Where did it go?

Martin

Quote by Reality_Patrol

... I have no idea what you're referring to. Care to clarify it?

I would, if only I knew what you were referring to.

Martin

Quote by SGT

With a finite R, energy is dissipated as heat, so we can account for the difference between initial and final energy. What happens with the missing energy in a lossless circuit? I understand the maths, but I want a physical answer. Energy cannot be destroyed nor created.

Lumped circuit theory is a self-consistent model that approximates real circuits. There are a number of assumptions inherent in lumped circuit theory that do not hold 100% in “real” circuits. How well the theory models the actual behavior of any “real” circuit will depend upon how closely those assumptions hold for that particular “real” circuit. For example, circuit theory assumes that Kirchhoff’s Laws are valid; that “real” circuit components can be represented by “lumped” parameters (i.e., discrete R, L, and C); and that all associated electric and magnetic fields remain within the confines of the circuit. (Specifically, these assumptions require that electrical effects happen instantaneously throughout the circuit, that there is no accumulation of charges at any point within the circuit, and that there is no magnetic coupling among the various elements of the circuit). These are all approximations that work rather well for many—but not all—applications. Strictly speaking, for example, lumped circuit theory is not valid for time-varying voltages and currents; “real” circuit components are never actually “lumped” quantities; and the electric and magnetic fields associated with circuits never remain fully within the confines of a real circuit.

In a “real” circuit, energy would be lost due to heat dissipation in the finite resistances of the physical wires and components as well as due to radiation resulting from time variations in voltages and current as the circuit responds to the throwing of the switch. These are the only mechanisms that can account for energy loss. But if we assume that we can ignore energy loss due to radiation—which is one of lumped circuit theory’s inherent assumptions—then lumped circuit theory will be “forced” to account for the entire energy loss via resistance (yes, even in the limit of zero resistance!). It does so in a lossless (zero resistance) circuit by requiring a current impulse at the moment that the switch is thrown. It’s only via an “infinite” current that a finite energy dissipation can be created in circuit with zero resistance. Of course, impulses are non-physical—just as zero resistances are non-physical.

In reality (that is, were you close a “real” switch in a “real” circuit consisting of “real” wires and “real” capacitors), the circuit would respond with a large current pulse at the moment that the switch was closed. Some energy would be dissipated in the resistances of the “real” wires, capacitors, and switch. However, there also would be some radiation loss caused by the higher-frequency components of the current pulse. The relative amounts would depend on the dimensions of the circuit and the (physical) properties of the switch, wires, and capacitors.

SGT

Quote by Martin

...Strictly speaking, for example, lumped circuit theory is not valid for time-varying voltages and currents; ...

Not true! Lumped circuit theory is valid not only for time-varying voltages and currents but also with time-varying and nonlinear components. The only requirement for a lumped circuit is that it is composed by lumped components and that the overall dimensions of the circuit are small when compared with the shortest wavelength involved. If what you wrote was true, you could not use the theory with an impulsional current, since an impulse is a time-varying waveform.

In a “real” circuit, energy would be lost due to heat dissipation in the finite resistances of the physical wires and components as well as due to radiation resulting from time variations in voltages and current as the circuit responds to the throwing of the switch. These are the only mechanisms that can account for energy loss. But if we assume that we can ignore energy loss due to radiation—which is one of lumped circuit theory’s inherent assumptions—then lumped circuit theory will be “forced” to account for the entire energy loss via resistance (yes, even in the limit of zero resistance!). It does so in a lossless (zero resistance) circuit by requiring a current impulse at the moment that the switch is thrown. It’s only via an “infinite” current that a finite energy dissipation can be created in circuit with zero resistance. Of course, impulses are non-physical—just as zero resistances are non-physical.
In reality (that is, were you close a “real” switch in a “real” circuit consisting of “real” wires and “real” capacitors), the circuit would respond with a large current pulse at the moment that the switch was closed. Some energy would be dissipated in the resistances of the “real” wires, capacitors, and switch. However, there also would be some radiation loss caused by the higher-frequency components of the current pulse. The relative amounts would depend on the dimensions of the circuit and the (physical) properties of the switch, wires, and capacitors.

Do you have a reference where I can read about energy loss without a dissipative element, or are those assumptions a theory of your own?

SGT

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.

Martin

Quote by SGT

Not true! Lumped circuit theory is valid not only for time-varying voltages and currents but also with time-varying and nonlinear components. The only requirement for a lumped circuit is that it is composed by lumped components and that the overall dimensions of the circuit are small when compared with the shortest wavelength involved.

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. When I commented on its validity, I said strictly speaking it was not valid for time-varying voltages and currents—and that is a true statement.

If what you wrote was true, you could not use the theory with an impulsional current, since an impulse is a time-varying waveform.

I didn’t impose a current impulse; the model predicted it.

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 non-zero 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 resistance-free 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.

Do you have a reference where I can read about energy loss without a dissipative element, or are those assumptions a theory of your own?

I have no references, and it is not my “theory.” I simply solved a problem using standard lumped circuit theory, recognizing that sometimes the simplifying assumptions typically used to model a real circuit cannot blindly be applied.

Martin

Quote by SGT

... 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.

See my previous post.

SGT

Quote by Martin

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. When I commented on its validity, I said strictly speaking it was not valid for time-varying voltages and currents—and that is a true statement.

So, in your opinion, lumped circuit theory is valid only for constant voltages and currents?

I didn’t impose a current impulse; the model predicted it.

You have imposed a step voltage , that is a time varying waveform. The impulse appeared because it is the time derivative of the step.

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 non-zero 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 resistance-free 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.
I have no references, and it is not my “theory.” I simply solved a problem using standard lumped circuit theory, recognizing that sometimes the simplifying assumptions typically used to model a real circuit cannot blindly be applied.

The affirmations you made about modelling are true. The mistake is to use a model in a situation where it does not apply. In my mechanical example, using a rigid body model to study the behavior of an inelastic collision. In the electrical example, using ideal capacitors and switch to study the charging of one capacitor by another.
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.

Martin

Quote by SGT

So, in your opinion, lumped circuit theory is valid only for constant voltages and currents?

I said:

“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 “quasi-static” approximation—that is, when the circuit dimensions are “small” enough compared to the wavelength) that radiation can be neglected. Nevertheless, strictly speaking, time-varying 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 time-varying voltages and currents.

You have imposed a step voltage , that is a time varying waveform. The impulse appeared because it is the time derivative of the step.

I was responding to your statement:

“If what you wrote was true, you could not use the theory with an impulsional current, since an impulse is a time-varying 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.

The affirmations you made about modelling are true. The mistake is to use a model in a situation where it does not apply. ... In the electrical example, using ideal capacitors and switch to study the charging of one capacitor by another. ...Is this a failure of the theory? No, the failure is to use it to model things it is not supposed to do.

I doubt that that you would’ve found it objectionable had I proposed a “more-complete” model—one that incorporated a “small” series resistance. Doing so leads to the results I described previously.

If using the resistance-free circuit to model an actual circuit truly were a “failure,” we would not get results consistent with such a more-complete model. Both models predict the same total loss of energy. The “more-complete” 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 “more-complete” model and its results lead naturally to the “resistance-free” 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 more-comfortably respond with “It [the (1/4) (1/C) X Q^2 energy] was very rapidly dissipated in the very small resistance of the very low-resistance wires by a very short-duration current of very large amplitude.”

SGT

Quote by Martin

I said:
“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 “quasi-static” approximation—that is, when the circuit dimensions are “small” enough compared to the wavelength) that radiation can be neglected. Nevertheless, strictly speaking, time-varying 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 time-varying voltages and currents.

The motive lumped circuit theory demands that the dimensions of the circuit are small compared to the wavelengths involved has nothing to do with radiation. The limitation arises from the applicability of Kirchoff's laws. You can't predict voltages and currents along a distributed circuit with them, you must use Maxwell's laws. This has nothing to do with radiation, since waveforms in a waveguide or in a coaxial cable don't emit radiation, but they vary along the length of the component.
There is not in lumped circuit theory that forbids steps, ramps or sinusoids tobe analysed.

I was responding to your statement:
“If what you wrote was true, you could not use the theory with an impulsional current, since an impulse is a time-varying 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.
I doubt that that you would’ve found it objectionable had I proposed a “more-complete” model—one that incorporated a “small” series resistance. Doing so leads to the results I described previously.
If using the resistance-free circuit to model an actual circuit truly were a “failure,” we would not get results consistent with such a more-complete model. Both models predict the same total loss of energy. The “more-complete” 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 “more-complete” model and its results lead naturally to the “resistance-free” 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 more-comfortably respond with “It [the (1/4) (1/C) X Q^2 energy] was very rapidly dissipated in the very small resistance of the very low-resistance wires by a very short-duration current of very large amplitude.”

I think you did a very good analysis of the behaviour of the circuit. My objection is that your conclusions are not consistent with the accepted laws of Physics. Capacitors and inductors store energy in their electric and magnectic fields respectively and resistors dissipate energy as heat. You are postulating an extension to that theory allowing a zero resistence element to dissipate energy if you have an impulsional power waveform. I agree with you that this extension is useful in the sense that it eliminates the paradox of the missing energy.

Martin

Quote by SGT

The motive lumped circuit theory demands that the dimensions of the circuit are small compared to the wavelengths involved has nothing to do with radiation.

Lumped circuit theory assumes that all electrical effects happen instantaneously throughout the circuit, which is approximately realized if the dimensions of the circuit are small compared to the wavelength. This assumption is equivalent to ignoring the finite velocity of electromagnetic field propagation (i.e., equivalent to assuming “instantaneous propagation”), which amounts to ignoring radiation effects.

The limitation arises from the applicability of Kirchoff's laws. You can't predict voltages and currents along a distributed circuit with them, you must use Maxwell's laws.

When the dimensions of the circuit are not small compared to the wavelength, the “lumped-parameter” representation of a physical circuit can be replaced with a “distributed-parameter” representation, and the circuit analyzed by applying Kirchoff’s Laws to incremental sections of the circuit. The analysis results in predicting voltage and current waves that travel along (propagate in the direction of) the wires. This is the model used in Transmission Line Theory.

This has nothing to do with radiation, since waveforms in a waveguide or in a coaxial cable don't emit radiation, but they vary along the length
of the component.

At cross-sectional points of discontinuity (either within the circuit, or in the region exterior to the circuit) the electromagnetic fields associated with these “guided waves” of voltage and current can be scattered into electromagnetic waves that propagate away (i.e., radiate) from the circuit. Radiation does not occur with closed waveguides and coaxial cables because those structures provide an inherent shielding, trapping all of the electromagnetic energy within themselves.

Quote by SGT

I think you did a very good analysis of the behaviour of the circuit. My objection is that your conclusions are not consistent with the accepted laws of Physics. Capacitors and inductors store energy in their electric and magnectic fields respectively and resistors dissipate energy as heat. You are postulating an extension to that theory allowing a zero resistence element to dissipate energy if you have an impulsional power waveform. I agree with you that this extension is useful in the sense that it eliminates the paradox of the missing energy.

I am postulating nothing; I am presenting the results of the analysis.

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.