Higher Engineering Mathematics, Sixth Edition

272 Higher Engineering Mathematics

5. The voltage drops across two compo-
   nents when connected in series across
   an a.c. supply are: v 1 =200sin314. 2 t and
   v 2 =120sin( 314. 2 t−π/ 5 )volts respectively.
   Determine the:
   (a) voltage of the supply (given byv 1 +v 2 )
   in the formAsin(ωt±α).
   (b) frequency of the supply.
   [(a) 305.3sin( 314. 2 t− 0. 233 )V
   (b) 50 Hz]


6. If the supply to a circuit isv=20sin628. 3 t
   volts and the voltage drop across one of
   the components isv 1 =15sin( 628. 3 t− 0. 52 )
   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.
   [(a) 10.21sin( 628. 3 t+ 0. 818 )V
   (b) 100 Hz (c) 10 ms]


7. The voltages across three components in a
   series circuit when connected across an a.c.
   supply are:

v 1 =25sin

(

300 πt+

π

6

)

volts,

v 2 =40sin

(

300 πt−

π

4

)

volts,and

v 3 =50sin

(

300 πt+

π

3

)

volts.

Calculate the:

(a) supply voltage, in sinusoidal form, in the

formAsin(ωt±α).

(b) frequency of the supply.

(c) periodic time.

[(a) 79.83sin( 300 πt+ 0. 352 )V

(b) 150 Hz (c) 6.667 ms]

25.6 Determining resultant phasors

by complex numbers

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. 25.20,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. 25.21

andy 2 is joined to the end ofy 1 atπ/6 radians, i.e. 30◦

to the horizontal. The resultant is given byyR.

y 155

/6 or 30 8

y 254

Figure 25.20

y 155

y (^25)
4
(^0) 
yR
a
b
308
Figure 25.21
In polar form,yR= 5 ∠ 0 + 4 ∠−
π
6
= 5 ∠ 0 ◦+ 4 ∠− 30 ◦
=( 5 +j 0 )+( 4. 33 −j 2. 0 )
= 9. 33 −j 2. 0 = 9. 54 ∠− 12. 10 ◦
= 9. 54 ∠− 0 .21rad
Hence, by using complex numbers, the resultant in
sinusoidal form is:
y 1 +y 2 =5sinωt+4sin(ωt−π/ 6 )
= 9 .54sin(ωt− 0 .21)
Problem 12. Two alternating voltages are given
byv 1 =15sinωtvolts andv 2 =25sin(ωt−π/ 6 )
volts. Determine a sinusoidal expression for the
resultantvR=v 1 +v 2 by using complex numbers.
The relative positions ofv 1 andv 2 at timet=0are
shown in Fig. 25.22(a) and the phasor diagram is shown
in Fig. 25.22(b).