You might be tempted to now say that
P
in
=
V
in
I
in
P
in
V
in
I
in
, but that is incorrect for sinusoidal excitation.
V
in
V
in
and
I
in
I
in
are phasors! So let's digress for
a second to see (or review, I hope) how to find power when the
voltage and current are phasor quantities. What really matters
is not the absolute phase angle of the two quantities, but
rather the phase angle between them. Suppose we have a voltage
phasor, VV which has zero phase
angle and a complex impedance
Z=|Z|ⅇⅈ
θ
z
Z
Z
θ
z
. Obviously, the current is given by
I
˜
=
V
˜
Z
˜
=|V||Z|ⅇ-
θ
z
I
˜
V
˜
Z
˜
V
Z
θ
z
(1)
To find power, we can not work just with phasors, we have to go
back to the complete function of time as well so we write:
Vt=|V|cosωt
V
t
V
ω
t
(2)
It=|V||Z|cosωt−
θ
z
I
t
V
Z
ω
t
θ
z
(3)
It=|I|cosωt−
θ
z
I
t
I
ω
t
θ
z
(4)
The power as a function of time is given as
Pt=ItVt=|V||V|cosωtcosωt−
θ
z
P
t
I
t
V
t
V
V
ω
t
ω
t
θ
z
(5)
We remember a useful trig identity:
cosA−B=cosAcosB+sinAsinB
A
B
A
B
A
B
(6)
Hence:
cosωt−
θ
z
=cosωtcos
θ
z
+sinωtsin
θ
z
ω
t
θ
z
ω
t
θ
z
ω
t
θ
z
(7)
which makes
Pt
P
t
Pt=cos2ωtcos
θ
z
+cosωtsinωtsin
θ
z
P
t
ω
t
2
θ
z
ω
t
ω
t
θ
z
(8)
We are really interested in finding
average
power since energy which flows into and then back
out of the line does no work for us. Clearly the second term
in
Equation 8 (going as
cosωtsinωt
ω
t
ω
t
) has an average value of zero, and so we can forget
about it. Time for one more trig identity:
cos2A=12+12cos2A
A
2
1
2
1
2
2
A
(9)
cos2ωt
2
ω
t
has zero average value as well, so we are left with
the following for the average value of the power
Pt¯
P
t
Pt¯=|V||I|2cos
θ
z
=|V|22|Z|cos
θ
z
P
t
V
I
2
θ
z
V
2
2
Z
θ
z
(10)
Note that one useful way that people sometimes use to express
this is to say
Pt¯=12
V
˜
V
˜
*
P
t
1
2
V
˜
V
˜
*
(11)
Back to our example:
V
in
=4.18∠38
V
in
4.1838
and
I
in
=0.144∠21
I
in
0.14421
Thus
P
in
t¯=124.18×0.144cos59Watts=0.155
P
in
t
1
2
4.18
0.144
59
Watts
0.155
(12)
As an alternative way of calculating the power into
the line note that we
know the magnitude of
the current through both the capacitor and the resistor of the
apparent
Z
in
Z
in
. They are just two elements in series, and so they both
have the same current flowing through them, namely,
I
in
I
in
. No power is dissipated in the capacitor, so we could
just as well have said
P
in
t¯=12|I|2R=120.144215=0.155
P
in
t
1
2
I
2
R
1
2
0.144
2
15
0.155
(13)
and gotten the answer in an even easier fashion!
(Note that we still have to keep the factor of "1/2" to account
for the time average of a sinusoidal product.) For reasons I do
not understand, students have always had an aversion to finding
power. It is not that hard, and in the end, is usually the
"bottom line" with regard to how a system will perform. Go back
over this section until it makes sense, as you may see power
crop up someplace else one of these days!
