# Energy/Potential Loss for current through a 90 degree bend wire/resisitor

• bigreat
In summary: You seem to be arguing that it does, but you haven't actually demonstrated this. Until you have, your argument is invalid. In summary, the analogy between water and electricity does not hold up. A resistor is full of scatterers, meaning the current does not really pass through it in a straight line even without a bend. There is no resistance when a bend is made in a wire or resistor.
bigreat
Hi All,

I know that when water pass through a 90 degree bend pipe, there will be a pressure lose. However, how to calculate the energy/potential loss for electrical current pass through a 90 degree bend wire/resistor.

THanks

You want an academic paper to discuss a very flawed analogy?
Just as soon as the water analogy for electricity starts to fail you need to drop it and move on.
Just study elementary circuit theory and you won't go wrong.

sophiecentaur said:
You want an academic paper to discuss a very flawed analogy?
Just as soon as the water analogy for electricity starts to fail you need to drop it and move on.
Just study elementary circuit theory and you won't go wrong.

Come on dude, at least you can somewhat prove it doesn't happen. If everybody just drops, I don't see anything will actually move on.

The bend won't make any difference whatsoever. A resistor is full of scatterers, meaning the current does not really pass through it in a straight line even without a bend.

The only time a "bend" can make a difference is if you really have ballistic transport, but that never happens in an ordinary conductor and definitely not a resistor.

bigreat said:
Come on dude, at least you can somewhat prove it doesn't happen. If everybody just drops, I don't see anything will actually move on.
Au contraire - it's up to you to prove that a resistive effect exists. I'm not prepared to trawl through the Science to try and find a loophole that would make your idea right - and I don't think anyone else would be either.
Until you have looked thoroughly at EM theory and measurements and have found such an effect then you are not really in a position to insist that your analogy applies here.
I can only ask "Why should it?". Things can only "move on" if people don't waste time without learning the present state of things.

Bends in microwave strip lines have to be compensated (mitered): See "mitered bends" in

http://www.microwaves101.com/encyclopedia/mitered_bends.cfm

This miter is actually an impedance correction (I think by adding extra inductance to correct for extra capacitance).

Bob S

Bob S said:
Bends in microwave strip lines have to be compensated (mitered): See "mitered bends" in

http://www.microwaves101.com/encyclopedia/mitered_bends.cfm

This miter is actually an impedance correction (I think by adding extra inductance to correct for extra capacitance).

Bob S

I think that ,unless there were some actual radiation from the bend, there could be no 'resistance' in addition to the existing dielectric and copper losses.
I don't think the OP was considering anything as esoteric as this, though.

It's a well known fact that high frequencies (UHF and up) are impeded by corners.

Yes, but that is because corners mess up the impedance of the line which is relevant at hight frequencies, it is a geometric effect which has to do with the field distribution in and around the line; it has nothing as such to do with the current having to "bend".

Also it is -as Bob S has already pointed out- perfectly possible to make sharp corners at high frequencies as long as you use a miter; the change in dimensions basically compensates for the extra coupling which in turn means that the impedance does not change.

Averagesupernova said:
It's a well known fact that high frequencies (UHF and up) are impeded by corners.
Mismatch is a different thing from Resistance. It's an entirely separate issue, involving reflected (coherent) power and it is certainly nothing like analogous to water going round corners. Why won't this thing just DIE?

To both of the posters who posted after I did (#8). Notice that nowhere did I say resistance. In fact I said IMPEDE which implies impedance. I may be wrong about this, but with the level of understanding that the OP seems to have (not have) I thought it was worth mentioning.
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I've heard the old mitered corner thing many times. And to me it's like saying: "We can't build this building on this site because it is too low and water will collect there. So to fix that, we will bring in enough fill to raise it 15 feet. Now we can build on this low-laying site." Ummm, duh, it's no longer low. Same thing with the mitered corner, it's no longer a sharp corner.

It is true that more or less 'anything' different on an AC transmission line will produce a mismatch and reflect some of the power flow. The word "impede" is not one which is defined in Electrical Engineering (afaik) so using it in an argument doesn't achieve much. If you place the appropriate matching element in front (or even after) a right angle bend in a microstrip, you can eliminate the mismatch perfectly at a particular frequency. I don't know of an equivalent in plumbing. . . . . . ?
I don't see what the remotest connection this has with the resistance to a continuous flow of water around a bend.
I think we should have a special Forum for people who want to invent / sustain analogies rather than try to get to grips with the real thing. What's the point in this one?

Averagesupernova said:
.
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I've heard the old mitered corner thing many times. And to me it's like saying: "We can't build this building on this site because it is too low and water will collect there. So to fix that, we will bring in enough fill to raise it 15 feet. Now we can build on this low-laying site." Ummm, duh, it's no longer low. Same thing with the mitered corner, it's no longer a sharp corner.

But the difference is that this corner IS still sharp; the miter just compensates for the fact that the EM field now can couple to a different part of the waveguide; if you don't want to use a miter you can instead change the layout of the ground plane, perhaps use stubs etc...
Also, a bend at these frequencies will results an impedance mismatch that in turn results in reflections, not resistive losses which is what the OP was referring to.
The coplanar waveguides I design and use are all superconducting so they are very nearly lossless, but sharp bends etc will of course still mess up the impedance if I do not compensate for them.

sophiecentaur said:
It is true that more or less 'anything' different on an AC transmission line will produce a mismatch and reflect some of the power flow. The word "impede" is not one which is defined in Electrical Engineering (afaik) so using it in an argument doesn't achieve much. If you place the appropriate matching element in front (or even after) a right angle bend in a microstrip, you can eliminate the mismatch perfectly at a particular frequency. I don't know of an equivalent in plumbing. . . . . . ?
I don't see what the remotest connection this has with the resistance to a continuous flow of water around a bend.
I think we should have a special Forum for people who want to invent / sustain analogies rather than try to get to grips with the real thing. What's the point in this one?

I asked a question, and very grateful for the answers. That's the point.

## 1. What is energy/potential loss for current through a 90 degree bend wire/resistor?

The amount of energy or potential loss for current through a 90 degree bend wire/resistor depends on the resistance of the material, the length of the wire/resistor, and the current flowing through it. The higher the resistance and the longer the wire/resistor, the greater the energy/potential loss will be.

## 2. Why does a 90 degree bend in a wire/resistor cause energy/potential loss?

A 90 degree bend in a wire/resistor causes the electrons to change direction abruptly, which results in increased collisions between the electrons and the atoms in the material. These collisions dissipate energy, leading to energy/potential loss.

## 3. How does the material of a wire/resistor affect energy/potential loss through a 90 degree bend?

The material of a wire/resistor affects energy/potential loss through a 90 degree bend because different materials have different resistances. Materials with higher resistance will experience more energy/potential loss compared to materials with lower resistance.

## 4. Is there a way to reduce energy/potential loss through a 90 degree bend in a wire/resistor?

Yes, there are a few ways to reduce energy/potential loss through a 90 degree bend in a wire/resistor. One way is to use materials with lower resistance. Another way is to increase the diameter of the wire/resistor, which reduces the resistance. Additionally, using smoother bends in the wire/resistor can also help reduce energy/potential loss.

## 5. How does current affect energy/potential loss through a 90 degree bend in a wire/resistor?

The higher the current flowing through a wire/resistor, the greater the energy/potential loss will be through a 90 degree bend. This is because a higher current means more electrons are flowing through the material, leading to more collisions and therefore more energy/potential loss.

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