OpenStax-CNX module: m11391
1
Power
∗
Bill Wilson
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 1.0†
Abstract
How to calculate the power that ows into and out of a line, as well as average power.
You might be tempted to now say that Pin = Vin Iin , but that is incorrect for sinusoidal excitation. Vin
and Iin are phasors! So let's digress for a second to see (or review, I hope) how to nd 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, V which has zero
phase angle and a complex impedance Z = |Z|eiθz . Obviously, the current is given by
I˜ =
=
Ṽ
Z̃
|V | −θz
|Z| e
(1)
To nd power, we can not work just with phasors, we have to go back to the complete function of time as
well so we write:
V (t) = |V |cos (ωt)
(2)
|V |
cos (ωt − θz )
|Z|
(3)
I (t) = |I|cos (ωt − θz )
(4)
I (t) =
The power as a function of time is given as
P (t)
=
I (t) V (t)
=
|V ||V |cos (ωt) cos (ωt − θz )
(5)
We remember a useful trig identity:
cos (A − B) = cos (A) cos (B) + sin (A) sin (B)
(6)
cos (ωt − θz ) = cos (ωt) cos (θz ) + sin (ωt) sin (θz )
(7)
P (t) = cos2 (ωt) cos (θz ) + cos (ωt) sin (ωt) sin (θz )
(8)
Hence:
which makes P (t)
∗ Version
1.1: Jun 20, 2003 12:00 am -0500
† http://creativecommons.org/licenses/by/1.0
http://cnx.org/content/m11391/1.1/
OpenStax-CNX module: m11391
2
We are really interested in nding average power since energy which ows into and then back out of the
line does no work for us. Clearly the second term in (8) (going as cos (ωt) sin (ωt)) has an average value of
zero, and so we can forget about it. Time for one more trig identity:
cos2 (A) =
1 1
+ cos (2A)
2 2
(9)
cos (2ωt) has zero average value as well, so we are left with the following for the average value of the power
−
P (t)
−
P (t)
=
=
|V ||I|
2 cos (θz )
(|V |)2
2|Z| cos (θz )
(10)
Note that one useful way that people sometimes use to express this is to say
−
P (t)=
1
Ṽ Ṽ ∗
2
(11)
Back to our example: Vin = 4.18∠38 and Iin = 0.144∠21 Thus
−
Pin (t)
(4.18 × 0.144) cos (59) Watts
=
1
2
=
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 Zin . They are just two elements in series, and so
they both have the same current owing through them, namely, Iin . No power is dissipated in the capacitor,
so we could just as well have said
−
Pin (t)
=
2
1
2 (|I|) R
2
1
15
2 0.144
=
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 nding 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!
http://cnx.org/content/m11391/1.1/