OpenStax-CNX module: m21475
1
Phasors: Sinusoidal Steady State
∗
and the Series RLC Circuit
Louis Scharf
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Phasors may be used to analyze the behavior of electrical and mechanical systems that have reached a kind of
equilibrium called sinusoidal steady state. In the sinusoidal steady state, every voltage and current (or force
and velocity) in a system is sinusoidal with angular frequency
ω.
However, the amplitudes and phases of
these sinusoidal voltages and currents are all dierent. For example, the voltage across a resistor might lead
◦ π
π
2 radians) and lag the voltage across an inductor by 90
2 radians .
In order to make our application of phasors to electrical systems concrete, we consider the series RLC
the voltage across a capacitor by
90◦
(
circuit illustrated in Figure 1. The arrow labeled
i (t)
denotes a current that ows in response to the voltage
applied,and the + and - on the voltage source indicate that the polarity of the applied voltage is positive on
the top and negative on the bottom. Our convention is that current ows from positive to negative, in this
case clockwise in the circuit.
∗ Version 1.6: Sep 17, 2009 11:35 am -0500
† http://creativecommons.org/licenses/by/3.0/
http://cnx.org/content/m21475/1.6/
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Figure 1: Series RLC Circuit
We will assume that the voltage source is an audio oscillator that pro- duces the voltage
V (t) = Acos (ωt + φ) .
(1)
We represent this voltage as the complex signal
V (t) ↔ Aejφ ejωt
(2)
V (t) ↔ V ; V = Aejφ .
(3)
and give it the phasor representation
We then describe the voltage source by the phasor V and remember that we can always compute the actual
voltage by multiplying by
ejωt
and taking the real part:
V (t) = Re{V ejωt }.
Exercise 1
Show that
Circuit Laws.
Re V ejωt = Acos (ωt + φ)
when
(4)
V = Aejφ .
In your circuits classes you will study the Kirchho laws that govern the low frequency
behavior of circuits built from resistors
(R),
inductors
(L),
and capacitors
(C).
In your study you will learn
that the voltage dropped across a resistor is related to the current that ows through it by the equation
VR (t) = Ri (t) .
(5)
You will learn that the voltage dropped across an inductor is proportional to the derivative of the current
that ows through it, and the voltage dropped across a capacitor is proportional to the integral of the current
that ows through it:
di
(t)
VL (t) = L dt
R
1
VC (t) = C i (t) dt.
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(6)
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Phasors and Complex Impedance.
Now suppose that the current in the preceding equations is sinu-
soidal, of the form
We may rewrite
where
I
i (t)
(7)
i (t) = Re{Iejωt }
(8)
as
is the phasor representation of
Exercise 2
Find the phasor
i (t) = B cos (ωt + θ) .
I
B
in terms of
i (t).
θ
and
in (8).
The voltage dropped across the resistor is
VR (t)
=
Ri (t)
=
RRe{Iejωt }
=
Thus the phasor representation for
VR (t)
Re{RIe
jωt
(9)
}.
is
VR (t) ↔ VR ; VR = RI.
(10)
R
We call R the impedance of the resistor because R is the scale constant that relates the phasor voltage V '
to the phasor current I.
The voltage dropped across the inductor is
d
di
(t) = L Re{Iejωt }.
dt
dt
Re [ ] operator (see Exercise
VL (t) = L
The derivative may be moved through the
VL (t)
Thus the phasor representation of
=
LRe{jωIejωt }
=
Re{jωLIejωt }.
(11)
) to produce the result
VL (t)
VL (t) ↔ VL ; VL = jωLI.
jωL the impedance of the
VL ' to phasor current I .
We call
voltage
Exercise 3
Prove that the operators
(12)
inductor because
d
dt and
Re []
jωL
(13)
is the complex scale constant that relates phasor
commute:
d
d
Re{ejωt } = Re{ ejωt }.
dt
dt
(14)
The voltage dropped across the capacitor is
VC (t) =
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1
C
Z
i (t) dt =
1
C
Z
Re{Iejωt }dt.
(15)
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Re [ ]
The integral may be moved through the
VC (t)
operator to produce the result
I jωt
1
}
C Re{ jω e
I
Re{ jωC
ejωt }.
=
=
Thus the phasor representation of
VC (t)
(16)
is
VC (t) ↔ VC ; VC =
I
jωC
(17)
1
1
jωC the impedance of the capacitor because jωC is the complex scale constant that relates phasor
voltage VC " to phasor current I .
We call
Kirchho's Voltage Law.
nation of
R, L,
and
C
Kirchho 's voltage law says that the voltage dropped in the series combi-
illustrated in Figure 1 equals the voltage generated by the source (this is one of two
fundamental conservation laws in circuit theory, the other being a conservation law for current):
V (t) = VR (t) + VL (t) + VC (t) .
(18)
If we replace all of these voltages by their complex representations, we have
Re{V ejωt } = Re{(VR + VL + VC ) ejωt }.
(19)
An obvious solution is
V
=
=
VR + VL + VC
R + jωL +
1
jωC
(20)
I
where I is the phasor representation for the current that ows in the circuit. This solution is illustrated in
Figure 2, where the phasor voltages
RI, jωLI ,
and
1
jωC I are forced to add up to the phasor voltage
Figure 2: Phasor Addition to Satisfy Kirchho's Law
Exercise 4
Redraw Figure 2 for
R = ωL =
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1
ωC
= 1.
V.
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Impedance.
5
We call the complex number
1
R+jωL+ jωC
the complex impedance for the series RLC network
V
because it is the complex number that relates the phasor voltage
to the phasor current
I:
V = ZI
Z = R + jωL +
The complex number
(C),
Z
(21)
1
jωC .
depends on the numerical values of resistance
but it also depends on the angular frequency
(ω)
(R),
inductance
(L),
and capacitance
used for the sinusoidal source. This impedance may
be manipulated as follows to put it into an illuminating form:
Z
1
R + j ωL − ωC
q √
L
ω LC − ω√1LC .
= R+j C
=
(22)
√1
is a parameter that you will learn to call an "undamped natural frequency" in
LC
your more advanced circuits courses. With it, we may write the impedance as
The parameter
ω0 =
Z = R + jω0 L
ω
ω0
−
ω0
ω
.
(23)
ν.
Then the impedence, as a function of
1
Z (ν) = R + jω0 L ν −
.
ν
(24)
ω
ω0 is a normalized frequency that we denote by
normalized frequency, is
The frequency
When the normalized frequency equals one
(ν = 1),
then the impedance is entirely real and
Z = R.
The
circuit looks like it is a single resistor.
h
|Z (ν) | = R 1 +
ω0 L 2
R
i1/2
1 2
ν
ν−
0L
ν−
argZ (ν) = tan−1 ωR
The impedance obeys the following symmetries around
1
ν
.
(25)
.
ν = 1:
Z (ν) = Z ∗
|Z (ν) | = |Z
1
ν
1
ν |
arg Z (ν) = − arg Z
(26)
1
ν
.
In the next paragraph we show how this impedance function inuences the current that ows in the circuit.
Resonance.
The phasor representation for the current that ows the current that ows in the series
RLC circuit is
I
=
=
1
|Z(ν)| e
V
Z(ν)
−jargZ(ν)
V
(27)
1
H (ν) = Z(ν)
displays a "resonance phenomenon." that is, |H (ν) | peaks at ν = 1 and decreases
ν = 0 and ν = ∞:
The function
to zero and
|H (ν) | = {
http://cnx.org/content/m21475/1.6/
0,
ν=0
1
R
ν=1
0,
ν = ∞.
(28)
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|H (ν) | = 0, no current ows.
|H (ν) | is plotted against the normalized frequency ν = ωω0 in Figure 3.14. The resonance
1
occurs at ν = 1, where |H (ν) | =
R meaning that the circuit looks purely resistive. Resonance phe-
When
The function
peak
nomena underlie the frequency selectivity of all electrical and mechanical networks.
Figure 3: Resonance in a Series RLC Circuit
Exercise 5
(MATLAB) Write a MATLAB program to compute and plot
|H (ν) |
argH (ν) versus ν for ν
L
ω0 R
= 10, 1, 0.1, and 0.01,
and
ranging from 0.1 to 10 in steps of 0.1. Carry out your computations for
and overplot your results.
Circle Criterion and Power Factor.
1
Z(ν)
brings insight into the resonance of an RLC circuit and illustrates the equency selectivity of the circuit.
Our study of the impedance
Z (ν)
and the function
H (ν) =
But there is more that we can do to illuminate the behavior of the circuit.
1
V = RI + j ωL −
ωC
I.
(29)
This equation shows how voltage is divided between resistor voltage RI and inductor-capacitor voltage
j ωL −
1
ωC
I.
V = RI + jω0 L
ω
ω0
−
ω0
ω
I
(30)
RI.
(31)
or
V = RI +
jω0 L
R
ν−
1
ν
In order to simplify our notation, we can write this equation as
V = VR + jk (ν) VR
where
VR
is the phasor voltage
RI
and
k (ν)
is the real variable
k (ν) =
(32) brings very important geometrical insights.
circuit is complex, the terms
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VR
and
(32)
jk (ν) VR
ω0 L
R
ν−
1
ν
.
(33)
VR in the RLC
π
radians. This means that, for every
2
First, even though the phasor voltage
are out of phase by
OpenStax-CNX module: m21475
allowable value of
V.
of
7
VR , the corresponding jk (ν) VR
must add in a right triangle to produce the source voltage
This is illustrated in Figure 4(a). As the frequency
VR
and
jk (ν) VR
that sum to
V.
ν
k (ν) changes, producing other values
jk (ν) VR are illustrated in Figure
phasor voltage V>R lies on a circle of radius
changes, then
Several such solutions for
3.15(b). From the gure we gain the clear impression that the
VR
and
V
V
2 centered at 2 Let's try this solution,
VR
V
2
=
=
V
2
V
2
ejψ
1 + ejψ ,
+
(34)
and explore its consequences. When this solution is substituted into (32), the result is
V
V
1 + ejψ + jk (ν)
1 + ejψ
2
2
V =
(35)
or
2 = 1 + ejψ [1 + jk (ν)] .
(a)
(36)
(b)
Figure 4: The Components of V ; (a) Addition of VR and jk (ν) VR to Produce V, and (b) Several Values
of VR and jk (ν) VR that Produce V
If we multiply the left-hand side by its complex conjugate and the right-hand side by its complex conjugate, we obtain the identity
4 = 2 (1 + cosψ) 1 + k 2 (ν) .
This equation tells us how the angle
The number
cosψ
lies between
http://cnx.org/content/m21475/1.6/
−1
ψ
and
depends on
+1,
k (ν)
and, conversely, how
(37)
k (ν)
depends on
ψ:
cos ψ =
1 − k 2 (ν)
1 + k 2 (ν)
(38)
k 2 (ν) =
1 − cosψ
1 + cosψ
(39)
so a circular solution does indeed work.
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Exercise 6
Check
and
ψ
−1 ≤ cosψ ≤ 1
k.
for
−∞ < k < ∞
and
−∞ < k < ∞
for
−π ≤ ψ ≤ π .
Sketch
k
versus
ψ
versus
The equation
VR =
V
2
1 + ejψ
is illustrated in Figure 5. The angle that
VR
makes with
V
is determined
from the equation
2φ + π − ψ = π ⇒ φ =
ψ
2
(40)
Figure 5: The Voltages V and VR , and the Power Factor cosφ
In the study of power systems,
cosφ
is a "power factor" that determines how much power is delivered to
the resistor. We may denote the power factor as
ψ
η = cosφ = cos .
2
But
cosψ
may be written as
η = cosφ = cos
But
cosψ
(41)
ψ
2
(42)
may be written as
cosψ = cos (φ + φ)
=
cos2 φ − −sin2 φ
= cos2 φ − 1 − cos2 φ
Therefore the square of the power factor
η
2cos2 φ − 1
=
2η 2 − 1.
(43)
is
η2 =
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=
cosψ + 1
1
=
2
1 + k 2 (ν)
(44)
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k (ν) = 0, corresponding
ν = 0, ∞(ω = 0, ∞).
The power factor is a maximum of 1 for
for
k (ν) = ±∞,
Exercise 7
With
k
corresponding to
dened as
k (ν) =
Exercise 8
Find the value of
ν
ω0 L
R
ν−
1
ν , plot
k 2 (ν), cosψ ,
that makes the power factor
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and
η = 0.707.
to
ν = 1(ω = ω0 ).
η2
versus
ν.
It is a minimum of 0