This is Mur 
🇳🇱
@mur2501 I presume the terminal strips they mention but dont show are where the circuit that **is** drawn (the one with the ammeter) connects to the board, rather than the wire connecting directly? In other words, the mention of the terminal strips isnt particularly important and the diagram is effectively complete as is?
@mur2501 im also slightly confused because it mentioned it wants me to find the current when a voltage is applies across the terminals.. yes doesnt say where on the board exactly those terminals connect (though i might assume its at the points on the illustration, but maybe not.
For example if the terminal leads are applied closer to the opposite corners I'd get a higher resistance than if it were applied to adjacent corners and a different current if applied to the middle of opposite sides... or are they implying by the terminal strip that the opposite sides are electrically connected across the entire side?
@freemo
The metal strips run on the sides may have a significance here then
@mur2501 Yea my gut tells me it is worded poorly.. terminal strips is not a good word to describe what I think they meant (as those are not a continuous electrical connection but lots of tiny posts insulated from each other. I think they meant something more like a BUS wire running down opposite sides.. Only reason that makes more sense is i feel like they are just trying to get you to use the equation for a resistor and making it seem complicated (which is all it would boil down to if opposite ends hav ea bus on each of them).
@freemo
To say, their solution doesn't really considers the contact points 
@mur2501 cant imagine how it can be a valid solution if it doesnt either a) consider the contact points or b) consider opposite sides as having a bus (at which point it wont matter where the contact point is)...
so likely b then, which would make the question kinda what its expected to be.
I'm also wondering if maybe the official answer oversimplifies the problem too and just considers the whole board as the average of the two conductivities (which strictly speaking would give an incorrect answer).
@mur2501 hmm, ill have to read through it when i get the chance.. looks legit at first glance though and doesnt appear they cheated as I thoughtt hey might.
@freemo
I don't know
You can see the solution yourself