Krichhoff's law & Conservation of Energy

[person_random_normal]

more precisely
how do we explain the the preferential loss of energy of electrons in a case when we have a battery and 2 resistors serially connected ?

 

[person_random_normal]

please reply

 

[Drakkith]

more precisely
how do we explain the the preferential loss of energy of electrons in a case when we have a battery and 2 resistors serially connected ?

Imagine you have a hundred blocks arranged in a circular track, with each block in physical contact with both the block in front of and behind itself. Frictionless rails are added to keep the blocks on the track. Now, imagine that the entire track is coated with a substance that has very little friction except for a single spot , three blocks long, that is roughed up and has a very high amount of friction.

So, if you push the blocks, it takes energy to get them moving and to keep them moving, as friction is constantly stealing energy away from all of the blocks. However, the rough patch steals much more energy per unit of distance than the rest of the track does. Since the blocks are in physical contact with each other, it doesn't matter which blocks are currently in contact with the rough patch, the entire circle of blocks feels the effect. If the track's coating is very close to being frictionless, then almost all of the energy provided to the blocks is being used to move the blocks past the rough patch.

This is analogous to a resistor in an electrical circuit. A voltage source provides energy to move the charges. To move a charge through a resistor takes much more energy that it does to move a charge through a conductor. Yet, because charges respond to each others electric fields, just like the blocks respond to the contact forces between them, the effect of the resistor is felt by all of the charges in the circuit.

If you add a second resistor then it's analogous to adding a second rough patch. If the rough patch is of equal size and has the same friction as the first patch, then it requires just as much energy to move the blocks past it. This means that unless you put more energy into moving the blocks, they will slow down and you will have less 'current'.

 

[Dale]

please reply

That is a little pushy. I had to go to a meeting, you should usually not "bump" a thread in less than 24 hours.

more precisely
how do we explain the the preferential loss of energy of electrons in a case when we have a battery and 2 resistors serially connected ?

I am not completely certain what you mean by "preferential". I assume that you mean that if one resistor is larger than the other then there will be more energy lost in the large resistor than in the small resistor, and you want to know why.

Again, the E field is conservative, which means that it has a potential which is a function of position, this is the voltage. The energy of an electron is proportional to the voltage. Since the larger resistor has a larger voltage drop the energy loss of an electron is therefore larger.

Think of the analogy I proposed above. If a rock rolls down a steep part of a hill and then a shallow part of a hill then it loses more energy on the steep part because the drop in height is also greater. The potential energy of the rock at a given position on the hill depends only on the height at that position. Similarly the potential energy of the electron depends only on the voltage at that position.

 

[atyy]

that solves my doubt partly !
but
how do those electrons understand where to loose how much of energy so that not energy gained and lost sums to zero

In general, the global conservation laws of classical physics can always be enforced by local laws, so that an electron only needs information about things that are near it. (Actually this point is a bit tricky, because the notion of locality in Newtonian physics and electromagentism is not the same, but let's skip this here.)

As an example, how is momentum conservation enforced in Newtonian mechanics? How can objects know to move so that momentum is always conserved? They just need to know Newton's third law, which says if you push me hard, I will push you back just as hard. So they only need local information.

There is a tricky point (another one). Energy is always conserved in classical physics. However, Kirchoff's voltage law is not as general. It breaks down when there is a time-varying magnetic field through the circuit. This more general phenomenon is called induction and the more general law is called Faraday's law.

Faraday's law of induction
https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-loop-rule-540-5636/
https://courses.cit.cornell.edu/ece303/Lectures/lecture11.pdf

 

[person_random_normal]

Imagine you have a hundred blocks arranged in a circular track, with each block in physical contact with both the block in front of and behind itself. Frictionless rails are added to keep the blocks on the track. Now, imagine that the entire track is coated with a substance that has very little friction except for a single spot , three blocks long, that is roughed up and has a very high amount of friction.

So, if you push the blocks, it takes energy to get them moving and to keep them moving, as friction is constantly stealing energy away from all of the blocks. However, the rough patch steals much more energy per unit of distance than the rest of the track does. Since the blocks are in physical contact with each other, it doesn't matter which blocks are currently in contact with the rough patch, the entire circle of blocks feels the effect. If the track's coating is very close to being frictionless, then almost all of the energy provided to the blocks is being used to move the blocks past the rough patch.

This is analogous to a resistor in an electrical circuit. A voltage source provides energy to move the charges. To move a charge through a resistor takes much more energy that it does to move a charge through a conductor. Yet, because charges respond to each others electric fields, just like the blocks respond to the contact forces between them, the effect of the resistor is felt by all of the charges in the circuit.

If you add a second resistor then it's analogous to adding a second rough patch. If the rough patch is of equal size and has the same friction as the first patch, then it requires just as much energy to move the blocks past it. This means that unless you put more energy into moving the blocks, they will slow down and you will have less 'current'.

as you said - if the analogy is perfectly correct then - on a circular track with one rough patch , and the blocks moving around (and assuming a place on the tack which provides energy to the blocks on the track to move around)
we can say or rather , i think that as in the analogy you mentioned- when the block just encounters the rough patch , it cannot loose whole of its energy (otherwise it will stay where it is) so it partially looses its energy due to friction , some fraction of energy is transferred to the block ahead and some fraction is sustained and
so goes on with the blocks throughout the rough patch ! so that all the blocks on the track keep moving

this implies -the energy provided the blocks on our track is not lost completely in that rough patch, some energy escapes from there and needs to lost in some other form , thereafter otherwise conservation of energy is violated !
So if the analogy is perfectly correct for electrical circuits Krichhoff's voltage law is certainly wrong !
in sense if we have a simple circuit like a battery connected across a resistor, the energy provided to the electrons in the wire by the battery is not completely lost in the resistor, it needs to be lost after having escaped the resistor , some other form

I WANT YOU TO REPLY THIS
PLEASE !

 

[Drakkith]

this implies -the energy provided the blocks on our track is not lost completely in that rough patch, some energy escapes from there and needs to lost in some other form , thereafter otherwise conservation of energy is violated !

How so? The energy is still contained in the motion of the blocks unless it is dissipated by friction. If it's not dissipated by friction, then the blocks don't slow down. Energy conservation isn't violated.

we can say or rather , i think that as in the analogy you mentioned- when the block just encounters the rough patch , it cannot loose whole of its energy (otherwise it will stay where it is) so it partially looses its energy due to friction , some fraction of energy is transferred to the block ahead and some fraction is sustained and
so goes on with the blocks throughout the rough patch ! so that all the blocks on the track keep moving

Remember that I'm pushing on the blocks and supplying energy to them. The blocks on the rough patch only lose energy equal to the amount of energy being supplied by myself, so they don't accelerate or decelerate. There is nothing that says they need to lose their kinetic energy until I stop pushing.

 

[person_random_normal]

How so? The energy is still contained in the motion of the blocks unless it is dissipated by friction. If it's not dissipated by friction, then the blocks don't slow down. Energy conservation isn't violated.
Remember that I'm pushing on the blocks and supplying energy to them. The blocks on the rough patch only lose energy equal to the amount of energy being supplied by myself, so they don't accelerate or decelerate. There is nothing that says they need to lose their kinetic energy until I stop pushing.

then what is Krichhoff' voltage law in the light of the analogy you mentioned ??

 

[Drakkith]

then what is Krichhoff' voltage law in the light of the analogy you mentioned ??

It's just an analogy. Don't try to read too far into it.

 

[Dale]

then what is Krichhoff' voltage law in the light of the analogy you mentioned ??

Do you really even need an analogy? The sum of the voltage changes around any closed loop is equal to 0. How much simpler could it be?

Again, KVL is a statement about the voltage, it holds even when current is not flowing. Do you understand that?

I think you are making this way more complicated than it needs to be. For your battery-resistor circuit, the voltage at the - terminal is 0 the voltage at the + terminal is V. So if you draw a loop starting at ground and going in the direction of the current you get a change of V across the battery and a change of -V across the resistor. V + (-V) = 0. You can also draw your loop the opposite direction of the current so that the change of voltage across the resistor is V and the change of voltage across the battery is -V. Again V + (-V) = 0.

 

[person_random_normal]

It's just an analogy. Don't try to read too far into it.

very very precisely I want to understand -
after switching on the circuit - at that very moment we have some kind of chaotic situation for few nanoseconds , before a steady state (following all rational rules like conservation of energy) establishes !
what happens in that time, which leads to the steady state ??

 

[Dale]

very very precisely I want to understand -
after switching on the circuit - at that very moment we have some kind of chaotic situation for few nanoseconds , before a steady state (following all rational rules like conservation of energy) establishes !
what happens in that time, which leads to the steady state ??

https://en.wikipedia.org/wiki/Maxwell's_equations

 

[person_random_normal]

i want that to be elucidated in a simple way
hence i am here !

please !

 

[Drakkith]

very very precisely I want to understand -
after switching on the circuit - at that very moment we have some kind of chaotic situation for few nanoseconds , before a steady state (following all rational rules like conservation of energy) establishes !
what happens in that time, which leads to the steady state ??

Obviously current begins to flow, but beyond that I'm not sure what you want to know. There are many different things happening, all of which have their own description, so you're going to have to give us a more specific question.

 

[Dale]

i want that to be elucidated in a simple way
hence i am here !

please !

Maxwell's equations are the simplest model for the time scales that you are now asking about. The alternative is quantum electrodynamics.

In order to avoid Maxwell's equations you need to look at time scales which are large compared to the speed of light so that the "steady state" is already established. Then you get KVL and KCL. Specifically, one of the key simplifying assumptions of circuit theory is that all EM effects propagate instantaneously. Sometimes it is called the "small circuit" assumption meaning that the circuit is small compared to the wavelengths of interest.

You are literally going out of your way to make things more complicated than they need to be. Simply use KVL. What could be easier?

 

[maline]

Think about a simpler case: a bunch of charge (let's say negative) is added to one end of a large, funny-shaped conducting object. Within a tiny fraction of a second, the extra electrons will arrange themselves all over the surface of the object in one very specific pattern: the one that gives the whole object the same electric potential. Do you understand how the electrons "know" how to find this arrangement? The general principle is the same.

 

[maline]

In the case of a conductor with no emf, the way the electron find the steady state solution is that whenever the voltage in one region is too low (say), this creates, or rather implies the existence of, a electric field that pushes electrons away from that region. This continues until we have a constant potential. That's what "steady-state solution" means: an arrangement such that the forces created do not change the arrangement.
In the case of a circuit of emf's and resistors, the potential at each point still depends on how the charge densities are arranged. Charges end up in stationary positions along the surfaces of the wires, and especially at the ends of each conductor, in the arrangement that creates the steady state voltages at each point in the circuit. The difference is what the steady state looks like. With the conductor, a steady solution means no current. In a circuit, steady means that the current is the same at all points along the loop, so that the charge density at any point will not change. This is achieved when the voltage drop across each resistor in the loop is proportional to the resistance.

 

[vanhees71]

very very precisely I want to understand -
after switching on the circuit - at that very moment we have some kind of chaotic situation for few nanoseconds , before a steady state (following all rational rules like conservation of energy) establishes !
what happens in that time, which leads to the steady state ??

In the quasistationary limit (which is justified for not too quickly changing fields) you can write down the differential equation for switching on the circuit. In order to make this sensible, you have to take into account the self-induction of the circuit. So we assume we have a series of a resistor and and ideal coil with self-inductance $L$. Then the equation of motion reads
$$L \dot{I}+R I=U_0,$$
where $U_0$ is the constant voltage of an ideal voltage source.

The solution of this linear first-order ode with constant coefficients is straight-forward. First you solve the homogeneous equation:
$$L \dot{I}+R I=0.$$
This you do with the standard ansatz
$$I(t)=I_0 \exp(\lambda t).$$
Plugging this into the homogeneous equation, leads to the algebraic equation
$$L \lambda+R=0 \; \Rightarrow \; \lambda=-\frac{R}{L}.$$
The general solution of the homogeneous equation thus reads
$$I(t)=I_0 \exp(-R t/L).$$
Now all you need in addition is a special solution of the inhomogeneous equation, which you can get by the ansatz of the type of the right-hand side, i.e., by
$$I=I_{\infty}=\text{const}.$$
Plugging this ansatz into the inhomogeneous equation, you get
$$R I_{\infty}=U_0 \; \Rightarrow \; I_{\infty}=\frac{U_0}{R}.$$
Now the general solution of the inhomogeneous equation is the sum of the general solution of the homogeneous and the just found particular solution of the inhomogeneous equation
$$I(t)=\frac{U_0}{R} + I_0 \exp \left (-\frac{R}{L} t \right).$$
Finally, we fix the integration constant $I_0$ by fulfilling the initial condition, $I(0)=0$, which leads to $I_0=-I_{\infty}$. Thus the final solution of your problem is given by
$$I(t)=\frac{U_0}{R} \left [1-\exp \left (-\frac{R}{L} t \right) \right ].$$
For a derivation of the laws underlying this derivation, see my Texas A&M Lecture notes (part III)

http://fias.uni-frankfurt.de/~hees/physics208.html

 

[Dale]

I just saw this very nice figure on the wiki page

https://en.m.wikipedia.org/wiki/Poynting_vector#Interpretation