EE201
Response of the basic elements to AC / Phasor Notation
Denard Lynch
Page 1 of 4
Oct 2009
Let’s go back and look at the voltage w.r.t. time in a RLC series circuit:
Take, for now, as a given:
i(t) = 2.48sin(377t)A
The voltage is also sinusoidal, of the
same frequency, and a magnitude of
170V
peak
, but unknown phase.
First, find R, X
L
and X
C
:
R = 47
Ω
(given)
X
L
=
ω
L = (377)(.1H) = 37.7
Ω
X
C
= 1/
ω
C = 1/{(377)(100X10
6
F)} = 26.5
Ω
Next, find the maximum voltages across the elements:
V
R
= I
M
R = (2.48A)(47
Ω
) = 116.6V
V
L
= I
M
X
L
= (2.48A)(37.7
Ω
) = 93.5V
V
C
= I
M
X
C
= (2.48A)(26.5
Ω
) = 65.7V
If we simply add these up we get 275.8V!!
Clearly this would violate Kirchhoff’s
voltage law as we know it!
To help understand why, let’s plot these voltages:
We can now see, as we noted earlier, that the various voltages all have differing phases
with respect to each other.
The problem we face is how to add sine waves of differing
phases and frequencies?
Here are three options to consider:
1) Graphically:
Use the plotted waveforms and graphically add them to get the resultant waveform.
Looking at the plot above, we can easily check the validity of this method by testing a
few points.
For example, if we look at values of the component waves at
t=0: v
R
= 0V
,
v
L
= max = 93.5V
and
v
C
= min = 65.7V
.
If we add these we see the result in the ‘sum’
waveform:
27.8V
.
Also, if we look at the point in time when
v
R
= v
C
= 0V
, the ‘sum’ =
v
R
and we can see that the curves cross at that point.
We
could
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 Spring '10
 Linch
 Alternating Current, Series Circuit, Volt, Complex number, Sine wave, phasor notation

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