What guarantee Kirchoff's law of potential difference?

[Twukwuw]

Kirchoff's law states that, the potential difference along a closed circuit must be ZERO.

Now, let's do a simple question 1.
We have a simple circuit consists of INDUCTOR and Voltage source-->V=V.sin(wt)

At any instant, why must the voltage supplied equal the voltage generated in the inductor? (this is written in many textbooks)
Can the voltage supplied exceed the voltage of inductor and hence we have a NON_ZERO potential difference along a closed loop?

Now, simple question 2.
Suppose we have a capacitor which has been charged.
Next, we connect the 2 plates of the capacitor by a ZERO RESISTANCE wire. The capacitor will take a very short time to neutralise itself.
In this very short time, we can see that, the potential difference is not zero along a close loop, because the resistence is zero.

So, after these 2 examples,
I would like to ask, what GUARNTEE kirchhoffs law of potential difference?

Thanks,
Twukwuw.

 

[Galileo]

Kirchoff's law states that, the potential difference along a closed circuit must be ZERO.

Now, let's do a simple question 1.
We have a simple circuit consists of INDUCTOR and Voltage source-->V=V.sin(wt)

At any instant, why must the voltage supplied equal the voltage generated in the inductor? (this is written in many textbooks)
Can the voltage supplied exceed the voltage of inductor and hence we have a NON_ZERO potential difference along a closed loop?

The Kirchoff voltage law (KVL) simply says the electric field is conservative. It comes from Faraday's law. If there is no changing magnetic field, then:
$$\vec \nabla \times \vec E=\vec 0$$
This zero curl allows you to define a potential $$V(\vec r)=\int_o'^{\vec r} \vec E \cdot d\vec l$$ which is path independent (so for a closed loop it's zero -> KVL)

Incidentally, KVL is NOT true if there is a changing magnetic field:
$$\vec \nabla \times \vec E=-\frac{\partial}{\partial t}\vec B$$
Then the E-field is not conservative. This is also true in circuits with inductors. However, you can still use KVL whilst using the rule that LdI/dt is the potential drop across the inductor (this is not true, it just gives the right answer).

Gotta go, will post more later

 

[jtbell]

Next, we connect the 2 plates of the capacitor by a ZERO RESISTANCE wire.

Where would you find such a wire?

 

[vanesch]

Now, simple question 2.
Suppose we have a capacitor which has been charged.
Next, we connect the 2 plates of the capacitor by a ZERO RESISTANCE wire. The capacitor will take a very short time to neutralise itself.
In this very short time, we can see that, the potential difference is not zero along a close loop, because the resistence is zero.

Ha, this was an item I got on my exam by my old professor in electronics The point is that even a zero-ohm wire has a finite geometrical extension, and hence MAKES UP A SELF. It has a certain self-inductance.

So what you have now, is a perfect LC circuit, which will oscillate for ever. The current in the zero-ohm wire will be maximal, when the potential over it has dropped to 0, and then the self-induction will now charge the capacitor in the opposite direction. When the current is 0, there is indeed a voltage over the conductor, but even though it has 0 resistance, it doesn't have zero IMPEDANCE (because it is a self).