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 $0$?
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:
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:
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!
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.
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. :)
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.
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.
