274 Higher Engineering Mathematics
Hence, v 2 = v − v 1 = 30 sin 100 πt
− 20 sin(100 πt − 0.59)
= 30∠0 − 20∠ − 0.59 rad
= (30 + j 0) − (16.619 − j 11.127)
= 13.381 + j 11.127
= 17.40∠0.694 rad
Hence, by using complex numbers, the resultant
in sinusoidal form is:
v − v 1 = 30 sin 100 πt − 20 sin(100 πt − 0.59)
= 17.40 sin(ωt + 0.694) volts
(b) Supply frequency, f =
ω
2 π
=
100 π
2 π
= 50 Hz
(c) Periodic time, T =
1
f
=
1
50
= 0.02 s or 20 ms
(d) R.m.s. value of supply voltage, = 0.707 × 30
= 21.21 volts
Now try the following exercise
Exercise 111 Further problems on
resultant phasors by complex numbers
In Problems 1 to 4, express the combination of periodic functions in the form A sin(ωt ± α) by using
complex numbers:
1. 8 sin ωt + 5 sin
ωt +
π
4
[12.07 sin(ωt + 0.297)]
2. 6 sin ωt + 9 sin
ωt −
π
6
[14.51 sin(ωt − 0.315)]
3. v = 12 sin ωt − 5 sin
ωt −
π
4
[9.173 sin(ωt + 0.396)]
4. x = 10 sin
ωt +
π
3
− 8 sin
ωt −
3π
8
[16.168 sin(ωt + 1.451)]
5. The voltage drops across two components when connected in series across
an a.c. supply are: v 1 = 240 sin 314.2t and
v 2 = 150 sin(314.2t − π/5) volts respectively.
Determine the:
(a) voltage of the supply (given by v 1 + v 2 )
in the form A sin(ωt ± α).
(b) frequency of the supply.
[(a) 371.95 sin(314.2t − 0.239)V
(b) 50 Hz]
6. If the supply to a circuit is v = 25 sin200πt
volts and the voltage drop across one of
the components is v 1 = 18 sin(200πt − 0.43)
volts, calculate the:
(a) voltage drop across the remainder of
the circuit, given by v − v 1 , in the form
A sin(ωt ± α).
(b) supply frequency.
(c) periodic time of the supply.
[(a) 11.44 sin(200πt + 0.715)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 = 20 sin
300πt −
π
6
volts,
v 2 = 30 sin
300πt +
π
4
volts, and
v 3 = 60 sin
300πt −
π
3
volts.
Calculate the:
(a) supply voltage, in sinusoidal form, in the
form A sin(ωt ± α).
(b) frequency of the supply.
(c) periodic time.
(d) r.m.s. value of the supply voltage.
[(a) 79.73 sin(300π − 0.536) V
(b) 150 Hz (c) 6.667 ms (d) 56.37 V]
Hence, v 2 = v − v 1 = 30 sin 100 πt
− 20 sin(100 πt − 0.59)
= 30∠0 − 20∠ − 0.59 rad
= (30 + j 0) − (16.619 − j 11.127)
= 13.381 + j 11.127
= 17.40∠0.694 rad
Hence, by using complex numbers, the resultant
in sinusoidal form is:
v − v 1 = 30 sin 100 πt − 20 sin(100 πt − 0.59)
= 17.40 sin(ωt + 0.694) volts
(b) Supply frequency, f =
ω
2 π
=
100 π
2 π
= 50 Hz
(c) Periodic time, T =
1
f
=
1
50
= 0.02 s or 20 ms
(d) R.m.s. value of supply voltage, = 0.707 × 30
= 21.21 volts
Now try the following exercise
Exercise 111 Further problems on
resultant phasors by complex numbers
In Problems 1 to 4, express the combination of periodic functions in the form A sin(ωt ± α) by using
complex numbers:
1. 8 sin ωt + 5 sin
ωt +
π
4
[12.07 sin(ωt + 0.297)]
2. 6 sin ωt + 9 sin
ωt −
π
6
[14.51 sin(ωt − 0.315)]
3. v = 12 sin ωt − 5 sin
ωt −
π
4
[9.173 sin(ωt + 0.396)]
4. x = 10 sin
ωt +
π
3
− 8 sin
ωt −
3π
8
[16.168 sin(ωt + 1.451)]
5. The voltage drops across two components when connected in series across
an a.c. supply are: v 1 = 240 sin 314.2t and
v 2 = 150 sin(314.2t − π/5) volts respectively.
Determine the:
(a) voltage of the supply (given by v 1 + v 2 )
in the form A sin(ωt ± α).
(b) frequency of the supply.
[(a) 371.95 sin(314.2t − 0.239)V
(b) 50 Hz]
6. If the supply to a circuit is v = 25 sin200πt
volts and the voltage drop across one of
the components is v 1 = 18 sin(200πt − 0.43)
volts, calculate the:
(a) voltage drop across the remainder of
the circuit, given by v − v 1 , in the form
A sin(ωt ± α).
(b) supply frequency.
(c) periodic time of the supply.
[(a) 11.44 sin(200πt + 0.715)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 = 20 sin
300πt −
π
6
volts,
v 2 = 30 sin
300πt +
π
4
volts, and
v 3 = 60 sin
300πt −
π
3
volts.
Calculate the:
(a) supply voltage, in sinusoidal form, in the
form A sin(ωt ± α).
(b) frequency of the supply.
(c) periodic time.
(d) r.m.s. value of the supply voltage.
[(a) 79.73 sin(300π − 0.536) V
(b) 150 Hz (c) 6.667 ms (d) 56.37 V]