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When we short-circuit a voltage source, the current will be very high. The voltage across the wire is 0V. If we apply KVL, also the voltage across the voltage source is 0V. How can it be 0V if we have a voltage source?

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  • $\begingroup$ You absolutely have to always think of a voltage source as an ideal voltage in series with a resistor. A good voltage source has a low value for that resistor, but it is not zero. $\endgroup$ – DanielSank Feb 7 '16 at 20:40
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Even if you put a superconductor across the terminals of a voltage source the current would be finite as all real voltage sources have a resistance. A circuit with a voltage source with no resistance does not exist.

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The fundamental problem is that the Ohm's law is a simple mathematical approximation, neglecting the V-A characteristic of a real source, resistance of wires, or even nonlinear physical response of materials under extreme conditions.

Using this approximation outside of its scope leads to a "0/0" problem, which is obviously wrong. Some other physical approximations are not so clear indicating where they cease to be applicable.

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    $\begingroup$ Ohm's law isn't the problem. The problem is assuming that the source has zero output resistance. Ohm's law just says $V = IR$. There's nothing approximate about that. OP is making the mistake of putting $R=0$ into that equation. $\endgroup$ – DanielSank Feb 7 '16 at 21:07
  • $\begingroup$ @DanielSank You are right. $\endgroup$ – dominecf Feb 7 '16 at 21:54

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