How do resistors handle high currents without burning up?

[ramonegumpert]

Hello Experts

I have a question regarding high amperage going through strands of wires.

For example, when hundreds of amperes surge through a thick cable of wire containing multiple copper wires, why is it that the current did not choose to go through just 1 of these strands of thinner copper wires and thus burning it up ?

I ask because, each wire is not perfectly the same in resistance, am i right?
If so, then the surge current should go the path of least resistance instead of spreading its current across all the wires within the thick cable is it not?

For example, your car battery cables. Typically, it consists of multiple thinner wires bunched together with a lug. When you start the car, more than a hundred amperes can go through the wires in a split second.

Thanks for reading.

Have a nice day.

Regards
Ramone

 

[tiny-tim]

Hello Ramone!

… the surge current should go the path of least resistance …

Not exactly.

If there are two paths between the same two points A and B, the electric potential difference between A and B must be the same along either path.

Since potential difference (= voltage drop) = current x resistance (V = IR),

that means the current will be inversely proportional to the resistance.

eg, if you short a circuit with your body, and your body has 1/999th of the resistance of the circuit, then 999/1000ths of the current will go through you (the path of least resistance), but 1/1000th will still go through the original circuit. ​

 

[Baluncore]

If all the fibres are the same length and material then they will each carry a current proportional to their cross-section. When wire gets hotter it's resistance rises in proportion to it's absolute temperature. Since higher current leads through higher temperature, to higher resistance, the current is reduced. That negative feedback will keep all the wires carrying their appropriate components of the total current.

 

[ramonegumpert]

Dear Tiny-Tim and Baluncore :)

So, its spread out across all the cross-section available and dynamically the resistance will change with temperature and the logic V = I * R will adjust the currents flowing across each of the strands of wires accordingly.

Thanks so much for enlightening !

Very impressed by your clear explanations.

Thanks for reading.

Have a nice day.
Regards
Ramone

 

[skeptic2]

Let's say you have two conductors, a lower resistance conductor A, and a higher resistance conductor B. If all the current were to flow through A, the voltage across B would be zero. In practice the currents across all parallel paths will be such that the voltage across each conductor is equal.

 

[mfb]

So, its spread out across all the cross-section available and dynamically the resistance will change with temperature and the logic V = I * R will adjust the currents flowing across each of the strands of wires accordingly.

While temperature effects can be interesting, they are not the main reason why current flows through all strands.

The path of least resistance is current in all strands (and even a tiny fraction through the insulation).

 

[Baluncore]

Remember that the OP specified a current surge that increases the temperature of one fibre very significantly. We are talking about a current surge, without the time for temperatures to equilibrate across the cable. In a sense, I am considering here not so much what does happen, but why something else does not happen.

when hundreds of amperes surge through a thick cable of wire containing multiple copper wires, why is it that the current did not choose to go through just 1 of these strands of thinner copper wires and thus burning it up ?

Even if not applicable in the real world, my thermal argument is applicable to this unreal situation. The one wire that carried all the current would get hot, raise it's resistance and so the current would find the many lower resistance paths available.

Semiconductor junctions with a negative temperature coefficient will self destruct if some way of balancing the current across the junction area is not employed. Two PN junction diodes in parallel will not carry the same current. One will become warmer and so carry a higher current. That is a positive feedback effect that can lead to thermal runaway and destruction.

For current surges, metal wires in parallel, with a positive temperature coefficient, have a negative thermal feedback that balances the current variation across the section. Consider the analogy of wiring two or more filament light globes in parallel.

 

[mfb]

Remember that the OP specified a current surge that increases the temperature of one fibre very significantly.

Sure - based on the misconception that there would be a single path of lowest resistance where all the current would flow.

We are talking about a current surge, without the time for temperatures to equilibrate across the cable.

Or even without the time to get significant heating?

Semiconductor junctions with a negative temperature coefficient will self destruct if some way of balancing the current across the junction area is not employed.

That depends on the temperature coefficients and other details.

 

[russ_watters]

"The path of least resistance" is a sloppy way to describe where current goes. The current actually goes through all available paths, in [inverse] proportion to their respective resistances.

 

[Baluncore]

russ_watters is correct. The conduction of parallel currents follows ohms law.
Lightning follows the “path of least resistance” because ionisation reduces the resistance of it's path.

Semiconductor junctions with a negative temperature coefficient will self destruct if some way of balancing the current across the junction area is not employed.

That depends on the temperature coefficients and other details.

Those details are the conditional “if some way of balancing the current across the junction area is not employed”.

 

[ramonegumpert]

Dear Experts

Thanks so much for sharing your knowledge.

Science is so interesting!

One of reasons I asked this forum question is because I notice that when 2 or more resistors of the same resistance , from the same packet, are connected in parallel, current will go through one of them more causing it to even burnt up. The effect of resistors in parallel, as far as i understand, is that the total effective resistance dropped.

In this case, I wonder is it current also flow through the rest of resistors that are in parallel or all the current went through just the one that got burnt?

Enjoy!

regards
Ramone

 

[Baluncore]

If you connect several identical resistors in parallel then the effective resistance will be the resistor value divided by the number of resistors.

There are a couple of things that can cause resistors to fail. Firstly operating them at too high a power dissipation, they get too hot. Secondly operating them above their rated voltage, usually above about 100 volt.