Module 3B: AC Circuits II - Complex numbers, Phasors, Impedance and RLC networks
I.
Review so far
In Module 3A we considered what would happen when our familiar DC voltage source was replaced with
a source whose voltage varied with time. Though at first blush this would seem to make matters
hopelessly complicated, this turned out not to be so.
If we assume that our AC source is periodic, we can express the time dependence of the source with a
few parameters: the period T, the frequency f = 1/T, and the angular frequency ω= 2πf.
We considered three possible periodic waveforms: the square wave, the triangular wave, and the
sinusoidal wave, though from here onward we will concentrate only on the sinusoidal wave.
The amplitude can be expressed in a number of ways. The simplest are the amplitude V0 in V(t) =
V0cos(ωt), the “peak voltage” Vp = V0, and the “peak-to-peak voltage” Vpp = 2V0. We also defined the
“average voltage” as the mean value of V(t) over half a period, with Vavg = (2/π)V0, and the “root mean
square” voltage as the square Root of the Mean value of V2(t) (“V Squared”) over a period T, with Vrms =
V0/sqrt(2). The principal significance of Vrms is that an AC source with a given Vrms will deliver the same
time-averaged power Pavg to a resistive load as a DC source with VDC = Vrms (although the actual
instantaneous power delivered by the AC source varies between 0 and 2Pavg).
When hooked to a resistor, an AC source will deliver a current I(t) = V(t)/R, so that the time dependence
of the voltage and current are “in phase” with one another (i.e. both voltage and current reach their
zeroes and maxima at the same time), with the amplitudes related by our “AC Ohm’s Law:” I0 = V0/R,
which can also be expressed in terms of our other definitions of amplitude, e.g. Irms = Vrms/R or Ipp =
Vpp/R. Series and parallel combinations of resistors are replaced with an Req in just the same way we
used in DC circuits: series resistances add, and parallel resistances combine by the “one over” formula
(alternately, parallel conductances add). The instantaneous power delivered to a resistive load is given
by P(t) = V(t) I(t) = V2(t)/R = I2(t) R, and we can integrate over time to find Pavg = Vrms Irms = Vrms2/R = Irms2R.
Capacitors, it seems, are even simpler to deal with under AC than they were under DC. Rather than
dealing with charging and discharging circuits with exponential behavior and time constants, they obey
an Ohm-like law, in which I0 = V0/XC, where XC = 1/(ωC) is the capacitive reactance, measured in Ω.
Inductors work similarly. The current drawn by an inductor has an amplitude given by I0 = V0/XL, where
XL = ωL is the inductive reactance, also measured in Ω.
Capacitive reactances in series and parallel, and inductive reactances in series and parallel, combine just
as resistances (and inductances) do: in series they add, and in parallel they combine using the “one
over” formula. Alternately, one can compute the susceptances, BC = 1/XC and BL = 1/XL, and in parallel,
capacitive susceptances BC add, and inductive susceptances BL add.
However, we also noted a few crucial differences between capacitors/inductors and resistors, to which
we now return.
II.
Things get complicated… the differences between DC and AC circuits
The three points of difference we noted between resistors and capacitors/inductors:
1.
The capacitive reactance XC = 1/ωC and inductive reactance XL = ωL are frequency-dependent,
so even though capacitors and inductors “act like” resistors in a sense, as they obey an Ohmlike law relating V and I, the “resistance” (XL or XC) is a function of frequency.
2. With capacitors and inductors, there is a phase angle of 90° between the current through the
component, and the voltage across it, because the relation between current and voltage for
capacitors and inductors involves a derivative, and derivatives of sines and cosines return “each
other,” in other words, the same function with a 90° phase shift. This means that the current
through a capacitor or inductor is zero when the voltage reaches its minimum (or maximum)
values, and the current is maximum/minimum when the voltage is zero. For a capacitor, the
current “leads” the voltage by 90°, and for an inductor, the voltage “leads” the current by 90°.
3. Because the voltage and current are 90° out of phase with each other, the time-averaged
power Pavg = 0∫TV(t)I(t)dt / 0∫Tdt = 0: the AC source does no work on the capacitor or inductor,
since it is alternately storing and releasing electrical energy over each cycle.
Issue #1 sounds like little more than an accounting pain (having to keep track of changing X’s with ω),
but turns out to be enormously useful when we consider high-pass and low-pass filters.
We’ll leave aside Issue #3 for now, but return to it when we discuss power in AC circuits.
However, Issue #2 is quite profound, and is the essential property that causes AC circuits to behave in
ways that our DC-thinking minds will find astonishing.
But to deal with the issue of phase shifts, I will have to delve a bit into some mathematical formalism
which involves complex numbers.i
III.
How we used to deal with these things
Our modus operandi so far, when we could not perform direct decomposition using our series and
parallel relations, has been to set up loop and node equations using Kirchhoff’s Laws.
When we dealt with resistive networks, these generally led to linear algebraic equations (since V = IR, so
all of our terms were either constants – from our sources – or resistive terms of the form I*R =
I*constant), which we could readily solve. As the networks became more complicated (more than one
loop or node), we generally ended up with simultaneous algebraic equations, which we could also solve,
but we had to unleash some rather advanced linear algebra techniques (including Gaussian elimination
and Cramer’s Rule), and even then, the solutions were rather tedious!
When capacitors were added in, we ended up with first-order differential equations, since the V = IR
resistor terms and Q = ∫I dt = CV capacitor terms involve not just I, but ∫I dt as well. This limited us to
solving only rather simple circuits, or using “shortcuts” that enabled us to tease out some aspects of the
behavior without having to directly solve the differential equations. With RL circuits, V = IR and V = LdI/dt
terms led us to similar first-order differential equations, with similar analysis and results.
When we are dealing with resistors, capacitors and inductors together in the same circuit powered by an
AC source, we get a loop equation involving I (for resistors, V = IR), dI/dt (for inductors, V = LdI/dt), and ∫Idt
(for capacitors, V = Q/C). Applying KVL leads to a second-order differential equation, of which we can
take a derivative (to replace the ∫I(t)dt term with an ordinary I(t)) to get:
LI’’(t) + RI’(t) + I(t)/C = V’(t)
in which even the V’(t) term is nonzero due to the fluctuating voltage of our AC source. If you have
taken differential equations you’ll recognize this as a “driven (by V’(t)) constant-coefficient second-order
differential equation,” where the primes (‘) represent time derivatives. Taking our cue from how we
dealt with the differential equations developed for RC circuits (and realizing that if V(t) ~ cos(ωt) or
sin(ωt) then V’(t) also ~ sin(ωt) or cos(ωt), we have an equation in which a function I(t), its derivative
I’(t), and its second derivative I”(t), are all of the same (or similar) functional form. Therefore, we may
guess an “ansatz solution” to this equation of the form I(t) ~ A cos(ωt) + B sin(ωt), possibly multiplied by
some form of exponential damping (since ekt is another function which returns “itself” as a derivative).
While we could go through the mechanics of this, the algebra turns out to be rather tedious, so I’ll leave
it to your differential equations class. For now, we will seek out an alternate means of analyzing RLC
circuits under an AC voltage source, one which is more graphical in nature.
IV.
Phasors: Visualizing an AC source
So far, I have described AC sources as providing a voltage that varies as V(t) = V0cos(ωt), with a time
dependence that looks like the left panel in the graph on the next page.
However, there is a powerful way of visualizing this voltage source which abstracts out some of the time
dependence. Instead of picturing a time-dependent voltage, picture voltage as a vector-like quantity,
with magnitude V0, which sweeps around in a counter-clockwise circle (like the trace on a radar screen)
over the course of one period, as depicted in the right panel in the figure below. That is, its phase angle
with respect to the positive x-axis is given by φ = ωt. At t = 0, the phase angle is zero, and as t increases,
this voltage vector, or “phasor,” sweeps around in a circle and returns to the starting point at time T.
Thus we draw a “phasor diagram” which depicts the voltage phasor, with length V0, and its position at
time t=0, which means it lies along the positive x-axis.
Though this phasor maintains the same magnitude V0 at all times, a measurement at a given time t only
measures the x-component of this vector, which can be shown by trigonometry to be
V(t) = V0cos(ωt)
If our AC source is connected to a resistor, our time-dependent value of I(t) is given by I(t) = I0cos(ωt),
which also can be described by a phasor, this time of length I0 = V0/R, and with a phase angle given by φ
= ωt, so φ(t=0) = 0 and the phasor diagram depicts our I(t) phasor along the positive x-axis as well.
If our AC source is connected to a capacitor, our time-dependent I(t) = -I0sin(ωt). This phasor has length
I0 = V0/XC, but its phase angle is given by φ = ωt+90°, since -sin(ωt) = cos(ωt+90°). Therefore, at t=0, φ =
90°, so our phasor diagram depicts our I(t) phasor along the positive y-axis.
Finally, if our AC source is connected to an inductor, I(t) = I0sin(ωt), with I0 = V0/XL. Here, the phase angle
is given by φ(t)=ωt-90°, since sin(ωt) = cos(ωt-90°). Therefore, our phasor for I(t) has length I0, and
points along the negative y-axis.
So far, this discussion has been entirely graphical. Though it’s tempting to think of these phasors as
vectors, and apply the vector mathematics we learned last year to it (and in fact this is very useful in
many cases!), there is a very powerful way to analyze phasor diagrams using complex numbers.
V.
Complex numbers and AC Circuits
First, consider this simple question: What is the square root of -1?
Well, the answer to this question is not simple. Though you may instinctively think that it is -1, it’s not,
as -12 = +1. In fact there is no number anywhere on the real number line which, when squared, gives
you a negative anything (let alone -1) as a result. However, it’s clear that such a number must exist as
the solution to x2+1 = 0. (You may point out that the graph of y(x) = x2+1 never touches y = 0, which just
underscores how “odd” this √(-1) number must be.)
Mathematicians call this number i, where i = sqrt(-1). However, since we’re already making heavy use
of i to represent current, electrical engineers call it j instead. From here on in I will refer to j (and
occasionally refer to it as “i” out of habit) but remember that j = i = sqrt(-1).
j does not fall anywhere along the real number line (which is why y(x) = x2+1
does not reach zero for any real number!), and is usually thought of as lying
along an imaginary axis which runs perpendicular to the real number line. This
creates a “plane” of numbers, known as the “complex plane.” A “complex
number” is a number containing both a “real part” and an “imaginary part,” as
in z = 5 + 3i (or 5 + j3 in our “electrical engineering notation”) and is plotted on
the complex plane using the “real part” as the “x-coordinate” and the
“imaginary part” as the “y-coordinate.” So we can deal with the complex plane
in much the same way that we deal with points on a set of Cartesian axes.
j has a number of interesting properties:
j2 = -1,
j3 = -j,
j4 = +1,
j5 = j,
any jn = jn-4,
and 1/j = -j
Note that any real number (such as 5) lies along the positive x-axis in our complex plane, and multiplying
it by j gives us 5j, which lies the same distance away from zero, but along the positive y-axis. Multiplying
5j by j again gives us -5, which lies along the negative x-axis, and multiplying -5 by j gives -5j, along the
negative y-axis. Finally, multiplying -5j by j returns us to +5, on the positive x-axis. In other words,
Multiplying anything by j results in a counterclockwise rotation of 90° about the origin in
the complex plane.
Similarly,
Multiplying anything by –j (or dividing it by +j) results in a clockwise rotation of 90°.
This should get our brains thinking…
is there some relation between this strange j number, and the way that our voltage and
current phasors for capacitors and inductors are rotated 90° with respect to each other?
Yes! And as Billy Mays famously said, “But wait, there’s more!”
It turns out that we can use these same complex numbers to represent the time dependence of our
voltage and current phasors (which we previously represented by imagining that the phasors “sweep”
around our diagram). There is a mathematical identity due to Euler which states that
ejx = cos(x) + jsin(x)
In other words, ejx has a “real part” (or “horizontal component” in our vector diagram) equal to cos(x),
and an “imaginary part” (or “vertical component”) equal to sin(x) multiplied by j. A few other useful
facts which can be derived from Euler’s identity:
ej0 = 1
ej(π/2) = j
ejπ = -1
ej(-π/2) = -j
Putting this together which what we found above,
Multiplication of a phasor by ejφ represents a counterclockwise rotation φ in the complex plane.
If we replace x with ωt and multiply Euler’s equation by V0, we find that
V0ejωt = V0cos(ωt) + jV0sin(ωt)
whose “real part” (or horizontal component) is exactly the same V(t) = V0cos(ωt) we used earlier.
If this source with V(t) = V0ejωt is connected to a resistor with resistance R,
I(t) = V(t)/R = (V0/R)ejωt= I0ejωt
whose “real part” / “horizontal component” is the selfsame I(t) = I0cos(ωt) we found earlier.
This leads us to the notion that we can treat V and I as these complex-valued quantities for the purposes
of our calculations, and then as a last step use the fact that what we actually measure is the “real part”
of the complex-valued quantity. Now, can we treat our resistances and reactances in the same way?
In a flash of inspiration, let’s take the ω in our equations for capacitive and inductive reactance, and
replace them with jω, since the capacitor and inductor result in a current which is “rotated” by 90° with
respect to the voltage.
Let’s connect our source to a capacitor with a capacitive reactance given by XC = 1/ωC, but using a
complex capacitive reactance given by 1/(jωC) = -jXC. Now
I(t) = V(t)/(-jXC) = +j(V0/XC)ejωt = +jI0ejωt
(using 1/j = -j)
To find the “real part” of I(t), we’ll use Euler’s identity:
I(t) = -jI0ejωt = +jI0 [ cos(ωt) + jsin(ωt) ] = +jI0cos(ωt) – I0sin(ωt)
(using j2 = -1)
whose “real part” is –I0sin(ωt), exactly as we found before!
Next, we’ll consider our AC source wired to an inductor, and use a complex inductive reactance given by
jωL = +jXL. The current is given by
I(t) = V(t)/(+jXL) = -j(V0/XL)ejωt = -jI0ejωt = -jI0 [ cos(ωt) + jsin(ωt) ] = -jI0cos(ωt) + I0sin(ωt)
whose “real part” is +I0sin(ωt), as we expect!
We can represent our values of R, -jXC and +jXL in the complex plane, as
depicted at right. Our inductive reactance +jXL lies along the positive y-axis,
our capacitive reactance –jXC along the negative y-axis, and the resistance R
is simply represented as a real number R, which lies along the positive x-axis.
(As an aside, one may wonder if there is a “fourth” circuit element that has a “negative resistance” along
the negative x-axis. That would imply a device that pulls a current 180° out of phase with the applied
voltage, and the answer is technically “no.” However, there are resistors with variable resistance called
memristors which exhibit some of these properties. In particular, as the voltage increases, the effective
resistance increases, so the current drawn drops, and as the voltage drops, the effective resistance
drops even faster, so the current increases. They are nonlinear (non-Ohmic) devices, although they
have a “linear regime” in which there is a linear relationship (with negative slope) between V and I.
Memristors are sometimes referred to as the “fourth element” in circuit analysis, although they don’t
exactly fit the bill of what a device along the “negative x-axis” of our phase diagram would be.)
The current, in turn, is given by the ratio of V (which lies along the positive x-axis) divided by the
appropriate resistance or reactance for the circuit in question.
VI.
Combining Resistance and Reactance: Impedance
Now, given that these “complex reactances” and resistances are vector-like
quantities in the complex plane, we may scratch our heads and wonder how
they can be combined, if, for example, we have R, L and C in series in a circuit.
Taking inspiration from the fact that resistance, capacitive reactance and
inductive reactance all add when combined in series, we can compute that –jXC,
+jXL and R have a “vector sum” Z given by
Z = R + j(XL-XC)
where Z is known as the impedance, and is also measured in ohms. Though Z is actually a complex
number with both real and imaginary parts, we can apply trigonometry to find the magnitude of Z, and
its angle φZ with respect to the positive x-axis, as in the diagram at right:
Z = |Z| = sqrt(Z Z*) = sqrt(R2+(XL-XC)2)
φZ = tan-1[ (XL-XC) / R ]
We may recall that we could write vectors in either “rectangular” or “component” form, or in “polar”
(magnitude and direction) form. We can do the same for Z, using the fact that Re(Z) = R = ZcosφZ, and
Im(Z) = (XL-XC) = ZsinφZ to state that
Z = R + j(XL-XC)
= ZcosφZ + jZsinφZ = Zejφz
(known as “rectangular” or “component” form)
(known as “polar” form, where Z = sqrt[R2+(XL-XC2)] )
To make our lives even simpler, we can introduce a shorthand notation for the polar form, in which
Z = Zejφz = Z ∠ φz
(known as “phasor” form)
The geometric interpretation of the polar and phasor forms is that Z is a vector (phasor) in the complex
plane with a magnitude of Z, and with a phase angle φZ with respect to the positive real axis. But Z can
equivalently be considered, using the rectangular form, to be a vector composed of two orthogonal
components: a “horizontal” or real component R, and a “vertical” or imaginary component XL-XC. And
using the above relations, which apply what we already knew about vectors, we can readily switch back
and forth between rectangular and polar/phasor forms.
For individual electronic components, the complex impedances are given by
Z = R = R ej0 = R ∠ 0
Z = -jXC = XCej(-π/2) = XC ∠ -90°
Z = +jXL = XLej(+π/2) = XL ∠ +90°
(resistor)
(capacitor)
(inductor)
This enables us to use some very nice properties of exponentials, most specifically the fact that ratios of
exponentials are found by adding or subtracting the values in the exponent:
ex / ey = e x – y
→
(A ∠ φA) / (B ∠ φB) = (A/B) ∠ (φA-φB)
since the “magnitude” portion of the phase diagram is just a regular number, and the “angle” portion
comes from the exponent in ejφ, which multiplies (or divides) by adding (or subtracting) the exponent.
VII.
Solving circuits with complex numbers
First, we can write our voltage V(t) = V0cos(ωt) in complex form, with the phase angle φ in our phasor
notation represented by the value of V(t) at t = 0:
V(t) = V0ejωt,
represented at t=0 by a phasor
V0ej0 = V0 ∠ 0
(φ = 0 since ωt = 0 at t = 0)
Now, if this AC voltage source were connected to a resistor with impedance R ∠ 0, the current will be
I = V / R = (V0 ∠ 0) / (R ∠ 0) = (V0/R) ∠ 0 = I0 ∠ 0 with I0 = V0/R
In other words, the magnitude of the current comes from our usual Ohm’s Law relation, and the phase
angle is zero, which means that the voltage and current are in phase with each other.
If the AC voltage source were connected to a capacitor, the current will be
I = V / XC = (V0 ∠ 0) / (XC ∠ -90) = (V0/XC) ∠ +90= I0 ∠ +90 with I0 = V0/XC
Again, the magnitude of our current comes from our “AC Ohm’s Law,” and the current leads the voltage
by 90 degrees.
Connected to an inductor, the current will be
I = V / XL = (V0 ∠ 0) / (XC ∠ +90) = (V0/XL) ∠ -90= I0 ∠ -90 with I0 = V0/XL
Our “AC Ohm’s Law” gives us the magnitude of current once again, and this time the current leads the
voltage by -90 degrees (i.e. the voltage leads the current by 90 degrees).
So, our AC Ohm’s Law relations look good, and ELI the ICE man is happy.
If we have combinations of inductors in series or parallel, or capacitors in series in parallel, or resistors in
series or parallel, they can be replaced by an XLeq, XCeq, or Req, as described above: series R’s or X’s add,
and parallel R’s or X’s combine using the “one-over” formula (or, use G’s or B’s, which add in parallel!).
But what happens when I have components of different types strung together? Here, the answer is not
so simple because either the voltages or the currents will be out of phase with respect to each other,
and we will have to do something a little more than simple “adding” or “one-overing,” which
necessitates our use of the complex impedance Z. Let’s first consider what happens in series.
VIII.
The series RLC Circuit.
Consider the circuit at right, in which an AC voltage source is connected to
a resistor, a capacitor, and an inductor, in series.
The resistor has resistance R, the inductor has an inductive reactance
given by +jXL, and the capacitor has a capacitive reactance given by –jXC.
The three combine as complex numbers to yield the impedance Z:
Z = R + j(XL-XC)
Z is in fact a complex number, so we cannot express it as a single number, but either as a sum of real and
imaginary components, or as a magnitude
Z = |Z| = sqrt(R2 + (XL-XC)2)
which, like resistance and reactance, is measured in ohms, and a phase angle
φZ = tan-1((XL-XC)/R)
We can therefore write our impedance equivalently in rectangular, polar or phasor forms:
Z = R + j(XL-XC) = Z ejφz = Z ∠ φZ
and we can easily transform between rectangular and polar/phasor forms using:
Z = |Z| = sqrt(R2+(XL-XC)2)
φZ = tan-1((XL-XC)/R)
R = Z cosφZ
XL-XC = Z sinφZ
A memory aid is to think of R, XL and XC as “orthogonal components” of the “vector” Z, where the
magnitude of Z is found using the Pythagorean theorem, the angle of Z from trigonometry, and the “x”
(R) and “y” (XL-XC) “components” calculated from the magnitude and angle of Z using the cosine and
sine, respectively. That is, we can unleash all our “vector” tools on these phasor quantities!
It’s important to be able to readily transform between these forms because addition and subtraction are
most readily performed on complex quantities in rectangular form, whereas multiplication, division and
exponentiation are most efficiently done using complex quantities in polar form. As we’ve already seen,
circuit analysis involves extensive manipulation of our circuit quantities, so we may have to transform
back and forth several times in one problem.
Now our series RLC combination, with a complex impedance Z, is connected to a voltage source
V(t) = V0cos(ωt) = V0 ∠ 0
Since the components are in series, the current in all three components must be exactly the same, both
in magnitude and in phase, and is given by the AC Ohm’s Law:
I(t) = V(t)/Z = (V0 ∠ 0) / (Z ∠ φZ) = (V0/Z) ∠ ̶ φZ
That is, the current drawn by this circuit will have a magnitude I0 = V0/Z, and will have a phase φI = -φZ,
where the convention is that φI represents the angle by which the current leads the voltage. If φZ < 0,
then φI > 0, and thus the current leads the voltage by φI (which is expected for a “capacitive” circuit in
which XC > XL). On the other hand, if φZ > 0 (which is expected for an “inductive” circuit in which XL > XC),
then φI < 0, and the voltage leads the current by φI.
Once the magnitude of the current I0 and its phase angle φI are known, one can calculate the voltage
across each of the three components using our Ohm-like laws, again in phasor form:
VR = IR:
VC = IXC:
VL = IXL:
VR ∠ φR = (I0 ∠ φI) (R ∠ 0) = I0R ∠ φI
VC ∠ φC = (I0 ∠ φI) (XC ∠ -90°) = I0XC ∠ φI -90°
VL ∠ φL = (I0 ∠ φI) (XL ∠ +90°) = I0XL ∠ φI +90°
That is, as we would expect, the voltage across the resistor is in phase with the current through it,
whereas the voltage across the capacitor lags the current by 90°, and the voltage across the inductor
leads the current by 90°. The three component voltages have three different phases with respect to
the supply voltage, however.
IX.
Parallel RLC circuits
Next, let’s consider the circuit shown at right, in
which R, L and C are in parallel to one another.
In our series RLC circuit, we utilized the fact that resistances and reactances in series add (which in this
case was complicated by the fact that our resistances were real and our reactances were imaginary, so
we had to add them as complex numbers), and the fact that devices in series must all have the same
current (that is, magnitude and phase angle) going through them. This leads us to formulae for the
voltages across the various components in the circuit, which by Kirchoff’s Voltage Law add up to zero,
once the phase angles (or equivalently, real and imaginary components) are considered.
In the parallel RLC circuit we will use the fact that devices in parallel must have the same voltage
(magnitude and phase angle) across them, and that the currents through each of the devices must sum
up to the total current drawn by the circuit. When considering devices all of the same type, we used the
“one over formula” to combine parallel resistances or reactances. While we may be tempted to use the
same formula to combine resistances with reactances, we must be conscious that the sum will be
complex (the denominators will contain real and imaginary parts) and this is a bit tedious to deal with.
So let’s do a little bit of thinking. When we combined resistors in parallel, I pointed out in passing that
it is possible to define the conductance G = 1/R, and define an equivalent conductance Geq = G1 + G2 + …
In analogy, we can define the susceptance B = 1/X, where X represents either the capacitive or inductive
reactance, and B the analogous susceptance. However, we added in a new wrinkle when we discussed
complex reactances, so complex susceptances will have to be used. We can derive the complex
susceptance by using the same ω → jω substitution:
G = 1/R
BC = 1/XC = 1/(1/jωC) = +jωC
BL = 1/XL = 1/(jωL) = -j/ωL
which, in terms of phase angles, can be written in phasor form as
G = 1/R ∠ 0°
BC = ωC ∠ +90°
BL = 1/ωL ∠ -90°
Note that the magnitudes are inverted, and the phase angles are reversed from what we had with our
resistance and reactances:
R = R ∠ 0°
XC: 1/ωC ∠ -90°
XL: ωL ∠ +90°
Finally, we are prepared to combine our susceptances and conductance to derive not the impedance,
but the reciprocal of the impedance, which is known as the admittance, Y:
Y = G + j(BC-BL) = 1/Z
with
Y = |Y| = sqrt(G2 + (BC-BL)2)
φY = tan-1((BC-BL)/G)
We can use the admittance directly to find the current drawn from the source as
I=VY
If V = V0 ∠ 0, and Y = Y ∠ φY, it follows that
I = I ∠ φY
Since each component drops the same voltage (which is equal to the source voltage, V0 ∠ 0), the current
drawn by each component may be found using our AC Ohm’s Law relations:
I = V G = ( V0 ∠ 0 ) ( 1/R ∠ 0 ) = (V0/R) ∠ 0
I = V BC = ( V0 ∠ 0 ) ( ωC ∠ +90 ) = (V0/XC) ∠ +90°
I = V BL = ( V0 ∠ 0 ) ( 1/ωL ∠ -90 ) = (V0/XL) ∠ -90°
for the resistor
for the capacitor
for the inductor
We can verify that for the inductor, I lags V by 90 degrees, for the capacitor, I leads V by 90 degrees, and
for the resistor, I and V are in phase. So ELI the ICE man is happy.
We can also compute the impedance, which can be done in polar form:
Z = 1 / Y = (1/Y) ∠ -φY
So that the magnitude of the impedance Z = 1/Y, and the phase angle of the impedance φZ = -φY.
In rectangular form, the computation is not quite as straightforward, and we have to use the fact that
any complex number multiplied by its complex conjugate is a real number, in order to simplify the
denominator we multiply 1/Y by 1 in the form of Y*/Y*:
Z = 1/Y = 1/( G + j(BC-BL)) = Y*/YY* = ( G - j(BC-BL)) / ( G + j(BC-BL)) ( G - j(BC-BL))
= ( G - j(BC-BL)) / (G2 + jG(BC-BL) – jG(BC-BL) + (BC-BL)2) = ( G - j(BC-BL)) / |Y|2
If BC > BL (which is equivalent to XC < XL, since the B’s go as 1 over the X’s), then φY > 0, so the current
leads the voltage, and the circuit is “capacitive.” If BL > BC (which is equivalent to XL > XC), φY < 0, so the
voltage leads the current, and the circuit is “inductive.” This is the opposite of what we had in the series
RLC circuit, in which the “bigger” reactance dominated the circuit. But it is perhaps not so surprising, as
it’s the “smaller” reactance that draws the most current, so the current drawn from the source will be
most profoundly affected by the smaller reactance, which therefore dominates the circuit behavior.
X.
More complex RLC circuits: bringing back the DC toolbox
Now that we have determined how to deal with simple series and parallel RLC combinations, we may
consider how to deal with more complex RLC networks. Well, we may try to use a “decomposition”
strategy in which R’s, L’s and C’s which can clearly be identified as “in series” may be replaced by a Zeq,
and R’s, L’s and C’s which can clearly be identified as “in parallel” may be replaced by a Yeq. Then the
“equivalent impedances” and/or “equivalent admittances” can be written in phasor form as
Zeq ∠ φZ
or
which can be converted to rectangular form as
Yeq ∠ φY
Zeq = Zreal + jZimag
Yeq = Yreal + jYimag
with Zreal = ZeqcosφZ and Zimag = ZeqsinφZ
with Yreal = YeqcosφY and Yimag = YeqsinφY
Then we use the fact that series impedances add, and that parallel admittances add, taking advantage of
the fact that addition can be done readily in rectangular form (by separately adding the real and
imaginary components).
If we need to convert between impedance and admittance, this is best done in polar form:
Zeq = 1/Yeq = (1/Yeq) ∠ -φY
and
Yeq = 1/Zeq = (1/Zeq) ∠ -φZ
Though the mathematics gets quite tedious (dealing with magnitudes and angles, and having to convert
back and forth between rectangular and polar form), the principles are exactly the same as they were
for purely resistive networks:
If components are in series, compute their impedance by combining R’s and X’s.
If components are in parallel, compute their admittance by combining G’s and B’s.
To add impedances or admittances, use rectangular form.
To convert between impedance and admittance, use polar/phasor form.
Once you get Zeq or Yeq, you can compute the total current drawn. Then you can “re-compose” the
circuit, using the fact that V across all components in parallel is the same, and I through all components
in series is the same. Once you know two of V, I and R/X/G/B/Z/Y for a component, you can use V = IZ or
I = VY to find the third.
As for circuits which cannot be decomposed, we will have to fall back to our Kirchhoff-based methods,
although here we have to take into account not only the magnitudes of voltages, currents and
impedances, but their phase angles as well. (Or, equivalently, we may write our voltages, currents and
impedances in rectangular form, taking care to keep track of both the real and imaginary components.)
This makes the algebra even more tedious than it was in the case of resistive networks, but all of our
tools apply as they did before.
For example, in the case of our simple series RLC circuit, we found that
V
VR
VC
VL
= V0 ∠ 0
= I0R ∠ φI
= I0XC ∠ φI -90°
= I0XL ∠ φI +90°
which can be re-written in rectangular form, as:
V = V0cos(0) + jV0sin(0) = V0 + j0
VR = I0Rcos(φI) + jI0Rsin(φI)
VC = I0XCcos(φI-90) + jI0XCsin(φI-90)
VL = I0XLcos(φI+90) + jI0XLsin(φI+90)
= I0XCsin(φI) - jI0XCcos(φI)
= -I0XLsin(φI) + jI0XLcos(φI)
since cos(φ±90) = - ±sin(φ) and sin(φ±90) = ±cos(φ). Then we apply KVL:
V(t) = VR(t) – VC(t) – VL(t) = 0
Then we use the facts that the real and imaginary components must separately add up to zero, and that
I0 = V0/Z, tan(φI) = (XL-XC)/R, so sin(φI) = (XL-XC)/Z and cos(φI) = R/Z:
Real parts:
ΣRe(V) = V0 – I0Rcos(φI) - I0XCsin(φI) + I0XLsin(φI)
= V0 – V0R2/Z2 + V0(XL-XC)2/Z2
= V0 – V0(R2+(XL-XC)2)/Z2
= V0 - V0 (Z2/Z2)
=0
Imaginary parts:
ΣIm(V) = I0Rsin(φI) + I0XCcos(φI) - I0XLcos(φI)
= I0 ( R(XL-XC)/Z + XCR/Z – XLR/Z )
= I0 (R(XL-XC)/Z – R(XL-XC/Z)
=0
Also keep in mind that in a circuit with multiple sources, the sources may have different amplitudes (i.e.
V0’s), but may also have different frequencies (i.e. ω1 and ω2), and even if they have the same frequency,
may be out of phase with respect to each other!
However, if a network is linear – and under AC, resistors, capacitors and inductors are all linear
components – we can apply our network theorems as we did before.
Generally, voltages/currents of different frequencies are treated separately from one another, i.e. you
solve separately for the voltages/currents at each frequency, and consider the signals to be independent
of one another. However, voltages/currents of the same frequency but different phase must be
combined together using our phase angles and trigonometric identities. For our purposes we will
consider all sources to be operating at the same frequency, and defer until later consideration of what
happens when different frequencies are present at once.
If there are multiple sources, we can use the superposition theorem, and solve for the voltages and
currents in each “subcircuit,” with the actual voltages and currents being the linear sum of the voltages
and currents in each subcircuit. If all sources operate at the same frequency (even if phases are
different), then you can simply sum the voltages and currents. (To apply superposition, each
deactivated source is replaced with its internal impedance. For ideal voltage sources, Z = 0 ∠ 0, but real
AC sources typically have a nonzero impedance. That said, typically the phase angle of that impedance
is zero, so the source is typically replaced by a resistor when it is “deactivated.”)
Finally, our good friends Thevenin and Norton apply as well.
Thevenin’s theorem in AC tells us that any linear network (provided that all sources operate at the same
frequency ω) may be replaced, from the perspective of two terminals on that network, by a single
sinusoidal voltage source Vth at frequency ω, and a single “Thevenin impedance” Zth connected in series
to the source. In this case, there are four quantities to solve for: the magnitude of the Thevenin voltage
Vth, its phase angle φVth, the magnitude of the Thevenin impedance Zth = |Zth|, and its phase angle φZth.
So there are four unknowns (as opposed to two in the DC case).
Norton’s theorem allows us to replace a linear network with an AC current source IN at frequency ω
(with magnitude IN and phase angle φIN), in parallel with a Norton admittance YN (with magnitude YN and
phase angle φYN).
In both cases, the procedure is the same as for DC circuits, done by applying any two of three steps:
Measure the open-circuit voltage VOC (which will have a magnitude and phase angle). VOC is
then equal to Vth (in magnitude and phase angle).
2. Measure the short-circuit current ISC (which will have a magnitude and phase angle). ISC is then
equal to IN (in magnitude and phase angle).
3. Deactivate all sources (i.e. replace them with their internal impedances), and determine the
equivalent impedance Zeq (magnitude and phase angle, or real and imaginary components)
between the terminals. Then Zeq = Zth = 1/YN (in magnitude and phase angle).
1.
And, as before, we can readily convert back and forth between Thevenin and Norton equivalents for a
given circuit, using
Vth = IN/YN
IN = Vth/Zth
If our Thevenin circuit is connected to a load with impedance ZL, then the current drawn will be
I = Vth / Ztot = Vth / (Zth + ZL)
where the calculation must be done using our complex mathematics techniques. That means we must
convert Zth and ZL to rectangular form in order to compute Ztot, and then convert Ztot back to polar form
to solve for I. (But, this is usually a heck of a lot simpler than dealing with the actual circuit!)