Without prefix:What is the explanation for Kirchhoff's Voltage Rule?

[Bashyboy]

Hello,

I was reading the article given in this link http://www2.ignatius.edu/faculty/decarlo/kirchhoff.htm, of which pertains to Kirchhoff's Rules.

In this article, these claims are made:

"1. Voltage Rule. This is based on the conservation of energy: 'the sum of voltages around a closed conducting loop (that is, a circuit) must be zero'

In other words, since voltage and work are related, we are saying the net work done must be zero."

I understand that the battery provides the electrons with energy, that they may traverse the circuit; in addition, I understand that when the electrons (which make up the electric current) flow through the resistor collisions between the electrons and the resistor occur, which results in the electrons losing energy, corresponding to the electric potential drop. However, I don't understand why it's necessary that all of the energy is used up as the electrons flow through the circuit, isn't it possible that some (perhaps even all of them) have some remaining energy when they come to the threshold of the positive terminal?

As I attempted to answer this question, I came across this article: http://farside.ph.utexas.edu/teaching/302l/lectures/node59.html

They, too, make a claim that I am unsure of:

"This rule [voltage rule] is also easy to understand...zero net work is done in slowly moving a charge around some closed loop in an electrostatic field."

Why is zero net work done?

I'd massively appreciate it if someone could help me resolve these questions.

Thank you for your time.

 

[tiny-tim]

Hello Bashyboy!

However, I don't understand why it's necessary that all of the energy is used up as the electrons flow through the circuit, isn't it possible that some (perhaps even all of them) have some remaining energy when they come to the threshold of the positive terminal?
…
"This rule [voltage rule] is also easy to understand...zero net work is done in slowly moving a charge around some closed loop in an electrostatic field."

Why is zero net work done?

energy is conserved

where do you think the energy goes?

(if, as in your example, it isn't used up, then it's still conserved, isn't it?)

since energy is conserved, that means the electric force is conservative

and voltage = electric potential = potential energy per charge

which by definition of potential energy is:

minus the work done by a conservative force, per charge​

 

[Bashyboy]

Well, doesn't some of the charge go into heating up the resistor?

 

[tiny-tim]

Well, doesn't some of the charge go into heating up the resistor?

Yes, some of the energy goes into heating up the resistor.

(the charge of course is constant)

The potential drop across the resistor (V = IR) is a measure of the heat …

it tells you where the energy has gone.

The battery puts energy into the loop, the resistors take energy out of the loop, the whole thing balances.

All the energy is accounted for by KVL.​

 

[Bashyboy]

I still don't see how to answer my original problems. For my first question, isn't it possible that an electron is given energy by the battery, it flows through the resistor and loses some energy--but not all of it--, and goes through the entire circuit, coming to the positive terminal with some remaining energy?

 

[stevendaryl]

I still don't see how to answer my original problems. For my first question, isn't it possible that an electron is given energy by the battery, it flows through the resistor and loses some energy--but not all of it--, and goes through the entire circuit, coming to the positive terminal with some remaining energy?

Sure, that's possible. That would be a transient situation, though. When you first connect a battery to a circuit, the current through the battery isn't immediately equal to the steady-state value; instead, the current will increase from zero to the steady-state value. The steady-state value is the value for which the increase in an electron's energy caused by the battery is exactly equal to the decrease in its energy caused by the resistor. Kirkhoff's laws are approximations that ignore those sort of transient changes.

 

[tiny-tim]

… an electron is given energy by the battery … coming to the positive terminal with some remaining energy?

the battery is like the motor in a car

the resistances are like the friction in the bearings, the wind resistance, etc

when the battery, or the motor, has been running for a short time, equilibrium is reached

the speed of the car adjusts so that the energy lost to friction, wind resistance, etc, exactly equals the energy supplied by the motor

the current from the battery adjusts so that the energy lost in the resistances exactly equals the energy supplied by the battery

EDIT: stevendaryl already said it!

 

[Bashyboy]

Does this somehow explain why the net work is zero?

 

[DrZoidberg]

The battery isn't throwing the electrons into the wire. It's not the case that the electrons move into the wire at high speed and then gradually slow down on their way to the other end. Their speed is the same everywhere in the wire assuming the diameter of the wire is equal everywhere. On their journey the electrons are constantly giving of energy to the metal atoms. Despite of that their speed doesn't change because they also are constantly receiving energy from the electric field produced by the battery. The electrons move very very slowly, so their kinetic energy is virtually zero. They can't have any energy left when they reach the end of the wire because they never really had any energy to begin with. When you use a rope to pull a heavy object you have to put a lot of energy into overcome the friction but the rope never really had much energy since it's very light and moves slowly. So when you stop pulling will the rope have any energy left in it?

"This rule [voltage rule] is also easy to understand...zero net work is done in slowly moving a charge around some closed loop in an electrostatic field."

Why is zero net work done?

This rule does not apply to an electric field that's induced by a changing magnetic field e.g. a transformer. In that case charge that moves around the transformer core in a closed loop will have work done on it. Such a field is called "non conservative". There is a great video about that topic here


Most electric fields however are conservative fields. That means that in a closed loop no net work is done on a charge. That is because of conservation of energy. Imagine you had two objects that are statically charged and then you take a third charged object and let it move in a circle between the other two. If it was receiving energy on it's path it could continue to circle forever and you would get energy from nothing.

 

[xAxis]

"This rule [voltage rule] is also easy to understand...zero net work is done in slowly moving a charge around some closed loop in an electrostatic field."

Why is zero net work done?

I'd massively appreciate it if someone could help me resolve these questions.

Thank you for your time.

I honestly think that it is confusing to explain the voltage rule in circuits with the work by charge in electrostatic field. At least one shouldn't think they are literaly the same. Rather, the charge moving in closed loop in the electric field is just analogy to charge moving across the closed circuit.
But I might be wrong.

 

[sophiecentaur]

Sure, that's possible. That would be a transient situation, though. When you first connect a battery to a circuit, the current through the battery isn't immediately equal to the steady-state value; instead, the current will increase from zero to the steady-state value. The steady-state value is the value for which the increase in an electron's energy caused by the battery is exactly equal to the decrease in its energy caused by the resistor. Kirkhoff's laws are approximations that ignore those sort of transient changes.

Using the idea that electrons are flowing around a circuit (although there is a mean drift speed, of course) in conjunction with the idea of a "transient situation" can hardly make sense. During any 'switch on' settling time (as little as a few ns duration for some circuits) an electron would hardly have traveled by any significant distance (10^-12m, typically). This is just another example of how discussing electric currents in terms of electrons can be nonsensical. Just stick to Current and 'charges' and worry about electron movement when it is relevant - like inside solid state and thermionic devices (black boxes when doing circuit analysis).

 

[sophiecentaur]

Does this somehow explain why the net work is zero?

Yes it does. If all the energy supplied by the battery were not dissipated or transferred within the circuit, where would it go or where would any extra energy come from?

 

[sophiecentaur]

I honestly think that it is confusing to explain the voltage rule in circuits with the work by charge in electrostatic field. At least one shouldn't think they are literaly the same. Rather, the charge moving in closed loop in the electric field is just analogy to charge moving across the closed circuit.
But I might be wrong.

Absolutely. A 9V battery, 0.02m long would have a field of 9/0.02V/m across its terminals. Another, physically bigger, battery - say 0.1m long, would have 1/5 of the electric field across its terminals. How could anyone invent a system for circuit analysis that worked on the basis that different batteries of the same voltage would produce different resuts. (Not to mention the lengths of the connecting wires in a DC circuit)
We use Volts, Potential Difference and Energy transferred etc etc because it works and Volts per Metre wouldn't work.
Volts per Metre becomes relevant in many instances, such as electron tubes and particle accelerators but it's 'horses for courses' in Science.

 

[stevendaryl]

Using the idea that electrons are flowing around a circuit (although there is a mean drift speed, of course) in conjunction with the idea of a "transient situation" can hardly make sense.

I don't think you're really addressing his question. He's asking about the claim that conservation of energy implies that the changes in voltages around a circuit add up to zero. It's true that for the domain of applicability of Kirkhoff's laws that the voltages add up to zero, but it doesn't follow from conservation of energy.

 

[sophiecentaur]

I don't think you're really addressing his question. He's asking about the claim that conservation of energy implies that the changes in voltages around a circuit add up to zero. It's true that for the domain of applicability of Kirkhoff's laws that the voltages add up to zero, but it doesn't follow from conservation of energy.

But his question involves an assumption that there is a reasonable answer in terms of electrons buzzing through the wires. I don't think there is one in those terms.

 

[stevendaryl]

Yes it does.

No, it doesn't.

If all the energy supplied by the battery were not dissipated or transferred within the circuit, where would it go or where would any extra energy come from?

Conservation of energy just says that the energy drained from the battery must equal the energy in the circuit minus the energy dissipated away by the resistors. To get that the energy drained from the battery is equal to the energy dissipated away, you have to assume that the energy in the circuit doesn't keep getting greater.

Mathematically, energy is being supplied by the battery at the rate

$I V_{battery}$

Energy is being dissipated by the resistor at the rate

$I^2 R$

To get these two equal, you need the assumption that the total energy in the circuit doesn't change with time. That will not absolutely true for all times, but it will be true to a very good approximation soon after the battery is connected.

 

[sophiecentaur]

I don't understand why it's necessary that all of the energy is used up as the electrons flow through the circuit, isn't it possible that some (perhaps even all of them) have some remaining energy when they come to the threshold of the positive terminal?[/B]

I just read this part of your OP. If there were any resistance in the contact with this terminal then the voltage at this point would not be zero. (And, of course, all real batteries do have some internal resistance so, once you start to drain current from them, some energy is, in fact, dissipated internally, so there is some energy 'left' after the charge has 'passed around' a real circuit.
I wonder if part of your problem with this is that you are thinking in terms of the 'electrons' carrying the energy round the circuit in the form of Kinetic Energy? This is a common misconception which can be resolved once one considers their 1mm/s mean drift speed and the total mass of moving electrons in a wire (1/120,000 of the mass of a copper wire they are moving through).

 

[sophiecentaur]

SO the energy in the circuit turns up in the form of magnetic or electric fields and they need to be included in the Kirchoff analysis. But K2 only applies where there is a conservative field. The OP was referring to the simple model of a battery and some resistors. Adding extra factors can't help in explaining the basic concept, surely.

 

[stevendaryl]

But his question involves an assumption that there is a reasonable answer in terms of electrons buzzing through the wires. I don't think there is one in those terms.

His question was why does conservation of energy imply that the sum of voltage changes around a circuit must add up to zero? He was trying to understand this by using a picture of electrons moving through the circuit, but his question is perfectly valid, however you try to understand it. How does one prove, from conservation of energy alone, that the sum of voltage changes around a loop add up to zero? You can't prove it from conservation of energy alone. If you let $U_{battery}$ be the internal energy of the battery, and $U_{circuit}$ be the internal energy of the circuit, then conservation of energy tells us that:

$\dfrac{d}{dt} U_{total} = - I^2 R$

where $U_{total} = U_{battery} + U_{circuit}$.

The additional assumption that's needed is something along the lines that
$U_{circuit} = 0$ for a perfect conductor. There is no energy associated with having current flowing through a wire. It's not exactly obvious that that's true, and I don't think it is precisely true, but it's approximately true, because as you say, the charges in a conductor don't move very far at all, so there's really no way to build up kinetic or potential energy inside a conductor.

 

[stevendaryl]

I just read this part of your OP. If there were any resistance in the contact with this terminal then the voltage at this point would not be zero. (And, of course, all real batteries do have some internal resistance so, once you start to drain current from them, some energy is, in fact, dissipated internally, so there is some energy 'left' after the charge has 'passed around' a real circuit.
I wonder if part of your problem with this is that you are thinking in terms of the 'electrons' carrying the energy round the circuit in the form of Kinetic Energy? This is a common misconception which can be resolved once one considers their 1mm/s mean drift speed and the total mass of moving electrons in a wire (1/120,000 of the mass of a copper wire they are moving through).

It's not a "misconception", it's a question. He's essentially asking whether there can be energy in the form of kinetic energy of charges in a circuit that increases with time as the battery is connected. The answer is that the additional kinetic energy due to drift velocity is completely negligible in ordinary circuits, and this energy is therefore ignored in Kirchoff's laws.

 

[sophiecentaur]

I wouldn't argue with that at all. But trying to start an explanation (not yours) from a flawed basis (i.e. that the motion of the electrons is relevant) is not going to be helpful for anyone. Quasi mechanical pictures of circuit behaviour can only get in the way of a proper understanding and I think it's always worth while trying to steer people away from them.
It's this nagging problem of 'some energy being left over' which is so attractive - particularly if you have a picture of stuff flowing round a circuit like water (which, of course, always has some energy left after going through a water wheel etc - because it has to get out of the way to make room for the next lot coming out).

 

[sophiecentaur]

It's not a "misconception", it's a question. He's essentially asking whether there can be energy in the form of kinetic energy of charges in a circuit that increases with time as the battery is connected. The answer is that the additional kinetic energy due to drift velocity is completely negligible in ordinary circuits, and this energy is therefore ignored in Kirchoff's laws.

I think the OP can answer for himself, whether it is a question or a question based on a misconception. But, as we seem to be agreeing about the actual facts, what's the problem?

 

[stevendaryl]

I wouldn't argue with that at all. But trying to start an explanation (not yours) from a flawed basis (i.e. that the motion of the electrons is relevant) is not going to be helpful for anyone.

Sure, but to me, the appropriate response was something like: applying a voltage across a wire makes a negligible change in the kinetic energy of the charges in the wire, for such and reason.

In a certain sense, though, the reason drift velocity is so low is because the conductor isn't perfect, it has resistance.

 

[sophiecentaur]

@stevendary
OK, I'll use you as my sub-editor in future.

 

[sophiecentaur]

Sure, but to me, the appropriate response was something like: applying a voltage across a wire makes a negligible change in the kinetic energy of the charges in the wire, for such and reason.

In a certain sense, though, the reason drift velocity is so low is because the conductor isn't perfect, it has resistance.

The drift velocity is a consequence of the charge density and the total rate of charge transfer.
See this link (or any textbook)
The resistance is not relevant.

 

[Bashyboy]

I haven't read all of the posts, but stevendaryl was correct: my question was, "why does conservation of energy imply that the sum of voltage changes around a circuit must add up to zero?" In my query, I proposed the situation with the electron, hoping it was correct and that someone might be able to answer my questions in terms of that situation. But, if that wasn't possible, I was fine with that.

 

[stevendaryl]

The drift velocity is a consequence of the charge density and the total rate of charge transfer.
See this link (or any textbook)
The resistance is not relevant.

That doesn't make any sense to me. By "total rate of charge transfer", do you mean "current"? If so, then current is inversely proportional to the resistance.

In this article, the claim is made that drift velocity is proportional to the mean-free path d between collisions.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html

The conductivity (the inverse of the resistance) is similarly proportional to the mean free path, and so we have:

$v_d \propto \dfrac{1}{R}$

So what you said is, I believe, completely wrong.

 

[stevendaryl]

I think the OP can answer for himself, whether it is a question or a question based on a misconception. But, as we seem to be agreeing about the actual facts, what's the problem?

The problem I have is that after reading your post, one feels more confused than when he started, and yet feels embarrassed to ask clarifying questions, for fear of being ridiculed. That's just the impression I get.

 

[Bashyboy]

I haven't read all of the posts, but stevendaryl was correct: my question was, "why does conservation of energy imply that the sum of voltage changes around a circuit must add up to zero?" In my query, I proposed the situation with the electron, hoping it was correct and that someone might be able to answer my questions in terms of that situation. But, if that wasn't possible, I was fine with that.

stevendaryl, have you seen my message that I posted, just moments before you posted yours?

 

[stevendaryl]

stevendaryl, have you seen my message that I posted, just moments before you posted yours?

I think that the answer is that there is negligible energy in the circuit in the form of kinetic energy of charges, because the drift velocity is so small, typically.

Conservation of energy would tell us that:

$\sum_i P_i = \dfrac{dU}{dt}$

where $P_i$ is the power generated or consumed by component $i$, and $U$ is the energy in the circuit in the form of kinetic energy of charges. So since the kinetic energy of charges is negligibly changed by applying a voltage, we can approximate:

$\dfrac{dU}{dt} = 0$

So we get:

$\sum_i P_i = 0$

The power $P_i$ generated or consumed by component $i$ is just equal to $I_i V_i$, where $I_i$ is the current through component $i$, and $V_i$ is the voltage change from one end of the component to the other. So we have:

$\sum_i I_i V_i = 0$

For a circuit without junctions that split the current, the current is the same for all components, so we would have:

$\sum_i I_i V_i = I \sum_i V_i = 0$
$\Rightarrow \sum_i I V_i = 0$

To extend the argument to the case of circuits with junctions is more work, but I assume something similar applies.