Let's say I have an ideal open circuit (i.e. there's no resistance in the conductors) containing only a charged capacitor and then I close it. Now, the charge on one end of the capacitor will move to the other end until the voltage drop on the capacitor is zero. What confuses me is that if I try to apply Kirchoff's voltage law to the circuit, I just get $V_C=\frac{Q}{C}=0$, which is the steady state of the system after the discharge. So my question is what, at the fundamental level, makes Kirchoff's voltage law inapplicable to this circuit during the discharge, and how would you find the current in the circuit as a function of time before it reaches the steady state?

  • $\begingroup$ The wire has resistance. $\endgroup$
    – user253751
    Aug 19, 2020 at 11:18
  • $\begingroup$ My question is what will happen if there's no resistance in the wire. $\endgroup$
    – Ofek Aman
    Aug 19, 2020 at 11:19
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    $\begingroup$ Then it's impossible. What happens when an unstoppable force hits an immovable object? You can approximate forces as unstoppable, or objects as immovable, but not both at the same time. $\endgroup$
    – user253751
    Aug 19, 2020 at 11:21
  • $\begingroup$ But the charge would still move from one plate of the capacitor to the other, no? Like in an ideal conducting object where the charge arranges itself so that the electric potential on the face of the conductor will be constant. Why don't we have to assume resistance in that case while in this one we do? $\endgroup$
    – Ofek Aman
    Aug 19, 2020 at 12:09
  • $\begingroup$ @user253751 is a Luddite. It is perfectly reasonable to ask what happens at the extreme ends of a model, even if it’s unlikely to be physically realizable. $\endgroup$
    – cms
    Aug 19, 2020 at 12:15

1 Answer 1


Your question is a variant of many such question about situation where there is no resistance and the inductance of the circuit is ignored. You have a loop which has self-inductance and so you are dealing with a LC circuit in which the charge (current) oscillates at a frequency given by $\omega^2=\frac{1}{LC}$.

  • $\begingroup$ So what you're saying is that you must regard either the resistance or inductance of the wire in order for you to be able to find the current in the circuit? What happens though when you try to analyze the movement of particles on the face of a conductor before it reaches a steady state? Would you also have to consider the resistance and inductance of the material? $\endgroup$
    – Ofek Aman
    Aug 19, 2020 at 12:34
  • $\begingroup$ @What I am saying is that you cannot just consider the capacitor as the sole circuit element as that is an unphysical situation. What you have when you include the inductor is simple harmonic motion of the current/charge/voltage. With no resistance present and no dissipative processes this oscillatory motion is the steady state. $\endgroup$
    – Farcher
    Aug 19, 2020 at 14:13
  • $\begingroup$ @Farcher "no dissipative processes" Can you stop the circuit radiating e-m waves (albeit at very low power)? $\endgroup$ Aug 28, 2020 at 22:30
  • $\begingroup$ @PhilipWood I do not think that the radiation of em waves can be stopped but my answer was an attempt to keep things simple and neglect such a process as in most cases it is a very small effect. $\endgroup$
    – Farcher
    Aug 28, 2020 at 23:24
  • $\begingroup$ @Farcher Quite so, but I find it interesting that even without resistance an LC circuit loses energy in theory. $\endgroup$ Aug 29, 2020 at 7:46

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