Why is the voltage across a capacitor (as a function of time) continuous? Why is the current across an inductor (as a function of time) continuous?
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From a circuit standpoint, isn't the voltage across a charged capacitor discontinuous? The voltage of one plate compared to the opposite plate jumps discontinuously. On the other hand, if you want to adopt a microscopic viewpoint and consider the internal construction of the capacitor, you could say that the voltage potential from one plate going through the air (or whatever the capacitor medium is) to the opposite plate does rise at a large, though continuous rate if the capacitor is charged. Maybe you need to clarify exactly what you mean by your question. – Dec 28 '16 at 20:56
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@SamuelWeir The OP probably considers continuity with respect to time ;-) – Massimo Ortolano Dec 28 '16 at 21:10
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@MassimoOrtolano - "The OP probably considers continuity with respect to time ;-)". I guess that would make more sense. Again, I think that he should have really explained his question better. – Dec 28 '16 at 21:26
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If the voltage across a capacitor was discontinuous the charge on the capacitor would be discontinuous which would mean a transfer of charge in an infinitely short period of time i.e. an infinite current.
A similar argument applies for an inductor except this time a discontinuous change in the current would imply an infinite rate of change of current which for an inductor would mean an infinite induced emf opposing the change producing it.
Farcher
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Farcher's answer is correct, but I would like to expand a bit on a point that is frequently unclear to students.
Circuit theory deals with mathematical models of electric circuits and elements. For circuit theory, a capacitor is a two-terminal element defined by the constitutive relationship
$$i = C\frac{\mathrm{d} v}{\mathrm{d} t},$$
where $C$ is the capacitance, $v$ the voltage across the capacitor terminals, and $i$ the current crossing the capacitor.
In the definition above, there is something missing: what kind of functions are $v$ and $i$? The common, implicit, assumption is that $v$ and $i$ are real functions of real variable of whatever type needed for the above equation to make mathematical sense. In this case, if $v$ is differentiable at time $t$ – for the above equation to make sense – it should be continuous at $t$ too, for a a well-known theorem of mathematical analysis.
However, we can allow for possible discontinuous voltages if we extend the relationship to distributions. Such an extension can be convenient in certain cases, but of course one should be cautious in the interpretation of the results. Another generalization that is frequently employed in circuit theory is that of considering $v$ and $i$ as complex-valued functions (or distributions).
Massimo Ortolano
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I don't like this explanation because it's not clear why $v$ should be differrnriable in try first place (unless you define the capacitor as a circuit element which obeys that relation, as you said). It's basically math wizardry and lacks physical significance – math_lover Dec 29 '16 at 09:46
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@JoshuaBenabou In circuit theory it's exactly like that: the capacitor is defined in that way. Of course, the physical significance is that that relationships models the electrical behaviour of a real capacitor. The differentiability is implicit in the fact that there is a derivative. Saying that the current can't be infinite is just another way of saying that the voltage should be differentiable. – Massimo Ortolano Dec 29 '16 at 10:01
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I understand what you're saying. I guess at the fundamental reason why the voltage is continuous is because particles travel at finite speeds. – math_lover Dec 29 '16 at 10:03
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Capacitors and Inductors store energy into EM fields. The question - "Why are currents in inductors and (change in) voltage across capacitors change "continuously"?" could be rephrased as "Why can't they change discontinuously/abruptly?"
A straightforward answer is that it is physically impossible to do so. [Please, don't be misguided by the math of it; they are just models reflecting the physical truth within. The equations are definitely using calculus on single variable continuous functions. Devices like capacitors and inductors are engineered to leverage these physical phenomena.]
Let me explain.
Firstly, let's assume there is a non-uniform charge distribution. Then, it is imperative that there will be an associated Electric field. Unequivocally, an increase in charges leads to increase in this field's intensity (regardless of the geometric configuration). The proportionality with which it happens is 'capacitance'. It captures the geometric information of structure (and no more). $$Q = CV$$
Second, let's assume looping current flowing. With it comes Mag flux. (What's flux? say, a small current segment exudes a Mag field; these segments when arranged as a loop, their mag field "interlace" upon each other.) Similar to above, an increment to current will effect in more dense mag flux. This proportionality is captured in 'inductance'. Yet again, a purely geometrical concept.
$$\Phi = LI$$
These devices utilize EM fields. EM fields are causal and dynamic in nature. They undergo time retardation. To have "discontinuous" characteristics, one has to have infinite charge separation or send a charge moving at infinite velocity. It is possible to do this at a "very large" magnitude, but not exactly beyond it. Furthermore, physical laws basing such EM dynamics like conservation and continuum of charge prohibits any-other vacuous point discontinuities; charge cannot occur somewhere magically and must be physically transported.
PS: I admit this answer is still lacking some rigorous mathematical explanation. This is just due to my own lack of knowledge.
