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According to KVL, $\sum \epsilon = \sum RI$ in any closed loop. However, in an ideal circuit with no resistance, R is $0$, and hence $\epsilon$ must also be $0$. Does KVL not apply in this case?

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    $\begingroup$ This circuit cannot exist as drawn if the capacitor is charged. There are various ways to interpret (change) this drawing into different circuits that can be realized. $\endgroup$
    – HTNW
    Jul 22 at 10:13
  • $\begingroup$ There are no capacitors in this circuit, only a battery and a wire. $\endgroup$
    – Matan E.
    Jul 22 at 10:18
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    $\begingroup$ @MaranE Whoops, guess I'm tired. Same difference: this circuit drawing is illegal. If you actually connect a battery to a wire like this the circuit diagram that describes it will be different. $\endgroup$
    – HTNW
    Jul 22 at 10:22
  • $\begingroup$ $\epsilon $ is not zero in this case, this is a common missaplication of ohms Law, current is not constant here so does not apply. KVL is correct, the closed line integral is equal to zero, the pd across the wire is the -pd across the battery $\endgroup$ Jul 22 at 14:00

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This is a self-contradictory circuit model.

Remember, the things that we put in a circuit diagram are there to represent real objects. As such, they are generally approximations and the real behavior is more complicated. For many batteries over their typical operating range treating them as an ideal voltage source is a good approximation. But when the current is very high it is no longer a good approximation and the internal resistance (among other effects) must be considered.

When we put together our models we need to take care that we are doing so where the approximations are good. The fact that this model is self contradictory is simply an indication that there are no real circuits where this is a good approximation.

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