TRIGONOMETRIC WAVEFORMS 159B
(c)Whent=10 msthenv=340 sin(
50 π10
103− 0. 541)=340 sin(1.0298)=340 sin 59◦
=291.4 V(d) Whenv=200 volts
then 200=340 sin(50πt− 0 .541)
200
340=sin(50πt− 0 .541)Hence (50πt− 0 .541)=arcsin200
340
= 36. 03 ◦or 0.6288 rad
50 πt= 0. 6288 + 0. 541
= 1. 1698Hence whenv=200 V,time,t=1. 1698
50 π=7.447 ms(e) When the voltage is a maximum,v=340 V.
Hence 340 =340 sin (50πt− 0 .541)
1 =sin (50πt− 0 .541)
50 πt− 0. 541 =arcsin 1
= 90 ◦or 1.5708 rad
50 πt= 1. 5708 + 0. 541 = 2. 1118Hence time,t=2. 1118
50 π=13.44 msA sketch ofv=340 sin(50πt− 0 .541) volts is shown
in Fig. 15.30.Figure 15.30Now try the following exercise.Exercise 71 Further problems on the
sinusoidal formAsin(ωt±α)In Problems 1 to 3 find the amplitude, peri-
odic time, frequency and phase angle (stating
whether it is leading or laggingAsinωt)ofthe
alternating quantities given.1.i=40 sin (50πt+ 0 .29) mA
[
40, 0.04 s, 25 Hz, 0.29 rad
(or 16◦ 37 ′) leading 40 sin 50πt]2.y=75 sin (40t− 0 .54) cm
[
75 cm, 0.157 s, 6.37 Hz, 0.54 rad
(or 30◦ 56 ′) lagging 75 sin 40t]3.v=300 sin (200[ πt− 0 .412) V
300 V, 0.01 s, 100 Hz, 0.412 rad
(or 23◦ 36 ′) lagging 300 sin 200πt]- A sinusoidal voltage has a maximum value
of 120 V and a frequency of 50 Hz. At time
t=0, the voltage is (a) zero, and (b) 50 V.
Express the instantaneous voltagevin the
formv=Asin(ωt±α)
[
(a)v=120 sin 100πtvolts
(b)v=120sin(100πt+ 0 .43) volts
]- An alternating current has a periodic time of
25 ms and a maximum value of 20 A. When
timet=0, currenti=−10 amperes. Express
the currentiin the formi=Asin(ωt±α)
[
i=20 sin
(
80 πt−π
6)
amperes]- An oscillating mechanism has a maximum
displacement of 3.2 m and a frequency of
50 Hz. At time t=0 the displacement is
150 cm. Express the displacement in the gen-
eral formAsin(ωt±α).
[3.2 sin(100πt+ 0 .488) m]- The current in an a.c. circuit at any time
tseconds is given by:
i=5 sin(100πt− 0 .432) amperes
Determine (a) the amplitude, periodic time,
frequency and phase angle (in degrees) (b) the
value of current att=0 (c) the value of