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And your numbers are off. For starters, double the weight of your presumed electrode.
The resistivity goes up as the temperature increases, which would tend to happen if you plan on actually vaporizing anything.
115.0 x 10^-8 ohm meter @ 3600K. I have no idea what it hits at 5828K, which is the vaporization point.
Using 115 gives resistance of 0.00024403757940757282 ohms
There is a reason sintering is the common method of working with tungsten. The energy required just to melt it is enormous, let alone vaporize it.
You assert that the energy required to melt tungsten is enormous, but I have calculated what it would be in this case, rather than just making unsubstantiated assertions.
You really don't have a clue do you? OK, let's say the electrode weighs twice as much as in my order of magnitude calculation. So it's twice as long, 24mm instead of 12mm. All that does is to double the energy and power required from a modest 70kJ and 14kW to a still modest 140kJ and 28kW. And it changes the current required (ignoring, as I did before, the temperature coefficient of resistivity - see below) from 3,780 amps to 2,672 amps while increasing the voltage by a factor 2 x sqrt(2). Yes that's less current required to do the job for a longer electrode.
And now let's turn to your other objection, the temperature coefficient of resistivity (change in resistivity with temperature). For tungsten that's something like 4.5 x 10^-3. So at its melting point it would be, as you say, 16 times higher than at room temperature or about 100 x 10^-8 ohm meter. Why didn't I include it that in the first calculation? Because I was doing worst case for current capacity - increasing the resistance reduces the current required to produce the resistive losses in the electrode that heat it up. So at the melting point the current required to maintain the required power is reduced by a factor of the square root of the resistance increase (or about four times). So at this stage, we don't need 2,672 amps, but a distinctly modest 668 amps. At 42 volts. So the temperature coefficient of tungsten, and its increase in resistivity when it liquifies, reduces the current requirement over my worst case order of magnitude calculation.
The amperage necessary to do what was observed is off the charts, in the range of an industrial electric arc furnace powered by a megawatt sized power plant.
Complete and utter bovine excrement. Pure marketing woo. The current necessary isn't
"off the charts" but a rather modest 2,762 amps reducing to 668 amps and then falling even further as the tungsten heats up and its resistivity increases. A MEGAwatt? Very funny. As we have seen it is, using your physical dimensions for the electrode, 28kW without thermal losses and maybe 56kW with. You're out by a factor of 20.
Of course, megawatt power is exactly what this proves.
No - burning out this light bulb does not need a megawatt. It really doesn't. See the sums.
Go find me a portable generator that's throwing out 10,000 amps that Mills could have hooked into.
But we don't need 10,000 amps at 100 volts as we have seen. Rounding, we need 2,750 amps and 10 volts to start, 675 amps at 40 volts at the melting point.
And then, you know what? - we need 0 amps and 0 volts, because after one second of the putative five second process the electrode melts and becomes a puddle on the floor. You see, 4/5ths of the energy required to vaporise the electrode is the latent heat of vaporisation and we can't apply that energy after gravity has done its work.
You said earlier:
No commercial power source known to man can vaporise 6mm tungsten electrodes in seconds?
and we have seen that this is utter tosh. Really - comp;ete marketing disinformation. Why should anyone trust anything else that you say?
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