Higher Engineering Mathematics, Sixth Edition

270 Higher Engineering Mathematics

Hence, by cosine and sine rules,

iR=i 1 +i 2 = 26 .46sin(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=2sinA+4cosA.
   [4.5sin(A+ 63. 5 ◦)]


2. Givenv 1 =10sinωtvolts and
   v 2 =14sin(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) 20.88sin(ωt+ 0. 62 )volts
   (b) 12.50sin(ωt− 1. 33 )volts

]

In Problems 3 to 5, express the given expressions

in the formAsin(ωt±α)by using the cosine and

sine rules.

3. 12sinωt+5cosωt
   [13sin(ωt+ 0. 395 )]


4. 7sinωt+5sin

(

ωt+

π

4

)

[11.11sin(ωt+ 0. 324 )]

5. 6sinωt+3sin

(

ωt−

π

6

)

[8.73sin(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 0ais called the horizontal component

ofFandabis called the vertical component ofF.

From trigonometry (see Chapter 11),

cosθ=

0 a

0 b

from which,

0 a= 0 bcosθ=Fcosθ

F

F sin 

F cos  a

b

0



Figure 25.16

i.e. the horizontal component ofF,H=Fcosθ

and sinθ=

ab

0 b

from whichab= 0 bsinθ

=Fsinθ

i.e. the vertical component ofF, V=Fsinθ

Determiningresultant phasorsbyhorizontal andvertical

components is demonstrated in the following worked

problems.

Problem 9. Two alternating voltages are given by

v 1 =15sinωtvolts andv 2 =25sin(ωt−π/6)

volts. Determine a sinusoidal expression for the

resultantvR=v 1 +v 2 by finding horizontal and

vertical components.

The relative positions ofv 1 andv 2 at timet=0are

shown in Fig. 25.17(a) and the phasor diagram is shown

in Fig. 25.17(b).

The horizontal component ofvR,

H=15cos0◦+25cos(− 30 ◦)= 0 a+ab= 36 .65V

The vertical component ofvR,

V=15sin0◦+25sin(− 30 ◦)=bc=− 12 .50V

Hence, vR=^0 c=

√

36. 652 +(− 12. 50 )^2

by Pythagoras’ theorem

=38.72 volts

tanφ=

V

H

=

− 12. 50

36. 65

=− 0. 3411

from which,φ=tan−^1 (− 0. 3411 )=− 18. 83 ◦

or − 0 .329 radians.

Hence, vR=v 1 +v 2 =^38 .72sin(ωt−^0.^329 )V

Problem 10. For the voltages in Problem 9,

determine the resultantvR=v 1 −v 2 using

horizontal and vertical components.