Problem/Question I am trying to understand why the current in resistors that are wired in series is the same. This makes no sense to me. Attempts at answering I will state everything I hold to be true, my entire reasoning process. So if at any point I am mistaken please correct me. By definition of resistance: First of all...resistance is supposed to reduce the current. So if the current is reduced after it passes through one resistor, let's say from 10A to 7A...how in the world will it be the same current in that subsequent resistor? Nevermind the next resistor in series, and even the next one after that. Taking energy into account and thinking of it in terms of volts: There is a voltage drop after every resistor. If there is less voltage between two points, then since the potential difference is not as strong the current will not be as large...or the charges will not be moving as fast. So by reasoning that the voltage drops every single time, and that the charges will continue to move slower, meaning that the current will also decrease, there is no way that the current could remain the same from one resistor to the other. Thinking of it in terms of collisions: Current is defined by the movement of charge carriers, let's say electrons here. As they are whizzing along due to the electric field that causes a potential difference, they start to bump into other atoms inside the resistor and slow down. If their speed has decreased...how can the current remain the same? Thanks in advance guys. I tried to make the question follow the 'template' given as best as possible. It's not really a homework question...just a question in my own self-study.
My friend, you have stumbled upon one of the most confusing hurdles in circuitry. Once you understand this, circuits are no problem. You are mixing up the current that is drawn and the current that actually passes through the resistors. The bigger the resistance, the less current passes through that resistor (assuming constant voltage), but once the lessened current starts passing through the first resistor it will pass through the rest of the resistors. Nothing the resistors can do will lower the current that passes through, but the resistors lower the amount of current drawn. You are saying that resistors reduce current. Current is charge passing the wire per unit of time. Therefore, you are saying that charge is being destroyed in the resistors. But how can that be? From this website: This is correct. This is not correct. You are complicating matters and mixing things up. Think of Ohm's law at each resistor. You are correct - the voltage is being dropped. But by how much? It is dropped by [itex]IR[/itex]. Because current is conserved through the resistors, the voltage drops are permitted to change if the resistances change.
Thank you very much for the prompt response Saketh. I'm still trying to absorb this all in and make sense of it...but I figured I'd ask a quick question since you're still online. I am beginning to understand what you mean by the resistors simply changing the amount of current drawn, so it affects the current as a whole throughout the entire circuit. How in the world do the charge carriers or electrons know what 'speed' to go, or what current will be had when it first starts off? How does it 'see' these resistors and know to adjust their speed appropriately? Basically how, why, and when does this 'current regulation' take place that is due to the resistors? It's pretty mind boggling. Edited: Just to add so you can better see my source of confusion...I didn't think charge was destroyed but rather I saw the resistors as imposing new speed limits so to speak. Just making the charges slow down and thus the current decrease. But now it seems from what you tell me that the charges are all obeying all of the 'speed limits' imposed by the resistors at the same time, at a constant rate, which makes no sense.
If you think of current as flowing you can compare it to water. Imagine water pouring through a hose. Further imagine that one section of the hose is thinner than the rest. The water doesn't need to know that the hose gets thinner in advance... it simply slows down because the water in front of it is slowing down. Charges figure it out quickly because disturbances propagate at the speed of light. Think about what would happen if you had a wire 3x10^8 meters long and created a voltage difference at each end simultaneously.
The way I was taught current was to view it as a bunch of cars on a highway, where resistors are speed bumps. Pretend that each ohm represents one speed bump. Series resistors are like having only one route for cars to travel on, and even though they are going to be slowed down (the potential difference drops) that they are forced to still go on the same road. Now, if on the road you are traveling there are 1000 speed bumps, but there is another road that will get you to the same place with only 100 speed bumps you will take the 100 bump road, which shows why current takes the path of least resistivity. I dunno, good analogy for me at least because now just replace cars with electrons and number of bumps with resistance values. In other words, current is the same through resistors in series because the electrons have no other path to take.
Lol, yes. That will be another question for me later on. Why exactly would charge take the path of least resistance? Aye, that does make sense. And comparing this to a thought experiment I was doing...I see there is a lapse in my knowledge in terms of current. Once I figure this out properly I think it will help me understand the first question. New Question Water I can physically understand because the molecules are bumping into each other and thus will all be going the same speed due to thinner parts of the hose or thicker. But electrons are a bit different. The thought experiment I meant was this: Let's say that we apply a voltage across a long copper tube...which is basically a circuit. We will take the resistance of the tube into consideration though. Now, all a voltage is, is a potential difference between two points. Which comes directly from there being an electric field somewhere 'pushing' the electrons along. Now if the force of the field dies off as distance squared, and the mass obviously remains the same of the electrons, logically the acceleration of the electrons will be diminishing due to the field. So by this train of thought the electron's speed should be diminishing as it reaches the end of the tube. This is NOT so though, and they have a constant velocity. Why? My conjecture would be that although the field that provided the initial push is dying off, they are still being carried by the field from each other. In other words, the field emanating from the electrons themselves are responsible for the constant velocity...or constant 'current.' HOW? No idea. But someone please explain how it is so. I imagine once I understand this I can better tackle my first problem.
By "copper tube" do you mean a solid tube made of copper, like a wire? I will assume yes, for the time being. Why do you think the field in the wire dies off as distance squared? The field exists throughout the wire, due to the surface charges created by the application of the voltage source. This field accelerates the electrons within the wire, creating the current flow. Note that the electrons are not just floating freely, but are constantly smacking into the lattice of the wire. The applied voltage creates a small drift velocity of the electrons that is the current. Check this out: Microscopic View of Ohm's Law
Yep, you're right. I was going off of Coulomb's Law for each individual point charge. I'm kind of viewing it as they get an initial 'jolt' from the field source, and by Coulomb's Law they slowly start to run out of 'gas' because of their distance. I also wasn't sure how the individual electrons themselves affected each other with their individual fields. So the surface charges are what keeps the electrons moving at a constant velocity? Now that you speak of lattice, that is another problem that I have had. If you could please check out this topic, I'm sure you could shed light on my problem: https://www.physicsforums.com/showthread.php?t=149621 Thanks a lot Doc Al.
As for why current is constant in the wire... You know those toys that people who sit behind big desks at work like to have? Where there are 5 or so metal balls hanging from a horizontal support, and when you pick the ball up from one side and let it drop onto the others, the ball at the other end bounces off? If there was no friction, no non-conservative force of any kind, the balls would do this forever, with the opposite ball rising the same amount as the first ball every time. With friction, though, the ball doesn't get quite as high. If a wire had no resistance, the push on electrons at the back of the wire would travel down the length at the speed of light, pushing electrons at the front just as hard. With resistance, there's ... 'friction' ... between the electrons and nuclei on the way and the drift velocity slows down. The electrons at the front of the current are only going to travel as fast as the electrons behind them are pushing, and the electrons pushing will travel only as fast as the electrons being slowed by resistance will let them.
In this particular case, you're using Coulomb's law incorrectly. You have applied a voltage across a distance. Voltage across the wire will be equal to the product of the electric field by the length of the wire (if you don't understand why, go over the definition of voltage). The length of the wire is constant, as is the voltage. Therefore, the electric field across the wire is constant. It follows that the force on each electron is constant. I think you misunderstand what the electrons are doing. It turns out that the main thing driving the electrons is thermal energy, not the electric field, but the electric field gives the electrons direction. So even though they're bouncing all around, they're bouncing all around in one general direction, on the average. So each individual electron can be doing something different, but on the average they tend to move in a direction because of the electric field. I can't think of a good analogy for this concept.
Whoa....the E-field is constant? How the hell is that possible? A constant voltage just means that there is a constant potential difference, so there is a difference in Joules per Coulomb at the ends of the wire. Don't the electrons need to expend Joules in order to get to the end? Shouldn't their energy and thus their speed be going down by that idea? And I'm sorry but I really don't understand how the E-field remains constant...I thought by the very nature and definition of it it caused forces on charges to drop off by distance squared. Definitely not arguing with you brother, I'm eager for you to tell me why I am wrong in my thoughts. I'll go read a bit see if I can piece it together in the meantime. Thanks man.
what is the "current being drawn"? in terms of the speed bump analogy, do the cars speed back up to their original speed? if so, how fast do we imagine this happening? if not, how is the current the same in the next resistor?
I think we've found a problem. You don't understand why electric fields can be constant. Not always. Infinite planes that are charged have a constant electric field. If you don't know what I'm talking about, you need to go over infinite planes that are charged (parallel plate capacitors). I think you should go over the infinite plane and electric potential of the infinite charged plane. That will help you understand that electric fields do not always do what you think they do. Until you understand Gauss's Law for electromagnetism, you will be confused with the point charge definition of an electric field. Think of a river going downhill. The water needs energy to move, and where does it get that energy? From a constant gravitational field. Gravity pulls the water and gives it the energy needed to pass obstacles. Now take that analogy, and extend it to the electrons. The constant electric field is like gravity, pulling the electrons (but once again, don't forget thermal energy).
Let's say you have a circuit which is just a loop with a resistor and a battery. How much current passes through the resistor? [tex]\frac{V}{R} = I[/tex], where V is the potential difference of the battery, R is the resistance of the resistor, and I is the current. So we say that the resistor has drawn [itex]\frac{V}{R}[/itex] amps of current. Now let's short circuit the resistor - in other words, take a wire, connect one end to one end of the resistor, and the other to the other end. We have effectively given the electrons a less resistance path to travel, so now the wire will draw more current than the resistor does. The difference between current being drawn and actual current is a terminology difference - they are the same idea. If the resistor draws X amount of current, X amount of current will pass through the resistor. You can also just say that X amount of current will pass through the resistor, and people will understand that the resistor has drawn X amount of current. Read this page. It will help your understanding.
Thanks again Saketh for taking the time to write something up. I I do now remember using the Gaussian pillbox with Gauss's Law and actually proving that the field would be constant any distance away from the infinite plane. Mind boggling. I think I'm starting to understand. So are you saying we can approximate any particular cross section of the wire as an infinite plane since there are so many electrons? So is there still a slight approximation involved? Here's another realization I came to...I think I'm seeing this as a singular electron traveling by itself through the wire. I imagine, and please correct me if I'm wrong, if there were only one electron traveling through the wire then he would eventually experience slow down because of distance as he got farther away from the field source. But a singular electron would mean that there is NO current. So I need to think in terms of the flow of charge....and thus need to take all of the other electrons in transit into consideration as well. Now all of these other electrons have fields as well...which change everything. Are these other electrons the reason why the electric field is constant throughout the wire? And thus of course why the current remains the same through resistors in series (my original question)? By the way...one more question. This same idea that the E-field was constant ALWAYS bothered me in parallel plate capacitors. Is it once again an approximation to an infinite plane (I think this is what you meant, but just want to make sure)? If this is so...then of course there is a small degree of error here right? Thanks and I apologize for the barrage of questions....but I've got to thoroughly pick your brain if you are in the know right?
What happens when the battery is applied to the wire is that surface charges are set up on the wire. These surface charges are more dense closer to the battery terminals; less dense further away. Think of these as rings of charge creating a uniform electric field throughout the wire. (If the field were not uniform, the current would build up a net charge correcting the situation quickly.) Using Saketh's analogy with gravity, think of the electrons not as water flowing downstream and getting faster as it flows, but as marbles falling through a maze of pegs. They are constantly being pulled by gravity (or the electric field) but they don't have a net gain in macroscopic KE since they are constantly bashing into pegs (or the wire lattice in the conductor) which turns the energy into random thermal energy.
Basically, yes. Yes, but the error involved is insignificant, because the drift velocity of the electrons depends on more than just the electric field. Once again, the electric field is constant. This means that no matter how far away from the "infinite plane" of one end of the wire you get, there will be no decrease in electric field strength. You are correct in that having one electron only would change our model, because then many of our assumptions would be on weak footing. But having one electron, according to our model of current, would not eliminate the current. Why is this? Let us define n as the number of electrons per unit volume of the wire. Therefore, n * V electrons pass through a cross-sectional area of wire in time [itex]\Delta t[/itex], where [itex]V = \Delta x A = v_{drift} \Delta t A[/itex]. Take a moment to convince yourself that this is true within our approximation of flowing electrons. Current is nothing more than number of electrons per unit of time passing through A, so we have [itex]I = \frac{enV}{\Delta t} = env_{drift}A[/itex], where e is the elementary charge (charge on one electron). Now, e has some value, [itex]v_{drift}[/itex] has some value, and A has some value. n would be 1 electron per cubic meter, because we have only one electron. So we can see that, within our model, current still exists even though there is only one electron. Yes, but it's a very good approximation. If you don't believe me, do the integration and find the actual electric field values of a finite plane. Yes, but the error is insignificant in most cases.
Thank you so much for this response and italicizing rings, it led me to the answer I was searching for. The hint you also give in parenthesis helped me understand this 'feedback mechanism' that I finally found out about. Thank you very much for your help as well Saketh. It helped me clarify thoughts and understand the big picture. I posted the .pdf file that also goes into detail about my problem for any and all who traipse upon this post with the same question to understand.
I'm glad we could be of help. And thanks for posting that paper by Sherwood--he goes into more detail, but that's exactly what I was talking about. Unfortunately, these subtleties are often glossed over or missing from many introductory texts.