Exercicios resolvidos valor eficaz RMS

1. Exercícios resolvidos – aula 2:
1a) Calcular o valor médio e eficaz:
1b)
[ ] 5
2
1010
10
11
2
0
2
0
0
0
=





==== ∫∫
+
T
T
t
T
dt
T
dt)t(f
T
V
/T
/T
Tt
t
med
[ ] [ ] 07750
2
100100
10
11 2
0
2
0
22
0
0
,
T
T
t
T
dt
T
dt)t(f
T
V
/T
/TTt
t
rms ==





==== ∫∫
+
[ ] [ ]( ) ( )
( ) ( ) 00
10
1
5220510
10
1
25675820
2
10
1
2010
1
2010
1 758
256
2
0
758
256
2
0
==−=
=





−−=−=








−+= ∫∫
m
m,.m.
m
m,m,
T
.
T
tt
T
dtdt
T
V
m,
m,
/T
m,
m,
/T
med
( ) [ ] [ ]( )
( ) ( )
( ) 2512
10
51
524005100
10
1
524005100
10
1
256758400
2
100
1
400100
1
2010
1 758
256
2
0
2
0
758
256
22
,
m
,
m,.m.
m
m,.m.
m
m,m,
T
T
t.t.
T
dtdt
T
V
m,
m,
/T
/T m,
m,
rms
==+=
=+=





−+=
=+=








−+= ∫ ∫

2. 1c)
( )
[ ] [ ]( ) ( ) 15115215
2
1
515
2
1
515
2
1
515
2
1
2
0
2
0
2
0
2
0
2
0
=−−=−+=
=







+=







+= ∫ ∫∫
)(.)tcos(t
td)t(sentdtd)t(senVmed
π
π
ωω
π
ωωω
π
ωω
π
ππ
π ππ
( ) ( )
[ ] [ ]
( ) ( ) ( )
( ) 41151492
2
1
252225
2
1
252225
2
1
4
22
2
2
25111502225
2
1
4
2
2
25150225
2
1
25150225
2
1
25150225
2
1
515
2
1
2
0
2
0
2
0
2
0
2
0
2
2
0
2
0
2
2
0
2
,
....
.sen
.
tsent
)tcos(t
td)t(sentd)t(sentd
td)t(sen)t(sentd)t(senVrms
==
=+=+=











−+−−=
=












−+−+=
=







++=
=







++=







+=
∫ ∫∫
∫∫
π
ππ
π
ππ
π
ππ
π
π
ωω
ωω
π
ωωωωω
π
ωωω
π
ωω
π
π
ππ
π ππ
ππ