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I made my donation so you have to read this (just kidding :-).
This raises another observation about the brush. It must be narrow so it doesn't span two turns at a time else it will be shorting out one turn to the next turn and that will short them together. They would heat up as would the brush which is causing the short. In my example this would cause a short across two turns that differ by exactly 1VAC. But each winding is capable of providing from 1 to 10 Amps under normal load situations for the size Variacs we are likely to encounter. With a brush causing a shorted winding it isn't hard to see that more than10 Amps could flow. To minimize the likelihood of a short the brush comes to a chisel point. The point is slightly narrower than one turn of wire and it is as wide as the removed enamel area of each turn. <I studied this on variacs and came to a different conclusion. The maximum power transfer theorem says that the source and load impedances have to be equal for maximum power to be transferred. A shorted turn is far from a matching impedance for the "primary" side, the rest of the variac winding. In order for a large amount of power to be coupled into a shorted turn the resistance of the wire would have to be much lower than what copper provides in order for the reflected impedance to be equal to the source impedance. Think Weller soldering gun. In addition, the brush, overlapping the pair of contact points is a resistor. Intentionally, so it throws the impedance mismatch further off. The result is a very lossy mismatched transformer action, little power is transferred. Also, the brush is large enough to dissipate its heat, most of which is caused by the output load current. The losses are so small you don't realize they are there.