If you consider an ideal wire with no resistance that shorts an ideal battery, the only voltage drop that exists is the emf of the battery, with nothing to balance it.

Obviously in the real world such a scenario is impossible, for the wire will have some resistance, but in this ideal example, is KVL not violated?

  • 2
    $\begingroup$ No, it's not violated, you just have to use it in a limit with $R_{wire}\to 0$. This is generally true for all physical phenomena where something becomes infinite. While the limit may not exist, we can still extract valuable information from the asymptotic behavior. $\endgroup$ – CuriousOne Jan 2 '16 at 0:54
  • $\begingroup$ Voting to close as a duplicate but, in solidarity with the majority of newcomers, I wont make any effort to look up the duplicate. $\endgroup$ – Alfred Centauri Jan 2 '16 at 1:37

No, it doesn't break. All it means is that all the voltage is on the ideal wire. According to Ohm's law there will be infinite current to account for the zero resistance. That's what happens in a real circuit - if you short a battery the current is very high.


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