Energy conservation equation problem

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Homework Help Overview

The discussion revolves around the energy conservation equation in the context of a capacitor connected to a constant energy source, with an initial assumption of zero resistance in the circuit. The original poster attempts to reconcile the energy stored in the capacitor with the work done by the source, raising questions about energy conservation when resistance is introduced.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of assuming zero resistance in a circuit, questioning the validity of energy conservation in this scenario. Some suggest considering practical limitations, while others propose theoretical constructs involving superconductors.

Discussion Status

The discussion is ongoing, with participants providing insights into the theoretical aspects of resistance and energy transfer. There is a recognition of the complexities involved in the assumptions made, particularly regarding superconductors and the nature of energy dissipation.

Contextual Notes

Participants note the constraints of real-world applications, emphasizing that while theoretical scenarios may allow for zero resistance, practical implementations always involve some level of resistance and energy loss. The conversation also touches on the potential for electromagnetic radiation as a form of energy dissipation in idealized circuits.

panzer1234
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A capacity(C) connects directly to a constant source(E); total resistant is zero. When the capacity has been charged already, it has energy :1/2CE^2. The work done by the source
is CE^2. Comparing two of them leads to an error of the energy conservation equation. What's wrong with it?

When the total resistant is not zero, we can calculate that the energy (CE^2-1/2CE^2) will be the heat that come out from the resistant. This satisfy the energy conservation equation.

How can we explain it?
 
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welcome to pf!

hi panzer1234! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)

i'd just say that there's no such thing as a wire with zero resistance …

the wire must have some resistance, and as you say half the energy will be converted to heat in the wire :smile:

(btw, it's a capacitor and it has capacitance :wink:)
 


suppose that the wire is made from super conductor material, so its resistant is zero, or we can have sth like that. Generally, i just want to consider the problem in theorem, so what is the fault?
 
hmm …

there must be an ordinary wire connecting to each end of the superconductor, and every voltage source has an internal resistance anyway …

so I'm staying with my previous answer …

there's no such thing as a wire with zero resistance

and there's no such thing as a free lunch :biggrin:
 


I agree with u that there is no such thing as a wire with zero resistance, in practice. However, I just want to consider the problem in theorem, that means there is no resistance. I wonder if the energy come out from the circuit as an electromagnetics wave. Please take a serious look at it.
P/S: I'm not an English native speaker, so please forgive me if my language confuses you :biggrin:
 
panzer1234 said:
I agree with u that there is no such thing as a wire with zero resistance, in practice. However, I just want to consider the problem in theorem, that means there is no resistance.

(you mean "in theory" :wink:)

yes, but even in theory there's no such thing as a wire with zero resistance …

the energy will always go into heating the connecting wire :smile:
 


tiny-tim said:
(you mean "in theory" :wink:)

yes, but even in theory there's no such thing as a wire with zero resistance …

the energy will always go into heating the connecting wire :smile:


In theory one could construct the circuit from superconducting material, which really does have zero resistance. Not just nearly zero, but really and truly zero. In place of the battery, use another much larger capacitor also made from superconducting materials. Assume that its capacitance is so large that charging the smaller capacitor will not significantly change its stored charge (or voltage).

However, even a perfectly straight piece of wire fraction of an inch long has inductance... so you end up with an LC circuit that will oscillate at some frequency and its harmonics, and so radiate the "excess" energy in the form of electromagnetic radiation.
 

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