## Impedance and the Electrical Analogy for Fluid Flow

There are certain concepts in electrical circuits that bear a strong similarity to fluid flow in networks of compliant tubes.    In the study of physical hemodynamics, aspects of the circulation are often diagrammed using the very same schematic elements that are used in discussing electrical circuits.  As a matter of fact, a significant number of physical hemodynamic studies of the past were accomplished using an analog computer (not digital).  These are (were) devices designed to facilitate the implementation of electrical circuits that are analogous to physical systems; the behavior of the circulation was studied by analyzing the behavior of the circuits!  The output – the result of an analog computer model is typically an electrical signal(s) – voltages and currents that vary with time.  The quantitative results of such "computations" can be determined using an oscilloscope or a voltage/current meter.

This article includes an introduction to circuits and impedance that will be useful in subsequent discussions where some of the same topics are discussed, but from a different perspective.

### Impedance of a Resistor

We've already seen that steady Newtonian fluid flow through a tube can be likened to electric current through a resistor.

$$\Large \Delta v = i R$$

$$\Large R = \frac{\Delta v}{i}$$

In the above, $$\Delta v \equiv v_1-v_2$$.   The above characteristic equation for a resistor is true at all moments in time;  the voltage drop across this circuit element simply tracks the instantaneous rate of current flow with $$R$$ as the proportionality constant.  A resistor is a circuit element that dissipates electrical energy – converts it into heat.  Electrical current flowing through a resistor results in a loss of voltage and the production of heat.  The rate of heat production is actually equal to $$i \Delta v = i^2 R$$; there is no ambiguity in this idealized representation.  In the case of the circulation, fluid flow is analogous to electrical current and pressure is analogous to voltage.  Consequently the equations relating resistive fluid flow through a tube are:

$$\Large \Delta p = q R$$

$$\Large R = \frac{\Delta p}{q}$$

$$p$$ symbolizes pressure and $$q$$ flow rate, e.g. mL/sec.  We retain the use of the symbol $$R$$ to represent a resistance in hemodynamics; you may be familiar with the value that arises when a Newtonian fluid flows at a steady rate in a long cylindrical tube (Poiseuille resistance):

$$\Large R = \frac{\Delta p}{q} = \frac{8 \mu l}{\pi r_0^4}$$

$$\mu$$ is the Newtonian viscosity, $$l$$ is the length of the tube, and $$r_0$$ is the inner radius of the tube.  In other words, resistance in fluid flow derives from physical aspects of the tube and fluid; it also depends entirely on the velocity profile which engenders the way in which fluid lamina shear against each other.

Up until now the notation has included $$\Delta p$$ (or $$\Delta v$$) to be explicit about the fact that the pressure (or voltage) is a difference across the circuit element - from one side to the other.  However this is always the case and the $$\Delta$$ is omitted in much that follows.  Resistance is also an example of an impedance, a ratio of sinusoids (pressure over flow or  voltage over current).  We saw in the last article that it is mathematically acceptable to divide, multiply, add, and subtract sinusoids of the same frequency.  We are still working on the fact that it will make physical sense to do so, but here's the first example:

$$\Large Z_R(j\omega) = \frac{P(j\omega)}{Q(j\omega)} = R(j\omega)$$

While subtle, something else has happened to this equation representing resistance; the pressure and flow got capitalized and $$j\omega$$ got stuck in all over the place.  This is the clue that somebody has stepped in and substituted Fourier transforms in place of the pressure ($$p$$) and flow ($$q$$) from the previous equation.  $$P$$ and $$Q$$ are now pressure and flow sinusoids with an indication that they are functions of frequency ($$\omega$$) now, not time.  $$Z$$ is the symbol for impedance now and $$Z_R$$ has been used to designate the impedance due to a resistor.  We'll find subsequently that there are several different kinds or usages of this term, but for now this will refer to a spectrum of ratios, pressure sinusoid divided by flow sinusoid as a function of frequency.  The impedance due to a resistance ($$Z_R$$) is ... a resistance.  Hmmmm .. pretty boring.  What this actually means is that a sinusoidal voltage applied across a resistor results in a sinusoidal current through the resistor that is in phase with the voltage.  The peaks and troughs of the voltage cycle coincide in time exactly with the peaks and troughs of the current; $$R$$ is the proportionality constant between the 2 sinusoids.  Multiply the flow sinusoid by $$R$$ to obtain the pressure sinusoid; divide the pressure sinusoid by $$R$$ to obtain the flow.  For the electrical resistor, it's the same value for  $$R$$ at all frequencies.  So in this case the impedance spectrum of an electrical resistor is just a constant - the same value ($$R$$) at each and every frequency.  We'll see later that this is one aspect of the electrical analogy that doesn't transfer directly to fluid flow.  Resistance for a sinusoidal fluid flow oscillation will turn out to increase with frequency due to the fact that the velocity profile changes with frequency.  More on this later.

Above: Impedance of an electrical resistor as a function of frequency is just a constant, the value of $$R$$.  The impedance phase (not shown) is $$0$$ at all frequencies.

### Impedance of a Capacitor (Compliance)

Resistors are not the only kind of gadget that can appear in an electrical circuit.  Where a resistor converts electrical energy into heat, capacitors are circuit elements that store energy in the form of an electrical field.  Here's the schematic symbol for a capacitor:

It's given this form because the electrical element is typically constructed from 2 conductive "plates" separated by an insulting membrane.  However the plates don't have to be flat and the whole gadget might be made up of 2 foil surfaces separated by a piece of paper and all rolled up into a cylinder.  The equation for a capacitor is:

$$\Large v(t) = \frac{1}{C} \int_0^t i(\tau) d\tau$$

$$\tau$$ as used here is sometimes called a "dummy variable".  The integration results in a function with dependent variable $$\tau$$. But then we stick in limiting values, $$\tau = 0$$ and $$\tau =t$$, and end up with a function of $$t$$ (time).  The integral of electrical current with respect to time is electrical charge (e.g. Coulombs) and capacitance has physical units of electrical charge divided by voltage.  Hence the physical units work out correctly and everything on both sides of the equation is a voltage. This is the integral form of the characteristic equation for a capacitor.  We can differentiate the equation to obtain a differential form:

$$\Large \frac{dv(t)}{dt} = \frac{1}{C} i(t)$$

OR

$$\Large C \frac{dv(t)}{dt} = i(t)$$

The fluid analogy with pressure similar to voltage and fluid flow similar to electrical current leads to the following:

$$\Large p(t) = \frac{1}{C} \int_0^t q(\tau) d\tau$$

OR

$$\Large p(t) = \frac{1}{C} \left[V(t)-V_0\right]$$

The latter shows explicitly that we get volume (e.g. mL) when flow rate (e.g. mL/s) is integrated with respect to time.  The interpretation of the "arbitrary" integration constant, $$V_0$$, is easier to see in this form.  $$V_0$$ is the volume of the vessel at zero distending pressure.  A compliiance is a mechanical construct that stores energy in the form of material displacement; the term "elastic recoil" appears frequently in the medical literature but it wouldn't be a bad idea to think of a spring that can store energy in the form of tension or compression.  In a later section will figure out how $$C$$ is related to the physical characteristics of a vessel.

A capacitor has a gap between the 2 plates that's occupied by an insulator.  Technically speaking, electrical current does not pass through the capacitor.  Yet current flows in to the capacitor and charges the plates. Similarly, we don't usually think of a compliance (blood vessel, cardiac chamber) as a vessel that allows fluid to flow through the wall although the latter can certainly happen to some degree, depending on the situation.  This is going to be an important concept for understanding the behavior of a compliance in the frequency domain.  You can't simply continue to add fluid volume to the chamber; the distending pressure would simply continue to increase until the vessel exploded.  Keep it in mind for what follows.

As was done for the resistance, we can take the Fourier transform of the characteristic equation and obtain:

$$\Large P(j\omega) = \frac{1}{j\omega C} Q(j\omega)$$

$$\Large Z_C (j\omega) = \frac{P(j\omega)}{Q(j\omega)} = \frac{1}{j\omega C}$$

$$Z_C (j\omega)$$  is used here to represent the impedance of a compliance and, again, we obtain a spectrum – a function that depends on frequency: $$Z_C = 1/(j\omega C)$$.  Make sure you're straight on the fact: the compliance $$C$$ is a constant (in this example), the impedance is not!  Note that no matter what we stick in for the value of $$C$$, the impedance is infinite at zero frequency ($$\omega = 0$$), and decreases to zero as oscillation frequency goes to infinity.  I was trying to set you up for this in the last paragraph.  The average flow in to or out of a compliance must be zero.  We are talking about filling a structure with fluid ( or a capacitor with charge ); it simply can't be distended more and more forever.

One more thing about this before we move on. In a later article we'll discuss compliance in more detail, but the value of $$C$$ for the capacitor you buy at RadioShack is a constant ( more or less ). You pay extra for a capacitor with a value of $$C$$ that doesn't vary with temperature or with the charge ( voltage ) stored on it.  Blood vessels and cardiac chambers are nonlinear.  The pressure-volume relationship is not a straight line, but a curve. Vessels like the ventricles ( and atria ) make their living by cycling i.e. changing their compliance over a cardiac cycle and we'll find that this is one of the best ways to describe cardiac function, at least for clinical purposes.  So considering compliance of the blood vessels as a constant is one of the shortcomings of the models presented in this section. However there is much to be learned by considering these models even though we must keep this limitation in the back of our minds.  It turns out that we can get away with this analysis for blood vessels (arteries anyway) if the distentions are "small" enough ( and depending on the purpose of the analysis).

Above: The impedance of a capacitor (or linear compliance) is a function of frequency even though the value of $$C$$ is a constant.  The modulus of the impedance is $$1/(\omega C)$$, i.e. a hyperbola.  From a mathematical standpoint, the voltage across an ideal capacitor is the integral ($$\int$$) of the current  (multiplied by a constant, $$1/C$$). Hence the plot is also the Fourier domain representation of integration (think about it!).  The impedance phase of a capacitor (compliance) is $$-\pi/2$$ (all frequencies).

### Impedance of an Inductor (Inertance)

The last type of circuit element we'll consider here is called an inductor.  In the electrical world, an inductor is constructed using coils of wire and it constitutes another way of storing energy in the form of an electromagnetic field. Here's the schematic of an inductor:

Yup, looks just like a coil. And $$L$$ is the symbol used to represent an inductor. You've seen inductors on the electrical cords used to power a wide range of gadgets that you use ( e.g. computers).  The characteristic equation for an inductor is:

$$\Large v(t) = L \frac{di(t)}{dt}$$

The voltage generated across an inductor is related linearly to how rapidly the current through it is changing, i.e. $$di(t)/dt$$; the inductance $$L$$ is the proportionality constant of the relationship.  The equation shows that when we multiply an inductance by a current that is changing in time (e.g. Amperes/sec), we'd better get a voltage.  Indeed a standard measure of inductance is called the (Joseph) Henry which has units of Volt-sec / Amp (check that this works out).

The fluid analogy relating to inductance is due to the mass of the fluid which requires a force to change its velocity, i.e. $$\textbf{F} = m \textbf{a}$$.

$$\Large p(t) = L \frac{dq(t)}{dt}$$

Here we have an equation identical to the last but with the usual analogy between pressure and voltage, fluid flow rate and current.  $$L$$ is not called an inductance anymore but inertance, clearly having something to do with inertia and mass. An inertance stores energy in the form of moving fluid.  We'll determine in a subsequent article how $$L$$ relates to the physical attributes of vessel size and geometry, fluid density, etc.

Proceeding as before, we now take the Fourier transform of the characteristic equation:

$$\Large P(j\omega) = L j\omega Q(j\omega)$$

$$\Large Z_L(j\omega) = \frac{P(j\omega)}{Q(j\omega)} = j\omega L$$

Here, $$Z_L(j\omega)$$ is used to represent the impedance of an inertance. Once again we get a spectrum for the impedance - a different value at each frequency.  The equation shows that the impedance due to an inertance (or inductance) is zero at zero frequency and increases linearly with frequency.  The value of an inductance, $$L$$, is a constant -- that's what you pay for at the store.  Different from the electrical analog, we'll find that the value of the inertance in a cylindrical tube changes somewhat with frequency when we're dealing with oscillatory fluid flow.  That's because the velocity profile changes with frequency.

Inductance and capacitance are sometimes referred to as "duals" of each other:

$$\Large v(t) = L \frac{di(t)}{dt}$$

$$\Large i(t) = C \frac{dv(t)}{dt}$$

With the characteristic equations side by side you can appreciate the symmetry of function.  I don't know why the word "dual" was chosen.

Above: The impedance of an inductor (or linear inertance) is a function of frequency even though the value of $$L$$ is a constant.  The modulus of the impedance is $$\omega L$$, i.e. a line with slope $$L$$ if plotted against $$\omega$$ as shown.  From a mathematical standpoint, the voltage across an ideal inductor is the derivative ($$d/dt$$) of the current  (multiplied by a constant, $$L$$). Hence the plot is also the Fourier domain representation of  differentiation (think about it!)  The impedance phase of an inductor (inertance) is $$+\pi/2$$ (all frequencies).

### How is this Used to Model the Circulation?

The electrical analogy is applied to circulatory analysis rather extensively and in different ways.  I believe the term "resistance" is in common clinical use (sometimes inappropriately).  I also hear cardiologists sling the term "impedance" around whenever something fluidy is going on that may not be so easy to understand.

Here's a simple model of the systemic (or pulmonary) circulation that's in pretty widespread use:

We learn about the total peripheral resistance somewhere in our first year physiology course, computed as the time-averaged pressure loss (aorta to right atrium) divided by the cardiac output.  In this case, the resistance is due to an entire complex network of vessels -- arteries, arterioles, microcirculation, venules, and veins. Actually it's more like a clinical parameter than a model.  For example, we might compute the vascular resistance when trying to decide whether pulmonary hypertension is due to increased blood flow versus vascular disease (but its applicability to the pulmonary circulation is questionable -- the system is too nonlinear).  It doesn't explain anything really about the relationship between time-varying pressure and flow in the circulation.  (If it did, you would see that time-varying pressure and flow signals look exactly the same with $$R$$ as the proportionality constant.)  We could also use this approach to "model" any part of the circulation, e.g. a vascular bed.  This gives us a conceptual framework by which blood flow might be distributed and arterial pressure controlled.

Above the aorta acts  as a "bus" in circuit terminology -- having approximately the same average pressure along its length (vena cava too) .  Vascular beds are connected in parallel arrangement so that the resistance of each can be adjusted to control blood flow at need.   The total resistance, due to the equivalent parallel resistor for all the vascular beds, impacts the aortic pressure and hence the perfusion pressure of the individual beds.  The resistors here are termed "lumped parameter models" in that they are meant to embody the resistance of a largish segment of the circulation.  Inductors and capacitors can be used in this way also, e.g. representing the compliance of an entire vascular bed.  We'll look at some lumped parameter circulatory models a little later.

The triangular, multi-plated gadget at the lower right corner of the above circuit is a schematic for a "ground", usually taken to represent 0.0 voltage ( pressure ).  However we could specify a specific fixed voltage, or even a time-varying voltage at this point in the circuit. If we were going to specify a time varying voltage however, we would probably call it either a voltage source or a current source and there are schematic representations of those also.

For more detailed hemodynamic models, impedances are employed to represent smaller segments of the system.  An inductor and resistor in series, for example, could represent (model) the inertial and resistive properties of the blood flow in a single blood vessel:

The compliance of the vessel may be represented by a capacitor.  A resistor in series with the capacitor conveys (linear) viscoelastic behavior to the vessel wall ($$R_w$$); a resistor in parallel allows for the possibility of fluid leakage through the wall (conductance, $$R_G$$).

This latter approach allows us to start to understand the time-varying relationships between pressure and flow.  The next step would be to allow these impedance elements to represent a limited portion of a vessel.   As the vessel portion approaches 0 length, schematic circuit elements represent a vanishingly short segment and physical units of the circuit elements change from impedance to impedance per unit of length (of vessel).  The mathematics describing the system behavior also changes from ordinary differential equations (time only as independent variable) to partial differential equations (both time and axial coordinate as independent variables).  Then we have a "distributed model" where characteristics of the circulation emerge relating to transmission of pressure and flow waves.  All of these aspects will be addressed in due course.

### Kirchoff's Laws

Idealized electrical circuits are subject to analysis using Kirchoff's Laws which are an idealized expression of charge conservation.  The schematic of each circuit element suggests wires so that elements can be connected at a "node" where the wires meet.  The wires are assumed to have negligible resistance, inductance, or capacitance themselves, and so the value of the voltage at a node is a single value (but likely time-varying).

Here are 2 schematics of exactly the same thing ... A capacitor, resistor, and inductor met at a node .. (fill in your own punchline).

The orientation and sequence of the circuit elements has nothing to do with the behavior, but the fact that the 3 elements meet at a node is unambiguous.

Now I'm going to ask you to make a big leap of faith.  Each of the elements in the circuit has its own impedance representation.  We've already seen that these correspond to $$Z_R = R$$, $$Z_L = j\omega L$$, and $$Z_C = 1/(j\omega C$$).   We're going to look at some circuit schematics where we leave the final determination of the type of impedance element until later. So this thing:

can also be represented by the following where $$Z_1$$ will correspond to the resistor, $$Z_2$$ the capacitor, and $$Z_3$$ the inductor:

We just leave the type of circuit gadget out of the discussion for the time being.  If we were talking about resistors, equations  like $$P = Q \; R$$ and $$R = P/Q$$ wouldn't bother you.  Now I'm asking you to accept the fact that $$P = Q\;Z$$ and $$Z = P/Q$$.  Also we are going to work for sinusoidal voltages and currents ( pressures and flows).  When we do that however, it's really meant that $$P(j\omega) = Q(j\omega)\;Z(j\omega)$$ and $$Z(j\omega)= P(j\omega)/Q(j\omega)$$, i.e. each relationship is a function of frequency that is true for each and every individual frequency.   Furthermore we'll be able to perform conceptual manipulations where the $$Z$$ can represent any type of circuit element we choose. In fact, each impedance element might represent an entire complicated network of impedances. This will allow us to blast ahead without having to write down so much stuff on the page.

Here now is the first of Kirchoff's Laws - the current law.   A node cannot store any charge and is in essence an infinitesimal point in a circuit.  Consequently the sum of currents entering the node is exactly equal to the sum of currents leaving or entering the node.  Each node has a single ( but likely time-varying ) voltage value.  To describe this situation unambiguously, we resort to math.  In the schematic below, we'll call the voltage at the central node  $$V$$.  The voltages at the dangling end of the circuit elements will be called $$V_A$$ through $$V_D$$.  Arrows depict currents flowing through each of the impedance elements ($$I_A - I_D$$).

While the figure is drawn with all of the arrows pointing towards the inner node, the sum of these currents must add up to ZERO.   Hence at least one of the currents must be negative (directed opposite the arrow) if the others are positive (in the direction of the arrow).  In terms of the voltages in the schematic, each current is equal to a voltage (difference) divided by an impedance (using the characteristic equation for each impedance element).

$$\Large \frac{V_A-V}{Z_A}+\frac{V_B-V}{Z_B}+\frac{V_C-V}{Z_C}+\frac{V_D-V}{Z_D}=0$$

$$\Large \frac{V_A}{Z_A}+ \frac{V_B}{Z_B}+ \frac{V_C}{Z_C}+ \frac{V_D}{Z_D} = V \left[\frac{1}{Z_A}+\frac{1}{Z_B}+\frac{1}{Z_C}+\frac{1}{Z_D} \right]$$

$$\Large \frac{ \frac{V_A}{Z_A}+ \frac{V_B}{Z_B}+ \frac{V_C}{Z_C}+ \frac{V_D}{Z_D} }{\frac{1}{Z_A}+\frac{1}{Z_B}+\frac{1}{Z_C}+\frac{1}{Z_D}}= V$$

And these results could readily be generalized to a situation with any  number of impedance elements meeting at a node. Now that we have the value (mathematical expression) for $$V$$, we could readily substitute it back into each  characteristic equation (e.g. $$I_A = (V_A-V)/Z_A$$) to determine the current through each impedance.  We derive exactly 1 independent equation for each node in the circuit due to Kirchoff's current law.  As depicted, $$V_{A-D}$$ in the above are all unknowns and we would need more information to determine the actual current through each element.  I'll just remind you again that the "answer" for each current or voltage isn't a number but a spectrum - a function of frequency.  In general the value of each impedance changes with frequency (is a function of frequency), and so the currents and voltages do also.  Each sinusoidal frequency remains separate from every other in a linear time-invariant system.  That's what allows us to do solve these types of problems with "ease".  We can determine the results (voltages and currents) from any  set of inputs by separating the inputs into Fourier (frequency) components, calculating the impedance and outputs at each frequency, then adding the Fourier outputs back together to get the outputs in the time-domain (functions of time).

The next useful item is called Kirchoff's Voltage Law which states that the net (directed) voltage change around any closed loop in the circuit is $$0$$.  Suppose we have the following seemingly complicated network with red arrows depicting a pseudo-arbitrary path through the circuit:

Starting at the upper left, the application of the voltage law for the path depicted by red arrows looks like this:

$$\Large (V_1-V_2)+(V_2-V_3)+(V_3-V_7)+(V_7-V_{11})+(V_{11}-V_{10})+(V_{10}-V_6)+(V_6-V_5)+(V_5-V_1) = 0$$

Now take a look at this statement and imagine how you would place the parentheses differently to show off the absurd simplicity and obvious truth of the statement.  Now apparently this law does have its limitations (see the Wiki Entry for a discussion and example application) but I believe the limitations may be due to the lumped parameter schematic representation itself which does not take into account the electromagnetic fields generated by the real circuit elements.

### A Crash-Course in Circuit Analysis

This article started with a determination of the behavior of individual circuit elements.  What we find is that the combination of these elements into circuit networks results in complicated behavior that can be used to model a host of physical processes, including circulatory function.  Obviously you could spend years studying circuit analysis and behavior.  This isn't a course in electrical engineering, so we're going to speed up the process -- a LOT.  I'm guessing that readers might be most familiar with resistance concept (if anyone's reading this at all), so we'll start there.  Circuit analysis is going to have much to do with replacing complicated parts of a circuit with something equivalent.  By equivalent, I mean mathematically identical, i.e. we're not being slack here.

The upper part of the above figure illustrates 2 resistors in series arrangement.  Our task is to replace them with a single equivalent resistor ($$R_e$$ ) that exhibits the same characteristics, i.e. the same trans-resistance pressure (voltage) for a given flow rate (current).  For flow rate $$q$$, the pressure across $$R_1$$  is $$\Delta p_1 = q R_1$$. Kirchoff's current law tells us that the flow through the 2 resistors is the same so $$\Delta p_2 = q R_2$$.  So the total $$\Delta p$$ is $$\Delta p_1 + \Delta p_2 = q (R_1+R_2)$$.  Hence the total resistance can be replaced by a single resistor, $$R_e = R_1+R_2$$. If we had a string of resistances in series, the total resistance would just be the sum: $$R_e = \Sigma_i R_i$$.  $$R_e$$ for a series combination is always greater than  the single resistors involved  (it's the sum).  That's all there is to that!  Resisters in series behave just like a single resistor whose value (resistance) is the sum of the individual resistances.

For the parallel combination (lower part of above figure) we see that a given trans-resistance pressure (voltage) causes flow through both resistances: $$q_1 = p/R_1$$, and $$q_2 = p/R_2$$ where the $$p$$ (or voltage) across the 2 resistors is the same (Kirchoff's voltage law).  So the total flow is $$q_1+q_2 = p(1/R_1+1/R_2)$$.  Hence $$R_e = p/q= p/(q_1+q_2) = 1/( 1/R_1+1/R_2 )$$ and $$R_e = 1/( 1/R_1+1/R_2 ) = R_1 R_2/(R_1+R_2)$$.  While it may not be obvious from looking at the formula,  $$R_e = R_1 R_2/(R_1+R_2)$$ is less than either $$R_1$$ or $$R_2$$.  FYI it turns out that the fraction of flow through $$R_1$$ is $$R_2/(R_1+R_2)$$ and the fraction through $$R_2$$ is $$R_1/(R_1+R_2)$$.  Adding the 2 fractions is exactly 1.0 of course.

The equivalent resistor arising from multiple resistors in parallel is also readily determined.

$$\Large q = \Sigma_{i=1}^n q_i = p \Sigma_{i=1}^n \frac{1}{R_i}$$

$$\Large \frac{p}{q} = R_e = \frac{1}{\Sigma_{i=1}^n \frac{1}{R_i}}$$

The fraction of flow through $$R_i$$ is just $$R_e/R_i$$.

#### And Now, Impedances ...

Now we're now going to replace the resistances with impedances.  These impedances might be entire complicated circuits; just don't worry about that for the moment.   We're going to work in the Fourier (frequency) domain also so the currents and voltages (flows and pressures) are all sinusoidal.   The equivalent impedance for this thing (series arrangement) is:

$$\Large Z_{eq}(j\omega) = Z_1(j\omega)+Z_2(j\omega)$$

Yup, just like the resistors.  And the equivalent impedance of this thing?

$$\Large Z_{eq}(j\omega) = \frac{Z_1(j\omega) Z_2(j\omega)}{Z_1(j\omega) + Z_2(j\omega)}$$

That's all there is to that!  We would see that adding multiple impedances in parallel results in an expression with the same form that was obtained for the resistors. This is going to turn out to be a quick and dirty shorthand for understanding impedance networks and we're going to put this to work, right now.

Here's an arbitrary example problem.  I'll warn you ahead of time that you won't see something like this in the circulation.  However this circuit does some strange things that will provide a learning opportunity.

So forget the fact that we've got capacitors, inductors, etc in this circuit for a moment.  The whole thing is really just:

where:

$$\Large Z_1 = R$$

$$\Large Z_2 =\frac{1}{j\omega C}$$

$$\Large Z_3 = j\omega L$$

So the first thing we'll do is replace the 2 impedances in parallel with an equivalent impedance, $$Z_{eq}$$.  I'm also going to stop writing $$j\omega$$ all over the place:

$$\Large Z_{eq} = \frac{Z_2 Z_3}{Z_2+Z_3}$$

We can see already that the impedance of the whole thing ($$Z_i$$ i.e. Input impedance) is just:

$$\Large Z_{i} = Z_1 + \frac{Z_2 Z_3}{Z_2+Z_3}$$

where we've added the 2 impedances in series ($$Z_1+Z_{eq}$$).  Now, if we want to know more about what $$Z_{eq}$$ actually is, replace $$Z_2$$ with $$1/(j\omega C)$$ and $$Z_3$$ with $$j\omega L$$ from the original circuit:

$$\Large Z_{eq} = \frac{\frac{1}{j\omega C} j\omega L}{\frac{1}{j\omega C} + j\omega L} = \frac{j\omega L}{1 +(j\omega)^2 LC} = \frac{j\omega L}{1 -\omega^2 LC}$$

Now we've just got 2 impedances in series, $$Z_1$$ and $$Z_{eq}$$, that can be added algebraically:

The final $$Z_{eq}$$ for the whole circuit is just:

$$\Large Z_{eq} = \frac{V(j\omega)}{I(j\omega)} = R + \frac{j\omega L}{1 +(j\omega)^2 LC} = \frac{R + j\omega L + (j\omega)^2 RLC}{1 +(j\omega)^2 LC} = \frac{R[1-\omega^2 LC] + j\omega L}{1 -\omega^2 LC}$$

Don't forget that $$j^2$$ is just $$-1$$.  It only took a little bit of algebra to convert the impedance to the expression on the far right.  Hey, we're done!  (really?)  This is the input mpedance spectrum (a function of $$\omega$$) of the whole circuit diagrammed previously.  What does it mean? What do we do with it?  Well, the expression can be evaluated at any (every) value of angular frequency ($$\omega$$) we choose.  That value will be a complex number, but don't let that bother you.  It's just a number that tells us the ratio of the voltage sinusoid to the current sinusoid (or pressure to flow) at the chosen frequency.  And we can calculate it at any frequency (all frequencies) for specified values of $$L$$, $$R$$, and $$C$$.  Furthermore, we can calculate the input current (flow) given the input voltage (pressure) and vice versa.  Suppose we're given the input current.  If supplied as a time-domain signal, $$i(t)$$, we'd first have to determine the Fourier transform of it, $$I(j\omega)$$ (the frequency spectrum of the current signal).  Here's an answer:

$$\Large V(j\omega) = I(j\omega) \frac{R[1-\omega^2 LC] + j\omega L}{1 -\omega^2 LC}$$

Given the current as a function of frequency ($$I(j\omega)$$), we multiply the current at each frequency by the impedance value at the corresponding frequency (value of $$\omega$$). That gives us the magnitude and phase of the voltage - at that frequency.  Then we'd have to compute the inverse Fourier transform of $$V(j\omega)$$ to get the time-domain voltage, $$v(t)$$.  Given the voltage, we have an equally straightforward problem, but we're multiplying by the reciprocal of the expression for the impedance:

$$\Large I(j\omega) = V(j\omega) \frac{1 -\omega^2 LC}{R[1-\omega^2 LC] + j\omega L}$$

Next we'll find the differential equation that relates the time-domain voltage and current signals at the input.  The expression for the impedance was:

$$\Large \frac{V(j\omega)}{I(j\omega)} = \frac{R + j\omega L + (j\omega)^2 RLC}{1 +(j\omega)^2 LC}$$

If we simply multiply these fractions out - "rationalize" them:

$$\Large [R + j\omega L + (j\omega)^2 RLC] I(j\omega) = [1 +(j\omega)^2 LC] V(j\omega)$$

Now multiplication by $$j\omega$$ in the frequency-domain is the same thing as a derivative with respect to time in the time domain:

$$\Large \left[R + \frac{d}{dt} L + \frac{d^2}{dt^2} RLC\right] i(t) = \left[1 +\frac{d^2}{dt^2} LC\right] v(t)$$

$$\Large R\; i(t) + L \frac{di(t)}{dt} + RLC \frac{d^2 i(t)}{dt^2} = v(t) + LC \frac{d^2 v(t)}{dt^2}$$

Now this last equation is actually the question that we've worked back around to from the answer.  Starting with the circuit diagram, we could apply Kirchoff's laws and eventually derive the last equation.  Then the next problem would be to solve this differential equation - potentially not much fun if you don't like math.  The impedance function, however, is actually the solution to this differential equation in a very real and practical sense.  We've already seen that we can use it (the impedance) to calculation the voltage (pressure) given the current (flow), and vice verse.  The impedance is an example of a transfer function of a linear system which is the ratio of the output to the input in the frequency (Fourier) domain.   Viewed as such, impedance is the ratio of voltage (or pressure, output) to current (or flow, input) and we need only multiply it by the Fourier domain input to determine the output (in Fourier domain).

The impedance spectrum amounts to a complex number that is a function of frequency.  Using the example we've started, let's see what is meant by this.  Looking back at the previously derived expression for the impedance, we had:

$$\Large Z_{eq} = \frac{V(j\omega)}{I(j\omega)} = R + \frac{j\omega L}{1 +(j\omega)^2 LC} = \frac{R + j\omega L + (j\omega)^2 RLC}{1 +(j\omega)^2 LC} = \frac{R[1-\omega^2 LC] + j\omega L}{1 -\omega^2 LC}$$

It can be a little difficult to separate a function into real and imaginary parts.  However this one is rather simple.  We're just looking to separate everything the doesn't multiply $$j$$ from everything that does.

$$\Large Re[Z_{eq}] = R$$

$$\Large Im[Z_{eq}] = \frac{\omega L}{1 +(j\omega)^2 LC} = \frac{\omega L}{1 -\omega^2 LC}$$

It's often preferable to express a complex function as a modulus (magnitude) and phase (angle).  The magnitude is readily determined: a complex number amounts to a right angle triangle where the 2 sides are made up of the real and imaginary parts.  The magnitude is just the length of the hypotenuse that's determined from the Pythagoras formula:

For our particular circuit, we just stick in the previously determined real and imaginary parts:

$$\Large \left| Z_{eq}\right| = \sqrt{[Re(Z_{eq})]^2 + [Im(Z_{eq})]^2 } = \sqrt{R^2 + \left[ \frac{\omega L}{1 -\omega^2 LC}\right]^2}$$

The triangle has an associated angle whose tangent is the imaginary part divided by the real part:

$$\Large \angle Z_{eq} = \tan^{-1}\left[{\frac{Im(Z_{eq}) }{Re(Z_{eq}) }}\right] = \tan^{-1} \left[ \frac{\omega L}{R -\omega^2 RLC} \right]$$

Sticking in somewhat arbitrary values for the circuit elements ($$R=100$$, $$L=10$$, $$C = 0.00025$$), and computing these functions over a range of $$\omega$$ yields these plots:

\

Hmmm.... What went wrong (if anything)? The impedance modulus of this circuit soars off to $$\infty$$ around $$\omega = 20$$; the phase looks like it's got a discontinuity in it at the same frequency.

So there's nothing wrong with this; or at least the math is correct and we could expect this circuit to behave something like this if we were silly enough to construct it.  Here again are the real and imaginary parts of the input impedance:

$$\Large Re[Z_{eq}] = R$$

$$\Large Im[Z_{eq}] = \frac{\omega L}{1 +(j\omega)^2 LC} = \frac{\omega L}{1 -\omega^2 LC}$$

While the real part is a constant ($$R$$), the imaginary part tends to infinity at a particular frequency, $$\omega = 1/\sqrt{LC}$$.  For the values of $$L$$ and $$C$$ noted, this corresponds to $$\omega = 20$$.  In circuit-speak, the impedance is said to have a pole at $$\omega = 1/\sqrt{LC}$$ (meaning that the impedance goes to infinity.  Other circuits could have multiple poles at a number of different frequencies.)  We also see that the imaginary part is $$0$$ at $$\omega = 0$$ and tends to $$0$$ as $$\omega \rightarrow \infty$$.  Consequently the input impedance has the same value as $$R$$ at $$\omega = 0$$ and tends back to that value for $$\omega \rightarrow \infty$$.  This is telling us that any attempt to drive this circuit with a current of frequency $$\omega = 1/\sqrt{LC}$$ would require an infinite voltage.   Why does the circuit behave this way?  We're going to dig a little deeper into this, and to do so I'm introducing a couple of tricks of the trade - the concepts of a voltage and current dividers.

#### Voltage Divider

This is going to be a recap of something we already looked at for resistors in series and in parallel arrangement.  If we place 2 impedances in series with each other and a sinusoidal voltage is applied, the voltage at the node between the 2 impedances is the input voltage multiplied by a fraction:

$$\Large V(j\omega) = V_{in}(j\omega) \frac{Z_2}{Z_1+Z_2}$$

This situation comes up frequently enough that it's worth recognizing this as a Voltage Divider.  The result is all worked out so it's just a good thing to be able to recognize it at a glance, not that you couldn't work it out for yourself.  For our particular circuit, we already determined that $$Z_1$$ is the resistor and $$Z_2$$ is due to the parallel combination of the inductor and capacitor:

$$\Large Z_1 = R$$

$$\Large Z_2 = \frac{j\omega L}{1 +(j\omega)^2 LC} = \frac{j\omega L}{1 -\omega^2 LC}$$

Using the voltage divider formula, the voltage $$V$$ at the intervening node is:

$$\Large V = V_{in}(j\omega) \frac{Z_2}{Z_1+Z_2} = V_{in} \frac{\frac{j\omega L}{1 - \omega^2 LC}}{R +\frac{j\omega L}{1 - \omega^2 LC}} = V_{in} \frac{j\omega L}{R[1-\omega^2 LC]+j\omega L}$$

At $$\omega = 0$$ we find that the voltage at the intervening node is ... $$0$$, since we have $$j\omega L = 0$$ the numerator.  Also true as $$\omega \rightarrow \infty$$ since we'll have:

$$\Large V = V_{in} \frac{j\omega L}{R[1-\omega^2 LC]+j\omega L} \approx V_{in} \frac{j\omega L}{R[-\omega^2 LC]} \approx V_{in} \frac{-j}{\omega RC}$$

(Note: the approximation process just shown is dependent on your understanding of how terms in a formula or equation dominate the behavior.  An introduction was given previously. )  This behavior shouldn't surprise you.  At $$\omega = 0$$ the inductor's impedance is $$j\omega L = 0$$ and the circuit reduces to this:

The voltage across the capacitor is $$0$$  and the resistor is just connected to ground ($$V_1 = 0$$).  As $$\omega \rightarrow \infty$$, the circuit starts to look like this:

and we have the same thing - the resistor connected to ground and the whole circuit looks like the resistor alone.  This is an important aspect of circuit and equation analysis, i.e. to check the behavior at limiting values of the independent variables.

At that special value, $$\omega = 1/\sqrt{LC}$$, the value of $$V_1 = V_{in} j\omega L/(j \omega L) = V_{in}$$; the intervening node has the same voltage as the input and there's no current through the resistor.  WHY?!  Well we could have expected this by looking a little closer at the impedance of the capacitor - inductor combination before proceeding. A few lines up the page it was:

$$\Large Z_2 = \frac{j\omega L}{1 +(j\omega)^2 LC} = \frac{j\omega L}{1 -\omega^2 LC}$$

And we can see that this thing has infinite impedance at $$\omega = 1/\sqrt{LC}$$.  That seems rather odd (to me anyway).  Let's try to figure out why this occurs.

#### Current Divider

The final installment for this circuits short course is the current divider which was already alluded to above.

$$\Large I_1(j\omega) = I_{in}(j\omega) \frac{Z_2}{Z_1+Z_2}$$

$$\Large I_2(j\omega) = I_{in}(j\omega) \frac{Z_1}{Z_1+Z_2}$$

Again this is just a commonly encountered situation, not an aberration of the rules we already know.  The equations show that the fraction of current through a branch in the circuit is simply expressed in terms of the impedances.  For our particular circuit, $$Z_1$$ and $$Z_2$$ correspond to the capacitor and inductor, respectively:

$$\Large Z_1 = \frac{1}{j\omega C}$$

$$\Large Z_2 = j\omega L$$

Using these expressions for the 2 impedances in parallel (the current divider), we can determine the current in each branch:

$$\Large I_1 = I_{in} \frac{j\omega L}{j\omega L + \frac{1}{j\omega C}} = I_{in} \frac{j\omega L}{\frac{1-\omega^2 LC}{j\omega C}} = I_{in}\frac{-\omega^2 LC}{1-\omega^2 LC}$$

$$\Large I_2 = I_{in} \frac{\frac{1}{j\omega C}}{j\omega L + \frac{1}{j\omega C}} = I_{in} \frac{\frac{1}{j\omega C}}{\frac{1-\omega^2 LC}{j\omega C}} = I_{in}\frac{1}{1-\omega^2 LC}$$

We see that the sum of the 2 currents adds up to the total, $$I_{in}$$, but that it begins to look like current is infinite in both branches around the value of $$\omega = 1/\sqrt{LC}$$; Can that be true ?!  (yup).  Even though they both approach $$\infty$$, the ratio of the 2 currents is $$-1$$, i.e. the 2 currents are 180° out of phase.  Attempting to stimulate the circuit at the frequency $$\omega = 1/\sqrt{LC}$$ causes an infinite current that bounces back and forth between the capacitor and the inductor and also results in infinite impedance of the circuit as a whole.  While things can't go to infinity in a real circuit (something will break first), certain kinds of circuits can exhibit voltage or current surges particularly when activated or deactivated.  That's a situation where a the circuit receives a wide range of input frequencies and there is bound to be something in the critical frequency range to cause a problem.  That's why there are circuit breakers and fuses.

This was just an exercise to demonstrate some of the aspects of circuit analysis that will come up in the future.  This type of circuit or physics is not going to come up in models of the circulation which is why people don't intermittantly explode.  Had this been a real circuit you would have been electrocuted in the process of reading the article.  For safety's sake, this circuit was presented only in this article that will never be read by anyone.