272 Higher Engineering Mathematics
5. The voltage drops across two components when connected in series across
an a.c. supply are: v 1 = 200 sin314.2t and
v 2 = 120 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) 305.3 sin(314.2t − 0.233)V
(b) 50 Hz]
6. If the supply to a circuit is v = 20 sin 628.3t
volts and the voltage drop across one of
the components is v 1 = 15 sin(628.3t − 0.52)
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) 10.21 sin(628.3t + 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 = 25 sin
300 πt +
π
6
volts,
v 2 = 40 sin
300 πt −
π
4
volts, and
v 3 = 50 sin
300 πt +
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.
[(a) 79.83 sin (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 functions may be found from their relative positions when
the time is zero. For example, if y 1 = 5 sinωt and y 2 =
4 sin(ωt − π/6) then each may be represented by phasors as shown in Fig. 25.20, y 1 being 5 units long and
drawn horizontally and y 2 being 4 units long, lagging
y 1 by π/6 radians or 30 ◦ . To determine the resultant of
y 1 + y 2 , y 1 is drawn horizontally as shown in Fig. 25.21
and y 2 is joined to the end of y 1 at π/6 radians, i.e. 30 ◦
to the horizontal. The resultant is given by y R .
y 1 5 5
/6 or 308
y 2 5 4
Figure 25.20
y 1 5 5
y 2 5 4
0
y R
a
b
308
Figure 25.21
In polar form, y R = 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 = 5 sinωt + 4 sin(ωt − π/6)
= 9.54 sin(ωt−0.21)
Problem 12. Two alternating voltages are given
by v 1 = 15 sin ωt volts and v 2 = 25 sin(ωt − π/6)
volts. Determine a sinusoidal expression for the
resultant v R = v 1 + v 2 by using complex numbers.
The relative positions of v 1 and v 2 at time t = 0 are
shown in Fig. 25.22(a) and the phasor diagram is shown
in Fig. 25.22(b).
5. The voltage drops across two components when connected in series across
an a.c. supply are: v 1 = 200 sin314.2t and
v 2 = 120 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) 305.3 sin(314.2t − 0.233)V
(b) 50 Hz]
6. If the supply to a circuit is v = 20 sin 628.3t
volts and the voltage drop across one of
the components is v 1 = 15 sin(628.3t − 0.52)
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) 10.21 sin(628.3t + 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 = 25 sin
300 πt +
π
6
volts,
v 2 = 40 sin
300 πt −
π
4
volts, and
v 3 = 50 sin
300 πt +
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.
[(a) 79.83 sin (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 functions may be found from their relative positions when
the time is zero. For example, if y 1 = 5 sinωt and y 2 =
4 sin(ωt − π/6) then each may be represented by phasors as shown in Fig. 25.20, y 1 being 5 units long and
drawn horizontally and y 2 being 4 units long, lagging
y 1 by π/6 radians or 30 ◦ . To determine the resultant of
y 1 + y 2 , y 1 is drawn horizontally as shown in Fig. 25.21
and y 2 is joined to the end of y 1 at π/6 radians, i.e. 30 ◦
to the horizontal. The resultant is given by y R .
y 1 5 5
/6 or 308
y 2 5 4
Figure 25.20
y 1 5 5
y 2 5 4
0
y R
a
b
308
Figure 25.21
In polar form, y R = 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 = 5 sinωt + 4 sin(ωt − π/6)
= 9.54 sin(ωt−0.21)
Problem 12. Two alternating voltages are given
by v 1 = 15 sin ωt volts and v 2 = 25 sin(ωt − π/6)
volts. Determine a sinusoidal expression for the
resultant v R = v 1 + v 2 by using complex numbers.
The relative positions of v 1 and v 2 at time t = 0 are
shown in Fig. 25.22(a) and the phasor diagram is shown
in Fig. 25.22(b).