule on triangle 0ab of Fig. 25.13 gives:
y
2
R = 5
2
+ 4
2
− [2(5)(4) cos 150
◦ ]
= 25 + 16 − (−34.641)
= 75.641
from which, y R =
√
75.641 = 8.697
Using the sine rule,
8.697
sin 150 ◦ =
4
sin φ
from which,
sin φ =
4 sin 150 ◦
8.697
= 0.22996
and
φ = sin
−1 0.22996
= 13.29
◦ or 0.232 rad
Hence, y R = y 1 + y 2 = 5 sinωt + 4 sin(ωt − π/6)
= 8.697 sin(ωt − 0.232)
Problem 7. Given y 1 = 2 sinωt and
y 2 = 3 sin(ωt + π/4), obtain an expression, by
calculation, for the resultant, y R = y 1 + y 2 .
When time t = 0, the position of phasors y 1 and y 2
are as shown in Fig. 25.14(a). To obtain the resultant, y 1 is drawn horizontally, 2 units long, y 2 is drawn
3 units long at an angle of π/4 rads or 45 ◦ and joined to
the end of y 1 as shown in Fig. 25.14(b).
From Fig. 25.14(b), and using the cosine rule:
y
2
R = 2
2
+ 3
2
− [2(2)(3) cos 135
◦ ]
= 4 + 9 − [−8.485] = 21.49
Hence,
y R =
√
21.49 = 4.6357
y 1 5 2
y 1 5 2
y 2 5 3
y 2 5 3
y R
␲/4 or 458
1358
458
␾
(a)
(b)
Figure 25.14
Using the sine rule:
3
sin φ
=
4.6357
sin 135 ◦
from which,
sin φ =
3 sin 135 ◦
4.6357
= 0.45761
Hence,
φ = sin
−1 0.45761
= 27.23
◦ or 0.475 rad.
Thus, by calculation, y R = 4.635 sin(ωt + 0.475)
Problem 8. Determine
20 sin ωt + 10 sin
ωt +
π
3
using the cosine
and sine rules.
From the phasor diagram of Fig. 25.15, and using the
cosine rule:
i
2
R = 20
2
+ 10
2
− [2(20)(10) cos 120
◦ ]
= 700
Hence, i R =
√
700 = 26.46 A
i 2 5 10 A
i 1 5 20 A
i R
608
␾
Figure 25.15
Using the sine rule gives :
10
sin φ
=
26.46
sin 120 ◦
from which,
sin φ =
10 sin 120 ◦
26.46
= 0.327296
and
φ = sin
−1 0.327296 = 19.10
◦
= 19.10 ×
π
180
= 0.333 rad