268 Higher Engineering Mathematics
Problem 5. Two alternating currents are
given by:i 1 =20sinωtamperes and
i 2 =10sin
(
ωt+
π
3
)
amperes. Determinei 1 +i 2
by drawing phasors.
Therelativepositionsofi 1 andi 2 attimet=0areshown
as phasors in Fig. 25.9, where
π
3
rad= 60 ◦.
The phasor diagram in Fig. 25.10 is drawn to scale with
a ruler and protractor.
i 15 20 A
i 25 10 A
608
Figure 25.9
i 25 10 A
i 15 20 A
iR
608
Figure 25.10
The resultantiRis shown and is measured as 26 A and
angleφas 19◦or 0.33 rad leadingi 1. Hence, by drawing
and measuring:
iR=i 1 +i 2 =26sin(ωt+ 0. 33 )A
Problem 6. For the currents in Problem 5,
determinei 1 −i 2 by drawing phasors.
At timet=0, currenti 1 is drawn 20 units long hor-
izontally as shown by 0ain Fig. 25.11. Currenti 2 is
shown, drawn 10 units long in broken line and lead-
ingby60◦. The current−i 2 is drawn in the opposite
direction to the broken line ofi 2 ,shownasabin
Fig. 25.11. The resultantiRis given by 0blagging by
angleφ.
By measurement, iR=17A and φ= 30 ◦ or
0.52 rad
Hence, by drawing phasors:
iR=i 1 −i 2 =17sin(ωt− 0. (^52) )
20 A
a
10 A
iR
b
0
2 10A
608
Figure 25.11
Now try the following exercise
Exercise 108 Further problems on
determining resultant phasors by
drawing
- Determine a sinusoidal expression for
2sinθ+4cosθby drawing phasors.
[4.5sin(A+ 63. 5 ◦)] - Ifv 1 =10sinωtvoltsandv 2 =14sin(ωt+π/3)
volts, determine by drawing phasors
sinusoidal expressions for (a) v 1 +v 2
(b)v 1 −v 2. [
(a) 20.9sin(ωt+ 0. 62 )volts
(b) 12.5sin(ωt− 1. 33 )volts
]
- Express 12sinωt+5cosωt in the form
Rsin(ωt±α)by drawing phasors.
[13sin(ωt+ 0. 40 )]
25.4 Determining resultant phasors
by the sine and cosine rules
As stated earlier, the resultant of two periodic func-
tions may be found from their relative positions when
the time is zero. For example, ify 1 =5sinωtandy 2 =
4sin(ωt−π/ 6 )then each may be represented by pha-
sors as shown in Fig. 25.12,y 1 being 5 units long and
drawn horizontally andy 2 being 4 units long, lagging
y 1 byπ/6 radians or 30◦. To determine the resultant of
y 1 +y 2 ,y 1 is drawn horizontallyas shown in Fig. 25.13
andy 2 is joined to the end ofy 1 atπ/6 radians, i.e. 30◦
to the horizontal. The resultant is given byyR.