Note that his circuit arrangement is exactly like mine, he went from a high potential to a low potential. The current, electric field in the wire, and the travelling direction are all the same, yet he gets +IR instead of -IR

is he still using this "E dot dl" [tex]-\int_{a}^{b} \mathbf{E}\cdot d\mathbf{s}= \Delta V[/tex]? How does Faraday's Law apply for non-closed paths? Because it seems like Lewin is using [tex]\int_{a}^{b} \mathbf{E}\cdot d\mathbf{s}= \Delta V[/tex] (no minus sign)

The same confusion goes when he talks about the electric field in the battery.

Second link wasn't very helpful and I can't pinpoint my answer in the first link because it seems like it derailed into some argument on Lewin's teaching ability.