266 Higher Engineering Mathematics
amplitude, of the resultant is 3.6. The resultant waveform leads y 1 = 3 sin A by 34 ◦ or 34 ×
π
180
rad = 0.593
rad.
The sinusoidal expression for the resultant waveform is:
y R = 3.6 sin(A + 34
◦ ) or
y R = 3.6 sin(A + 0.593)
Problem 2. Plot the graphs of y 1 = 4 sinωt and
y 2 = 3 sin(ωt − π/3) on the same axes, over one
cycle. By adding ordinates at intervals plot
y R = y 1 + y 2 and obtain a sinusoidal expression for
the resultant waveform.
y 1 = 4 sinωt and y 2 = 3 sin(ωt − π/3) are shown plotted in Fig. 25.2.
908
y
y 1 5 4 sin t
y 2 5 3 sin(t 2 /3)
0
26
24
22
6
6.1
4
2
258
258
y R 5 y 1 1 y 2
t
1808
2708
3608
/2
3/2
2
Figure 25.2
Ordinates are added at 15 ◦ intervals and the resultant is shown by the broken line. The amplitude
of the resultant is 6.1 and it lags y 1 by 25 ◦
or 0.436 rad.
Hence, the sinusoidal expression for the resultant waveform is:
y R = 6.1 sin(ωt − 0.436)
Problem 3. Determine a sinusoidal expression
for y 1 − y 2 when y 1 = 4 sinωt and
y 2 = 3 sin(ωt − π/3).
y 1 and y 2 are shown plotted in Fig. 25.3. At 15 ◦ intervals
y 2 is subtracted from y 1 . For example:
at 0
◦
, y 1 − y 2 = 0 − (−2.6) = +2.6
at 30
◦
, y 1 − y 2 = 2 − (−1.5) = +3.5
at 150
◦
, y 1 − y 2 = 2 − 3 = −1, and so on.
908
y
y 1
0
24
22
4
2
3.6
458
y 2
y 1 2 y 2
t
1808
2708
3608
/2
3/2
2
Figure 25.3
The amplitude, or peak value of the resultant (shown by
the broken line), is 3.6 and it leads y 1 by 45 ◦ or 0.79
rad. Hence,
y 1 − y 2 = 3.6 sin(ωt + 0.79)
Problem 4. Two alternating currents are given by:
i 1 = 20 sin ωt amperes and
i 2 = 10 sin
ωt +
π
3
amperes.
By drawing the waveforms on the same axes and
adding, determine the sinusoidal expression for the
resultant i 1 + i 2 .
i 1 and i 2 are shown plotted in Fig. 25.4. The resultant
waveform for i 1 + i 2 is shown by the broken line. It has
the same period, and hence frequency, as i 1 and i 2 .
2 angle t
198
198
i 1 5 20 sin t
908
1808
2708
3608
230
220
210
10
20
26.5
30
3
2
2
i R 5 20 sin t 110 sin (t 1 )
3
i 2 5 10 sin(t 1 )
3
Figure 25.4
The amplitude or peak value is 26.5 A.
The resultant waveform leads the waveform of
i 1 = 20 sin ωt by 19 ◦ or 0.33 rad
amplitude, of the resultant is 3.6. The resultant waveform leads y 1 = 3 sin A by 34 ◦ or 34 ×
π
180
rad = 0.593
rad.
The sinusoidal expression for the resultant waveform is:
y R = 3.6 sin(A + 34
◦ ) or
y R = 3.6 sin(A + 0.593)
Problem 2. Plot the graphs of y 1 = 4 sinωt and
y 2 = 3 sin(ωt − π/3) on the same axes, over one
cycle. By adding ordinates at intervals plot
y R = y 1 + y 2 and obtain a sinusoidal expression for
the resultant waveform.
y 1 = 4 sinωt and y 2 = 3 sin(ωt − π/3) are shown plotted in Fig. 25.2.
908
y
y 1 5 4 sin t
y 2 5 3 sin(t 2 /3)
0
26
24
22
6
6.1
4
2
258
258
y R 5 y 1 1 y 2
t
1808
2708
3608
/2
3/2
2
Figure 25.2
Ordinates are added at 15 ◦ intervals and the resultant is shown by the broken line. The amplitude
of the resultant is 6.1 and it lags y 1 by 25 ◦
or 0.436 rad.
Hence, the sinusoidal expression for the resultant waveform is:
y R = 6.1 sin(ωt − 0.436)
Problem 3. Determine a sinusoidal expression
for y 1 − y 2 when y 1 = 4 sinωt and
y 2 = 3 sin(ωt − π/3).
y 1 and y 2 are shown plotted in Fig. 25.3. At 15 ◦ intervals
y 2 is subtracted from y 1 . For example:
at 0
◦
, y 1 − y 2 = 0 − (−2.6) = +2.6
at 30
◦
, y 1 − y 2 = 2 − (−1.5) = +3.5
at 150
◦
, y 1 − y 2 = 2 − 3 = −1, and so on.
908
y
y 1
0
24
22
4
2
3.6
458
y 2
y 1 2 y 2
t
1808
2708
3608
/2
3/2
2
Figure 25.3
The amplitude, or peak value of the resultant (shown by
the broken line), is 3.6 and it leads y 1 by 45 ◦ or 0.79
rad. Hence,
y 1 − y 2 = 3.6 sin(ωt + 0.79)
Problem 4. Two alternating currents are given by:
i 1 = 20 sin ωt amperes and
i 2 = 10 sin
ωt +
π
3
amperes.
By drawing the waveforms on the same axes and
adding, determine the sinusoidal expression for the
resultant i 1 + i 2 .
i 1 and i 2 are shown plotted in Fig. 25.4. The resultant
waveform for i 1 + i 2 is shown by the broken line. It has
the same period, and hence frequency, as i 1 and i 2 .
2 angle t
198
198
i 1 5 20 sin t
908
1808
2708
3608
230
220
210
10
20
26.5
30
3
2
2
i R 5 20 sin t 110 sin (t 1 )
3
i 2 5 10 sin(t 1 )
3
Figure 25.4
The amplitude or peak value is 26.5 A.
The resultant waveform leads the waveform of
i 1 = 20 sin ωt by 19 ◦ or 0.33 rad