Why, when conductors are connected in series, their current strength is equal, because the resistivity of the conductors can be different?
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5 Answers
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Not only conductors. This logic applies to anything: insulators, vacuum, plasma, whatever.
The reason lies in the conservation of charge, which is a principle deeply rooted in physics. The charges carrying the current through a surface will come out the other side of that surface.
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Perhaps a water pipe analogy is in line.
If you have two pipes with different diameters connected in series, will the flow rate (liters/minute, gal/minute) be different in each section if there are no leaks? The flow rate will be the same in each section of the pipes.
Same goes for electrical conductors in series. The current, \$ I = {Q \over t} \$. in each conductor will be the same since the amount of charge transferred over time will be the same in each section of the conductors.
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Because the electrons have nowhere else to go.
If you think about electricity as the physical movement of electrons, then in a DC circuit all of the mobile electrons enter at one end of the two joined conductors, go through the first one, go through the junction of the two wires (wire nut, solder, connector pins, whatever) and go through the second conductor returning to the power source. If one of the two conductors is significantly smaller (higher resistance), that will throttle back the electron flow for the entire circuit.
In a series circuit, no matter how many components, or types of components, or wires, 100% of the current goes through 100% of the components 100% of the time. The component types have a major effect on the electron flow, such as a capacitor blocking DC, but that is included in the rule. Even if the capacitor has a very small leakage current, that tiny current moves through the other components.
Also, in a parallel circuit, 100% of the voltage appears across 100% of the components 100% of the time.
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1 amp is equal to 1 colomb's worth of electrons passing a given point in a second. 1 coulomb is about 6x10¹⁸ electrons.
If the two wires are in series, then the current in the two must be the same. If it wasn't, then either electrons would be piling up at the point where the wires meet, or else electrons would be magically appearing out of nowhere. This is expressed more formally as Kirchhoff's current law.
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\$\begingroup\$ "piling up" does indeed occur, that's what a capacitor describes \$\endgroup\$Ben Voigt– Ben Voigt2024-07-12 19:45:04 +00:00Commented Jul 12, 2024 at 19:45
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\$\begingroup\$ @BenVoigt yes, but it's accompanied by an "emptying out" of charges on the other plate, so to an observer with access only to the capacitor's terminals, it seems as if there's no piling up. \$\endgroup\$Simon Fitch– Simon Fitch2024-07-13 03:43:17 +00:00Commented Jul 13, 2024 at 3:43
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The current through the conductors is going to be determined by the total resistance, which you get by adding together the resistance of each conductor.
Moreover, what we see using Ohm's law, is that the total applied voltage is distributed across each conductor in proportion to its resistance. So the conductor with more resistance will have a higher voltage across it, and that higher voltage will "push harder against" the higher resistance resulting in the same flow of current as the others.
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\$\begingroup\$ Kirchoff's Current Law isn't restricted to ohmic devices. \$\endgroup\$Ben Voigt– Ben Voigt2024-07-12 19:45:56 +00:00Commented Jul 12, 2024 at 19:45
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\$\begingroup\$ I like to think that once we can understand simple electric circuits, then we're ready to move on to more advanced concepts like plumbing :) \$\endgroup\$gbarry– gbarry2024-07-12 20:00:27 +00:00Commented Jul 12, 2024 at 20:00
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\$\begingroup\$ Circuits with only ohmic devices are generally not very interesting. In even my "simple" circuits, I usually like to include a voltage or current source. \$\endgroup\$Ben Voigt– Ben Voigt2024-07-12 20:04:24 +00:00Commented Jul 12, 2024 at 20:04
sea of conduction band chargesin both copper wires (unimaginable numbers) would start to move. But their surfaces would experience just the right charge gradients so as to propel the central, inner charges through both wires to yield a constant current. This would NOT be at a constant linear velocity in both wires, though!!! Get that straight. \$\endgroup\$