Higher Engineering Mathematics

VECTORS, PHASORS AND THE COMBINATION OF WAVEFORMS 235

D

Using the sine rule:

3

sinφ

=

4. 64

sin 135◦

from which

sinφ=

3 sin 135◦

4. 64

= 0. 4572

Henceφ=sin−^10. 4572 = 27 ◦ 12 ′or 0.475 rad.

By calculation,

yR= 4 .64 sin (ωt+ 0 .475)

Problem 13. Two alternating voltages are

given byv 1 =15 sinωtvolts and

v 2 =25sin (ωt−π/6) volts. Determine a sinu-

soidal expression for the resultantvR=v 1 +v 2

by finding horizontal and vertical components.

The relative positions ofv 1 andv 2 at timet=0 are
shown in Fig. 21.25(a) and the phasor diagram is
shown in Fig. 21.25(b).

Figure 21.25

The horizontal component ofvR,

H=15 cos 0◦+25 cos (− 30 ◦)

=oa+ab= 36 .65 V

The vertical component ofvR,

V=15 sin 0◦+25 sin (− 30 ◦)

=bc=− 12 .50 V

Hence vR(=oc)=

√

[(36.65)^2 +(− 12 .50)^2 ]

by Pythagoras’ theorem

= 38 .72V

tanφ=

V

H

(

=

bc

ob

)

=

− 12. 50

36. 65

=− 0. 3411

from which, φ=tan−^1 (− 0 .3411)=− 18 ◦ 50 ′

or−0.329 radians.

HencevR=v 1 +v 2 = 38 .72 sin(ωt− 0. 329 )V.

Problem 14. For the voltages in Problem 13,

determine the resultantvR=v 1 −v 2.

To find the resultantvR=v 1 −v 2 , the phasorv 2 of

Fig. 21.25(b) is reversed in direction as shown in

Fig. 21.26. Using the cosine rule:

v^2 R= 152 + 252 −2(15)(25) cos 30◦

= 225 + 625 − 649. 5 = 200. 5

vR=

√

(200.5)= 14 .16 V

Figure 21.26

Using the sine rule:

25

sinφ

=

14. 16

sin 30◦

from which

sinφ=

25 sin 30◦

14. 16

= 0. 8828

Henceφ=sin−^10. 8828 = 61. 98 ◦or 118.02◦. From

Fig. 21.26,φis obtuse,

henceφ= 118. 02 ◦or 2.06 radians.

HencevR=v 1 −v 2 = 14 .16 sin (ωt+ 2 .06) V.