Problem understanding this graphic.

[Dsg319]

I do not understand why the current is dropping among this circuit. Can anyone explain. I’m sure it’s a simple reason I’m just not getting it lol.Thanks

 

[ptonsparky]

Available fault current due to conductor size, distance, impedance, etc.

That's my SEWAG.

 

[Dsg319]

Available fault current due to conductor size, distance, impedance, etc.

That's my SEWAG.

Certainly makes sense being that afc decreases the farther and smaller conductor you go.

 

[ptonsparky]

Certainly makes sense being that afc decreases the farther and smaller conductor you go.

I did that exercise, calculation, to determine if there was enough current to blow a fuse on a long circuit, last year. It didn't end well.

 

[Dsg319]

I did that exercise, calculation, to determine if there was enough current to blow a fuse on a long circuit, last year. It didn't end well.

Ha! At least now you know.

 

[don_resqcapt19]

Get Bussmann's free FC² app and play with it to see how much difference adding wire length will make to the available fault current.

 

[Dsg319]

Get Bussmann's free FC² app and play with it to see how much difference adding wire length will make to the available fault current.

I’ve actually got that app and played around with it lol. I’m guessing I was having a moment. It seems like a pretty handy app can even print labels with it too I believe.

 

[Carultch]

I do not understand why the current is dropping among this circuit. Can anyone explain. I’m sure it’s a simple reason I’m just not getting it lol.Thanks

Current really isn't different throughout the path, in the snapshot of any given fault. If the 561A fault occurs, and all other loads are shut off, the 561A fault current will be the same in every location along the line conductor, assuming no other loads are drawing current (fault or otherwise). It is a consequence of Kirchhoff's current law, since charge is ultimately conserved. However, when the fault occurs upstream, closer to the source, the fault current will be higher. This is due to the fact that the impedance of the conductor along the way, contributes to reducing the fault current.

 

[Dsg319]

Current really isn't different throughout the path, in the snapshot of any given fault. If the 561A fault occurs, and all other loads are shut off, the 561A fault current will be the same in every location along the line conductor, assuming no other loads are drawing current (fault or otherwise). However, when the fault occurs upstream, closer to the source, the fault current will be higher. This is due to the fact that the impedance of the conductor along the way, contributes to reducing the fault current.

In any given example of a fault, the fault current will be the same thorughout the circuit, assuming no other loads are drawing current or are faulted. It is a consequence of Kirchhoff's current law.

A theoretical ideal voltage source has infinite fault current. That is, the kind of voltage source that would always output same voltage, regardless of what load is connected. If used with superconductor wires to short-circuit this source, the charges would accelerate indefinitely with no limiting value of current (neglecting relativistic effects). A real voltage source has an unintended resistance/impedance, that limits the maximum current it can deliver to a fault. An ordinary wire conductor also has its own resistance/impedance, that adds up with the source resistance/impedance, and continues to limit the fault current.

Awesome explanation.Thanks for clearing things up !I understand the concept behind the closer to the source the higher the available fault current, and the father away and smaller the conductors the higher impedance and lower fault current. But for some reason the graphic really threw me off as it looking like the current was dropping along the way, rather than showing the available faults current at different lengths along the way.

Thanks everyone!