I was reading about LC Oscillations in my book and came upon this:

In actual circuits,there is always some resistance and some energy is therefore transformed to internal energy.

Why internal energy? I had read that resistances only lose energy in the form of heat. Is that what the authors are talking about? Do they mean to say that this internal energy is first stored in the resistor in the form of internal energy and then dissipated as heat?

I did a google search on Internal energy of resistors and got no specific results. So I assumed that the book was wrong. But then it is mentioned many a times in the book.

  • $\begingroup$ This is a good question. By the way, in English, each sentence is separated from the previous one by a period or question mark and then one space. I edited the post to fix it. $\endgroup$ – DanielSank Jan 13 '16 at 17:47
  • $\begingroup$ Yes, resistors will transform electrical energy to heat, which is considered "internal", however, you will not find many treatments of electrical circuits in terms of thermodynamics. The reason for that is because electrical circuits are extremely far away from thermal equilibrium and thermodynamics has very little useful things to say about that. The correct way to treat this is with the en.wikipedia.org/wiki/Fluctuation-dissipation_theorem and by learning how to calculate noise power/voltage and current for resistive components. I think either is a bit too advanced at your stage. $\endgroup$ – CuriousOne Jan 13 '16 at 18:18
  • $\begingroup$ @CuriousOne Nice, although I still don't understand why so many of your contributions never make it to the answer box. $\endgroup$ – DanielSank Jan 13 '16 at 18:24
  • 1
    $\begingroup$ @CuriousOne I see. By the way, you may find this other post amusing. $\endgroup$ – DanielSank Jan 13 '16 at 18:30
  • 1
    $\begingroup$ @DanielSank: The pretty thing about the second law of thermodynamics is that one can get all these answers for absolutely free without having to understand any of this and without even mentioning entropy once. It's just so trivial and so broad that people don't like to hear it. They are longing for some man behind the curtain, even a century and a half after the curtain has been removed by Clausius. $\endgroup$ – CuriousOne Jan 13 '16 at 19:01

I had read that resistances only lose energy in the form of heat.


Is that what the authors are talking about?

Yes, the author is just using confusing language.

Do they mean to say that this internal energy is first stored in the resistor in the form of internal energy and then dissipated as heat?

The author means that electrical energy is converted to heat in the material of the resistor. This is a bit confusing because heat has some different properties as compared with the electrostatic charge energy stored in a capacitor or the magnetostatic flux energy stored in an inductor. The electrostatic or magnetostatic energies are still in a form which can do work on other parts of the circuit. Heat, on the other hand, is for the most part lost. It's not entirely lost because you can build an engine based on heat differences; for example you can boil water to spin a turbine like in a nuclear power plant. However, in most cases the heat energy is no longer a major part of the dynamics of the circuit itself.$^{[a]}$ This is the case because heat energy is the motion of the atomic ion cores in the material of the resistor vibrating around, whereas the circuit operates on the electrical energy of the electrons.

You can verify that resistors generate heat easily by observing an electric stove. The stove top is just a big resistor, and it works by putting current through it to generate heat.

$[a]$: As mentioned by CuriousOne in the comments, heat in resistive elements actually does interact with the electrical dynamics of the circuit in the form of noise. This was first discovered by J.B. Johnson and then theoretically explained by H. Nyquist. The phenomenon is called Johnson-Nyquist noise. Johnson-Nyquist noise is fascinating and critical in understanding the role of electrical circuits in thermodynamic systems, as explained in another SE post.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.