When we short-circuit a voltage source, the current will be very high. The voltage across the wire is $0\rm\ V$. If we apply Kirchhoff voltage law, also the voltage across the voltage source is $0\rm\ V$. How can it be $0\rm\ V$ 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. $\endgroup$– DanielSankCommented Feb 7, 2016 at 20:40
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3 Answers
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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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2$\begingroup$ 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. $\endgroup$ Commented Feb 7, 2016 at 21:07
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Your setup is self contradictory:
When we short-circuit a voltage source
That means the voltage source is no more - its shorted after all. You can't short two terminals and still say they have a potential difference. Applying KVL then gives, $0_{\text{sum must be zero}}=0_{\text{no potentials}}$ which is trivial but consistent.
How can it be $0 V$ if we have a voltage source?
It can't. But again, a shorted voltage source isn't one at all.
What about real life batteries which involve internal resistances? Surely there is some contradiction there?
No. Applying KVL gives $0=V_{ideal}-I R_{internal}$ where $V_{ideal}$ is the emf of the battery. All the short does is drive a current through the resistor which consumes the potential in KVL. Internally, the voltage source itself stays at $V_{ideal}$ while between its external terminals there seems no potential difference.
But still at the instant of the short, for an ideal battery, if the battery's potential has gone to zero, why is there a current? If there is a current, then the battery hasn't gone to zero and KVL is violated. What now?
This is a very subtle point. During the initial transient phase when the battery potential hasn't become zero, there is a changing current (since the potential, as per our earlier discussion, is changing to comply with being called shorted). Since we have shorted the circuit, there now exists a current loop. The changing current causes a changing magnetic flux in this loop. As per Faraday's law this induces an emf in the circuit which is equal and opposite to the battery's potential, at that time. Its not just that Kirchhoff's law is violated it outright fails, in this transient phase. When electromagnetic induction is involved, its inadequate and Faraday's law needs to be used.
A real life (and simultaneously theoretic) case of this subtlety is shorting of a charged linear capacitor. Even though the cap's voltage isn't zero over a finite time, KVL is violated at all times.
There's a video about it.
