284 Basic Engineering Mathematics
Hence, vR=√
36. 652 +(− 12. 50 )^2
by Pythagoras’ theorem=38.72 voltstanφ=V
H=− 12. 50
36. 65=− 0. 3411from which φ=tan−^1 (−^0.^3411 )
=− 18. 83 ◦or− 0 .329 radians.Hence,vR=v 1 +v 2 = 38 .72sin(ωt− 0. 329 )VProblem 10. For the voltages in Problem 9,
determine the resultantvR=v 1 −v 2 using
horizontal and vertical componentsThe horizontal component ofvR,
H=15cos0◦−25cos(− 30 ◦)=− 6 .65V
The vertical component ofvR,
V=15sin0◦−25sin(− 30 ◦)= 12 .50VHence, vR=√
(− 6. 65 )^2 +( 12. 50 )^2
by Pythagoras’ theorem
=14.16 voltstanφ=V
H=12. 50
− 6. 65=− 1. 8797from which φ=tan−^1 (−^1.^8797 )
=118.01◦or2.06 radians.Hence, vR=v 1 −v 2 = 14 .16sin(ωt+ 2. 06 )VThe phasor diagram is shown in Figure 30.18.v 15 15 V2 v 25 25 Vv 25 25 VvR308308Figure 30.18Problem 11. Determine20sinωt+10sin(
ωt+π
3)using horizontal and vertical componentsi 15 20 Ai 25 10 A608Figure 30.19From the phasors shown in Figure 30.19,
Total horizontal component,
H=20cos0◦+10cos60◦= 25. 0
Total vertical component,
V=20sin0◦+10sin60◦= 8. 66
By Pythagoras, the resultant,
iR=√[
25. 02 + 8. 662]
= 26 .46APhase angle,φ=tan−^1(
8. 66
25. 0)= 19. 11 ◦or 0 .333rad
Hence, by using horizontal and vertical components,20sinωt+10sin(
ωt+π
3)
= 26 .46sin(ωt+ 0. 333 )Now try the following Practice ExercisePracticeExercise 121 Resultant phasors by
horizontal and vertical components (answers
on page 353)In problems 1 to 5, express the combination of
periodic functions in the formAsin(ωt±α)by
horizontal and vertical components.- 7sinωt+5sin
(
ωt+π
4)- 6sinωt+3sin
(
ωt−π
6)- i=25sinωt−15sin
(
ωt+π
3)- v=8sinωt−5sin
(
ωt−π
4)- x=9sin
(
ωt+π
3)
−7sin(
ωt−3 π
8)- The voltage drops across two components
when connected in series across an a.c.
supply arev 1 =200sin314. 2 tand
v 2 =120sin( 314. 2 t−π/ 5 )volts
respectively. Determine
(a) the voltage of the supply (given by
v 1 +v 2 ) in the formAsin(ωt±α).