Why do transmission line equations use current through inductor rather than capacitor?

I could not fit the title in I actually wanted which was this: In the transmission line equations why is the current difference either side of the inductor used in the capacitative current drop instead of the current difference either side of the capacitor?

In many references, eg. equation 2 from here:

The rate of change of voltage with x at a particular time is a function of the rate of change of current with time and the current itself.

$$I(x)-I(x+\delta X) = \frac{\partial V(x+\delta x)}{\partial t} C \delta x \\ -\frac{\partial I(x)}{\partial x} \delta x = \frac{\partial V(x)}{\partial t}C \delta x + \frac{\partial^2 V(x)}{\partial x \mathrm d t}\delta x C \delta x \\ \lim_{\delta x \to 0} \frac{\partial I}{\partial x} = -C \frac{\partial V}{\partial t}$$

the current drop through the inductor and resistor is used in the equation that relates the current drop through a capacitor to the capacitance and rate-of-change of voltage across a capacitor.

Surely the capacitance tells you about the current either side of the capacitor, in which case the currents used should be at the top and bottom of the capacitor?

EDIT: I will try to make this clearer as people ask questions. Normally I am used to $CdV/dt$ being given by the current through the capacitor (represented by $C$). But in the equation I linked to, $CdV/dt$ is being given by the difference in the current either side of the resistor-inductor pair. Why?

• Current isn't across something. – Olin Lathrop Dec 1 '14 at 20:03
• I changed "current across" to "current through", hopefully that is what you wanted. – kotozna Dec 1 '14 at 21:10
• OK, I have undone my downvote. – Olin Lathrop Dec 1 '14 at 22:17
• What do you mean by "current drop through a capacitor"? Ditto for "current drop through the inductor and resistor". Not terminology that helps me (at least!). – Andy aka Dec 1 '14 at 22:34
• @GeorgeHerold there is a "current drop" also, across some infinitesimally small section of the transmission line. Some of the current that went in, instead of being passed on to the next section, went through the capacitor (to change its voltage) and returned to the voltage source. Charge is still conserved in the system as a whole, but the current along the length of say, a coax center conductor (modeled by a infinitesimal series inductors and resistors) need not be uniform, so it must be "dropping" somewhere. – Phil Frost Dec 2 '14 at 1:42

Surely the capacitance tells you about the current either side of the capacitor, in which case the currents used should be at the top and bottom of the capacitor?

First, I think you might have a misconception, suggested by these words. In a capacitor, even when it is charging, the current is equal on both leads. That is, at every instant, for every unit of charge you put in one lead, you get exactly that much charge out the other lead in that same instant. This might seem counterintuitive, because we know a capacitor is made from two plates that are insulated from each other. How can a current flow through a capacitor, if there is no path for electrons to move from one side to the other?

But it doesn't matter: if you put an extra electron on one plate, it will repel exactly one electron from the other plate. It doesn't matter that the electrons you put in on one side are not the electrons you get out on the other.

Bill Beaty has a great article on capacitors, which has this pretty great image: So now, what about transmission lines? We know that a transmission line has some distributed capacitance. If we want to vary the voltage on a transmission line, we need some current, somewhere, to "charge" this capacitance, because you can't change the voltage across a capacitor without a current (by definition of what capacitors do: $i = C\:\mathrm dv/\mathrm dt$).

Here's another question: we know that at any instant, the current along a transmission line is not uniform. But how can that be? According to the lumped element model and Kirchoff's current law, if we put some charge in one end of a wire, we should get that same quantity of charge out the other. So, how can we have nonuniform current? There must be a "current drop" across something, but how is this possible?

With those two questions in mind, look at this again: simulate this circuit – Schematic created using CircuitLab

This represents some infinitesimally small segment of the transmission line. There's some current in. This is a lumped model, so all the current that enters must exit, but it can do so through two paths: it can continue to the next segment of the transmission line, or it can go through the capacitor and return to the source.

This resolves those two questions: when there's nonuniform current on the transmission line, it's because some of the current is being shunted by the distributed capacitance. The current "dropped" by each section of the transmission line is exactly the current required to charge (or discharge) the capacitance of that section of the line.

• Thanks, a very nice answer. I think if the diagram had shown I(x) and I(x+dx) either side of the capacitor's upper connection, I would not have been confused about the role of the other components. – kotozna Dec 2 '14 at 13:02

Remember that although we can have a voltage drop across an element, the current through a device is a single value. On the other hand, at a node (wire) the voltage is the same for each path, but the currents can differ as long as they sum to 0.

In the diagram you reference: there are two labeled currents. $I(x)$ is the current entering the differential element of transmission line and going through the inductor and resistor, and $I(x+\delta x)$ is the current leaving the differential element of transmission line.

Since charge is conserved, the amount of current entering a node must be equal to the amount leaving it. The current through the resistor and inductor cannot be different from one side to another since there is nowhere else for the current to go.

However, at the node above the capacitor the current splits. We know that $I(x)$ enters from the left and $I(x+\delta x)$ leaves to the right. That difference in current must come from the capacitor, therefore the current entering the capacitor is simply the difference, $I(x)-I(x+\delta x)$.

Applying the capacitor differential equation, $i(t)=C\frac{\mathrm dV(t)}{\mathrm dt}$, we get equation 2:

$$I(x;t) - I(x+\delta x;t) =(C\delta x)\frac{\mathrm dV(x;t)}{\mathrm dt}$$

In other words, your intuition was correct: we are actually using the current through the capacitor, not the "current drop across the inductor."

• Also a nice answer, thanks, I simply missed the fact the current I(x) could be placed after the inductor/resistor (i.e. further to the right, just before the capacitor). A basic error :0 – kotozna Dec 2 '14 at 13:05

Hmm, they are doing the change in the current as one moves along the transmission line.
The current at (x) Ix is different from the current at (x+dx). the only place that current can come from is the cap. (the current through the L and R are the same.) I think that's what is going on.

• Yes. But my question is why are they not using the current either side of the capacitor? – kotozna Dec 1 '14 at 21:50
• @kotozna, OK you seem to be a bit confused about current 'through' a capacitor. (True, there is current along the ground side of the cap, and also current from the L on the top side. The current through the cap (which can only flow at AC frequencies) links these two. – George Herold Dec 2 '14 at 1:26

In the picture you gave as reference, they are modelling a transmission line. For a given length dx you will get some leakage, or loss of signal.
The current through the equivalent series resistor (and inductor) remain constant. However, signal leakage is through current in the capacitor: C dV/dt as you mentioned.

You may be confused by the term I(x + dx) which implies the current is increasing... If the waveguide is a dipole (antenna) you could say the current is increasing as signal is picked up. Normally, however, dx would be a negative value as the signal is attenuated.