Kirchhoff law and Inductors I am struggling with using Kirchhoff's rule in circuits with Inductors. Looks like if you have an inductor, we have Electric field (E) that is created using a time varying magnetic field and that E is non conservative and hence you will have to use Faraday's and not Kirchhoff's rule. I get that. But when you use Faraday's rule, Electric field within inductor is taken to be zero as it has zero resistance. I don't get the connection of zero resistance and zero electric field.
Below is link where Prof Lewin uses Faraday's law. I get everything in this video other than why E should be zero within the inductor
https://www.youtube.com/watch?v=cZN0AyNR4Kw
 A: 
Looks like if you have an inductor, we have Electric field (E) that is created using a time varying magnetic field and that E is non conservative and hence you will have to use Faraday's and not Kirchhoff's rule.

You should not be thinking about E field when analyzing an inductor. There will be an EMF across the inductor, due to the magnetic field through the coil. But this EMF is not related to an E field in the wire (and this is why I am calling it an EMF and not just a "voltage", which might be confused with an electrostatic potential difference). 
You can certainly apply Kirchoff's laws to circuits containing inductors, provided the other requirements defining lumped circuits apply, and all the important magnetic fields are confined to the inductors, rather than interacting (significantly) with the wires joining the circuit. The "voltage" difference across the inductor will depend on the rate of change of the current, rather than the instantaneous value of the current as it does with a resistor.

I get everything in this video other than why E should be zero within the inductor

Practically, the E-field will not be exactly 0 in the wires of the inductor. But if we need to model its effect, we'll usually do that by including a separate series resistor element in our circuit model, rather than by including the E-field effect in the resistor model.
A: Kirchhoff's law in variable regime is not an easy subject, and the lectures on the subject are not always clear. One approach that seemed consistent to me is that of Anuparm Garm in "Classical electromagnetism in a nutshell" (Princeton)
(This is not the only possible approach)
He considers that the lumped circuit is formed of black boxes with an input $A$ and an output $B$ such that the circulation of the electric field on a path outside the black box does not depend on the path followed. This assumes that the magnetic field is negligible outside the box. He then defines the voltage $V\left(t\right)$ as the circulation from $A$ to $B$ on a path outside the box:
$V\left(t\right)=\int_{A\ Exterior\ path}^{B}{\vec{E}\vec{dl}}$ on an exterior path.
For an inductor formed from a perfect conductor, the electric field is zero within the conductor. If we define a closed path which goes from $A$ to $B$ through the conductor and returns from $B$ to $A$ through the exterior, Faraday's law tells us that $\int_{A\ in\ the\ inductor}^{B}{\vec{E}\vec{dl}}+\int_{B\ Exterior\ path}^{A}{\vec{E}\vec{dl}}=-\frac{d\emptyset}{dt}$
The first integral is zero (zero electric field) and the second is $-V\left(t\right)$
Therefore $V(t)=\frac{d\emptyset}{dt}=L\frac{di}{dt}$
He also find $V (t) = Ri$ for a resistor and $V (t) = Q / C$ for a capacitor.
To find Kirchhoff's law, it suffices to write that the circulation of the electric field on a closed path outside the black boxes is zero.
Sorry for my poor english. It is not my native language.
A: The electric field is always zero inside the thickness of an ideal conductor (check out my answer here to see why). Now for an inductor, which is usually shaped as a solenoid, the electric field inside the wires of the solenoid (so on the surface of the cylinder) is zero (assuming it's an ideal conductor). The induced electric field you're talking about does not occur inside the wires of the solenoid, but in the "hole" inside (the inner volume of the cylinder).
