I'm working at a subway (metro) company and knowing the risk of electrocution from the traction supply of trains posed some questions regarding the current flow.
So our trains run at around 800VDC. Their supply comes from an extra rail which is known as the "+" (or 3rd rail), and the other (normal) rails are known as the "-" (or return). We're instructed not to get close to the 3rd rail (+) but we can even touch the normal rails (-) if necessary.
If the actual current flow is from the negative to the positive, why is the positive side more dangerous?
We're also instructed not to separate any return conductor from the return circuit while the supply is live, since the current could find a way through the body and get electrocuted. But if we take the conventional direction of current, from the positive to the negative, and the train is the only consumer, so all the voltage should drop on it, there shouldn't be any current left in the return circuit.
Going back tot the humble diode, if the notation shows the flow being from the positive to the negative, is it actually made to only conduct from the negative to the positive? And then our positive rail is actually the negative one that carries the current (electrons)?
Also, is it correct to say that in DC the positive wire should be thicker because it carries all the current, and the negative can be thinner because it's just a return path? And obviously, in AC both wires should have the same thickness since the current flows in both directions.
Is it all just a big lie?
So our trains run at around 800VDC. Their supply comes from an extra rail which is known as the "+" (or 3rd rail), and the other (normal) rails are known as the "-" (or return). We're instructed not to get close to the 3rd rail (+) but we can even touch the normal rails (-) if necessary.
If the actual current flow is from the negative to the positive, why is the positive side more dangerous?
We're also instructed not to separate any return conductor from the return circuit while the supply is live, since the current could find a way through the body and get electrocuted. But if we take the conventional direction of current, from the positive to the negative, and the train is the only consumer, so all the voltage should drop on it, there shouldn't be any current left in the return circuit.
Going back tot the humble diode, if the notation shows the flow being from the positive to the negative, is it actually made to only conduct from the negative to the positive? And then our positive rail is actually the negative one that carries the current (electrons)?
Also, is it correct to say that in DC the positive wire should be thicker because it carries all the current, and the negative can be thinner because it's just a return path? And obviously, in AC both wires should have the same thickness since the current flows in both directions.
Is it all just a big lie?
The normal rails are safe to touch because they have a potential near ground. They will have some voltage on them but not very much. Hence only a small current will flow from them through you to ground. The positive side is dangerous not because it is positive but because it is not near ground potential. A subway system in another city could have a dangerous live rail which is negative with safe return rails which are positive; it all depends on potential away from ground, not the sign of that potential.
Please assure us that you are not an electrician at the subway company!!
The conventional current flow is from positive to negative. This is completely irrelevant to the question in the second part of your sentence.
You need to do a lot of reading about electricity! You may be confusing voltage and current. All the voltage only drops across the train if the train is the only thing in the circuit and all other wiring has zero resistance. In this case there would be zero voltage on the return, but non-zero current.
No, complete nonsense. Both carry the same current.
Please assure us that you are not an electrician at the subway company!!
The reason why it's dangerous is that it has a high voltage to ground. It doesn't matter here if it's positive or negative. You don't want any voltage across your body, as it screws up your nervous system. Stops hearts and such.
You've seen birds perching on high voltage wires, but they get away with it because they don't form a circuit path to ground, and thus feel nothing. But don't go standing on the 3rd rail...
Because Ben Franklin guessing wrong about the way current flows as compared to what the electrons actually do (he had 50-50 odds and got it backwards) we are forever stuck with current going the opposite direction to the flow of electrons. It's too late to fix this, as it would require changing every electronics book in the world. It's easier to just get used to it.
The reason for the positive wire being thicker is it has more insulation on it. And if there's a high voltage negative wire around, it will also have thick insulation. And also for AC.
The reason for not breaking the return ground if there's a load and the high voltage is on, is that the high voltage will appear at the break. And people may not realize that it's there before they get zapped.
Electricity is dangerous, but mankind has learned to manage it and to be safe. That's what the work rules around it are for.
I am, but that doesn't mean that I'm completely stupid to put myself or others in danger. I know what I'm doing and I respect the work rules, I've also built and repaired circuits from small line level audio to mains circuits. I asked these questions so that I have a correct picture in my head of what's actually happening and fill possible voids. It's good to sometimes go back to the basics.
I think this explains everything very well. I was wrongfully assuming an ideal circuit (like in Kirchoff's second law), where the return circuit is actually the ground and has 0V potential. But in this situation, where the actual return point of the supply is maybe kilometers away, there is a lot of resistance on the way, so there's a current flowing.
But I still don't understand how there can be zero current without any voltage.
Hi,
Current flows from positive to negative. As electrons are negative
they actually flow from negative to positive to give positive current.
3 rail tracks are best looked at as the live rail, the one with all the
insulators, and the two normal tracks with no insulation, known
as earth, they are not remotely dangerous, except in combination
with the live rail. I believe the live rail is universally positive, but
the other rails are not remotely negative, they are at earth, 0V.
The supplies along the track AFAIK are only joined to the live
rail and the earth rail it is next to as an earth return, but it is
possible to use both rails as an earth return with connections.
The return point is not kilometers away, its the spacing of
the power supplies, all locally grounded to local earth.
(Which are AC to DC and thus fully locally floating.)
So no live voltage no local current, whats not to understand ?
rgds, sreten.
Current flows from positive to negative. As electrons are negative
they actually flow from negative to positive to give positive current.
3 rail tracks are best looked at as the live rail, the one with all the
insulators, and the two normal tracks with no insulation, known
as earth, they are not remotely dangerous, except in combination
with the live rail. I believe the live rail is universally positive, but
the other rails are not remotely negative, they are at earth, 0V.
The supplies along the track AFAIK are only joined to the live
rail and the earth rail it is next to as an earth return, but it is
possible to use both rails as an earth return with connections.
The return point is not kilometers away, its the spacing of
the power supplies, all locally grounded to local earth.
(Which are AC to DC and thus fully locally floating.)
So no live voltage no local current, whats not to understand ?
rgds, sreten.
In our case, the tracks are not directly connected to the ground, but through separators that only let DC flow back to the supply, but block AC. The AC component is a small voltage used for detecting the train and it's part of a separate detection circuit that has nothing to do with the supply. The rails are insulated from the ground and thick wires ensure continuity where the rails join. These are the connections I was talking about.
The separators ensure a return path to the supply (called substation, which is also the point where the live rail is supplied from).
The separators ensure a return path to the supply (called substation, which is also the point where the live rail is supplied from).
Hi,
Well you tell me how something that allows DC can block half of an AC waveform.
Your just spouting some details you don't really seem to understand.
Of course butted rails need thick wires for continuity, are
they on both sides or just the rail next to the live rail ?
No way do the common rails have the same insulation as the live rail.
In fact continuity is so poor its simply likely not needed, for anything.
rgds, sreten.
Well you tell me how something that allows DC can block half of an AC waveform.
Your just spouting some details you don't really seem to understand.
Of course butted rails need thick wires for continuity, are
they on both sides or just the rail next to the live rail ?
No way do the common rails have the same insulation as the live rail.
In fact continuity is so poor its simply likely not needed, for anything.
rgds, sreten.
bogdan₂₀₁₁;4766410 said:
So far, you have received a lot of pretty good advice. The most important part though (especially for your health and safety) is … conventional current flows from a higher (positive) potential to a lower (relatively less positive, more negative) one. That the electrons themselves actually flow the other way is for the most part immaterial.
The next idea is, current flows in loops: If 1 amp from each of 2 positive sources flows into a Y junction (with the ground, or negative on the bottom), then 1+1=2 amps must flow out the bottom to the negative supply rail.
Next, is that you have an +800 VDC 'hot' rail, and something called “negative rail” which is safe. The “negative rail” need only be less positive than the positive rail for current to flow through the rail-car's motors, and be returned to the power supplies (“substations”). In the SAFETY sense, the negative rails can be very close to ground potential, and current from the +800 will find a way to the other “negative” rails.
Nominally that “way” is via the subway car contacting rails, both to the +800 and to the 'negative' (ground potential) rail(s). Abnormally, should you touch both the 800 and anything that might be 'ground', you become the conductor, light up like a light bulb, and shortly after are buried with honors to live out eternity pushing up daisies.
Likewise, the 'hot' rail could be +400, and the rolling-stock could be –400. The difference is 800, so the motors wouldn't seen anything different. But consider: how likely would it be to avoid touching the –400 VDC rolling stock? Nearly impossible. And any ground would then be a ground-return 'center' conductor. You'd again light up spectacularly. The funeral would be just as sad.
Continuing on that line, the 'positive' rail could be +0.001 V, and the rolling stock at –800 VDC. The motors see nothing different. Touching a naked rolling-stock rail would likely be fatal.
So, it makes no rational sense to ground the 'hot' rail and light up the cold rolling stock. It only makes sense the way you described.
Finally (yep!) remember [E = I R] … voltage (in a circuit) equals current (I) times the resistance (R) of all the conductors combined from source thru the loop and back. The current is a constant thru the loop (assuming a simple loop), at whatever level it achieves. The voltage drop thru all the resistances adds up to equal the supply voltage.
People, apart from being electric-current sensitive, are also inconveniently HIGH resistors, so when bridging the hot rail to ground, they take all the voltage. Which ruins Mother Nature's delicate nerve signaling, which results in assorted problems, the least of which is shutting down the heart. The rest has a bad outcome.
GoatGuy
The return resistance isn't the cause of the current. If a car is present to complete the circuit, a certain current will flow (serially, the same current) through each rail and the car, and no current will flow when there is no car present.
The supply voltage should be measureable across the rails whether the circuit is completed or not, with the exception that a relatively small portion of this voltage will be dropped on the resistance of the rails when current does flow.
The current flow depends on the resistance of the car plus the rails. The resistance of the rails is relatively insignificant in practice and for this reason the current flow is primarily determined by the car, not the rails.
Last edited:
No. From school, through high school, through university, to qualification courses, you just learn a lot of theory, take some things as they are because books say so, learn some safety rules and do everything "because reasons". Sorry if I'm the kind of guy that actually has questions.
I started asking questions at school, then continued at university; eventually I had to answer my own questions (which is what research is). My aim was to understand theory, as I find I can't learn anything without understanding it. I hate learning rules; when forced to do so I usually annoy people by pointing out errors and inconsistencies in the rules.
I am still astonished that someone can qualify as an electrician and get a job as an electrician on a public transportation system without understanding electricity. Please don't tell us you studied EE at university?
I am still astonished that someone can qualify as an electrician and get a job as an electrician on a public transportation system without understanding electricity. Please don't tell us you studied EE at university?
We don't have the details of its construction. It's just a box, cables come from the rails, cables go back to the power station. I only know that it doesn't let the high frequency signal leak along with the tracktion return. Most likely a filter.
Both rails have the wires, since the return current goes through both of them. They are insulated from the tunnel with rubber pads (under the rail) and everything connected to the rail is also insulated (continuity cables, the rail switches etc).
The live rail is completely separate and has its own insulation.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.