Why aren't resistance loads shorts?

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So instead of using reality, you're proffering theory. I mentioned this already. I work in the real world, not a theoretical one.
 
480sparky said:
So instead of using reality, you're proffering theory. I mentioned this already. I work in the real world, not a theoretical one.
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That just means that instead of truly infinite current in an "idealized" short circuit, you get the available fault current feeding a real world fault instead. Available fault current, usually measured in kiloamps, which is a function of the voltage and impedance of the source and distribution system. The heat generation ends up being inside the source and wires to an extent they aren't meant to be heated, rather than as usable power delivered to the load where we intend to use the power.
 
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Current through a circuit depends on the applied voltage and the resistance. Ideally, all of the applied voltage is seen at the load terminals, and none lost across other parts of the circuit.

The designed resistance of the load limits the current to the designed level for a given load, and the supplying circuit should be designed to approach the goal I just described above.

When the actual current exceeds the circuit's designed capacity, due to too much load or a load defect, the circuit's over-current device should trip to prevent damage to the circuit.

A short circuit occurs when an extremely high current occurs due to a unintentional, direct electrical connection occurs either between circuit conductors or a conductor and ground.

You could say that the difference between an overcurrent and a short circuit is the current level, where in the circuit it happens, or the conductive parts involved, but there is a difference.
 
infinity said:
I had been taught that as well, simple yes but that definition would also include a ground fault which IMO is not the same as a short circuit. I think of a short circuit as a shortcut between normal current carrying conductors.

So I guess we call both a ground fault and a short circuit a short?
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With respect, I don't agree with that. You can have a fault between live and neutral. That isn't a the same as ground fault.
 
Besoeker3 said:
With respect, I don't agree with that. You can have a fault between live and neutral. That isn't a the same as ground fault.
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A fault between any two live conductors, or the live conductor and the neutral, would be what the NEC would term a short circuit. A fault between any two conductors that are built for carrying current under normal non-fault circumstances, even if no loads are set up to put current on the neutral.

A ground fault works just like a short circuit, the way the term is defined in Physics, but with the added hazard of normally non-energized metal, becoming electrically energized. There also may be more resistance in a ground-fault current path, than in the return-path between current-carrying conductors, so the ground fault current could be a lot less than the short circuit fault current on the same circuit. The NEC doesn't include a ground fault in its vocabulary of a short circuit, because a ground fault is a separate safety issue with its own requirements for mitigating.
 
Carultch said:
A fault between any two live conductors, or the live conductor and the neutral, would be what the NEC would term a short circuit. A fault between any two conductors that are built for carrying current under normal non-fault circumstances, even if no loads are set up to put current on the neutral.

A ground fault works just like a short circuit, the way the term is defined in Physics, but with the added hazard of normally non-energized metal, becoming electrically energized. There also may be more resistance in a ground-fault current path, than in the return-path between current-carrying conductors, so the ground fault current could be a lot less than the short circuit fault current on the same circuit. The NEC doesn't include a ground fault in its vocabulary of a short circuit, because a ground fault is a separate safety issue with its own requirements for mitigating.
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Thank you, sir. Pretty as I see it also.
 
retirede said:
An intended load with “very low resistance” is not a short.
A short is an unintended path causing current to bypass the intended load.
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But shorting jumpers can be added when you want to work on things safely (e.g. CTs). I'd say a fault is an untended pathway. It could have high or low resistance.
 
suemarkp said:
But shorting jumpers can be added when you want to work on things safely (e.g. CTs). I'd say a fault is an untended pathway. It could have high or low resistance.
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If it has a high resistance, it is not a "short". Similarly, if it has a high impedance in other forms (e.g. inductance), it is also not a short.

CT shorting is the rare exception where a short circuit is intentional, and can be a little counterintuitive for a person learning about it for the first time. The reason is that voltage would be what builds up to unsafe levels, if you neglect to short the CT output when the meter is disconnected. The current in CT output circuit will be no greater than the CT ratio can generate from the primary current. So the CT current will remain at safe levels when the output is shorted. You need to shut off the load when you go to open the CT output circuit, and connect it to the meter, but while you have the meter disconnected, shorting the CT's is what you do by design to keep the secondary de-energized as much as possible.
 
suemarkp said:
But shorting jumpers can be added when you want to work on things safely (e.g. CTs). I'd say a fault is an untended pathway. It could have high or low resistance.
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The CT is the source. The load is the attached measuring device. So by shorting a CT, you are bypassing the intended load.
I think the definition still fits this case.
 
An opened CT secondary under load allows the CT to behave like a PT, with the inverse ratio producing a high voltage.
 
JoeNorm said:
On a similar thought, why doesn't a motor short through its brushes(I admittedly have a narrow understanding of motors)?



thanks
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In a typical DC motor, the armature is a bar and a short between the brushes. That high current flow gives DC motors typical high torque at starting, limited by the resistance through the brushes and the supply system inpedance.

As the motor starts to turn, the armature bar cuts through the magnetic flux of the field coils and this induces a back emf, or an effective voltage drop, across the armature bar. The bar sees effectively less voltage across it and resulting less current flow as it rotates and increases rotation. At locked rotor or zero rotation, the DC motor does have its armature shorted to the supply through the brushes.

It is fair to say that at locked rotor the DC motor has fault current flowing through the armature bar. Locked rotor current flow is not the intended load.

AC three phase induction motors, only the field coils are connected to the supply, which would not be considered a short. The field coil winding is the normal load. Length of the field coil winding and the smaller coil winding wire, limit fault current flow at starting to a number smaller than what the supply has available. Current flow in the rotor bars is induced, with no direct connection to the supply voltage. Machine design in this instance limits current flow in the rotor bars to below fault current levels.
 
480sparky said:
We can talk about 'there is no such thing as infinite resistance' and 'there is no such thing as a perfect conductor'. Yes, technically all insulation is in reality a semi-conductor. But that's theory, not in actual practice.
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Well if you had a perfect conductor with absolutely zero resistance, and jumped it across a voltage source current is still limited by the impedance of the source.

It doesn't take resistance to allow current to flow it takes continuity which is at the other end of the spectrum.
 
Carultch said:
A fault between any two live conductors, or the live conductor and the neutral, would be what the NEC would term a short circuit. A fault between any two conductors that are built for carrying current under normal non-fault circumstances, even if no loads are set up to put current on the neutral.

A ground fault works just like a short circuit, the way the term is defined in Physics, but with the added hazard of normally non-energized metal, becoming electrically energized. There also may be more resistance in a ground-fault current path, than in the return-path between current-carrying conductors, so the ground fault current could be a lot less than the short circuit fault current on the same circuit. The NEC doesn't include a ground fault in its vocabulary of a short circuit, because a ground fault is a separate safety issue with its own requirements for mitigating.
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Then throw in an ungrounded system, a ground fault there generally doesn't cause any significant abnormal current, it just turns makes an unintended ground reference. A second ground fault is where you get unintended current on this system.
 
480sparky said:
So instead of using reality, you're proffering theory. I mentioned this already. I work in the real world, not a theoretical one.
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In your real world, if a circuit has NO resistance, does current flow?:unsure:

480sparky said:
If a load doesn't have some resistance, current cannot flow.
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Hmm, in my REAL world No resistance is zero (0) Ohms, and can flow lots of current.o_O

Hence your statement is wrong, as if there is no resistance, lots of current will flow. That's why posters are pointing out the flaw.

Infinite resistance (Insulator) is quite a LOT of resistance! Not Zero. And no current flow.

Also if there is NO resistance, you get a divide by zero error.... in I = E/R so use .00000000001 ohms instead.;)

On the other topic, when people say "it has a short" when describing a lamp that flickers as the cord/switch is moved, I correct them and say "Nope, it has a long" and they don't understand....:D
 
Frank DuVal said:
In your real world, if a circuit has NO resistance, does current flow?:unsure:



Hmm, in my REAL world No resistance is zero (0) Ohms, and can flow lots of current.o_O

Hence your statement is wrong, as if there is no resistance, lots of current will flow. That's why posters are pointing out the flaw.

Infinite resistance (Insulator) is quite a LOT of resistance! Not Zero. And no current flow.

Also if there is NO resistance, you get a divide by zero error.... in I = E/R so use .00000000001 ohms instead.;)

On the other topic, when people say "it has a short" when describing a lamp that flickers as the cord/switch is moved, I correct them and say "Nope, it has a long" and they don't understand....:D
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Though it may be impossible to achieve resistance of zero, isn't that the concept behind "superconductors" is to achieve as little resistance possible to increase current carrying ability?

If you have zero resistance or even super low resistance, and no load in the circuit to add resistance to the circuit then the only thing limiting current in that circuit is the impedance of the source.
 
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