KVL in short circuit with 0 resistance [duplicate]

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?

• 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.
– HTNW
Jul 22 at 10:13
• There are no capacitors in this circuit, only a battery and a wire. Jul 22 at 10:18
• @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.
– HTNW
Jul 22 at 10:22
• $\epsilon$ 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 Jul 22 at 14:00