Alternating Current Analysis 229
Note that
AVm t dt A dt A T
TT
sin() 0 ,
0
22
∫∫ (^0)
and
V t dt V t dt
V
dt
V
T
m
T
m
T
mm
22
0
2
0
22
0
1
2
1
2
2
22
∫∫sin () cos( )
TT
∫
Then
V
T
AT
V
T
A
V
A
V
RMS
m
mm
1
2
2 2
2
2
2
2
2
2
Note that
V
m VVmRMSAC
2
∗0 707.
Then
VAVRMS^22 RMS AC ,()forvtAVtmsin()
If
v( )tAVmm11 2 2sin( tV) sin( t V)mnnsin( t)
then
VA
V
RMS mk AV
k
n
RMS
n
2
2
1
22
(^21)
∑∑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
tt
Ri t dt
t
t
1
21
2
1
2
∫ ()