What's the reason behind the current remaining the same after passing by a resistance?
I've been wondering why does this really happen, I mean by intuition if electrons are driven by EMF (ignoring wire's resistance), $n$ coulombs would pass by a point per second, until they encounter something that slows them down thus the rate of flow would change. Why does current remain the same?
One answer I saw somewhere that made sense to me is that it indeed slows electrons down, but electrons lose some of their energy to compensate the loss of velocity in a way that would bring the current back to the constant current, and this lose of energy is called drop in voltage and that's why voltage decreases when running over some resistance, is this true?
electric-circuits electric-current charge electrical-resistance voltage
add a comment |
I've been wondering why does this really happen, I mean by intuition if electrons are driven by EMF (ignoring wire's resistance), $n$ coulombs would pass by a point per second, until they encounter something that slows them down thus the rate of flow would change. Why does current remain the same?
One answer I saw somewhere that made sense to me is that it indeed slows electrons down, but electrons lose some of their energy to compensate the loss of velocity in a way that would bring the current back to the constant current, and this lose of energy is called drop in voltage and that's why voltage decreases when running over some resistance, is this true?
electric-circuits electric-current charge electrical-resistance voltage
1
Because otherwise they'd build up in the resistor? How come the amount of water before a half-closed valve is the same as the amount of water after it?
– immibis
Nov 24 at 7:25
not necessarily building up on the resistor, but the current before entering the resistor or the velocity of electrons before entering the resistor is more than the velocity of them after leaving the resistor. Or maybe to elaborate my point more: 2 coulombs flow through a cross section of the wire in a second before entering the resistor, after entering it, 1 coulomb (for e.g) would flow through a cross section of that wire in a second, and the other coulomb wouldn't build up on the resistor, rather it would leave it but with a slower velocity. isn't my assumption true?
– Just_Cause
Nov 24 at 16:00
If it leaves with a slower velocity, then it's still leaving, isn't it? Coulombs per second is not velocity, it's amount of electrons per second. If the amount of electrons coming out is less than the amount of electrons going in, then where are the electrons going?
– immibis
Nov 25 at 2:54
let's forget about velocity, i used it because it plays a role in current's formula. we agree to some point that the number of coulombs that flow across a point before entering the resistor in a second, is the same amount that does the same after leaving the resistor. However, what I want to get to is that why does that really happen? electrons face obstacles inside the resistor and there must be something that played a role in getting their rate of flow back the same. V.F and Steeven for example pointed that out in different ways.
– Just_Cause
Nov 25 at 13:16
add a comment |
I've been wondering why does this really happen, I mean by intuition if electrons are driven by EMF (ignoring wire's resistance), $n$ coulombs would pass by a point per second, until they encounter something that slows them down thus the rate of flow would change. Why does current remain the same?
One answer I saw somewhere that made sense to me is that it indeed slows electrons down, but electrons lose some of their energy to compensate the loss of velocity in a way that would bring the current back to the constant current, and this lose of energy is called drop in voltage and that's why voltage decreases when running over some resistance, is this true?
electric-circuits electric-current charge electrical-resistance voltage
I've been wondering why does this really happen, I mean by intuition if electrons are driven by EMF (ignoring wire's resistance), $n$ coulombs would pass by a point per second, until they encounter something that slows them down thus the rate of flow would change. Why does current remain the same?
One answer I saw somewhere that made sense to me is that it indeed slows electrons down, but electrons lose some of their energy to compensate the loss of velocity in a way that would bring the current back to the constant current, and this lose of energy is called drop in voltage and that's why voltage decreases when running over some resistance, is this true?
electric-circuits electric-current charge electrical-resistance voltage
electric-circuits electric-current charge electrical-resistance voltage
edited Nov 23 at 18:36
Qmechanic♦
101k121831142
101k121831142
asked Nov 23 at 10:57
Just_Cause
634
634
1
Because otherwise they'd build up in the resistor? How come the amount of water before a half-closed valve is the same as the amount of water after it?
– immibis
Nov 24 at 7:25
not necessarily building up on the resistor, but the current before entering the resistor or the velocity of electrons before entering the resistor is more than the velocity of them after leaving the resistor. Or maybe to elaborate my point more: 2 coulombs flow through a cross section of the wire in a second before entering the resistor, after entering it, 1 coulomb (for e.g) would flow through a cross section of that wire in a second, and the other coulomb wouldn't build up on the resistor, rather it would leave it but with a slower velocity. isn't my assumption true?
– Just_Cause
Nov 24 at 16:00
If it leaves with a slower velocity, then it's still leaving, isn't it? Coulombs per second is not velocity, it's amount of electrons per second. If the amount of electrons coming out is less than the amount of electrons going in, then where are the electrons going?
– immibis
Nov 25 at 2:54
let's forget about velocity, i used it because it plays a role in current's formula. we agree to some point that the number of coulombs that flow across a point before entering the resistor in a second, is the same amount that does the same after leaving the resistor. However, what I want to get to is that why does that really happen? electrons face obstacles inside the resistor and there must be something that played a role in getting their rate of flow back the same. V.F and Steeven for example pointed that out in different ways.
– Just_Cause
Nov 25 at 13:16
add a comment |
1
Because otherwise they'd build up in the resistor? How come the amount of water before a half-closed valve is the same as the amount of water after it?
– immibis
Nov 24 at 7:25
not necessarily building up on the resistor, but the current before entering the resistor or the velocity of electrons before entering the resistor is more than the velocity of them after leaving the resistor. Or maybe to elaborate my point more: 2 coulombs flow through a cross section of the wire in a second before entering the resistor, after entering it, 1 coulomb (for e.g) would flow through a cross section of that wire in a second, and the other coulomb wouldn't build up on the resistor, rather it would leave it but with a slower velocity. isn't my assumption true?
– Just_Cause
Nov 24 at 16:00
If it leaves with a slower velocity, then it's still leaving, isn't it? Coulombs per second is not velocity, it's amount of electrons per second. If the amount of electrons coming out is less than the amount of electrons going in, then where are the electrons going?
– immibis
Nov 25 at 2:54
let's forget about velocity, i used it because it plays a role in current's formula. we agree to some point that the number of coulombs that flow across a point before entering the resistor in a second, is the same amount that does the same after leaving the resistor. However, what I want to get to is that why does that really happen? electrons face obstacles inside the resistor and there must be something that played a role in getting their rate of flow back the same. V.F and Steeven for example pointed that out in different ways.
– Just_Cause
Nov 25 at 13:16
1
1
Because otherwise they'd build up in the resistor? How come the amount of water before a half-closed valve is the same as the amount of water after it?
– immibis
Nov 24 at 7:25
Because otherwise they'd build up in the resistor? How come the amount of water before a half-closed valve is the same as the amount of water after it?
– immibis
Nov 24 at 7:25
not necessarily building up on the resistor, but the current before entering the resistor or the velocity of electrons before entering the resistor is more than the velocity of them after leaving the resistor. Or maybe to elaborate my point more: 2 coulombs flow through a cross section of the wire in a second before entering the resistor, after entering it, 1 coulomb (for e.g) would flow through a cross section of that wire in a second, and the other coulomb wouldn't build up on the resistor, rather it would leave it but with a slower velocity. isn't my assumption true?
– Just_Cause
Nov 24 at 16:00
not necessarily building up on the resistor, but the current before entering the resistor or the velocity of electrons before entering the resistor is more than the velocity of them after leaving the resistor. Or maybe to elaborate my point more: 2 coulombs flow through a cross section of the wire in a second before entering the resistor, after entering it, 1 coulomb (for e.g) would flow through a cross section of that wire in a second, and the other coulomb wouldn't build up on the resistor, rather it would leave it but with a slower velocity. isn't my assumption true?
– Just_Cause
Nov 24 at 16:00
If it leaves with a slower velocity, then it's still leaving, isn't it? Coulombs per second is not velocity, it's amount of electrons per second. If the amount of electrons coming out is less than the amount of electrons going in, then where are the electrons going?
– immibis
Nov 25 at 2:54
If it leaves with a slower velocity, then it's still leaving, isn't it? Coulombs per second is not velocity, it's amount of electrons per second. If the amount of electrons coming out is less than the amount of electrons going in, then where are the electrons going?
– immibis
Nov 25 at 2:54
let's forget about velocity, i used it because it plays a role in current's formula. we agree to some point that the number of coulombs that flow across a point before entering the resistor in a second, is the same amount that does the same after leaving the resistor. However, what I want to get to is that why does that really happen? electrons face obstacles inside the resistor and there must be something that played a role in getting their rate of flow back the same. V.F and Steeven for example pointed that out in different ways.
– Just_Cause
Nov 25 at 13:16
let's forget about velocity, i used it because it plays a role in current's formula. we agree to some point that the number of coulombs that flow across a point before entering the resistor in a second, is the same amount that does the same after leaving the resistor. However, what I want to get to is that why does that really happen? electrons face obstacles inside the resistor and there must be something that played a role in getting their rate of flow back the same. V.F and Steeven for example pointed that out in different ways.
– Just_Cause
Nov 25 at 13:16
add a comment |
5 Answers
5
active
oldest
votes
Yes, it is true.
Charge cannot disappear. Or appear. This is Kirchhoff's current law. For a steady current, all charge that enters any point each second must also leave that point each second:
$$sum i_{in}=sum i_{out}$$
Otherwise charge would accumulate at that point. And eventually the enormous total charge that is accumulated will be large enough to repel any further entering charge - large enough to balance out the battery voltage - and all current would stop flowing. Since this doesn't happen, charge isn't accumulating anywhere, so Kirchhoff's current law must be true.
So what is happening?
When turning on your battery / voltage source, charges are being "pushed" forward by the battery voltage. The first charge hurries with almost the speed of light. Very, very quickly it reaches the resistor. Here it is slowed down. All the charges behind it now have to queue up and wait in line - they slow down as well. Soon they all move at the exact same speed.
When exiting the resistor the charge continues with whatever speed it came out with - which naturally is the same as the speed of all those waiting in line.
So, now, suddenly, all charge moves at the same speed. In other words, the current (amount of charge passing per second) is the same everywhere.
The push on the charges, on the other hand, is large before the resistor and zero (relatively) after the resistor. In the same way that the pressure in a water hose is large, but as soon as the water is out, there is no pressure any more. The pressure is released. Likewise, the voltage is dropped.
The supplied voltage is "spent" across the resistor, so to speak, and we talk about it as a voltage drop. And the full "pressure" / voltage supplied by the battery will be spent. In other words, the entire supplied voltage must be dropped and spread out over all components along the way in the circuit. Which leads to Kirchhoff's other law, his voltage law or loop law: all "spent" voltage in any circuit loop must necessarily equal what is being supplied somewhere else: $$sum v_{added}=sum v_{dropped}$$
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Nov 27 at 14:16
add a comment |
What's the reason behind the current remaining the same after passing
by a resistance?
Because, due to the redistribution of charges (and voltages), the electric field inside a resistor is stronger than the electric field inside a wire.
This redistribution happens automatically. Initially, the electric field is evenly distributed and the electrons in the wire move faster than the electrons inside the resistor. As a result, electrons accumulate in front of the resistor and positive ions accumulate behind it. This increases the voltage and the electric field across the resistor, causing the electrons inside the resistor to move faster.
Since the voltage of the battery stays the same, the increase of the voltage across the resistor leads to the decrease of the voltage and, therefore, the decrease of the electric field in the wire, causing the electrons in the wire to move slower.
As the electrons in the resistor move faster and faster and the electrons in the wire move slower and slower, at some point, their speeds will equalize. At this point, the redistribution of charges will stop and the current in all parts of the circuit will be the same.
"at this point, their speeds will equalize" , you mean the drift speed, right? Individual electrons must be changing their speed all the time, because of collisions with other electrons and vibrating atoms.
– physicsguy19
Nov 23 at 15:50
2
@physicsguy19 Yes, I am talking about the drift speed - the speed that contributes to the current.
– V.F.
Nov 23 at 16:23
add a comment |
Current us a measure of how much charge is passing a given point (or cross section) of a wire.
If the currents were not equal at all points in a simple circuit, there would have to be charges entering or exiting the circuit. This however does not happen.
Water pipe analogy: current is something like liters per minute that pass through a certain point. If there are no leaks or additional pipes joining, at each point there has to be the same water flow in liters per minute.
1
This implies some "non-compressability" of the "electron gas" which is not entirely intuitive (and actually not even entirely true, or we wouldn't have capacitors ;-) ).
– Peter A. Schneider
Nov 23 at 18:27
Well, it's the reason you use a water pipe analogy and not an air pipe analogy.
– Peter A. Schneider
Nov 23 at 19:34
@Harry Johnston: No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out (incl. for a capacitor) are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero.
– Peter Mortensen
Nov 24 at 23:17
@Peter A. Schneider: non-compressibility and capacitors are actually compatible (Kirchhoff's current law still holds for a time-varying current (the current in and current out are exactly equal, incl. for a capacitor) - at sufficient low frequency). The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in only one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:31
add a comment |
Compare the conductor with a water pipe: the flow rate in the water pipe is the same everywhere (assuming there are no branches or leaks or whatever). Now introduce a constriction in the pipe, or a valve that's half closed. In this new situation the flow rate in the pipe is lower, but it is lower everywhere in the pipe: both before and after the constriction. The flow rate is still the same everywhere in the pipe (but not the same as before).
The same is true in a conductor: in a conductor with a resistance the current (flow rate) will be lower than in a conductor without (given the same voltage), but each conductor will have the same current everywhere.
Re "... a difference between current in and current out": No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero . The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:11
@PeterMortensen: I stand corrected. You are right, that section of my answer was wrong, so I deleted it.
– Roel Schroeven
Nov 24 at 23:34
add a comment |
I have found a nice explanation called the Drude Model, a simple mechanical model proposed by the German physicist Paul Drude in 1900. It is based on the idea that the charge carriers bounce around randomly at high speeds, with an average momentum of
$$langle vec{p} rangle = qvec{E}tau$$
where $q$ is the charge of a carrier (i.e. the electron charge), $vec{E}$ is the electric field vector and $tau$ is the average time between bounces.
Quoting from the wikipedia article, use the following substitutions
$$langle vec{p} rangle = mlangle vec{v} rangle,; vec{J} = nqlanglevec{v}rangle$$
where $m$ is the electron mass, $n$ is the number density of the charge carriers, $langle vec{v} rangle$ is the average velocity, and $vec{J}$ is the current density. This gives us:
$$vec{J} = left(frac{nq^2tau}{m}right)vec{E}$$
Now consider a resistor with the oriented cross-section area $vec{S}$. Multiply the above formula together with $vec{S}$ to get the current through $vec{S}$:
$$I = vec{J}cdotvec{S} = left(frac{nq^2tau}{m}right)vec{E}cdotvec{S}$$
For simplicity, assume $vec{E}$ and $vec{S}$ are constant and parallel, and that $|vec{E}|=Delta U/L$, where $Delta U$ is the voltage difference and $L$ is the length of the resistor. This gives us the current in terms of the voltage difference times a factor:
$$I = left(frac{nq^2tau}{m}right)frac{Delta U,S}{L} = left(frac{nq^2tau S}{mL}right)Delta U$$
This is Ohm's law if we identify $left(dfrac{nq^2tau S}{mL}right)=dfrac{1}{R}$.
Note: the above only shows the steady-state situation. It doesn't really explain how the local voltages and Kirchoff conditions reach this steady state after the voltage is applied initially. But the Drude model can also handle time-varying dynamics. Please see the Wikipedia article for details.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f442719%2fwhats-the-reason-behind-the-current-remaining-the-same-after-passing-by-a-resis%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes, it is true.
Charge cannot disappear. Or appear. This is Kirchhoff's current law. For a steady current, all charge that enters any point each second must also leave that point each second:
$$sum i_{in}=sum i_{out}$$
Otherwise charge would accumulate at that point. And eventually the enormous total charge that is accumulated will be large enough to repel any further entering charge - large enough to balance out the battery voltage - and all current would stop flowing. Since this doesn't happen, charge isn't accumulating anywhere, so Kirchhoff's current law must be true.
So what is happening?
When turning on your battery / voltage source, charges are being "pushed" forward by the battery voltage. The first charge hurries with almost the speed of light. Very, very quickly it reaches the resistor. Here it is slowed down. All the charges behind it now have to queue up and wait in line - they slow down as well. Soon they all move at the exact same speed.
When exiting the resistor the charge continues with whatever speed it came out with - which naturally is the same as the speed of all those waiting in line.
So, now, suddenly, all charge moves at the same speed. In other words, the current (amount of charge passing per second) is the same everywhere.
The push on the charges, on the other hand, is large before the resistor and zero (relatively) after the resistor. In the same way that the pressure in a water hose is large, but as soon as the water is out, there is no pressure any more. The pressure is released. Likewise, the voltage is dropped.
The supplied voltage is "spent" across the resistor, so to speak, and we talk about it as a voltage drop. And the full "pressure" / voltage supplied by the battery will be spent. In other words, the entire supplied voltage must be dropped and spread out over all components along the way in the circuit. Which leads to Kirchhoff's other law, his voltage law or loop law: all "spent" voltage in any circuit loop must necessarily equal what is being supplied somewhere else: $$sum v_{added}=sum v_{dropped}$$
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Nov 27 at 14:16
add a comment |
Yes, it is true.
Charge cannot disappear. Or appear. This is Kirchhoff's current law. For a steady current, all charge that enters any point each second must also leave that point each second:
$$sum i_{in}=sum i_{out}$$
Otherwise charge would accumulate at that point. And eventually the enormous total charge that is accumulated will be large enough to repel any further entering charge - large enough to balance out the battery voltage - and all current would stop flowing. Since this doesn't happen, charge isn't accumulating anywhere, so Kirchhoff's current law must be true.
So what is happening?
When turning on your battery / voltage source, charges are being "pushed" forward by the battery voltage. The first charge hurries with almost the speed of light. Very, very quickly it reaches the resistor. Here it is slowed down. All the charges behind it now have to queue up and wait in line - they slow down as well. Soon they all move at the exact same speed.
When exiting the resistor the charge continues with whatever speed it came out with - which naturally is the same as the speed of all those waiting in line.
So, now, suddenly, all charge moves at the same speed. In other words, the current (amount of charge passing per second) is the same everywhere.
The push on the charges, on the other hand, is large before the resistor and zero (relatively) after the resistor. In the same way that the pressure in a water hose is large, but as soon as the water is out, there is no pressure any more. The pressure is released. Likewise, the voltage is dropped.
The supplied voltage is "spent" across the resistor, so to speak, and we talk about it as a voltage drop. And the full "pressure" / voltage supplied by the battery will be spent. In other words, the entire supplied voltage must be dropped and spread out over all components along the way in the circuit. Which leads to Kirchhoff's other law, his voltage law or loop law: all "spent" voltage in any circuit loop must necessarily equal what is being supplied somewhere else: $$sum v_{added}=sum v_{dropped}$$
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Nov 27 at 14:16
add a comment |
Yes, it is true.
Charge cannot disappear. Or appear. This is Kirchhoff's current law. For a steady current, all charge that enters any point each second must also leave that point each second:
$$sum i_{in}=sum i_{out}$$
Otherwise charge would accumulate at that point. And eventually the enormous total charge that is accumulated will be large enough to repel any further entering charge - large enough to balance out the battery voltage - and all current would stop flowing. Since this doesn't happen, charge isn't accumulating anywhere, so Kirchhoff's current law must be true.
So what is happening?
When turning on your battery / voltage source, charges are being "pushed" forward by the battery voltage. The first charge hurries with almost the speed of light. Very, very quickly it reaches the resistor. Here it is slowed down. All the charges behind it now have to queue up and wait in line - they slow down as well. Soon they all move at the exact same speed.
When exiting the resistor the charge continues with whatever speed it came out with - which naturally is the same as the speed of all those waiting in line.
So, now, suddenly, all charge moves at the same speed. In other words, the current (amount of charge passing per second) is the same everywhere.
The push on the charges, on the other hand, is large before the resistor and zero (relatively) after the resistor. In the same way that the pressure in a water hose is large, but as soon as the water is out, there is no pressure any more. The pressure is released. Likewise, the voltage is dropped.
The supplied voltage is "spent" across the resistor, so to speak, and we talk about it as a voltage drop. And the full "pressure" / voltage supplied by the battery will be spent. In other words, the entire supplied voltage must be dropped and spread out over all components along the way in the circuit. Which leads to Kirchhoff's other law, his voltage law or loop law: all "spent" voltage in any circuit loop must necessarily equal what is being supplied somewhere else: $$sum v_{added}=sum v_{dropped}$$
Yes, it is true.
Charge cannot disappear. Or appear. This is Kirchhoff's current law. For a steady current, all charge that enters any point each second must also leave that point each second:
$$sum i_{in}=sum i_{out}$$
Otherwise charge would accumulate at that point. And eventually the enormous total charge that is accumulated will be large enough to repel any further entering charge - large enough to balance out the battery voltage - and all current would stop flowing. Since this doesn't happen, charge isn't accumulating anywhere, so Kirchhoff's current law must be true.
So what is happening?
When turning on your battery / voltage source, charges are being "pushed" forward by the battery voltage. The first charge hurries with almost the speed of light. Very, very quickly it reaches the resistor. Here it is slowed down. All the charges behind it now have to queue up and wait in line - they slow down as well. Soon they all move at the exact same speed.
When exiting the resistor the charge continues with whatever speed it came out with - which naturally is the same as the speed of all those waiting in line.
So, now, suddenly, all charge moves at the same speed. In other words, the current (amount of charge passing per second) is the same everywhere.
The push on the charges, on the other hand, is large before the resistor and zero (relatively) after the resistor. In the same way that the pressure in a water hose is large, but as soon as the water is out, there is no pressure any more. The pressure is released. Likewise, the voltage is dropped.
The supplied voltage is "spent" across the resistor, so to speak, and we talk about it as a voltage drop. And the full "pressure" / voltage supplied by the battery will be spent. In other words, the entire supplied voltage must be dropped and spread out over all components along the way in the circuit. Which leads to Kirchhoff's other law, his voltage law or loop law: all "spent" voltage in any circuit loop must necessarily equal what is being supplied somewhere else: $$sum v_{added}=sum v_{dropped}$$
edited Nov 23 at 13:36
answered Nov 23 at 13:04
Steeven
26k560107
26k560107
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Nov 27 at 14:16
add a comment |
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Nov 27 at 14:16
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Nov 27 at 14:16
Comments are not for extended discussion; this conversation has been moved to chat.
– rob♦
Nov 27 at 14:16
add a comment |
What's the reason behind the current remaining the same after passing
by a resistance?
Because, due to the redistribution of charges (and voltages), the electric field inside a resistor is stronger than the electric field inside a wire.
This redistribution happens automatically. Initially, the electric field is evenly distributed and the electrons in the wire move faster than the electrons inside the resistor. As a result, electrons accumulate in front of the resistor and positive ions accumulate behind it. This increases the voltage and the electric field across the resistor, causing the electrons inside the resistor to move faster.
Since the voltage of the battery stays the same, the increase of the voltage across the resistor leads to the decrease of the voltage and, therefore, the decrease of the electric field in the wire, causing the electrons in the wire to move slower.
As the electrons in the resistor move faster and faster and the electrons in the wire move slower and slower, at some point, their speeds will equalize. At this point, the redistribution of charges will stop and the current in all parts of the circuit will be the same.
"at this point, their speeds will equalize" , you mean the drift speed, right? Individual electrons must be changing their speed all the time, because of collisions with other electrons and vibrating atoms.
– physicsguy19
Nov 23 at 15:50
2
@physicsguy19 Yes, I am talking about the drift speed - the speed that contributes to the current.
– V.F.
Nov 23 at 16:23
add a comment |
What's the reason behind the current remaining the same after passing
by a resistance?
Because, due to the redistribution of charges (and voltages), the electric field inside a resistor is stronger than the electric field inside a wire.
This redistribution happens automatically. Initially, the electric field is evenly distributed and the electrons in the wire move faster than the electrons inside the resistor. As a result, electrons accumulate in front of the resistor and positive ions accumulate behind it. This increases the voltage and the electric field across the resistor, causing the electrons inside the resistor to move faster.
Since the voltage of the battery stays the same, the increase of the voltage across the resistor leads to the decrease of the voltage and, therefore, the decrease of the electric field in the wire, causing the electrons in the wire to move slower.
As the electrons in the resistor move faster and faster and the electrons in the wire move slower and slower, at some point, their speeds will equalize. At this point, the redistribution of charges will stop and the current in all parts of the circuit will be the same.
"at this point, their speeds will equalize" , you mean the drift speed, right? Individual electrons must be changing their speed all the time, because of collisions with other electrons and vibrating atoms.
– physicsguy19
Nov 23 at 15:50
2
@physicsguy19 Yes, I am talking about the drift speed - the speed that contributes to the current.
– V.F.
Nov 23 at 16:23
add a comment |
What's the reason behind the current remaining the same after passing
by a resistance?
Because, due to the redistribution of charges (and voltages), the electric field inside a resistor is stronger than the electric field inside a wire.
This redistribution happens automatically. Initially, the electric field is evenly distributed and the electrons in the wire move faster than the electrons inside the resistor. As a result, electrons accumulate in front of the resistor and positive ions accumulate behind it. This increases the voltage and the electric field across the resistor, causing the electrons inside the resistor to move faster.
Since the voltage of the battery stays the same, the increase of the voltage across the resistor leads to the decrease of the voltage and, therefore, the decrease of the electric field in the wire, causing the electrons in the wire to move slower.
As the electrons in the resistor move faster and faster and the electrons in the wire move slower and slower, at some point, their speeds will equalize. At this point, the redistribution of charges will stop and the current in all parts of the circuit will be the same.
What's the reason behind the current remaining the same after passing
by a resistance?
Because, due to the redistribution of charges (and voltages), the electric field inside a resistor is stronger than the electric field inside a wire.
This redistribution happens automatically. Initially, the electric field is evenly distributed and the electrons in the wire move faster than the electrons inside the resistor. As a result, electrons accumulate in front of the resistor and positive ions accumulate behind it. This increases the voltage and the electric field across the resistor, causing the electrons inside the resistor to move faster.
Since the voltage of the battery stays the same, the increase of the voltage across the resistor leads to the decrease of the voltage and, therefore, the decrease of the electric field in the wire, causing the electrons in the wire to move slower.
As the electrons in the resistor move faster and faster and the electrons in the wire move slower and slower, at some point, their speeds will equalize. At this point, the redistribution of charges will stop and the current in all parts of the circuit will be the same.
answered Nov 23 at 14:10
V.F.
10.9k21028
10.9k21028
"at this point, their speeds will equalize" , you mean the drift speed, right? Individual electrons must be changing their speed all the time, because of collisions with other electrons and vibrating atoms.
– physicsguy19
Nov 23 at 15:50
2
@physicsguy19 Yes, I am talking about the drift speed - the speed that contributes to the current.
– V.F.
Nov 23 at 16:23
add a comment |
"at this point, their speeds will equalize" , you mean the drift speed, right? Individual electrons must be changing their speed all the time, because of collisions with other electrons and vibrating atoms.
– physicsguy19
Nov 23 at 15:50
2
@physicsguy19 Yes, I am talking about the drift speed - the speed that contributes to the current.
– V.F.
Nov 23 at 16:23
"at this point, their speeds will equalize" , you mean the drift speed, right? Individual electrons must be changing their speed all the time, because of collisions with other electrons and vibrating atoms.
– physicsguy19
Nov 23 at 15:50
"at this point, their speeds will equalize" , you mean the drift speed, right? Individual electrons must be changing their speed all the time, because of collisions with other electrons and vibrating atoms.
– physicsguy19
Nov 23 at 15:50
2
2
@physicsguy19 Yes, I am talking about the drift speed - the speed that contributes to the current.
– V.F.
Nov 23 at 16:23
@physicsguy19 Yes, I am talking about the drift speed - the speed that contributes to the current.
– V.F.
Nov 23 at 16:23
add a comment |
Current us a measure of how much charge is passing a given point (or cross section) of a wire.
If the currents were not equal at all points in a simple circuit, there would have to be charges entering or exiting the circuit. This however does not happen.
Water pipe analogy: current is something like liters per minute that pass through a certain point. If there are no leaks or additional pipes joining, at each point there has to be the same water flow in liters per minute.
1
This implies some "non-compressability" of the "electron gas" which is not entirely intuitive (and actually not even entirely true, or we wouldn't have capacitors ;-) ).
– Peter A. Schneider
Nov 23 at 18:27
Well, it's the reason you use a water pipe analogy and not an air pipe analogy.
– Peter A. Schneider
Nov 23 at 19:34
@Harry Johnston: No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out (incl. for a capacitor) are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero.
– Peter Mortensen
Nov 24 at 23:17
@Peter A. Schneider: non-compressibility and capacitors are actually compatible (Kirchhoff's current law still holds for a time-varying current (the current in and current out are exactly equal, incl. for a capacitor) - at sufficient low frequency). The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in only one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:31
add a comment |
Current us a measure of how much charge is passing a given point (or cross section) of a wire.
If the currents were not equal at all points in a simple circuit, there would have to be charges entering or exiting the circuit. This however does not happen.
Water pipe analogy: current is something like liters per minute that pass through a certain point. If there are no leaks or additional pipes joining, at each point there has to be the same water flow in liters per minute.
1
This implies some "non-compressability" of the "electron gas" which is not entirely intuitive (and actually not even entirely true, or we wouldn't have capacitors ;-) ).
– Peter A. Schneider
Nov 23 at 18:27
Well, it's the reason you use a water pipe analogy and not an air pipe analogy.
– Peter A. Schneider
Nov 23 at 19:34
@Harry Johnston: No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out (incl. for a capacitor) are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero.
– Peter Mortensen
Nov 24 at 23:17
@Peter A. Schneider: non-compressibility and capacitors are actually compatible (Kirchhoff's current law still holds for a time-varying current (the current in and current out are exactly equal, incl. for a capacitor) - at sufficient low frequency). The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in only one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:31
add a comment |
Current us a measure of how much charge is passing a given point (or cross section) of a wire.
If the currents were not equal at all points in a simple circuit, there would have to be charges entering or exiting the circuit. This however does not happen.
Water pipe analogy: current is something like liters per minute that pass through a certain point. If there are no leaks or additional pipes joining, at each point there has to be the same water flow in liters per minute.
Current us a measure of how much charge is passing a given point (or cross section) of a wire.
If the currents were not equal at all points in a simple circuit, there would have to be charges entering or exiting the circuit. This however does not happen.
Water pipe analogy: current is something like liters per minute that pass through a certain point. If there are no leaks or additional pipes joining, at each point there has to be the same water flow in liters per minute.
answered Nov 23 at 13:02
Jasper
7861415
7861415
1
This implies some "non-compressability" of the "electron gas" which is not entirely intuitive (and actually not even entirely true, or we wouldn't have capacitors ;-) ).
– Peter A. Schneider
Nov 23 at 18:27
Well, it's the reason you use a water pipe analogy and not an air pipe analogy.
– Peter A. Schneider
Nov 23 at 19:34
@Harry Johnston: No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out (incl. for a capacitor) are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero.
– Peter Mortensen
Nov 24 at 23:17
@Peter A. Schneider: non-compressibility and capacitors are actually compatible (Kirchhoff's current law still holds for a time-varying current (the current in and current out are exactly equal, incl. for a capacitor) - at sufficient low frequency). The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in only one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:31
add a comment |
1
This implies some "non-compressability" of the "electron gas" which is not entirely intuitive (and actually not even entirely true, or we wouldn't have capacitors ;-) ).
– Peter A. Schneider
Nov 23 at 18:27
Well, it's the reason you use a water pipe analogy and not an air pipe analogy.
– Peter A. Schneider
Nov 23 at 19:34
@Harry Johnston: No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out (incl. for a capacitor) are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero.
– Peter Mortensen
Nov 24 at 23:17
@Peter A. Schneider: non-compressibility and capacitors are actually compatible (Kirchhoff's current law still holds for a time-varying current (the current in and current out are exactly equal, incl. for a capacitor) - at sufficient low frequency). The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in only one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:31
1
1
This implies some "non-compressability" of the "electron gas" which is not entirely intuitive (and actually not even entirely true, or we wouldn't have capacitors ;-) ).
– Peter A. Schneider
Nov 23 at 18:27
This implies some "non-compressability" of the "electron gas" which is not entirely intuitive (and actually not even entirely true, or we wouldn't have capacitors ;-) ).
– Peter A. Schneider
Nov 23 at 18:27
Well, it's the reason you use a water pipe analogy and not an air pipe analogy.
– Peter A. Schneider
Nov 23 at 19:34
Well, it's the reason you use a water pipe analogy and not an air pipe analogy.
– Peter A. Schneider
Nov 23 at 19:34
@Harry Johnston: No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out (incl. for a capacitor) are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero.
– Peter Mortensen
Nov 24 at 23:17
@Harry Johnston: No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out (incl. for a capacitor) are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero.
– Peter Mortensen
Nov 24 at 23:17
@Peter A. Schneider: non-compressibility and capacitors are actually compatible (Kirchhoff's current law still holds for a time-varying current (the current in and current out are exactly equal, incl. for a capacitor) - at sufficient low frequency). The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in only one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:31
@Peter A. Schneider: non-compressibility and capacitors are actually compatible (Kirchhoff's current law still holds for a time-varying current (the current in and current out are exactly equal, incl. for a capacitor) - at sufficient low frequency). The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in only one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:31
add a comment |
Compare the conductor with a water pipe: the flow rate in the water pipe is the same everywhere (assuming there are no branches or leaks or whatever). Now introduce a constriction in the pipe, or a valve that's half closed. In this new situation the flow rate in the pipe is lower, but it is lower everywhere in the pipe: both before and after the constriction. The flow rate is still the same everywhere in the pipe (but not the same as before).
The same is true in a conductor: in a conductor with a resistance the current (flow rate) will be lower than in a conductor without (given the same voltage), but each conductor will have the same current everywhere.
Re "... a difference between current in and current out": No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero . The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:11
@PeterMortensen: I stand corrected. You are right, that section of my answer was wrong, so I deleted it.
– Roel Schroeven
Nov 24 at 23:34
add a comment |
Compare the conductor with a water pipe: the flow rate in the water pipe is the same everywhere (assuming there are no branches or leaks or whatever). Now introduce a constriction in the pipe, or a valve that's half closed. In this new situation the flow rate in the pipe is lower, but it is lower everywhere in the pipe: both before and after the constriction. The flow rate is still the same everywhere in the pipe (but not the same as before).
The same is true in a conductor: in a conductor with a resistance the current (flow rate) will be lower than in a conductor without (given the same voltage), but each conductor will have the same current everywhere.
Re "... a difference between current in and current out": No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero . The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:11
@PeterMortensen: I stand corrected. You are right, that section of my answer was wrong, so I deleted it.
– Roel Schroeven
Nov 24 at 23:34
add a comment |
Compare the conductor with a water pipe: the flow rate in the water pipe is the same everywhere (assuming there are no branches or leaks or whatever). Now introduce a constriction in the pipe, or a valve that's half closed. In this new situation the flow rate in the pipe is lower, but it is lower everywhere in the pipe: both before and after the constriction. The flow rate is still the same everywhere in the pipe (but not the same as before).
The same is true in a conductor: in a conductor with a resistance the current (flow rate) will be lower than in a conductor without (given the same voltage), but each conductor will have the same current everywhere.
Compare the conductor with a water pipe: the flow rate in the water pipe is the same everywhere (assuming there are no branches or leaks or whatever). Now introduce a constriction in the pipe, or a valve that's half closed. In this new situation the flow rate in the pipe is lower, but it is lower everywhere in the pipe: both before and after the constriction. The flow rate is still the same everywhere in the pipe (but not the same as before).
The same is true in a conductor: in a conductor with a resistance the current (flow rate) will be lower than in a conductor without (given the same voltage), but each conductor will have the same current everywhere.
edited Nov 24 at 23:33
answered Nov 23 at 15:25
Roel Schroeven
1412
1412
Re "... a difference between current in and current out": No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero . The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:11
@PeterMortensen: I stand corrected. You are right, that section of my answer was wrong, so I deleted it.
– Roel Schroeven
Nov 24 at 23:34
add a comment |
Re "... a difference between current in and current out": No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero . The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:11
@PeterMortensen: I stand corrected. You are right, that section of my answer was wrong, so I deleted it.
– Roel Schroeven
Nov 24 at 23:34
Re "... a difference between current in and current out": No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero . The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:11
Re "... a difference between current in and current out": No, this is not correct. Kirchhoff's current law still holds (for a sufficient low frequency) - the current in and current out are exactly equal. The electrical energy in a capacitor is by a positively charged plate and a negatively charged plate - the net charge is zero . The water pipe analogy is a rubber membrane that completely blocks the pipe (some flow is possible, but there cannot be flow in one direction indefinitely).
– Peter Mortensen
Nov 24 at 23:11
@PeterMortensen: I stand corrected. You are right, that section of my answer was wrong, so I deleted it.
– Roel Schroeven
Nov 24 at 23:34
@PeterMortensen: I stand corrected. You are right, that section of my answer was wrong, so I deleted it.
– Roel Schroeven
Nov 24 at 23:34
add a comment |
I have found a nice explanation called the Drude Model, a simple mechanical model proposed by the German physicist Paul Drude in 1900. It is based on the idea that the charge carriers bounce around randomly at high speeds, with an average momentum of
$$langle vec{p} rangle = qvec{E}tau$$
where $q$ is the charge of a carrier (i.e. the electron charge), $vec{E}$ is the electric field vector and $tau$ is the average time between bounces.
Quoting from the wikipedia article, use the following substitutions
$$langle vec{p} rangle = mlangle vec{v} rangle,; vec{J} = nqlanglevec{v}rangle$$
where $m$ is the electron mass, $n$ is the number density of the charge carriers, $langle vec{v} rangle$ is the average velocity, and $vec{J}$ is the current density. This gives us:
$$vec{J} = left(frac{nq^2tau}{m}right)vec{E}$$
Now consider a resistor with the oriented cross-section area $vec{S}$. Multiply the above formula together with $vec{S}$ to get the current through $vec{S}$:
$$I = vec{J}cdotvec{S} = left(frac{nq^2tau}{m}right)vec{E}cdotvec{S}$$
For simplicity, assume $vec{E}$ and $vec{S}$ are constant and parallel, and that $|vec{E}|=Delta U/L$, where $Delta U$ is the voltage difference and $L$ is the length of the resistor. This gives us the current in terms of the voltage difference times a factor:
$$I = left(frac{nq^2tau}{m}right)frac{Delta U,S}{L} = left(frac{nq^2tau S}{mL}right)Delta U$$
This is Ohm's law if we identify $left(dfrac{nq^2tau S}{mL}right)=dfrac{1}{R}$.
Note: the above only shows the steady-state situation. It doesn't really explain how the local voltages and Kirchoff conditions reach this steady state after the voltage is applied initially. But the Drude model can also handle time-varying dynamics. Please see the Wikipedia article for details.
add a comment |
I have found a nice explanation called the Drude Model, a simple mechanical model proposed by the German physicist Paul Drude in 1900. It is based on the idea that the charge carriers bounce around randomly at high speeds, with an average momentum of
$$langle vec{p} rangle = qvec{E}tau$$
where $q$ is the charge of a carrier (i.e. the electron charge), $vec{E}$ is the electric field vector and $tau$ is the average time between bounces.
Quoting from the wikipedia article, use the following substitutions
$$langle vec{p} rangle = mlangle vec{v} rangle,; vec{J} = nqlanglevec{v}rangle$$
where $m$ is the electron mass, $n$ is the number density of the charge carriers, $langle vec{v} rangle$ is the average velocity, and $vec{J}$ is the current density. This gives us:
$$vec{J} = left(frac{nq^2tau}{m}right)vec{E}$$
Now consider a resistor with the oriented cross-section area $vec{S}$. Multiply the above formula together with $vec{S}$ to get the current through $vec{S}$:
$$I = vec{J}cdotvec{S} = left(frac{nq^2tau}{m}right)vec{E}cdotvec{S}$$
For simplicity, assume $vec{E}$ and $vec{S}$ are constant and parallel, and that $|vec{E}|=Delta U/L$, where $Delta U$ is the voltage difference and $L$ is the length of the resistor. This gives us the current in terms of the voltage difference times a factor:
$$I = left(frac{nq^2tau}{m}right)frac{Delta U,S}{L} = left(frac{nq^2tau S}{mL}right)Delta U$$
This is Ohm's law if we identify $left(dfrac{nq^2tau S}{mL}right)=dfrac{1}{R}$.
Note: the above only shows the steady-state situation. It doesn't really explain how the local voltages and Kirchoff conditions reach this steady state after the voltage is applied initially. But the Drude model can also handle time-varying dynamics. Please see the Wikipedia article for details.
add a comment |
I have found a nice explanation called the Drude Model, a simple mechanical model proposed by the German physicist Paul Drude in 1900. It is based on the idea that the charge carriers bounce around randomly at high speeds, with an average momentum of
$$langle vec{p} rangle = qvec{E}tau$$
where $q$ is the charge of a carrier (i.e. the electron charge), $vec{E}$ is the electric field vector and $tau$ is the average time between bounces.
Quoting from the wikipedia article, use the following substitutions
$$langle vec{p} rangle = mlangle vec{v} rangle,; vec{J} = nqlanglevec{v}rangle$$
where $m$ is the electron mass, $n$ is the number density of the charge carriers, $langle vec{v} rangle$ is the average velocity, and $vec{J}$ is the current density. This gives us:
$$vec{J} = left(frac{nq^2tau}{m}right)vec{E}$$
Now consider a resistor with the oriented cross-section area $vec{S}$. Multiply the above formula together with $vec{S}$ to get the current through $vec{S}$:
$$I = vec{J}cdotvec{S} = left(frac{nq^2tau}{m}right)vec{E}cdotvec{S}$$
For simplicity, assume $vec{E}$ and $vec{S}$ are constant and parallel, and that $|vec{E}|=Delta U/L$, where $Delta U$ is the voltage difference and $L$ is the length of the resistor. This gives us the current in terms of the voltage difference times a factor:
$$I = left(frac{nq^2tau}{m}right)frac{Delta U,S}{L} = left(frac{nq^2tau S}{mL}right)Delta U$$
This is Ohm's law if we identify $left(dfrac{nq^2tau S}{mL}right)=dfrac{1}{R}$.
Note: the above only shows the steady-state situation. It doesn't really explain how the local voltages and Kirchoff conditions reach this steady state after the voltage is applied initially. But the Drude model can also handle time-varying dynamics. Please see the Wikipedia article for details.
I have found a nice explanation called the Drude Model, a simple mechanical model proposed by the German physicist Paul Drude in 1900. It is based on the idea that the charge carriers bounce around randomly at high speeds, with an average momentum of
$$langle vec{p} rangle = qvec{E}tau$$
where $q$ is the charge of a carrier (i.e. the electron charge), $vec{E}$ is the electric field vector and $tau$ is the average time between bounces.
Quoting from the wikipedia article, use the following substitutions
$$langle vec{p} rangle = mlangle vec{v} rangle,; vec{J} = nqlanglevec{v}rangle$$
where $m$ is the electron mass, $n$ is the number density of the charge carriers, $langle vec{v} rangle$ is the average velocity, and $vec{J}$ is the current density. This gives us:
$$vec{J} = left(frac{nq^2tau}{m}right)vec{E}$$
Now consider a resistor with the oriented cross-section area $vec{S}$. Multiply the above formula together with $vec{S}$ to get the current through $vec{S}$:
$$I = vec{J}cdotvec{S} = left(frac{nq^2tau}{m}right)vec{E}cdotvec{S}$$
For simplicity, assume $vec{E}$ and $vec{S}$ are constant and parallel, and that $|vec{E}|=Delta U/L$, where $Delta U$ is the voltage difference and $L$ is the length of the resistor. This gives us the current in terms of the voltage difference times a factor:
$$I = left(frac{nq^2tau}{m}right)frac{Delta U,S}{L} = left(frac{nq^2tau S}{mL}right)Delta U$$
This is Ohm's law if we identify $left(dfrac{nq^2tau S}{mL}right)=dfrac{1}{R}$.
Note: the above only shows the steady-state situation. It doesn't really explain how the local voltages and Kirchoff conditions reach this steady state after the voltage is applied initially. But the Drude model can also handle time-varying dynamics. Please see the Wikipedia article for details.
answered Nov 24 at 19:52
Cuspy Code
36615
36615
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f442719%2fwhats-the-reason-behind-the-current-remaining-the-same-after-passing-by-a-resis%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Because otherwise they'd build up in the resistor? How come the amount of water before a half-closed valve is the same as the amount of water after it?
– immibis
Nov 24 at 7:25
not necessarily building up on the resistor, but the current before entering the resistor or the velocity of electrons before entering the resistor is more than the velocity of them after leaving the resistor. Or maybe to elaborate my point more: 2 coulombs flow through a cross section of the wire in a second before entering the resistor, after entering it, 1 coulomb (for e.g) would flow through a cross section of that wire in a second, and the other coulomb wouldn't build up on the resistor, rather it would leave it but with a slower velocity. isn't my assumption true?
– Just_Cause
Nov 24 at 16:00
If it leaves with a slower velocity, then it's still leaving, isn't it? Coulombs per second is not velocity, it's amount of electrons per second. If the amount of electrons coming out is less than the amount of electrons going in, then where are the electrons going?
– immibis
Nov 25 at 2:54
let's forget about velocity, i used it because it plays a role in current's formula. we agree to some point that the number of coulombs that flow across a point before entering the resistor in a second, is the same amount that does the same after leaving the resistor. However, what I want to get to is that why does that really happen? electrons face obstacles inside the resistor and there must be something that played a role in getting their rate of flow back the same. V.F and Steeven for example pointed that out in different ways.
– Just_Cause
Nov 25 at 13:16