Higher Engineering Mathematics, Sixth Edition

Methods of adding alternating waveforms 273

v 15 15 V v 1

v 25 25 V vR

/6 or 30 8  1508

(a) (b)

Figure 25.22

In polar form,vR=v 1 +v 2 = 15 ∠ 0 + 25 ∠−

π

6

= 15 ∠ 0 ◦+ 25 ∠− 30 ◦

=( 15 +j 0 )+( 21. 65 −j 12. 5 )

= 36. 65 −j 12. 5 = 38. 72 ∠− 18. 83 ◦

= 38. 72 ∠− 0 .329 rad

Hence, by using complex numbers, the resultant in

sinusoidal form is:

vR=v 1 +v 2 =15sinωt+25sin(ωt−π/ 6 )

= 38 .72sin(ωt− 0. 329 )

Problem 13. For the voltages in Problem 12,

determine the resultantvR=v 1 −v 2 using complex

numbers.

In polar form,yR=v 1 −v 2 = 15 ∠ 0 − 25 ∠−

π

6

= 15 ∠ 0 ◦− 25 ∠− 30 ◦

=( 15 +j 0 )−( 21. 65 −j 12. 5 )

=− 6. 65 +j 12. 5 = 14. 16 ∠ 118. 01 ◦

= 14. 16 ∠ 2 .06 rad

Hence, by using complex numbers, the resultant in

sinusoidal form is:

y 1 −y 2 =15sinωt−25sin(ωt−π/ 6 )

= 14 .16sin(ωt− 2. 06 )

Problem 14. Determine

20sinωt+10sin

(

ωt+

π

3

)

using complex

numbers.

From the phasors shown in Fig. 25.23, the resultant may

be expressed in polar form as:

i 25 10 A

i 15 20 A

608

Figure 25.23

iR= 20 ∠ 0 ◦+ 10 ∠ 60 ◦

i.e. iR=( 20 +j 0 )+( 5 +j 8. 66 )

=( 25 +j 8. 66 )= 26. 46 ∠ 19. 11 ◦A or

26. 46 ∠ 0 .333rad A

Hence, by using complex numbers, the resultant in

sinusoidal form is:

iR=i 1 +i 2 = 26 .46sin(ωt+ 0. 333 )A

Problem 15. If the supply to a circuit is

v=30sin100πtvolts and the voltage drop across

one of the components is

v 1 =20sin( 100 πt− 0. 59 )volts, calculate the:

(a) voltage drop across the remainder of the

circuit, given byv−v 1 , in the form

Asin(ωt±α)

(b) supply frequency

(c) periodic time of the supply

(d) r.m.s. value of the supply voltage

(a) Supply voltage,v=v 1 +v 2 wherev 2 is thevoltage

across the remainder of the circuit.