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

Kirchhoff's law states that the sum of voltages around any closed loop sum to zero. The law is true as the electric field is conservative in circuits. Why can we not apply the law here?

Why doesn't the law hold here despite the fact that the electric field is conservative and the voltages should add up to $$0$$?

• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. – David Z May 15 '20 at 0:28
• @FedericoPoloni What about capacitors in that case? – Alan Whitteaker May 17 '20 at 8:58
• @AlanWhitteaker All circuit elements should have a voltage drop / increase. So in retrospect, my comment above sort of misses the main point better highlighted in the answers. I'd better delete it. :) – Federico Poloni May 17 '20 at 9:21
• @FedericoPoloni Atleast you acknowledged. : ] – Alan Whitteaker May 17 '20 at 9:24

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 $$V$$.
• @hezizzenkins - there are superconducting wires, but no superconducting battery to my knowledge. – MaxW May 15 '20 at 4:14
• The question mentioned an ideal wire - but not an ideal battery. The ideal wire will discharge the battery over time (no infinite current, ad the battery has an internal resistance). Voltage drop occurs inside the battery. – Klaws May 15 '20 at 5:47
• The update makes this a great answer, +1 – Jeffrey May 15 '20 at 17:39
• @hezizzenkins don't forget that superconductors are also not quite ideal: after the current becomes so large that its associated magnetic field exceeds the critical field, the material stops being superconductive. This leads to sudden increase of resistivity and thus dissipation of a lot of power ($P=I^2R$). This can be quite violent, you shouldn't be near this device at such a moment. – Ruslan May 15 '20 at 23:09
• Just to extend: One can kick the can down the figurative road quite far and still find limiting cases. As said, there are really no superconducting batteries, but even if there were, and even if the wire had no critical field density, the current would create an immense magnetic field, whose induced voltage would cancel the 5V in short order. After that, lorenz forces would rip the device to pieces, or finally the energy density would collapse into a singularity. – Elmore May 16 '20 at 6:43

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.

• Or to say this without the phrasing of “tending to 0/infinity” (which is accurate, but often misunderstood/misused): realistically, the resistance of the will be very small but nonzero, and the smaller it is, the larger the current will be, in order to maintain a voltage drop of 5V. – PLL May 14 '20 at 17:32
• In the limit of $r \to 0$, then $I \to \infty$ and the quasi-equilibrium state you are describing (in which the current $I$ flows) persists for a vanishingly short time. This then violates the assumption of quasi-equilibrium to start with, leaving you with a contradiction. I would not call this holding up perfectly. – ComptonScattering May 15 '20 at 14:22

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!

• Or, put another way, the question is equivalent to asking why Newton's laws of motion don't work when an irresistable force is applied to an immovable object. – occipita May 15 '20 at 10:31

There are a number of points here.

First if you are saying that there is no resistance in the circuit and nothing else is present then the situation is unphysical and as such you cannot apply Kirchhoff's laws.

However, as drawn the circuit is a loop and therefore has a self inductance $$L$$.

Once inductance is considered there is a problem because there is a non-conservative electric field generated by the inductor if the current changes so some would say that Kirchhoff's laws cannot be used.

In the end and assuming that there is no resistance in the circuit, by whatever route you take you end up with an equation of the form $$V= L\dfrac {dI}{dt}$$ where $$\dfrac {dI}{dt}$$ is the rate of current in the circuit.

So suppose that you have a switch in the circuit at close it at time $$t=0$$ so the initial current is zero.

Integration of the equation yields $$I=\dfrac VL \,t$$ with the current increasing linearly with time for ever, again not a very realistic situation.

• "First if you are saying that there is no resistance in the circuit and nothing else is present then the situation is unphysical" Superconducting wires? – jamie May 14 '20 at 12:11
• Everything you say about inductance and the transient behavior when a switch is closed is true, but none of that addresses the OP's question. Only your second sentence hints at the answer, which is that Kirchoff's Law would work if there was resistance in the circuit, but without resistance, the voltage and current are undefined. – Solomon Slow May 14 '20 at 13:27
• @jamie You can form a superconducting loop with current flowing through it and then a constant current will flow for ever. – Farcher May 14 '20 at 15:12
• Kirchoff's law only makes sense when you apply it to ideal circuit elements. An ideal voltage source, by definition, maintains a constant voltage between its terminals, and any two points in a circuit that are connected to the same node, by definition, must be at the same voltage. Unless the voltage of source in the OP's circuit is zero (unlikely), then by definition the voltage between its terminals must be both zero and not zero (i.e., it's undefined.) – Solomon Slow May 14 '20 at 17:06
• @SolomonSlow: The ideal circuit includes an inductor in series. Typically it is ignored since circuit theory would rather not have to include R_{wire} and L_{wire}. After all, the whole point of circuit theory is to loose details to ease finding a solution. But, while R_{wire} and R_{source} are material consequences, and therefore potentially avoidable with judicious choice of materials, L_{wire} is not: it is a direct consequence of Maxwell's equations. Therefore, without L_{wire} the question cannot exist in the discipline of E&M; it may be ignored only when it's irrelevantly small. – user16035 May 17 '20 at 12:35

We can apply Kirchhoff's law here; it works just fine.

Suppose we design a circuit consisting of a 5 V battery shorted with an ideal wire. Then Kirchhoff's law is applicable, and it tells us that the voltage across the battery will be 0. This makes sense because a battery with a lot of current through it fails to act like an ideal voltage source.

Suppose we design a circuit consisting of an ideal 5 V voltage source shorted with a real wire. Then Kirchhoff's law is applicable, and it tells us that the voltage across the wire will be 5 V. This makes sense because a wire with a lot of current through it fails to act like an ideal wire.

Suppose we design a circuit consisting of an ideal 5 V voltage source shorted with an ideal wire. Then Kirchhoff's law is applicable, and it tells us that such a circuit cannot be built.

When we draw a resistor in a circuit diagram, typically we are thinking of some thing which has a resistance, not necessarily an actual resistor. Just as when we use point masses, we aren't actually thinking of a point mass, but of some object which can be modelled by a point mass.

When you connect a battery to itself using a wire, the wire itself has a resistance. So strictly speaking, if you wanted to draw a circuit diagram representing a battery connected to itself, you should include a resistor. That resistor would represent the resistance of the wire itself.

Then in circuit diagram terms, the p.d. across that virtual resistor would be the voltage of the battery, and Kirchhoff's Law would hold.

You might ask why we don't do this for every circuit. The answer is that we would if we wanted to work to a high degree of precision. But typically the resistance of a wire is very small compared to whatever we are measuring, so that our virtual resistor can be safely 'set to zero resistance', i.e. ignored.

We can't apply Kirchhoff's law here as it states that the sum of the voltage drop in a closed loop in any circuit is zero and we know that since their is no any circuit element so their is no voltage drop.

NOTE: assuming the wire to be resistance less .

Thanks for asking. Hope it helps.

It still does apply.

The ideal wire has a resistance of zero, and it has a 5V drop across it. Simple maths says that anything divided by zero is infinity, so you have an infinite current.

In a practical circuit, the voltage source and wire both have resistances, so the current will be finite. As the resistances drop though, the current increases, and mathematically it "tends to" infinity as the resistances drop.

So there's no mystery here. You've just discovered why dividing by zero is a problem. :)

• "Simple maths says that anything divided by zero is infinity" it isn't. Division by zero is undefined and limit of a function as the variable approaches zero is a different other matter (and in most cases depends on how the limit is calculated, still not resulting in zero necessarily). – gented May 15 '20 at 14:18
• In this case it's true though: $\lim_{x\to0}5/x=\infty$. – MannyC May 15 '20 at 16:32
• What voltage drop does Ohm's law predict for an infinite current running though a zero resistance ? – my2cts May 15 '20 at 16:36
• @MannyC No, it isn't even true, because the function $1/x$ does not possess a limit in 0 (the left limit and the right limit are different, therefore the limit doesn't exist because if it existed they would to be the same). – gented May 15 '20 at 22:49
• True. But the resistance is positive. I should have said $\lim_{x\to0^+}$. – MannyC May 15 '20 at 22:54

Nice question. If you short an ideal battery with an ideal conductor, without contact resistance it will discharge in zero time. All of the stored energy will be released at once in the form of electronic kinetic energy, as the battery and the wire have no resistance. A huge current comes into existence, as the total electron kinetic energy equals the energy stored in the battery. Something will blow up. Just before that happens The voltage will be zero, Kirchhoff's law trivially applies but Ohm's law does not. This is because Ohm's law does not take into account the kinetic energy of the electrons, which in this case is the only contribution.

In practice the battery has an internal resistance so even if it is shorted the current is limited. Because of this finite resistance Ohm's law applies once a (quasi-) stationary state current is reached, before something will blow up. Nevertheless, don't try this at home as even in the non-ideal case something may blow up.

It is possible to have wires with zero resistance (to a point). But it is not possible to have wires with zero inductance.

The (ideal) voltage source will maintain the 5V, and the wire will have a 5V drop due to the changing current.

$$V = L \frac{di}{dt}$$

If we assume the inductance and the voltage are both constant, you can solve for the change in current over time.

The voltage in this ideal transient situation will be 5V, even though the resistance is zero.

There's no loop. That entire figure's just one point.

Sure, you're probably imagining an ideal wire as a good conductor with very low resistance; but that's only an approximation. A real ideal wire is direct physical contact, i.e. the endpoints are literally the same point in physical space.

The above depicts an ideal battery where the anode and cathode are literally the exact same point in physical space.

The ideal battery will have to have a $$\Delta V$$ of $$0 ,$$ as any other value would be contradictory. As an ideal battery with joined terminals with no voltage drop is indistinguishable from a non-component, the entire circuit can be redrawn as a single point.

• To state the obvious: if we retreat from the ideal wire being literally ideal, then the problem disappears automatically. – Nat May 17 '20 at 10:22

The Short version:

Any ideal circuit must have an L since any current flow will set up a magnetic field.

This L, while usually ignored, is very important in situations where the source and wire resistance are smaller than this L.

The Long version:

An ideal wire and an ideal voltage source appears to creates a paradox; namely you have two potential differences at the same pair of nodes. Which is like saying A = B and A =/= B at the same time. So an ideal source and ideal wire is nonsensical; but actually there is a solution to a circuit with an ideal source and wire - an implied L that is nearly always neglected.

Let's say at t = 0s I close the switch of my ideal circuit. Current violently starts to flow and current creates a magnetic field. More importantly, this magnetic field is changing, therefore it creates a back emf! Basically $$L*dI/dt = V_{source}$$. Note that the L term is not a flaw of your ideal conductor, but a fundamental property of current flowing in any circuit.

So in the most idealized of circuits, you still have an emf along the wire that perfectly matches that of your source. But how do you get L?

That's a much more difficult question, to solve it you need E&M, not just circuit theory (you could measure it if you can set up a circuit that is ideal enough). Instead, typically, this inductance is negligible and therefore neglected. There are geometries which minimize L. So what happens as L is minimized (say by shaping your source and wire as a Möbius strip)? The electron still has a finite mass, and therefore inertia. The electron's inertia is still an L. Therefore L can never be zero, and the paradox is solved.