082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

476 Chapter 15 Applications of integration
Example15.4 Calculate the r.m.s.value of f (t) =Asin(ωt + φ) across [0,2π/ω].
√ ∫ 2π/ω
A
0
Solution 2 sin 2 (ωt + φ)dt
r.m.s. =
2π/ω
√
∫
A
=
2 ω 2π/ω
1−cos2(ωt +φ)dt
4π 0
√
[
]
A
=
2 ω sin2(ωt + φ) 2π/ω
t −
4π 2ω
0
=
√
(
A 2 ω 2π
4π ω
−
sin2(2π + φ)
2ω
+ sin2φ )
2ω
Now sin2(2π + φ) = sin(4π + 2φ) and since sin(t + φ) has period 2π we see that
sin(4π +2φ) = sin2φ. Hence,
√ √
A2 ω 2π A
r.m.s. =
4π ω = 2
2 = √ A = 0.707A
2
Notethatsin(ωt + φ)hasperiod2π/ω.TheresultofExample15.4illustratesageneral
result:
The r.m.s. value of any sinusoidal waveform taken across an interval of length one
periodis
0.707 ×amplitudeofthe waveform
Root mean square value is an effective measure of the energy transfer capability of a
time-varyingelectricalcurrent.Toseewhythisisso,considerthefollowingEngineering
application.
Engineeringapplication15.3
Average power developed across a resistor by a time-varying
current
Consideracurrenti(t)whichdevelopsapowerp(t)inaloadresistorR.Thiscurrent
flowsfromtimet =t 1
totimet =t 2
.LetP av
betheaveragepowerdissipatedbythe
resistor during the time interval [ ]
t 1
,t 2 . We require that total energy transfer,E,be
the same inboth cases. So wehave
E =P av
(t 2
−t 1
)=
∫ t2
t 1
p(t)dt