On my E&M test I was given the circuit diagram below and asked to find the charge on the capacitor after a long time or as $ t \to \infty $. In class we had used Kirchhoff’s voltage rule on the same circuit except with a AC power supply instead of a DC one so I thought it would still apply. Making that assumption, I drew in the currents into the diagram and put a charge of CV on the capacitor and everything was consistent with Kirchoff’s voltage rule. However, apparently I was supposed to recognise that Kirchhoff’s rule doesn’t apply and get a different answer. If anyone could explain to me why and how this works and where my circuit analysis was wrong, I would greatly appreciate it.

circuit diagram I was given on my test minus the currents and charge on the capacitor that I drew in.

  • $\begingroup$ Just looking at your sketch you wrote $I(t) = Vt/L$ which we are supposed to evaluate at $t\rightarrow\infty$. What does that tell us? $\endgroup$ – noah Apr 13 at 18:33
  • $\begingroup$ @noah I think it tells us that this is a badly posed problem, because that IS the current through the inductor as a function of time if you take this circuit diagram seriously. Maybe OP was supposed to recognize that wires always have internal resistance, which is negligible at small times (and small currents) but non-negligable for larger times (and larger currents). But overall, OP drew a completely correct conclusion for the circuit drawn. And even taking internal resistance into account, he still got the correct capacitor charge! $\endgroup$ – Jahan Claes Apr 13 at 18:37
  • $\begingroup$ @JoshMessing I would chalk this up to a poorly written problem, unless you have copied the circuit diagram incorrectly. $\endgroup$ – Jahan Claes Apr 13 at 18:37
  • $\begingroup$ @JahanClaes Yeah I was more pointing at it as very poorly stated. The most sense I can make of the question would be that students are supposed to recognize that there's clearly some problem. $\endgroup$ – noah Apr 13 at 18:59
  • $\begingroup$ I take it that 𝑉 is the emf of the battery or cell. Are you to assume that its internal resistance is zero? $\endgroup$ – Philip Wood Apr 13 at 19:22

There are various versions of Kirchhoff's voltage rule. A useful one is this...

In any closed loop the algebraic sum of the emfs is equal to the sum of the potential drops.

I see no reason why this rule shouldn't be applied to your circuit: the capacitor and resistor both have potential drops across them; the cell has an emf and so does the inductor – but only if the current through it is changing.

If the internal resistance of the cell is to be taken as zero, the question itself is trivially easy' as the pd across the capacitor is always equal to the source emf, whatever the current in the other components. If the internal resistance of the cell isn't to be taken as zero, the question is still easy but very odd if the inductor has no resistance of its own. The current through the inductor will rise and rise, causing more and more of a potential drop across the cell's internal resistance until eventually there is no pd across the inductor and therefore none across the capacitor!

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    $\begingroup$ All Stack Exchange posts are version controlled, so adding content like "ANSWER MODIFIED" at the top of your post is useless and goes against the spirit of a VC'd system (which is to just seemlessly integrate the new material into the original post). $\endgroup$ – Kyle Kanos Apr 13 at 22:17
  • $\begingroup$ I thought that $I$ had done the seamless integration! Having done so I thought it only right to warn anyone who'd looked at the old version (including the person who voted for it) that it had been modified. Could you tell me what 'version controlled' means? $\endgroup$ – Philip Wood Apr 13 at 22:28
  • $\begingroup$ There's an "Edited {date}\\name" at the bottom of the post that is a link to the edit history (here's yours). Interested users can just check out the link and see what changed. For version control, here's a Wikipedia entry for you $\endgroup$ – Kyle Kanos Apr 13 at 22:31
  • $\begingroup$ Well! I never knew that! Thank you so much for the enlightenment. I'd still argue that 'ANSWER MODIFIED' alerts anyone who doesn't think to look at the edit history, but I bow to your superior wisdom on this matter. $\endgroup$ – Philip Wood Apr 13 at 22:36
  • $\begingroup$ Sorry for lag in response, got suspended yesterday for saying the S-word twice. There's a Meta post about the subject of "revision tables" in posts physics.meta.stackexchange.com/q/5886/25301 $\endgroup$ – Kyle Kanos Apr 15 at 11:24

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