# Why doesn't Kirchhoff's Law work when a battery is shorted with an ideal wire?

Just to complement the other answers: This isn't really about Kirchhoff's law. Rather, it is about an idealised situation that does not have a solution at all.

When you draw such a diagram, you can think of it in two ways:

- As a sketch of a real circuit. Then the voltage source is, e.g. a battery or a power supply, and the line is a wire. You can connect them this way, and something will happen (possibly, something will break or catch fire).
- As an idealised circuit. Then the voltage source maintains a fixed (presumably nonzero) voltage $V$ between the poles and supplies whatever current is necessary. The wire has no resistance, inductance or capacitance -- it will carry any current and produce zero voltage drop. You immediately see that you cannot satisfy both conditions. Hence, this idealised circuit does not admit a solution.

UPDATE

To extend this a bit: You can approximate the behaviour of real devices with combinations of ideal circuit element. For a battery, a common way is a series conection of an ideal volatge source and a resistor (see e.g. wikipedia), and a real wire would be an ideal wire with, again, a resistor (and possibly inductance and capacitance, see wikipedia again).

So in your case, you would have to include two resistors: An internal resistance $R_\text{int}$, which you can think of as part of the battery, and a wire resistance $R_\text{w}$, which really is distributed along all of the real wire and not a localised element.

The you wil have a current$$I=\frac{V}{R_\text{int}+R_\text{w}}\,$$ and an "external voltage", i.e. the voltage aong voltage source and internal resistance, of $$U_\text{ext}=V-I\cdot R_\text{int}=V\left(1-\frac{R_\text{int}}{R_\text{int}+R_\text{w}}\right)\,.$$ In the fully idealised case $R_\text{int}=R_\text{w}=0$, these expressions are ill-defined.

You can look at two posible limiting cases:

- "Superconducting wire": If $R_\text{w}=0$ but $R_\text{int}\neq0$, i.e. superconducting ideal wire shorting a real battery, current is limited by internal resistance and external voltage is zero (and the battery will likely overheat).
- "Real wire on ideal battery": If, on the other hand, $R_\text{int}=0$ but $R_\text{w}\neq0$, current is limited by the wire resistance, and the external voltage is just $V$.

The law does hold up perfectly here. There's a battery, with `v`

volts. Let's use 5v.

Then, there's a wire. In the circuit above, there will be some (high) current going through the wire and by ohm's law, some voltage drop will appear. -5v, actually.

5v + -5v = 0. Solved.

The 5v for the battery is a fixed value. If you want to solve for the current, you can do:

`v = rI`

`5 = rI`

`r`

might tend to 0, and `I`

might tend to infinite. But that's not a problem. `rI`

still is 5, and you still get a 5v voltage drop.

Kirchoff's law only applies to consistent circuits. It is possible to write a circuit which is not self-consistent using ideal wires and ideal batteries, but any tool which gives you a solution for the circuit will have to fail because there is no such solution in the first place.

In this case, if you work out the equations, you see that you have an overdefined system with 1 unknown and 2 equations.

In a similar vein, there are many rules you will learn in physics class (and even math class!) which MC Escher broke with gusto!