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So I have been mulling over Ohm's law for hours today. It makes sense to me that the current gets bigger as voltage rises, since the voltage is supplying more electrons and current is the number of electrons flowing. However, I'm confused by the theory that as resistance drops, current increases, even when voltage stays constant. Such as:

$$V/R = I$$ $$\frac{1V}{0.0001 \Omega} = 10,000,000 mA!!$$

That's a big current from such a small voltage!

My mental picture of a current is more electrons flowing. Electrons are supplied by the voltage, right? If we start off with a small voltage, say 1V, but infinitely decrease the resistance, we'll get a bigger and bigger current. How can this be? Isn't current the number of electrons flowing? With fewer electrons (because of less voltage), how can less resistance infinitely increase the current?

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    \$\begingroup\$ "because of less voltage" WUT. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 3 '13 at 21:21
  • \$\begingroup\$ Google "internal resistance" and you will see why it doesn't actually happen that way. \$\endgroup\$ – AJMansfield Aug 4 '13 at 1:15
  • \$\begingroup\$ "voltage stays constant" [...] "because of less voltage". You have a contradiction in your text, making it pointless to answer. \$\endgroup\$ – Kaz Aug 4 '13 at 15:50
  • \$\begingroup\$ @Kaz, I think that this was the crux of my misunderstanding, that the voltage supplied all of the electrons to the current and that the conductor was only a sort of pipe that they went through. \$\endgroup\$ – NickRamirez Aug 4 '13 at 16:40
  • \$\begingroup\$ Note that while a voltage (potential difference) can cause a current, a current does not cause a voltage. There are billions of ions flowing past the earth every second, from the sun and interstellar space, these are charged carriers, moving at an enormous velocity, and produce a massive (terra-amps) current. However, there is no voltage, they are propelled by gravity (orbiting ion streams) and by momentum gained from the nuclear reactions that produced them. \$\endgroup\$ – crasic Aug 4 '13 at 18:27
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Your basic misconception is that voltage is "supplying the electrons". The electrons (or whatever the charge carriers are in whatever material you are using) are always there. Voltage is the push to get them moving at the macro level. It is this motion that we call current.

Current therefore is a function of two things, how hard you push on the charge carriers and how much the material resists the movement of these carriers. Double the voltage, and you get double the current at the same resistance. Halve the resistance, and you get double the current at the same applied voltage.

In theory, the current would go infinite as the resistance becomes zero. In practise, the voltage source won't be able to support more than some current before it can no longer provide the voltage.

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  • \$\begingroup\$ Copied from my comments to RedgrittyBrick - 2) Your first sentence implies that the voltage source need not supply any charge, that is bound to be very confusing to the OP. In fact this runs counter to an understanding of ohms law. \$\endgroup\$ – placeholder Aug 4 '13 at 15:23
  • \$\begingroup\$ This may be a silly question, but may help me to visualize it: if a conductor had very, very low resistance--nearly infinitely low, would a small push basically disintegrate the conductor as its electrons are pushed away? I am visualizing a sand castle being blown away by the the wind (or in this case, a gentle breeze). \$\endgroup\$ – NickRamirez Aug 4 '13 at 16:21
  • \$\begingroup\$ @rawb: Voltage isn't so much a supply of electrons as a push for the existing electrons. We talk about a voltage source as "supplying current" from a circuit point of view. That can be a useful abstraction, but note that this is actually not correct. All the current that a voltage source "supplies" out one lead is returned to is via the other lead. This is like the water pump in a forced hot water heating system. It doesn't add water, just move around the existing water. \$\endgroup\$ – Olin Lathrop Aug 5 '13 at 13:14
  • \$\begingroup\$ If a voltage source does not supply current the output voltage will drop. What you describe is a charged capacitor instantaneously connected to a resistor, not a power supply. In your pump example it is as if the pump has generated a head and then stopped, as the water flows the head will drop and the flows stops. What is being conflated is the total number of charge carriers in the system (the working volume of water) vs. the supply and (re)collection at individual components points in the circuit. cont'd \$\endgroup\$ – placeholder Aug 5 '13 at 17:13
  • \$\begingroup\$ When you turn on your faucet people understand that the tap doesn't magically materialize water into your cup, it's understood that the water flows through your system, the plumbing, evaporates, rains, etc. etc. is collected and eventually flows in a loop. The OP didn't write "creates" he wrote "supplies". Don't be misleading to prove nerd points. \$\endgroup\$ – placeholder Aug 5 '13 at 17:14
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Electrons are supplied by the voltage right?

No, electrons are "supplied" by the conductor (in the case of typical metal conductors). In fact they are naturally present. Note that in other types of conductor the charge carriers may be something other than electrons.

The voltage just "supplies" a force that pushes those charge carriers slowly along.

The source of the voltage, for example a battery, does supply charge carriers, but it is the battery supplying the carriers, it is not the voltage supplying the carriers.

The drift velocity in metals (and hence current in coulombs of charge per second) depends on electron mobility and electric field (which is measured in volts per metre)

Conductivity is proportional to the product of mobility and carrier concentration. Conductivity is a property of a material, it's inverse is resistivity which is measured in ohm metres. For a given length and cross-sectional area of a material you can calculate resistance in Ohms from resistivity of the material.

So a lower resistance (higher conductance) implies higher mobility (carriers can move faster) or higher carrier concentration (more carriers) - either lead to more current.

Isn't current the number of electrons flowing?

Not exactly. It's the number of electrons (or other charge carriers) flowing past a specific point per second. It's a rate not a number.

Maybe an analogy will help: Think of cars travelling along a highway as you stand on a bridge watching. The number of cars passing per minute does depend on the number of cars per mile of highway (the carrier concentration) but it also depends on the speed of the cars (related to carrier mobility).

Say you have a uniform density of cars along a 120 mile circular highway like the M25 around London. You are standing on a bridge counting cars passing each minute. If all those cars are travelling at 30 MPH (because the speed-limit signs have been set to 30) maybe you measure 10 cars passing a minute. If the cars are later all travelling at 60 MPH, you would measure 20 cars per minute without any change in the number of cars on the highway and no change in their density (concentration).

The speed of the cars depends on the propelling force (like voltage) but also depends on rolling resistance, air resistance, the need to negotiate junctions and around obstacles etc. Of course charge-carriers don't have built in engines for propulsion, maybe they are more like wind-driven or solar-powered batteryless vehicles.

Like all analogies, you don't get far before it becomes misleading. At the atomic level, the charge carriers are all in perpetual random motion but don't move far on average unless there is an additional force provided by an electric field (measured in volts per metre) that causes them to, on average, drift in a specific direction. Their movement is impeded by the characteristics of the material, this is a resistance to their movement.

Note that charge carriers in motion are carrying energy from one place to another, this is somewhat more important than the carrying of charge because the net movement† of charge is zero (in the car analogy, there are always as many cars in sight in each direction from the bridge no matter how many cars are in motion or how fast they move)

†What I mean here is that after a period of time, there are still the same number of charge carriers in a portion of conductor as there were at the start. There has been movement of carriers around the circuit - so there is charge movement - but the net effect (on number of carriers, and charge, in the portion of conductor) is as if there had been none. On the other hand, energy is dissipated in the conductor due to this motion, so you can tell from the elevated temperature of the portion of conductor that there has been motion of charge carriers.

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  • \$\begingroup\$ "Note that in other types of conductor the charge carriers may be something other than electrons." - Are you talking about this? Or...? \$\endgroup\$ – BlueRaja - Danny Pflughoeft Aug 4 '13 at 1:20
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    \$\begingroup\$ In a salt solution for example it is Na+ and Cl- ions, not free electrons. \$\endgroup\$ – ithisa Aug 4 '13 at 1:36
  • \$\begingroup\$ Several flaws with this that simply will confuse a beginner. 1) *-ivity is different than *-ance, resitance and resitivity are similar but different, same with conductivity etc. look it up, fix it. 2) Your first sentence implies that the voltage source need not supply any charge, that is bound to be very confusing to the OP. I fact this runs counter to an understanding of ohms law. 3) Your last para, is totally wrong. If you have no movement of charge you have no current - full stop. You are confusing charge balance with charge flow. But that is consistent, your doesn't supplying charge.. \$\endgroup\$ – placeholder Aug 4 '13 at 15:18
  • \$\begingroup\$ The car analogy helps I think. Now I'm seeing that electrons are present in the conductor, so a small push may get them going only if there is little resistance. Going with the car analogy, if some materials are better conductors because they have more "cars" packed into them, does this mean that there are conductors that have a high concentration of electrons, and maybe they're not very mobile, but given a big enough push you'll get a lot of charge flowing? You'd just need more voltage when using that conductor? \$\endgroup\$ – NickRamirez Aug 4 '13 at 16:16
  • \$\begingroup\$ @rawbrawb: Thanks for the corrections, I've edited the answer and attempted to address those points. I may be making things less clear instead of clearer - so please feel free to dive in and edit it further (convert to wiki if you wish). \$\endgroup\$ – RedGrittyBrick Aug 4 '13 at 19:53
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That's a BIG current from such a small voltage!

While I understand why that might astonish you, it is, in fact, unastonishing.

Consider: a current can exist in the absence of any voltage whatsoever.

A voltage is not required for there to be a current. A current is simply electric charge in motion.

Yes, according to Ohm's Law, the current through a resistor is proportional to the voltage across.

But, for example, for an (ideal) inductor, the current can be any non-zero constant value for zero voltage!

Thus, what you must do is to refine your understanding of the relationship between voltage and current. There's much more to it than just Ohm's Law.

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  • \$\begingroup\$ "But, for example, for an (ideal) inductor, the current can be any non-zero constant value for zero voltage!" - I don't think that's true - an ideal inductor is still an ohmic device. Any magnetic flux through an inductor would cause a potential-difference (voltage) between the two ends of the inductor. The current is still described by Ohm's law in that case, though the resistance is a result of the impedance, rather than any physical property of the inductor. \$\endgroup\$ – BlueRaja - Danny Pflughoeft Aug 4 '13 at 1:11
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    \$\begingroup\$ @BlueRaja-DannyPflughoeft: Alfred Centauri is correct. The instantaneous voltage across the inductor in a typical switching voltage regulator is exactly zero thousands of times a second. The instantaneous voltage across the inductor of a typical mains power filter is exactly zero dozens of times a second. However, just as Alfred Centauri says, at those instants, the current through that inductor is not zero. (In both situations, that current is very small when the load has recently pulled very little power, and that current is much higher when the load has recently pulled a lot of power). \$\endgroup\$ – davidcary Aug 4 '13 at 2:51
  • \$\begingroup\$ And you can have current flow in a vacuum (in the absence of even a conductor) ... how does introducing another device/concept (inductor) help him understand Ohms law when he is focusing on a resistor and simpler concepts? \$\endgroup\$ – placeholder Aug 4 '13 at 15:53
  • \$\begingroup\$ I haven't studied inductors yet, but the idea that you could have a current without voltage pushing it forward is intriguing. \$\endgroup\$ – NickRamirez Aug 4 '13 at 16:23
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As Olin pointed out, I think you have a few misconceptions regarding what those magnitudes mean. When you say "the voltage supplies electrons" I'm guessing you're imagining a typical DC voltage supply connected to a resistor, and it's reasonable to think that. It's clearly supplying the energy, because a resistor just sitting there won't do anything, then it must be supplying electrons, sort of kicking them out into the circuit and thus making everything circulate. But this is not the best way to see it.

You can imagine a DIFFERENCE in voltage between two points, as a difference in potential energy (it's potential energy per unit charge, but the analogy works). To compare with gravity, a difference in voltage is a difference in altitude, a slope so to speak. It is the case, both with gravity and voltage that, masses or electrons will go from the point with higher potential to the point with lower potential, because there's a force pushing them in that direction (gravity or the electric field). And that's the main idea, the electric field is aligned with the direction in which the potential changes and pushes them in that direction.

Now if you connect a supply to the resistor, it will create a difference in voltage between both its terminals, sort of a "drop". This difference implies there's an electric field, and it will push and pull EVERY electron in the wire and the resistor. So it's more accurate to think that the voltage supply provides a uniform force across the resistor, rather than kicking electrons to the circuit.

On the other hand, current in a wire or resistor is defined as the quantity of charge (number of electrons if you will) that go through the cross section of the wire every second. We model a resistor as a piece of wire that sort of pushes back a little bit on electrons, or like a pipe filled with floating debris.

Electrons move through erratically, crashing into the debris and stopping, and then gaining speed with the stream of water again. The push of the stream is the electric field here, you can picture a vertical pipe so gravity will be the electric field if you want. The weak point of the analogy is that debris should be fixed in place, since they represent atoms, and the electric field is always too weak to rip a whole atom out of the molecular structure.

The more a filled with debris a wire is, the more resistive it is. If you have little debris, the electrons can catch a higher speed before they crash into something, hence they average a higher speed than in a debris-filled pipe. And that's why you have a higher current if R is low... the force might be low, but if there's nothing to fight it, the electrons will end up moving really fast. The analogy breaks down when you put too much current because probably you can't model a wire as a partially obstructed pipe... the debris might start to do weird stuff, but I'm not sure. The friction will also melt the wire and maybe you can no longer say that V = RI along a pool of boiling copper.

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  • \$\begingroup\$ I'm thinking that boiling copper isn't too bad a conductor, so why not? :-) \$\endgroup\$ – Anindo Ghosh Aug 4 '13 at 3:54
  • \$\begingroup\$ Okay, so this make that whole "drift velocity" thing make more sense. Electrons move slowly as individuals (bounce around in a confined area?), but together large numbers of them are being pushed past a point. I was having trouble before understanding how charge could have been moving fast, if electrons just drift along at a slow pace. A bit like herding cats along a hallway? \$\endgroup\$ – NickRamirez Aug 4 '13 at 16:34
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Here are the two forms of ohms law that your are talking about:

1) \$ V=IR\$

2) \$ I = \dfrac{v}{r} \$

Eqn #1 was says, if I push "I" current through a resistor it will generate "V" volts across it. My power supply has to be able to supply that current regardless of the voltage needed.

Eqn #2 says that if I force "V" volts across a resistor, my power supply has to be able to supply "I" current.

There is obviously limits to what real world power supplies can do, as you note in your 10,000 A example.

The conductor/resistor has what is known as "a sea of electrons" in it. These power supplies push the charge into one end and charge comes out the other end. The charge is balanced in the resistor, but current flows, every carrier pushed on by your power supply is matched by another electron flowing out the other end. This is now work is accomplished.

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Suppose one has a vertical water pipe which is one hundred meters tall and has a 100cm2 cross-sectional area, and it has a small hole in the bottom. The size of the hole is such that when the water is one meter high (a volume of 10 liters) , one liter per second will flow out through the hole. When there is more water in the pipe, it will flow out faster; when there is less water, slower.

If water is added to the tank at a rate of 1 liter per second and there are less than 10 liters of water in the pipe, the rate at which water is leaving the pipe will be less than the rate at which it is entering, and thus the amount of water in the pipe will increase. If there are more than 10 liters of water, the amount of water entering will be insufficient to keep up with the water that's leaving, and so the amount of water will decrease. With the indicated size hole, an equilibrium will be reached with ten liters of water in the pipe.

If one were to reduce the size of the hole, the depth required to put enough pressure on the hole to get a liter per second through it would increase. As the water level rose, the amount of power required to keep adding a liter per second would increase, but if hole were small enough, the only limit on the equilibrium depth would be the point where the tank overflowed or otherwise failed, or if pump was no longer able to pump a liter per second against the increasing back pressure.

Conversely, if one were to vastly increase the size of the hole, the rate at which water would have to be supplied to maintain a one meter depth would vastly increase. If the hole were large enough, the rate at which water would need to be added to maintain that depth could increase beyond the ability of any water source to supply.

Note that from a practical perspective, the effect of resistance that approaches infinity (the tiny hole) is not to say that voltage will also reach infinity, but rather that the supply won't be able to produce enough voltage to produce a desired amount of current. Likewise, as resistance approaches zero, current doesn't approach infinity, but instead the supply will be unable to feed enough current to produce a desired amount of voltage.

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Interesting question! Here's the response that I hope you will like:

So, a resistor is a sea of positive and negative charges, right? So, imagine applying an electric field across this sea. Electrons flow. Now, lets look at the length of the resistor. If we add to the length, we will have a larger sea, but the rate will be the same, so it won't help increase current at all. Also, resistance increases with length, so the rate will actually decrease!

So, let's add to the sea by increasing the cross-sectional area of the wire. We will have a bigger sea, and the electric field that before was able to move along a current can now move along a greater current, and for seemingly no extra cost! In physics, this is described as decreasing the resistance of the sea, although in reality the rate of collisions experienced by individual electrons is unchanged. The fact that there are more electrons able to respond is what increases the rate/current.

So now we can see that if we decrease the resistance to an infinitely small amount we can view this simultaneously as creating a huuuge cross section, a vast sea of responsive electrons ready to be influenced by the field. This makes a huge current by no extra-large voltage source. Hope you get to read this. Have a good day.

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  • \$\begingroup\$ That analogy didn't help simplify the explanation in the slightest. \$\endgroup\$ – Funkyguy Sep 30 '15 at 18:28
  • \$\begingroup\$ Haha in retrospect this still isn't bad, voltage=E*l, and if the cross sectional area increases the E field suffers none, and so by symmetry whatever cross section is added shall have the same proportional current as the original wire, thus adding current at no cost. \$\endgroup\$ – Andres Salas Nov 5 '15 at 11:02

protected by Olin Lathrop Jul 14 '14 at 21:22

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