Alternating Current Analysis
229
Note that
AV
t dt
A dt A T
m
T
T
sin( )
,
ϭ
ϭ
0
0
2
2
0
∫
∫
and
V
t dt
V
t dt
V dt
V T
m
T
m
T
m
m
2
2
0
2
0
2
2
0
1
2
1
2
2
2
2
sin ( )
cos(
)
∫
∫

 

 
ϭ
ϭ
ϭ
Ϫ
T T
∫
Then
V
T
A T
V T
A
V
A
V
RMS
m
m
m
ϭ
ϩ
ϭ
ϩ
ϭ
ϩ
1
2
2
2
2
2
2
2
2
2







 

 
Note that
V
V
V
m
m
R M S A C
2
0 707
ϭ
ϭ
∗ .
Ϫ
Then
V
A
V
t
A V
t
RMS
RMS AC
m
ϭ
ϭ
2
2
ϩ
ϩ
Ϫ
,
()
sin( )
for v
If
v( )
sin( )
sin( )
sin( )
t
A V
t V
t
V
t
m
m
m n
n
ϭ ϩ
ϩ
ϩ
1
1
2
2
then
V
A
V
A
V
RMS
mk
k
n
RMS
n
ϭ
ϭ
ϭ
ϭ
2
2
1
2
2
1
2
ϩ
ϩ
Ϫ
∑
∑
k
k
for k AC sources and one DC source (A).
R.3.12 The power dissipated by a resistor R, having a current i(t) over an interval of time
[t 1 , t 2 ], is given by
P
t
t
Ri t dt
t
t
ϭ Ϫ
1
2
1
2
1
2
( )
∫
CRC_47760_CH003.indd 229
CRC_47760_CH003.indd 229
7/23/2008 1:27:29 PM
7/23/2008 1:27:29 PM