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.