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Where did it go?by Martin
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#37
Nov905, 04:33 AM

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#38
Nov905, 09:58 AM

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This problem has a mechanical analog. The inelastic collision between two bodies, where momentum is conserved and kinectic energy not.
Before collision the total momentum is [tex]p = mv + 0 = mv[/tex] and the kinectic energy is [tex]W = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2[/tex] After collision the total momentum is [tex]m\frac{v}{2} + m\frac{v}{2} = mv[/tex] while the kinectic energy is [tex]W = \frac{1}{2}m\left(\frac{v}{2}\right)^2 + \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{4}mv^2 [/tex] We consider that work is done in deforming the bodies and this accounts for the missing energy, that is turned into heat. If the two bodies are rigid there can be no inelastic collision. The problem with the two capacitors is that there is no mechanism that can account for the missing energy. The model just does not work, in the same way that the rigid body model is unsuitable to an inelastic collision. 


#39
Nov905, 09:03 PM

P: 31

The model is just that: A model. It is self consistent. It has a set of rules (Kirchoff’s Laws) that must be satisfied at every node and around every loop. Each lumped circuit element is specified by unambiguous terminal relationships. In the circuit presented, the model predicts an impulse of current at t=0, and an instantaneous energy dissipation. The model doesn’t “know” how well its predictions mirror the corresponding “real” circuit that we are using it to represent—it predicts what it predicts, and does so such that all of its “rules” are obeyed. The fact that the model predicts an impulse of current (i.e., we didn’t apply an impulse of current to the circuit) and that an impulse of current (or an impulse of any physical quantity) cannot occur in the “real world,” and that the model predicts an energy loss even though there is no nonzero resistance in the circuit to account for dissipation tells us that the model is attempting to predict the behavior of a real circuit in an extreme situation. We have pushed the use of the model as an accurate representation of a “real” circuit up against its limits. Specifically, we have ignored the small inductance and resistance that would exist in a configuration of “real” capacitors, a switch, and wires. Why did we do that? We did that because in most “real” configurations that involve “real” circuit elements, ignoring the “incidental” resistances, inductances, and capacitances is legitimate: They are indeed very small quantities compared to the “real” circuit elements themselves. Apparently, then, we can’t directly make such an assumption for this particular configuration. However, we may be able to indirectly do so. We have to ask: What is this “ideal” circuit trying to tell us? To answer that, we do the simplest thing that we can do: Insert a resistance in the circuit, and do the analysis to determine the relationship between the current and resistance and between the energy dissipated and the resistance; and then we examine how the current and energy dissipation change as we allow the resistance to shrink to zero. In other words, we take as the solution to our resistancefree problem the solution to an analogous problem that includes a resistance, in the limit as that resistance goes to zero. Applying the concept of limits to arrive at solutions that otherwise would escape our grasp is pretty standard stuff. 


#40
Nov905, 09:12 PM

P: 31




#41
Nov1005, 03:19 AM

P: n/a

Back to the mechanical example, the collision makes appear a force [tex]f = m\frac{dv}{dt}[/tex] in the same way that the charge and discharge of the capacitors give origin to a current [tex]i = C\frac{dV}{dt}[/tex]. In both cases, if dt tends to zero, the derivative becomes an impulse. The difference is that in the mechanical case dt is only zero in an elastic collision. In this case both momentum and kinectic energy are conserved. Lumped circuit theory has no correspondent element to a rigid body. Is this a failure of the theory? No, the failure is to use it to model things it is not supposed to do. 


#42
Nov1005, 08:13 PM

P: 31

“Lumped circuit theory is valid in the sense that it provides useful approximate results whose accuracy depends upon the degree to which its underlying assumptions and restrictions are met in any particular application.” Inherent to the theory’s applicability is the assumption that there is no radiation—that the only way energy can be lost from the circuit is via heat dissipation in resistances. The moment that you allow time varying fields, you necessarily allow for energy loss via radiation. Of course, we can apply the theory when the voltages and currents vary in time “slowly” enough (the “quasistatic” approximation—that is, when the circuit dimensions are “small” enough compared to the wavelength) that radiation can be neglected. Nevertheless, strictly speaking, timevarying voltages and currents do give rise to radiation, and there is no mechanism to account for radiation in lumped circuit theory. Consequently, strictly speaking, the theory is not valid for timevarying voltages and currents. “If what you wrote was true, you could not use the theory with an impulsional current, since an impulse is a timevarying waveform.” I pointed out that I didn’t impose a current impulse; the model predicted it. While I agree that closing the switch leads to the appearance of a step voltage across the ideal capacitor (across both ideal capacitors, actually), the appearance of that step voltage, like the appearance of the current impulse, is predicted by the model (via KVL and the voltage/current relationship required by the ideal capacitors), not imposed by me. If using the resistancefree circuit to model an actual circuit truly were a “failure,” we would not get results consistent with such a morecomplete model. Both models predict the same total loss of energy. The “morecomplete” model enables us to see how the charge, voltage, current, power, and energy dissipated depend on the size of the resistance. It shows that, regardless of the size of the resistance, both the total charge transferred and the total energy dissipated does not change. And the “morecomplete” model and its results lead naturally to the “resistancefree” model and its results as limits. But, if you prefer, place a small—infinitesimal even—series resistance in the circuit to account for the inevitable resistance that must exist in a “real” circuit. Then, instead of the answer that I offered (“It [the (1/4) (1/C) X Q^2 energy] was instantly dissipated in the zero resistance of the ideal wires by an instantaneous current of infinite amplitude”), you can morecomfortably respond with “It [the (1/4) (1/C) X Q^2 energy] was very rapidly dissipated in the very small resistance of the very lowresistance wires by a very shortduration current of very large amplitude.” 


#43
Nov1105, 04:29 AM

P: n/a

There is not in lumped circuit theory that forbids steps, ramps or sinusoids tobe analysed. 


#44
Nov1205, 08:41 PM

P: 31

The analysis tells us that finite energy is dissipated by a zero resistance. This seemingly paradoxical result is a consequence of the model’s insistence that conservation laws (charge and energy) not be violated (i.e., by requiring that Kirchoff’s Laws be obeyed), regardless of the size of the resistance. The conservation of charge and the conservation of energy are the accepted laws of physics; the mechanisms by which they are satisfied are not. The only way that energy dissipation can occur when the resistance is zero (a physical impossibility) is if an infinite current (another physical impossibility) flows through it. Infinity times zero is meaningless, unless a finite result is arrived at (as it is in this instance) as a limiting process. 


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