My previous post was long and tedious, and all it did was presenting the three (passive) circuit elements as well as the concept of impedance. It show the *inner workings *of these little devices are actually quite complicated. Fortunately, the conclusions were very neat and short: for all circuit elements, we have a very simple relationship between (a) the voltage *across *the terminals of the element (**V**) and (b) the current that’s going *through *the circuit element (**I**). We found they are always in some ratio, which is referred to as the** impedance**, which we denoted by

**Z**:

**Z** = **V**/**I ⇔ V **=** I∗Z**

So it’s a ‘simple’ ratio, indeed. But… Well… Simple and not simple. It’s a ratio of two complex numbers and, therefore, it’s a *complex *number itself. That’s why I use the ∗ symbol when re-writing the **Z** = **V**/**I** formula as **V **=** I∗Z**, so it’s clear we’re talking a product of two complex numbers). This ‘complexity’ is best understood by thinking of the voltage and the current as *phase vectors* (or* **phasors *as engineers call them). Indeed, instead of using the sinusoidal functions we are used to, so that’s

- V = V
_{0}·cos(ωt + θ_{V}), - I = I
_{0}·cos(ωt + θ_{I}), and - Z = Z
_{0}·cos(ωt + θ) = (V_{0}/I_{0})·cos(ωt + θ_{V}− θ_{I}),

we preferred the *complex *or *vector *notation, writing:

**V**= |**V**|*e*^{i}^{(ωt + }^{θV}^{) }= V_{0}·*e*^{i}^{(ωt + }^{θV}^{)}**I**= |**I**|*e*^{i}^{(ωt + }^{θI}^{) }= I_{0}·*e*^{i}^{(ωt + }^{θI}^{)}**Z**= |**Z**I*e*^{i}^{(ωt + }^{θ}^{) }= Z_{0}·*e*^{i}^{(ωt + }^{θ}^{) }= (V_{0}/I_{0})·*e*^{i}^{(ωt + θV − θI}^{)}

For the three circuit elements, we found the following *solution *for **Z **in terms of the previously defined *properties *of the respective circuit elements, i.e. their *resistance *(R), *capacitance *(C), and *inductance* (L) respectively:

- For a
**resistor**, we have**Z**(resistor) =**Z**_{R }= R - For an
**capacitor**, we have**Z**(capacitor) =**Z**_{C }= 1/*i*ωC = –*i*/(ωC) - For an
**inductor**, we have**Z**(inductance) =**Z**_{L}=*i*ωL

We also explained what these formulas meant, using graphs like the ones below:

- The graph on the left-hand side gives you the ratio of the
*peak*voltage and*peak*current for the three devices as a function of C, L, R and ω respectively. - The graph on the right-hand side shows you the relationship between the
*phase*of the voltage and the current for a capacitor and for an inductor. [For a resistor, the phases are the same, so no need for a graph. Also note that the lag of the phase*inductor*, while it’s 270 degrees for a*capacitor*(which amounts to the current*leading*the voltage with a 90° phase difference).]

The inner workings of our circuit elements are all wonderful and mysterious, and so we spent a lot of time writing about them. That’s finished now. The summary about describes all of them in very simple terms, relating the voltage and current *phasor *through the concept of *impedance*, which is just a ratio—albeit a *complex *ratio.

As the graphs above suggest, we can build all kinds of crazy circuits now, and the idea of *resonance* as we’ve learned it when studying the behavior of waves will be particularly relevant when discussing circuits that are designed to *filter *certain frequencies or, the opposite, to *amplify *some. We won’t go that far in this post, however, as I just want to explain the basic rules one needs to know when looking at a circuit, i.e. *Kirchoff’s circuit laws*. There are two of them:

**1. Kirchoff’s Voltage Law (KCL): The sum of the voltage drops around any closed path is zero.**

The principle is illustrated below. It doesn’t matter whether or not we have other circuits feeding into this one: Kirchoff’s Voltage Law (KCL) remains valid.

We can write this law using the concept of circulation once again or, what you’ll probably like more, just using plain summation:

**2. Kirchoff’s Current Law (KCL): The sum of the currents into any node is zero.**

This law is written and illustrated as follows:

This Law requires some definition of a *node*, of course. Feynman defines a node as any set of terminals such as a, b, c, d in the illustration above which are *connected*. So it’s a set of connected terminals.

Now, I’ll refer you Feynman for some *practical* *examples*. The circuit below is one of them. It looks complicated but it all boils down to solving a set of linear equations. So… Well… That’s it, really. We’re done! We should do the exercises, of course, but then we’re too lazy for that, I guess. 🙂 So we’re done!

Well… Almost. I also need to mention how one can *reduce *complicated circuits by **combining parallel impedances**, using the following formula:

And then another powerful idea is the idea of ** equivalent circuits**. The rules for this are as follows:

- Any two-terminal network of passive elements is equivalent to (and, hence, can be replaced by) an
(*effective*impedance**Z**)._{eff} - Any two-terminal network of passive elements is equivalent to (and, hence, can be replaced by) a
*generator in series with an impedance*.

These two principles are illustrated below: (a) is equivalent to (b) in each diagram.

The related formulas are:

**I**=**Ɛ**/**Z**_{eff}**V**_{n }=**Ɛ**_{eff }−**I**_{n}∗**Z**_{eff}

Last but not least, I need to say something about the **energy** in circuits. As we noted in our previous post, the impedance will consist of a real and an imaginary part. We write:

**Z** = R + *i*·X

This gives rise to the following powerful equivalence: **any impedance is equivalent to a series combination of a pure resistance and a pure reactance**, as illustrated below (the ≡ sign stands for equivalence):

Of course, because this post risks becoming too short 🙂 I need to insert some more formulas now. If **Z** = R + *i*·X is the impedance of the whole circuit, then the whole circuit can be summarized in the following equation:

**Ɛ **= **I**∗**Z** = **I**∗(R + *i*·X)

Now, if we bring the analysis back to the *real *parts of this equation, then we may write our current as I = I_{0}·cos(ωt). This implies we chose a t = 0 point so θ_{I}** _{ }**= 0. [Note that this is somewhat different than what we usually do: we usually chose our t = 0 point such that θ

_{V}

**= 0, but it doesn’t matter.] The**

_{ }*real*emf is then going to be the real part of

**Ɛ**=

**I**∗

**Z**=

**I**∗(R +

*i*·X), so we’ll write it as Ɛ (no bold-face), and it’s going to be the real part of that expression above, which we can also write as:

**Ɛ **= **I**∗**Z** = I_{0}·*e ^{i}*

^{(ωt)}

^{ }∗(R +

*i*·X)

So Ɛ is the real part of this **Ɛ** and, you should check, it’s going to be equal to:

Ɛ = I_{0}·R·cos(ωt) − I_{0}·X·sin(ωt)

The two terms in this equation represent the voltage drops across the resistance R and the reactance X in that illustration above. […] Now that I think of it, in line with the -or and -ance convention for circuit elements and their properties, should we, perhaps, say *resistor *and *reactor *in this case? 🙂 […] OK. That’s a bad joke. [I don’t seem to have good ones, isn’t it?] 🙂

Jokes aside, we see that the voltage drop across the resistance is *in phase* with the current (because it’s a simple cosine function of ωt as well), while the voltage drop across the purely reactive part is *out of phase* with the current (as you know, the sine and cosine are the same function, but with a phase difference of π/2 indeed).

You’ll wonder where are we going with this, so let me wrap it all up. You know the power is the emf times the current, and so let’s integrate this thing *over one cycle* to get the *average* rate (and then I mean a *time rate of change*) of the energy that gets *lost *in the circuit. So we need to solve the following integral:

This may look like a monster, but if you look back at your notes from your math classes, you should be able to figure it out:

- The first integral is (1/2)I
_{0}^{2}·R.. - The second integral is zero.

So what? Well… Look at it! It means that the (average) energy loss in a circuit with impedance **Z** = R + *i*·X *only *depends on the real part of **Z**, which is equal to I_{0}^{2}·R/2. That’s, of course, how we want it to be: ideal inductances and capacitors *store *energy when being powered, and give whatever they stored when ‘charging’ back to the circuit when the current reverses direction.

So it’s a nice result, because it’s consistent with everything. Hmm… Let’s double-check though… Is it also consistent with the power equation for a resistor which, remember, is written as: P = V·I = I·R·I = I^{2}·R. […] Well… What about the 1/2 factor?

Well… Think about it. I is a sine or a cosine here, and so we want the average value of its *square*, so that’s 〈cos^{2}(ωt)〉 = 1/2.

** Done! **🙂