Current through two wires, does it split?

[geratheg]

I'm not sure if this scenario is an important one or not, but I'm curious, in an ideal circuit where a wire has no resistance (I know in practice wires have a little bit of resistance), if we had two wires in parallel would the current split?

For example, in the circuit shown below the total current would be 3A. Would each wire 1 and wire 2 each experience half of this current (1.5A each)? Or would all of the current go through the shortest path, which is wire 1 with no current through wire 2?
What if wire 1 and wire 2 were the same sized paths?

[suicidaleggroll]

The current will split based on the the relative resistance in the two paths.  If the resistances are equal, the current will be split evenly.  With "ideal" wires there is no resistance, but this is one of those cases where "ideal" wires should not be used as the equations break down.

[Ian.M]

True zero resistance implies superconductivity, so it gets *REALLY* messy and non-intuitive.

As far as I understand it, you have to consider the magnetic field the loop formed by the two wires is in at the moment they were connected.  The flux through the loop is pinned in place and cannot be changed even if the external source of the magnetic field removed.  A current will be induced in the loop by field removal sufficient to maintain the same flux.   As such, when you add a battery and resistor to impose an external current across two points on the perimeter of the loop it must divide so as to maintain the same flux through the loop.  This will obviously be highly dependent on the exact geometry of the loop.  If you connect one wire before the other one, then there will be no current in the second wire (unless the external field changes or the total current changes)  as the second wire completes the loop pinning the flux between it and the first wire generated by the current in the first wire in place.

See: http://web.pdx.edu/~pmoeck/lectures/312/supercon.pdf for the basic concepts.

 

[geratheg]

True zero resistance implies superconductivity, so it gets *REALLY* messy and non-intuitive.

As far as I understand it, you have to consider the magnetic field the loop formed by the two wires is in at the moment they were connected.  The flux through the loop is pinned in place and cannot be changed even if the external source of the magnetic field removed.  A current will be induced in the loop by field removal sufficient to maintain the same flux.   As such, when you add a battery and resistor to impose an external current across two points on the perimeter of the loop it must divide so as to maintain the same flux through the loop.  This will obviously be highly dependent on the exact geometry of the loop.  If you connect one wire before the other one, then there will be no current in the second wire (unless the external field changes or the total current changes)  as the second wire completes the loop pinning the flux between it and the first wire generated by the current in the first wire in place.

See: http://web.pdx.edu/~pmoeck/lectures/312/supercon.pdf for the basic concepts.

This went beyond what I asked lol, but for considering magnetic fields and flux . Other than that, I guess ideal wiring isn't so important and doesn't affect the important calculations.

[Rudane]

You are correct, in theory, when you say that each wire, being equal, will have 1.5 Amps of current. In the lab, because of manufacturing tolerances, you might measure 1.45 A and 1.55 A. If the wires are made to be unequal, each will carry an amount of current proportional to its conductance.

[Ian.M]

If you take two long insulated copper wires from the same reel, of exactly the same length, lightly twist them together so they remain at the same temperature, and terminate them at each end by crimping both into a single blind hole in copper blocks so that they are inserted to the same depth, you can easily get equal current to three significant figures.  If they are long enough you can get them to match to five figures.

To get a difference of over 6%, ("1.45 A and 1.55 A") either the wires must be short, which makes the terminations and length matching much more critical, or they must be at different temperatures or be made from really crappy wire of varying thickness or be made from different batches. e.g. the conductivity is highly dependent on how pure the copper is.

[AG6QR]

You are correct, in theory, when you say that each wire, being equal, will have 1.5 Amps of current. In the lab, because of manufacturing tolerances, you might measure 1.45 A and 1.55 A. If the wires are made to be unequal, each will carry an amount of current proportional to its conductance.

Yes, but be careful when translating the theory to the lab measurements.  This is a case where the measuring instrument could easily disturb the value of the measurement.  We usually are safe in assuming that a current meter has negligible resistance, because it usually IS very low compared to the other significant resistances in the circuit.  But it won't necessarily be negligible compared to the two wires in parallel.

If you have identical wires, and you measure the current through each with identical ammeters, the results ought to be close to identical.  But with those same wires, if you use two ammeters that have differing burden voltage (shunt resistance), you could easily see significantly different measurements, where the differences are caused solely by the differences in the meters.  If you don't have two identical meters, but you decide to do the measurements with one meter, swapping it between the two different wires, it's likely that the shunt resistance of the meter will cause lower current to be carried through whichever wire is being measured, while the unmeasured wire carries more current.


A completely separate interesting point which I haven't seen noted yet:  If you have two wires that are as close to identical as you can make them, but things weren't quite perfect and one is slightly lower resistance than the other, the lower resistance wire will draw more current, thus get hotter than its neighbor.  Most metal wires will have a positive temperature coefficient of resistance, meaning that the hotter wire increases its resistance.  This will tend to bring the two wires a little closer to being in balance than they were initially.  Contrast that with LEDs, where the hotter an LED gets, the more current it draws, causing it to get still hotter, in a vicious cycle.  With a bunch of LEDs in parallel, any minor imbalance is amplified by the effect of temperature, while with wires, any minor imbalance is lessened.

[tszaboo]

To all the above answerers: Please dont confuse people, the answer is two words.
To OP: Ohm's law.

That is all.

[German_EE]

Don't you mean Kirchoff's Law? https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

[codeboy2k]

See: http://web.pdx.edu/~pmoeck/lectures/312/supercon.pdf for the basic concepts.

Printed for the loo....

[tszaboo]

Don't you mean Kirchoff's Law? https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

Not alone. Kirchhoff's law alone doesn't answer the question. In fact if you only apply Kirchoffs law, there could be millions of amps spinning around in that circle, as long as it is only spinning around there.
Put a small resistor instead the wire, proportional to the length. Solve it with Ohms, and Kirchoffs law (yes you need both). This is like middle school electronics.

[Ian.M]

Ideal wire = ideal superconductor.  Millions of amps spinning round the loop is a valid solution that corresponds to a large change in the field the loop is in after it was closed/became superconducting.  It even corresponds to certain real life situations: http://en.wikipedia.org/wiki/Superconducting_magnet#Persistent_mode

[dannyf]

-if we had two wires in parallel would the current split-

For ideal conductor the current through each wire is undefined. If you assume that the two wires are identical, the current through them must be the same.

[xrunner]

To all the above answerers: Please dont confuse people, the answer is two words.
To OP: Ohm's law.

That is all.

Right. You have to place two resistors in the wires to even solve it, otherwise the current is v / 0.00000000000000000000... ohms which is undefined. Just place two perfect resistors in the wires to solve it.

Two wires like that generalizes to one wire. What if you looked at one wire under an electron microscope? How many paths are there in ONE wire? It's just one wire for the purposes of calculating paths.

[ConKbot]

To all the above answerers: Please dont confuse people, the answer is two words.
To OP: Ohm's law.

That is all.

Now now, how will people show off how smart they are, while obviously avoiding context clues about the prerequisite knowledge of the target audience.

[geratheg]

Right. You have to place two resistors in the wires to even solve it, otherwise the current is v / 0.00000000000000000000... ohms which is undefined. Just place two perfect resistors in the wires to solve it.

Two wires like that generalizes to one wire. What if you looked at one wire under an electron microscope? How many paths are there in ONE wire? It's just one wire for the purposes of calculating paths.

That equation makes it make sense. Undefined current.

Are you suggesting there are many paths in one wire?

[cosmicray]

If you have identical wires, and you measure the current through each with identical ammeters, the results ought to be close to identical.  But with those same wires, if you use two ammeters that have differing burden voltage (shunt resistance), you could easily see significantly different measurements, where the differences are caused solely by the differences in the meters.  If you don't have two identical meters, but you decide to do the measurements with one meter, swapping it between the two different wires, it's likely that the shunt resistance of the meter will cause lower current to be carried through whichever wire is being measured, while the unmeasured wire carries more current.

Without trying to run this into the ground, does measuring the current via a clamp style device cause any effect similar to putting a shunt-style meter in series (i.e. a minuscule drop caused by the meter or shunt) ?

[xrunner]

That equation makes it make sense. Undefined current.

Are you suggesting there are many paths in one wire?

Well, look at the schematic attached (forgive the poor drawing). Suppose SW1 is open, and the wires are perfect conductors. The current goes around the outside path because the batteries are in series.

Now, close SW1. What do you think happens in the center wire (assume Zero resistance of the wire and switch). Each battery is shorted out (except for the resistance), and each would try to push current through the center wire in two different directions. Since the wire is without any resistance, do you think it would allow current in two different directions?

[Ian.M]

Does measuring the current via a clamp style device cause any effect similar to putting a shunt-style meter in series (i.e. a minuscule drop caused by the meter or shunt) ?

Not at DC with non-ideal wires.  Any momentary effect from closing the core round the wire will damp out pretty quickly.  AC is another matter, though at typical power frequencies the effect is normally insignificant and less than the other sources of error in the measurement.

If you actually want to do the parallel wires experiment in practice, as affordable clamp on ammeters don't have enough accuracy or resolution, you need a matched pair of identical calibrated four terminal current shunts so that both paths continue to have exactly the same resistance even when you are switching the measuring instrument between them.  Its easiest to arrange by replacing the common termination at one end of the pair of wires with the two shunts bolted end to end, connecting the two wires to the outer ends and using the junction of the shunts as the common.   However most of us are content to check the resistances with a good milliohmmeter and trust that Ohm's law will continue to apply (neglecting thermal effects).