Cooked engine loom question, Looking for cause.

[m_c]

Well no, because the kettle is a resistive load, therefore what will actually happen is it just won't give you 3kw on the 'foreign 230v' - it'll give you 2.75kw instead.
Much the same as adding resistance to a 12v/24v DC circuit will reduce the current. If you have a pull in coil with the output hooked to the starter input, it's plausible to burn that out that way - but more resistance leads to less current.

It's not a fixed power value, it's a fixed resistance.
It's no good using the right equation if you can't use it the right way around.

I'm going to have to comment, but both of you are partly right and partly wrong.

The point @Ubique was trying to make, was due to the coil being an inductive load, as voltage drops (or technically the back emf drops), the effective resistance also drops, resulting in a higher current.

The kettle analogy wasn't a good example, as it's a resistive load, but the figures are a reasonable example of how an inductive load might act, although in practise it's probably not going to a linear response (but at 240V vs 230V it's not likely to be that non-linear)

 

[PhillipM]

As soon as the coil has stopped moving, which is very quickly indeed, it's pretty much just a pure resistive load.
Same as spinning up a radiator fan. They'll draw more* inrush current, but once up to speed they'll run at a lower speed with lower power.

*well, maybe, but I've never seen it, if you run them on 7v they draw *proportionally* more inrush current compared to the operating current, but the absolute inrush current is still lower than with 14v.