9
$\begingroup$

I'm reading through my E&M textbook (Physics for Scientists and Engineers 3rd edition, Knight) and watching as many lectures on YouTube (Shankar, Pomerantz, Lewin, etc) to prepare for next quarter. I have one question as I don't want to learn a wrong concept.

I've noticed when you charge a capacitor with a constant current you get a voltage slope. After watching Carver Mead talk on G4g, I later saw that for an inductor $V = L \frac{dI}{dt}$. Which looks strikingly similar to the equation I've learned up to this point for capacitors, $I = C \frac{dV}{dt}$. That's weird.

If you charge a capacitor with a constant current you get a slope of the voltage, then do you get the slope of the current if you charge an inductor at a constant voltage?

Are these mirrored concepts?

If this is the case it makes sense to me. So much of an electrons motion seems dependent on it's path or type of path. At least from what I've learned to this point. Conductors and inductors seem to be taking advantage of the same rules for different outcomes.

|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ What is your precise notion of "mirrored concepts"? They behave in many ways complementary/oppositely to each other, but what exactly is your question? $\endgroup$ – ACuriousMind♦ Dec 20 '15 at 20:33
  • 3
    $\begingroup$ en.wikipedia.org/wiki/Duality_(electrical_circuits) $\endgroup$ – user83548 Dec 20 '15 at 20:56
  • $\begingroup$ @ACuriousMind I'm surprised you don't seem to know about the duality transformation that Bruce mentions, given the interests you seem to have. Anyhow, if you don't check it out as I think you'll find it elegant. $\endgroup$ – WetSavannaAnimal Dec 20 '15 at 22:07
  • $\begingroup$ @WetSavannaAnimalakaRodVance: I never heard about these transformations before. I never particularly cared about circuits, though. It's interesting to see that the abstract duality transformation goes all the way down to the circuit level. $\endgroup$ – ACuriousMind♦ Dec 20 '15 at 22:19
  • 3
    $\begingroup$ The duality is much deeper than just on the circuit level. Do we have an answer about the symmetry properties of Maxwell's equations, possibly including the symmetry breaking between the existence of electric monopoles vs. the non-existence of magnetic monopoles? I think that would be a much better answer, in terms of physics, than the technically correct but somewhat engineering centric circuit transform result. $\endgroup$ – CuriousOne Dec 21 '15 at 0:07
16
$\begingroup$

Capacitors and inductors are images of one another under the self-inverse mapping that transforms a linear electrical network to its dual network.

The network duality transformation maps the network's graph to its topological dual graph,then all the impedances (either as lone-frequency complex scalars or as Laplace transfer functions) in the dual graph links become their reciprocals and current sources become voltage sources and contrariwise.

The physical meaning of this mapping is that we are finding a network where the voltages and currents in the network's state equations swap roles. So this is the reason for your observation: your two equations result from one another if you swap the roles of the voltage and current.

Some common examples: the Norton equivalent source is the dual of the Thévenin equivalent source and contrariwise. Likewise the star-delta transformation is an evocative example, showing how loops become links and contrariwise in the dual graph.

|cite|improve this answer
$\endgroup$
  • 2
    $\begingroup$ +1 but dang, you beat me to it. $\endgroup$ – Alfred Centauri Dec 20 '15 at 22:04
  • 1
    $\begingroup$ this is all true for networks without transformers, but if there are inductive coupled loops? $\endgroup$ – hyportnex Dec 20 '15 at 22:21
  • 1
    $\begingroup$ @jakemckenzie The simplest form of duality arises for linear networks, but the notion works for arbitrary elements as long as the set of nonlinear arbitrary elements is closed under the operation of swapping their port current and voltage state variables (e.g. if there were an element with $V = Iˆ2$, then there would need to be a dual element kind with the relationship $I = Vˆ2$ in the set of allowed elements. It is the symmetry in the state equations between the current and voltage state variables. $\endgroup$ – WetSavannaAnimal Dec 22 '15 at 2:38
  • 1
    $\begingroup$ @jakemckenzie ....The transformation is $\mathbb{R}ˆN\to\mathbb{R}ˆN$, where $N$ is the dimension of the state space. The other essential ingredient to making all this work is the fact that the current conservation law, when transformed, makes a valid statement about voltages - it indeed becomes Kirchoff's voltage law, a statement about conservation of energy. As with CuriousOne's comment, the notion applies to Maxwell's equations too, particularly if one allows magnetic monopoles: electric and magnetic fields swap roles as do electric and magnetic charge / currents. This duality technique.... $\endgroup$ – WetSavannaAnimal Dec 22 '15 at 2:40
  • 1
    $\begingroup$ @jakemckenzie .... has been used to analyze antennas: look up the Schelkunoff field equivalence method, where virtual magnetic currents are used to replace awkward boundary conditions $\endgroup$ – WetSavannaAnimal Dec 22 '15 at 3:06
2
$\begingroup$

For a slightly more prosaic elucidation, again using the electrical circuit paradigm, I find it helpful to rework everything into a (rather approximately, surely) "pseudo-Ohm's Law" or "impedance-oriented" configuration:

Inductance: $$V=\frac{\textrm{d}I}{\textrm{d}t}\cdot L$$

Resistance: $$V = I\cdot R$$

Capacitance: \begin{align}\Delta V& = \Delta q\cdot C \\&= \int_{\Delta t}{I\!\left(t'\right)\mathrm{d}t'}\cdot C\;.\end{align}

So, yes: inductance and capacitance mirror each other in that the former relates the voltage to the derivative of the current, whereas the latter relates the voltage to the integral of the current.

|cite|improve this answer
$\endgroup$

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.