In high school we were all taught that electricity only flows in a loop. But it seems that one can send a "pulse" down an open conductor. This is used in a time-domain reflectometer. How is this possible?

How can you send electricity down an open conductor? What didn't they explain in high school that allows this to be possible?

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    \$\begingroup\$ Imagine a device which cannot be explained by grade-school science books: a big wide capacitor with the terminals at one edge of the plates. If you discharge it, the whole thing can't discharge instantly. Instead, during fast discharge a WAVE goes at the speed of light across the plates. (Then the wave reaches the far edge of the plates, and bounces! It can bounce upon returning, bounce repeatedly, so during discharge, the capacitor rings like a bell.) TDR uses this "capacitor plate-wave" effect. \$\endgroup\$
    – wbeaty
    Commented May 18, 2021 at 6:20
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    \$\begingroup\$ Also, what does "electricity" really mean? "Electricity," is it a form of energy? But electrical energy always travels one-way, going from source to load (from dynamo to distant washing machines.) This "electricity" energy never flows in a circle. Even during AC, the energy goes in a single direction, source to load. So "Electricity" ...ISN'T a form of energy? Faraday would agree. JC Maxwell would too! In other words, Faraday and Maxwell would fail your high-school science test, because the test actually contradicts the physics they discovered. \$\endgroup\$
    – wbeaty
    Commented May 18, 2021 at 6:32
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    \$\begingroup\$ When you get to AC circuitry, capacitance and inductance, you will see that this IS a complete circuit involving the coax cable's inductance and capacitance up to the break. Thus it permits current flow ... just not DC current. \$\endgroup\$
    – user16324
    Commented May 18, 2021 at 10:38
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    \$\begingroup\$ Also eternally relevant an old Radio Yerevan joke: Radio Yerevan was asked: "How does telegraphy work?" Radio Yerevan answered: "Imagine a large dog. The tail is in Moscow, the snout in Yerevan. When somebody pulls the tail in Moscow, it barks in Yerevan." "Thanks. Then how does wireless telegraphy work?" "Exactly the same way, just without the dog." There is probably more truth in it than the original author intended. \$\endgroup\$ Commented May 18, 2021 at 11:53
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    \$\begingroup\$ Most things taught in high school aren't exactly how they work. Just a close enough approximation. \$\endgroup\$
    – Wesley Lee
    Commented May 18, 2021 at 14:38

8 Answers 8


How does radio work? There's certainly no loop between the transmitter and the receiver, and yet they communicate, somehow.

How does a circuit with a capacitor, which has two plates separated by an insulator, work? If you start pulling the two plates apart, at what point does it stop being a capacitor and start being an open circuit?

How does the current know how much to flow, before having seen the resistor?

In high school we were all taught that electricity only flows in a loop.

Rigorously, what does this mean? What "electricity" are we talking about, anyway? This statement doesn't mean much because it doesn't even really define what it's talking about.

Kirchoff's circuit laws are more rigorous. Specifically, Kirchoff's current law states:

The algebraic sum of currents in a network of conductors meeting at a point is zero.

If you can't draw a loop, it's because you have a node somewhere that connects to only one thing. And the only way for one thing to sum to zero is for that one thing to be zero. So, current can't flow through an open circuit.

But here's the thing: a schematic is a mathematical model: it is not a physical electrical device. The lines are not real wires and the capacitors are not real capacitors. Rather, the things on a schematic represent idealized components which obey simple mathematical rules which may or may not be sufficiently representative of the real world.

The paradox arises when you take a wire connected to a TDR and model it as a line on a schematic connected to a TDR and nothing else:


simulate this circuit – Schematic created using CircuitLab

According to the rules of schematic evaluation, no current can flow. Yet the TDR does indeed work.

The paradox is resolved when the schematic is updated to more completely model the real world. Every section of a real wire is a small inductor. Likewise, every section of that wire also has some capacitance to all the things around it, like Earth. So a more complete schematic looks something like this:


simulate this circuit

Now there is a loop in which current can flow, and the paradox is resolved. Physically, the circuit is still just a TDR and a wire, but now the schematic more accurately models the real-world behavior of a real wire.

Continuing down this path you might want to model the resistance of the wire and other things. Eventually you will arrive at a transmission line model.

The moral of the story is that schematics are only models, and the model must include all the real behaviors of the physical device it describes. If you're just powering a small light with a wire, neither the current nor the voltage change rapidly, and so the inductance of the wire and its capacitance to the surrounding environment or other things in the circuit aren't really interesting, so you can leave them out of the schematic. Assuming low current for the gauge of the wire, you can also neglect the resistance of the wire. But a TDR is explicitly designed to send a very fast step down the wire, and now that inductance and capacitance is relevant, so omitting it from the schematic means the model doesn't sufficiently capture the real behavior of the device.

  • \$\begingroup\$ Well done explaining it succinctly and still almost as simple as school-level education; specifically by reducing it to components probably known to pupils... \$\endgroup\$
    – AnoE
    Commented May 20, 2021 at 11:38
  • \$\begingroup\$ This is a very good explanation, but I don't think it quite explains the how of my question. How does electricity actually flow down an open conductor? In a DC system you need a loop between positive and negative sides of the battery, or you cannot have current flow (as far as I am aware). In an AC system the same seems to be true. So in a DC system if you connect a wire between positive and negative electrons are pulled from the nearest atom to the positive side (which lacks electrons), this causes a wave of electron movement to travel down the conductor. How does this work in TDR? \$\endgroup\$
    – Jason
    Commented May 20, 2021 at 22:40
  • \$\begingroup\$ @Jason it flows through the capacitors. \$\endgroup\$
    – Phil Frost
    Commented May 21, 2021 at 2:47
  • \$\begingroup\$ @Jason think of it this way: you can make an object charged, right? For example, you can rub a latex balloon on a cat, and you can tell the balloon is now charged because it will stick to a wall. So the balloon must have more or less charge than it did before, so there must have been some nonzero current at some point. Where did that current flow, if it was never part of any loop? The answer: the capacitance between the balloon, the cat, the Earth, you, and all the other matter in the universe completes the loop. \$\endgroup\$
    – Phil Frost
    Commented May 21, 2021 at 2:51
  • \$\begingroup\$ @Jason another way to think of it: it's not very different from connecting an initially discharged capacitor to a battery. At first the capacitor isn't charged. Then it is. The transition between not charged and charged involves some transient current. Say you're connecting a twisted pair to a TDR. What's a twisted pair but a capacitor with really long thin plates? \$\endgroup\$
    – Phil Frost
    Commented May 21, 2021 at 2:54

The statement that electricity flows only in complete circuits is a simplification most suitable for so-called lumped systems, where all of the elements (e.g. capacitors, resistors, etc) are connected by short and relatively ideal wires, and the wavelength of any waves are much longer than the physical dimensions of the circuit. For example, the wavelength of a 100 kHz radio wave is on the order of single-digit kilometers, so a lumped-circuit model is suitable for discussion the operation of e.g. a linear audio amplifier.

Lumped system modeling is not an adequate model for systems where your signals have wavelengths shorter than the circuit elements themselves -- in that situation, distributed models and electromagnetic theory are better descriptions and the lumped-circuit model falls apart. This theory is commonly seen in microwave and high-speed radio circuits, where even the shape and positioning of wires is key to achieving the necessary performance goals.

In the distributed-element model, a transmission line can be modeled as a medium where voltage and current waves travel under particular constraints. The key ones are: the propagation speed (how fast a disturbance moves down the line), characteristic impedance (the ratio of voltage to current waves in a disturbance traveling on the line), and loss tangent (how much the disturbance decays as it travels).

Under these assumptions, discontinuities (where the characteristic impedance of the line changes) must lead to a reflection as a result of mathematical boundary conditions at the discontinuity. Time-domain reflectometry relies upon this exact mechanism, transmitting a sharp pulse and noting when reflections return to the source. This is not unlike ultrasonic inspection or sonar detecting cracks or objects in an acoustic medium.


They omitted details and simplified things to teach basic concepts. Think of static electricity. That is only one way. Or an simple dipole antenna.

Nothing happens faster than the speed of light. Which means it takes time for all parts of any circuit to react and reach an equilibrium. The voltage source doesn't instantly and magically know what the value of the resistor is in the circuit to know how much current to push. When something changes in a circuit (such as when you first connect the power supply or hit a switch or anything), you get transient currents flowing back and forth and this is basically the components "talking" to each other to reach an equilibrium throughout the entire circuit. These may be very short, but they are there and can be measured with proper equipment and setup.

These are called transmission line effects and this is what TDR is looking at.

Here are some mechanical analogies:

  • A closed circuit is like a loop of pipe with a pump in the middle. it can pump water in one direction continuously. This is what you are first taught in school. Current can only flow CONTINUOUSLY in a closed loop.
  • An open-circuit on a wire is like pipe that is plugged on one end and a pump on the other end. But instead of a pump, let's just say it's you pouring water into the pipe. You might start pouring the water at some given rate and it will all enter the pipe smoothly... until the water reaches the plugged end. At which point, the water will start to "come back towards you" as the pipe gets full and when it arrives back at your end it is going to splash you in the face. So current can flow in an open-circuit but NOT CONTINUOUSLY.
  • An impedance discontinuity in a circuit (whether open or closed) is like you pouring water into a pipe that has a narrowing somewhere down the line. It can also be a widening but this doesn't work so well for mechanical analogies. Let's just stick with a narrowing. Like above, you start pouring water at some rate and it goes in smoothly but when it reaches the narrowing some water continues but since you were pouring in water too quickly there is an excess and the first, wider portion of the pipe begins to fill and water works its way back towards you. And when it gets to your end it splashes you in the face.

The water coming back and you and splashing is basically a reflection coming back at you telling you what is on the other end of wire, and that takes time. This is what TDR is using. If you were a voltage source, the splashing tells you that you need to adjust the how fast you are pushing water into the pipe. In the second case you would adjust the flow to zero, and in the third case you would reduce the flow so that water continuous to flow smoothly through the narrowing without splashing. And as you adjust this rate, you have to observe how much it continues to splash (or overflow) and this takes time back and forth between you and the narrowing. These are the transient signals bouncing back and forth in the wire communicating the difference parts of the circuit to each other to reach equilibrium.

You CAN send current into a dead end the same way you can send water into a dead end pipe...but not continuously. You will eventually reach an equilibrium where the accumulated charge in the antenna equally resists how hard you are trying to push charge into the antenna and current will stop flowing. If you then increase the voltage you can shove more charge into the antenna. Water is incompressible so doesn't have an equivlent. However, if imagine it with air, then you can increase the pressure to shove more air into the dead end pipe. Of course you can repeatedly inject and suck out water out of a dead end pipe and this is what an antenna is except with charge.

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    \$\begingroup\$ One thing that might help bridge the gap here: when you say "you will eventually reach an equilibrium", consider the time constants involved. For a water pipe, it might be measured in seconds or even tens of seconds. But for DC current flowing into a wire, we're talking about picoseconds to low nanoseconds to reach equilibrium, which is why we can often simplify it away. \$\endgroup\$ Commented May 18, 2021 at 20:16

what didn't they explain in high school that allows this to be possible?

A lot of what electricity truly is is not covered in high school, because at that time you have neither the physics nor the math foundations to discuss the concepts.

For example, waveguide is coaxial cable with no center conductor. Heaviside had to invent vector calculus to describe coaxial cable, which he invented, and to "adjust" Maxwell's equations to the form we know today.

As a concept, shielded cable is relatively straightforward. Far less so is how two conductors separated by a Teflon insulator can have a "characteristic impedance", just one of the consequences of the finite speed of light on how an electromagnetic field pulse propagates.

Transforms, tensors, complex vectors - there's a lot to learn and discover. Enjoy the ride.


  • \$\begingroup\$ The characteristic impedance of a coax conductor has NEVER made sense to me in a physics sense. I just trust that it's a thing...and it means something and "oh, by the way, make sure the last device on the end of this is terminated actual matching resistor. Otherwise, you'll have ALL sorts of unwanted noise." To think that the guy invented the math to accommodate the phenomena is fascinating. Really need to jump back in and finish my degree after 10 years of practical work, and screwing around with all these "rules of thumb". \$\endgroup\$
    – depwl9992
    Commented May 18, 2021 at 23:26
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    \$\begingroup\$ Heaviside didn't just invent the math - he patented the coaxial cable! waveform.com/blogs/main/… physicstoday.scitation.org/do/10.1063/PT.5.030967/full \$\endgroup\$
    – D Duck
    Commented May 19, 2021 at 14:50
  • \$\begingroup\$ @depwl9992 I think of it as a water pipe with different widths down the length and how smoothly water can be injected into the pipe. If it's a dead end (open-circuit, high-impedance) , you get water hammer. \$\endgroup\$
    – DKNguyen
    Commented May 19, 2021 at 21:10
  • \$\begingroup\$ Water pipes and HVAC ducts have a characteristic impedance. If you have a constant pressure source (fan) and a long duct (transmission line), the shape of the opening at the far end affects net air flow. Measure the air flow with the duct abruptly ending into a wide-open space, the equivalent of an open circuit. Now add a hyperbolically-curved flair to the end of the duct, like the curved body of a PA horn. This acts as an impedance matching transformer between the duct and the space. The airflow will increase. \$\endgroup\$
    – AnalogKid
    Commented May 19, 2021 at 22:21

were all taught that electricity only flows in a loop

This is a "lie to children": a simplification of something very complicated.

Broadly, "electricity" does not exist, or at least the word does not cleanly relate to specific things in the physical world.

Charge exists, as a property of charge carriers, which are real quantum-scale physical objects. But charge is not electricity. It can be observed as the phenomenon "static electricity", and anyone who's ever got a shock off their carpet will have noticed that these pulses don't require a loop - they flow from an accumulation of charge.

Electric fields exist, around every charge. They are as real as gravity and can be observed in the same way, because electric fields exert a force on charge.

The point at which it starts to get complicated is EM waves; changes in the field don't propagate instantly, they propagate as a wave does from a stone dropped into water. And they propagate through both air and metals, but in very different ways (see "dielectric").

A TDR pulse is an EM wave. Like a radio wave. It may help to think of it as a radar pulse telling you the range to a target, except because of the structure of the conductor it's (mostly) trapped inside. In the same way that a pulse of light is trapped in an optical fiber.


While both of the other answers are technically correct, I believe they are too technical based on your question.

If you have a coax cable with a fault in it, perhaps an open or a short, you could use a multimeter and measure the resistance and it would read infinity or close to 0 Ohms. But that would not tell you WHERE the problem lies.

This is where a tool like a TDR comes into play. The multimeter uses a DC voltage to make its measurements but a TDR uses a pulse. In high school they teach you basic AC and DC circuit fundamentals but these are simplifications of what's really going on. In nearly all cases with things like household wiring these concepts hold. But once you move to either long wires or high frequencies (RF) those simplified concepts break down.

If you take that same coax cable and use a pulse generator instead of a constant DC source and look at the voltage with an oscilloscope, you will see something very different from what you might expect. You'll see the original pulse but at some point later you'll see that pulse again either at the same polarity or reversed polarity. What's going on? It's a REFLECTION from the fault in the line. You can take a full semester course in a college level electrical engineering curriculum on traveling waves (in most programs it's required) and learn all about this but the bottom line is that when there is something different in the coax, it will reflect.

So this is the basis of a TDR. It injects a pulse and then measures the reflection. There is one other thing that needs to be known before it can determine how far down the line the fault is. That's the speed at which the pulse travels. This is a characteristic of the particular type of coax you are looking at. If you look at the specs for a type of coax, or any other transmission line type, you will see a propagation speed usually as some percentage of "c" the speed of light. So your coax may have a propagation speed of 0.75c or 75% of the speed of light.

The TDR measures the time for the reflection to return and then uses the propagation speed to calculate the distance (i.e. time * speed = distance) and you will know where the fault is.

So the bottom line is that while in the simplified case your line is either open or shorted but in reality it's more complex and looks to the driver as a distributed set of resistance, capacitance, and inductance values. If you search for "transmission line model" you'll see some examples of what can be used to model a transmission line or "T line".

  • \$\begingroup\$ But the question is why it's possible to send a pulse into an open-circuited cable in the first place. Introductory electricity teaches that since it is not a closed circuit, no current can flow at all. \$\endgroup\$ Commented May 18, 2021 at 14:53
  • \$\begingroup\$ As I said, what you've been taught is simplified and does not totally represent all real-world cases. Once you transition into cases where there are frequencies which are greater than about 1/4 of the wavelength, those simplifications break down and you must use a better model. This is analogous to Newton's Laws of Motion. For nearly every case we encounter in our everyday lives, they work with 100% accuracy. But if you were working at the Large Hadron Collider you would not be able to use Newton's, you must use Einstein's Relativity because it's a better model than Newton's. \$\endgroup\$
    – jwh20
    Commented May 18, 2021 at 15:00
  • \$\begingroup\$ From a DC circuit perspective, an open transmission line is simply an open circuit. Your voltmeter will show that. But for a pulse generator, it's not open at all. Your oscilloscope will show that as well a TDR. \$\endgroup\$
    – jwh20
    Commented May 18, 2021 at 15:03
  • \$\begingroup\$ and that's the question: why is it not open at all? You have not answered this \$\endgroup\$ Commented May 18, 2021 at 15:04
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    \$\begingroup\$ @jwh20, while all you're writing seems factually correct, I think OP is interested in the small sentence It injects a pulse. How (physically/technically) can a pulse (what "is" a pulse, physically) be injected into a open wire, by what mechanism does it happen? School only teaches about the mechanism where an anode or cathode "push" or "pull" electrons around in conjunction with each other, but this clearly seems to be to the case when talking about antennae, open coax cables and such. \$\endgroup\$
    – AnoE
    Commented May 20, 2021 at 11:35

In electrodynamics, current is more than just the flow of charge. Where there is a changing electric field, a displacement current flows. To apply the electricity flows in a loop rule (Kirchhoff's Current Law) to a dynamic situation, you must account for the displacement current. It is the displacement current that closes the circuit when you launch an electromagnetic wave down a cable in TDR.

  • \$\begingroup\$ This is the correct answer. \$\endgroup\$ Commented May 18, 2021 at 17:22

Your teacher was fundamentally correct, but the fine detail was omitted to be taught later to student that need to know on their particular career path, You are obviously on such a path and need to know. The TDR is normally used on cables with two conductors. The two conductors are often either a closely coupled pair of wires or a coaxial cable. I you place your Ohm meter on such a pair of wires, and the pair are long enough, you may notice a tiny flick of the needle on your meter. This will only be noticeable on an analogue meter off course. What is happening, is the cable is being charged like a tiny capacitor. The meter effectively applies a pulse, or more correctly a step onto the line, this step travels close to the speed of light down to the open cct. end and is reflected or bounces back to the meter, is reflected and bounces back to the open cct. end again. This bouncing continues until the line has reached the voltage of the meter terminals, the meter now indicates an open cct. The whole process lasts less than a microsecond for a cable in the 100 yard length range. The flick of the needle is small because it is incapable of response for these small duration's. The TDR is designed to operate at these speeds and shows the charging process in detail. The TDR may also apply an extremely narrow pulse rather than a step and only display the first bounce, the distance in time between the two pulse’s is the length of the line the, TDR is usually calibrated in distance rather than time for operator convenience.

Your high school teacher was demonstrating DC characteristics which assumes very long settling times. The very fast charging and reflection (bouncing events) are more akin to AC. Hope this helps, but even this is not the full story, pico second TDR’s can observe discontinuities in microwave structures down to distances of the order of millimetre's.


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