Let's say you have two really long, straight wires (extremely low resistance). Hook one of them to the + of a battery, and the other to a -. Would it conduct (for a limited time) as if there were a circuit?
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Let's say you have two really long, straight wires (extremely low resistance). Hook one of them to the + of a battery, and the other to a -. Would it conduct (for a limited time) as if there were a circuit?
at a few kV it will.
just go look at a schematic of a tesla coil.
Only if the wires are close enough. There are 4 variables that determine something called the characteristic impedance. These are (all per unit length): the resistance of the conductor, the inductance of the conductor, the conductivity of the dielectric (insulator) and the capacitance of the dielectric.
An ideal wire has zero resistance, inductance, capacitance and conductivity. Remember that resistance and inductance are along the length of the line and conductivity and capacitance are between the conductors.
To answer your question, the only way for current to flow is if the dielectric is conducting or breaks down. A spark would be an example of a breakdown, as would the small amount of current flowing from one conductor to another through the insulator.
Two ideal wires where the wires are seperated by an ideal dielectric will not result in current flow. Most two wire real world cables (coax, parallel wire etc.) don't have perfect dielectrics, so there is current flow even with the wires open on one end. This would not happen in an ideal transmission line. In short, there has to be some non-infinite impedance between the two wires. If the dielectric isn't perfect, then several factors will determine how much current flows in this otherwise open line including the line length and the frequency of the applied voltage.
Ah right, I forgot about the dielectric. How about a perfect insulator? My question should have been more along the lines of "if there is no circuit that actually exists, (not even through air), would physics allow current to flow?"
I figured that the battery has to do SOMETHING to the wires, since you can connect them in the middle (far, far away from the battery), and somehow it's supposed to "know" that it can start conducting now. But if the wires are, say, light-years in length, we don't know if they're connected in the middle or not, so I'd say it conducts until _something happens_. I have no idea what that something is.. Maybe it actually does not conduct, even if they are connected. I have no idea. :DCode:=======WIRE====== =======WIRE=======
+ -
-----
|bat|
|ter|
| y |
-----
I think that energy would flow into the wire until the initial impulse bounces back (at about the speed of light).
Consider the water hammer effect (its kinda the opposite but similar): when you flush a toilet in an old building, water flows through the pipes. When the valve slams shut, the water immediately next to the valve stops right away. The water a little further away slams into the stopped water and stops itself. And so on and so on and so on. This creates a single pressure wave traveling backwards from the valve at the speed of sound. When this wave hits obstructions like bends in the pipe or other valves, some of it bounces back, some of it just shakes the pipe. This is what causes the banging sound you hear after a valve slams shut. Sounds like someone is banging on the pipes with a hammer.
When you open a valve quickly, water right next to the valve begins to flow right away. Water further away doesn't know the valve is open until the wave of lower pressure gets to it. Its the opposite of the water hammer effect and similar to what you are describing.
Thinking of energy flowing in isn't the best way to describe an ideal lossless line. In a non ideal case, the capacitances would be charged etc., making the energy flow idea a better way of looking at things. Since you have currents and voltages, describing the line in terms of power flow is better.Quote:
Originally posted by russ_watters
I think that energy would flow into the wire until the initial impulse bounces back (at about the speed of light).
I think the most important clarification that needs to be made is between current and voltage. Current deals with "flow", wheras voltage deals with "potential". With perfect lines, the battery would do something to the end of the wire...they would be at the same voltage potential as the battery terminals. However, without the ends being connected, current cannot flow. This is why you shortly after shorting the ends, current would flow. That is to say, its instantaneous not because the battery knows per se, but because the ends of the wires were at the same potential as the battery terminals (after some time, explained below).Quote:
Originally posted by Strogian
Ah right, I forgot about the dielectric. How about a perfect insulator? My question should have been more along the lines of "if there is no circuit that actually exists, (not even through air), would physics allow current to flow?"
I figured that the battery has to do SOMETHING to the wires, since you can connect them in the middle (far, far away from the battery), and somehow it's supposed to "know" that it can start conducting now. But if the wires are, say, light-years in length, we don't know if they're connected in the middle or not, so I'd say it conducts until _something happens_. I have no idea what that something is.. Maybe it actually does not conduct, even if they are connected. I have no idea. :DCode:=======WIRE====== =======WIRE=======
+ -
-----
|bat|
|ter|
| y |
-----
I think what you're looking to get an answer to is what happens when the battery is connected...how can the voltage at the ends be the same as the voltage at the battery instantly? Well it can't. What happens is connecting a battery sends what is known as a transient down the line. A transiant is just a change in voltage or current, and in this case would be from 0V to whatever the battery's voltage is. This transiant travels at the phase velocity of electricity in the wire (some speed less than the speed of light,c, usually about .5-.9c). This transiant would eventually reach the end of the wire (if the wire were a light year long, after a year) and then depending on what was at the end, it could bounce back.
I hope I'm not confusing you. The distinctions to be made are between current and voltage and the concept of transiant propogation. You can take a whole class on the topic :). Hopefully Moridin will come by to put a more comprehensible spin on the topic.
russ_watters is correct, a current will flow then reflect off the end of the line. This is a classic transmission line problem. (Try googleing transmission line effects or time domain reflectometry)
As Ramuman said, the voltage/current at the end of the line cannot chance instantaneously, as this would violate relativity. This means what happens when you initially connect a voltage/current source to a wire must be completely independent of the resistance on the end of that wire.
When you first connect the wire to the source, rather then the resistance at the end of the wire (which in this case is infinite) the current/voltage is a function of the characteristic impedance of the wire. IIRC a bare wire in air will have the characteristic impedance of air which is 377 ohms.
When you initially connect the terminal a current of I = V/377 will flow into the wire. (You do not even need the matching pair, a single wire will do. IIRC, this will alter the characteristic impedance.) If the wire ends in an open circuit the voltage pulse (a negative of???) you first applied will be reflected back in the other direction.
Once it arrives back at the battery it will reflect once again, but this time it will not reflect completely. Instead the internal resistance of the battery will dampen the reflection and you will end up with a reflection factor of less then 1.0. If the battery had an output impedance of 100 ohms the reflection coefficient would be (377-100)/(377+100) = 0.58.
The initial signal will in fact reflect back and forth down the transmission line until the damping form the reflection coefficient and the resistance of the line reduce the voltage pulse to a vale near zero. The voltage/current in the line at any point will be the sum of all the pulses that have traveled down the line up until that point.
As you would expect, since it’s an open circuit, this sum will tend to zero for the current and Vbattery for the voltage. It’s been a while, and I suspect I have some signs wrong, and possibly a few other errors but the description is basically correct. (I now think I should have discussed a current pulse rather then a voltage pulse)
This is why it’s important to use matched transmission lines. For example the coax that goes into your TV is a 75ohm transmission line. Your TV terminated this with a 75 ohm load, so the reflection coefficient for the first reflection is (75-75)/(75+75) = 0 so there is no reflection.
If you are applying an AC voltage/current, the situation is even more complex, and what ultimate happens if you do not have a matched line is that you end up with standing waves on the line.
What Ramuman posted is also correct, but it’s more of a technical background for deriving the mathematical formulas surrounding the effects I described, as well as quantities like the characteristic impedance.
Discussions like these are why i fear magnetism/electrodynamics in general physics 2 next semester. Kinda interesting though, i did learn something from all this about the reflections.:)
It shouldn't be a problem if its anything like my Physics 2 was. In physics you shouldn't be modeling transmission lines etc. You'll deal more with particles in a magnetic or eletric field etc. What was a pain was the actual E&M class in EE. I took a class on just transmission lines last semester and that's really where I understood these topics. Like I said above, you can easily spend an entire class discussing this stuff and Smith charts etc. :)Quote:
Originally posted by speedyaxon
Discussions like these are why i fear magnetism/electrodynamics in general physics 2 next semester. Kinda interesting though, i did learn something from all this about the reflections.:)
Oh one final thing...I just realized that the title of this post was "conducting with no circuit". The way we're modeling this line, it is a circuit. The real transmission line itself is basically just a complex load (complex in the imaginary+real numbers = complex sense, not in the simple/complex sense) because it basically becomes an RCL or resistance/capacitance/inductance network.
Current was the word I was looking for - I do that sometimes.Quote:
Originally posted by Ramuman
Thinking of energy flowing in isn't the best way to describe an ideal lossless line.