# Krichhoff's law & Conservation of Energy

1. Jul 19, 2015

### Shreyas Samudra

Krichhoff's voltage law (kvl) is said to be conservation of energy but i couldn't get a satisfactory explanation for that,
i want to say -
say, we have a simple circuit consisting of a battery(of emf E) and a resistor(of resistance R), so having connected them by ideal wires, we have electrons in the wire which sense the potential difference of the battery, and hence get some sort of energy
then they move in the wire till they encounter the resistor,and then as krichhoff's law says the formerly energized electrons experience equal and -ve potential drop due to resistor so that net potential drop/ gain in the loop is zero.
so i think this can be interpreted as - the resistor consumes all of the energy of the electrons provided to them by the battery and converts that to heat !!!!
but the pitfall here in this logic is what happens to those electrons then , after they leave the resistor ??
i couldn't think of that
so is it correct or no ??
and another thing is - it cannot even capacitive circuits
i would like to have the answer on microscopic understanding , a classical point of view. i don't understand quantum mechanics

2. Jul 19, 2015

### Staff: Mentor

A force which conserves energy is called a conservative force, and such forces can be written as the gradient of a potential. In the case of electric circuits, that potential is called voltage.

Potentials have the property that the net change in the potential is 0 around any closed path. For voltage this gives KVL.

It is not necessary that any individual electron actually travel the whole closed path, only that the net change in the potential be 0 around the loop.

3. Jul 19, 2015

### vanhees71

Have a look at the lecture notes on (sorry for the hand-writing)

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

In Part III you find the circuit theory. There are also some extra worked-out examples.

4. Jul 19, 2015

### USeptim

The electrons lose the same amount of energy in the resistance plus wire that the energy they get in the battery. The electrons leaving the resistance may leave the loop being replaced by other electrons, anyway electrons are indistinguishable except by their quantum numbers.

With a capacitor the situation is the same but a time dependence is introduced.

5. Jul 20, 2015

### Shreyas Samudra

in nutshell i want to say -
battery gives energy to electrons
those electrons loose same energy in resistor (as heat)
so leaving the resistor what happens to those electrons(in general)
how do they go up to + ve terminal of the battery ??
what energy do they have to do that ?
as per KVL those electrons have same energy as they had when battery wasn't present to create potential difference (net change in energy is zero-KVL)

6. Jul 21, 2015

### vanhees71

As I've derived in another thread by a very simple argument (Drude model), the electrons drift due to the electric field. Together with some friction due to scattering they reach a constant limiting speed (in the DC case). This leads to a microscopic classical model for the derivation of the electric conductivity. A full treatment is very complicated. You'd need QED ad finite temperatures and linear-response theory to treat it on the most fundamental level.

7. Jul 21, 2015

### tommyxu3

In the whole circuit, with the push of the emf, all the electrons are keep going.

8. Jul 21, 2015

### Shreyas Samudra

so
can you please illustrate that in a simple with more of english than mathematics , please TRY IT
I am eager for that !!!

9. Jul 21, 2015

### vanhees71

This is utmost simple math. Write down the equation of motion for an electron with the force given by linear friction and the force due to the electromagnetic field
$$m\ddot{x}=-m \gamma \dot{x}-e E.$$
In the stationary limit, $\dot{x}=\text{const}$, i.e., $\ddot{x}=0$ you get
$$m \gamma \dot{x}=-e E \; \Rightarrow \; \dot{x}=-\frac{e}{m \gamma} E.$$
With the number density of the conduction electrons $n$, the current density is given by
$$j=-e n v=\frac{n e^2}{m \gamma} E \; \Rightarrow \; \sigma=\frac{n e^2}{m \gamma}.$$
More English spoils the clarity of the argument! :-)).

10. Jul 21, 2015

### Shreyas Samudra

i know this
i fear that you have not understood my question
i am simply endeavouring to prove/ visualize or feel KVL

11. Jul 21, 2015

### vanhees71

Hm, perhaps somebody else with more didactical experience can help better :-(.

12. Jul 21, 2015

### Shreyas Samudra

oh , please just try it , waiting for somebody might take too long
do i again state my doubt (more specifically)

13. Jul 21, 2015

### Shreyas Samudra

14. Jul 21, 2015

### vanhees71

But I don't understand the question!

15. Jul 21, 2015

### Shreyas Samudra

simply

how can we prove KVL

16. Jul 21, 2015

### vanhees71

The proof of Kirchhoff's Laws is as follows. You assume circuits with a spatial extension small against the wavelength of the electromagnetic fields (i.e., low frequencies) so that the quasistationary Maxwell equations are good enough, i.e., you neglect the Maxwell "displacement current" in the Ampere-Maxwell Law simplifying it to the Ampere Law. Then you integrate the Maxwell equations along the wire, using Stokes's Law.

See my Texas A&M Lecture Notes for the details. They were well received by 2nd semester engineering students. The only obstacle is that they are handwritten:

http://fias.uni-frankfurt.de/~hees/physics208/phys208-notes-I.pdf
http://fias.uni-frankfurt.de/~hees/physics208/phys208-notes-II.pdf
http://fias.uni-frankfurt.de/~hees/physics208/phys208-notes-III.pdf
http://fias.uni-frankfurt.de/~hees/physics208/phys208-notes-IV.pdf

Kirchhoff's laws for AC can be found in Part III. There are also some worked-out examples:

http://fias.uni-frankfurt.de/~hees/physics208/RL-circuit.pdf
http://fias.uni-frankfurt.de/~hees/physics208/RC-circuit.pdf
http://fias.uni-frankfurt.de/~hees/physics208/CL-circuit.pdf

17. Jul 21, 2015

### Staff: Mentor

What was wrong with post 2? It is hard to help if you don't even bother to comment meaningfully on the responses you have already received.

18. Jul 22, 2015

### Shreyas Samudra

i am very sorry for that

so you meant - moving a test charge in the circuit- net work done on it will be zero.
so does it mean that energy provided to electrons in the wire by battery is equal and opposite to that , which dissipated as heat in resistor ??
and if that is the case the energy left with electrons after leaving the resistor will be the same as it was, when the wires had no battery ,resistor connected across.(wires were just a bundle of wires- kept away from anything!)
so electrons, having left the resistor electrons have what energy that drives them back to the +ve terminal of the battery ??????

19. Jul 22, 2015

### Staff: Mentor

Yes, but note that a "test charge" is not an actual electron in the actual current. It is a hypothetical charge that can be moved around the circuit at will subject to a hypothetical external force. It's only purpose is in establishing the potential. Once the potential is determined you have no more need of the hypothetical test charge and you simply deal directly with the potential.

For example, consider a series RC circuit driven by a battery. In steady state there is no current, but KVL still holds. Even though no actual charges are moving, if you had a test charge on a stick you would find that it would take a certain amount of energy to move it to different points on the circuit. From that you would establish the potentials.

Then removing the test charge you still have KVL even though no current is flowing. KVL only says that the changes in potential (voltage) is 0 around any loop, not that charges need to be moving around the loop.

20. Jul 22, 2015

### Shreyas Samudra

i understand that but
what this mean -
''increase in electron energy in any closed loop in zero, if that were not true laws of thermodynamics might have been violated''
can you illustrate this using a simple battery-resistor circuit ??

21. Jul 22, 2015

### Staff: Mentor

If a different amount of energy were dissipated by the resistor than were provided by the battery, then the first law of thermodynamics (conservation of energy) would be violated.

22. Jul 22, 2015

### atyy

The wires are idealized as "superconducting", so the electrons don't need any energy to go along them. Real wires have resistance distributed along their entire length, and electrons lose energy along the entire length of the wire. But here an approximation is made in which all the resistance is lumped in one place, and the rest of the wires are "superconducting".

23. Jul 22, 2015

### stedwards

See if you can generalize my simple, following example to a more general network.

A constant voltage source applies $V$ volts, and supplies $I$ current to two restistors $R_1$ and $R_2$ in series. The supply power is $P=IV$. This better show up as the power dissipated in the two resistors or energy conservation is violated.

The dissipated power in $R_1$ and $R_2$ is $V^2/R_1 +V^2/R_2$

Apply the voltage divider rule to calculate the voltage across each resistor to find $V_1$ and $V_2$, then calculate the sum of the dissipated power from both, together. It should equal the amount delivered by the voltage source.

The power is constant, so the energy delivered or dissipated at some particular time is also constant.

Last edited: Jul 23, 2015
24. Jul 23, 2015

### Shreyas Samudra

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

Last edited: Jul 23, 2015
25. Jul 23, 2015

### Staff: Mentor

Huh? The electrons don't have to understand anything. They have potential energy by virtue of their location. More specifically, by virtue of the value of the potential at that location.

Would you say that a rock on a hill has to "understand" anything to have PE or to lose it as it rolls down the hill?