Task 1 V = 340 sin (50πt − 0.541) Answer Sine wave in form: V =

Task 1
V = 340 sin (50πt − 0.541)
Answer
Sine wave in form:
V = Vm sin (ωt − φ)
has an amplitude of Vm, angular velocity of ω and phase angle (lagging) of φ
radians.
a)
Comparing the given sine wave equation with the standard form, we have:
Amplitude = 340 V
Angular velocity : ω = 50π
frequency : f = ω
2π = 50π
2π = 25 Hz
Time Period : T = 1
f = 1
25 = 0.04 sec
Phase Angle : φ = 0.541 radians (lagging) or =⇒ 0.541 × 180
π = 31◦ (lagging)
b)
When t = 0:
V = 340 sin (50π · 0 − 0.541)
= 340 sin (−0.541) = −175.0978 V
c)
When t = 10 ms:
V = 340 sin (50π · 10−2 − 0.541)
= 340 sin (0.5π − 0.541) = 291.446 V
d)
1

When V = 200 V
200 = 340 sin (50πt − 0.541)
=⇒ sin (50πt − 0.541) = 200
340
50πt − 0.541 = sin−1
( 200
340
)
50πt − 0.541 = 0.6289
∴ t = 0.6289 + 0.541
50π = 0.007 sec
The sine wave first reaches 200 V at the time t = 0.007 sec.
e)
Maximum voltage occurs when the sine term of the sine wave is maximum, that
is sin (50πt − 0.541) = 1. Implies,
Vmax = 340 · max[sin (50πt − 0.541)]
︸ ︷︷ ︸
=1
=⇒ sin (50πt − 0.541) = 1
∴ 50πt − 0.541 = sin−1(1)
[sin−1(1) = π
2 + 2n, where, n = 0, 1, 2, . . . ]
∴ 50πt − 0.541 = π
2 + 2n
=⇒ t = 1
50π
( π
2 + 2n + 0.541
)
The sine wave first reaches it’s maximum value, when n = 0, that is:
t = 1
50π
( π
2 + 0.541
)
= 0.0134 sec
f )
The given sine function is plotted using a graphing calculator (Desmos). The
y-axis is scaled by a factor of 10000, so 1 unit which is seen to be 0.005 V
actually represents 50 V.
2

Figure 1: Voltage sine wave
Task 2
3 sin ωt + 4 cos ωt
Answer
R sin (ωt + α) = 3 sin ωt + 4 cos ωt
[Trig. formula: sin (a + b) = sin a cos b + cos a sin b]
=⇒ R cos α sin ωt + R sin α cos ωt = 3 sin ωt + 4 cos ωt
[Comparing LHS and RHS we get:]
R cos α = 3 (1)
R sin α = 4 (2)
[Squaring and adding (1) and (2)]
R2 sin2 α + R2 cos2 α = 42 + 32
=⇒ R2 = 25 or R = 5
[Dividing (2) by (1) :]
tan α = 4
3
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