Power consumption in voltage multipliers
What you have to realize is that the output voltage can be very high, but the output current is usually very low.
High voltage doesn't mean that you get a lot of power out of them.
They don't &quot;consume&quot; much power at all.  A typical Cockcroft-Walton multiplier may put out only a few milliamperes at several thousand volts.
Say you have a Cockcroft-Walton multiplier that puts out 3000VDC at 10 milliamperes.  You will have to supply (3000V* 0.01A = 30W) at the low voltage side.
You wouldn't normally need that much current, though.
You want the high voltage to drive a Marx generator.
A Marx generator slowly charges its capacitors through the resistors.  When the voltage gets high enough, the capacitors discharge through the spark gaps.  All of the energy is released at once, then the process starts again.
More current means a faster recharge of the capacitors in the Marx generator - less current just means that you get individual discharges (or discharges at longer intervals) rather than a constant &quot;buzz&quot; of Marx discharges.
Since you saw an example of a 9 volt battery driving the whole thing, it can't be really consuming much power - 9 volt batteries don't &quot;do&quot; high power.  If you push one really hard, you might get a couple of hundred milliampers out of it - at nine volts, maybe a couple of watts for a little while.
Your examples don't match what you would have seen in the videos, though.
The Marx generator shown would need about 45000 volts DC input to get to 180000 volts output.
The Cockcroft-Walton multiplier shown has a DC output far too low to drive the shown Marx generator to 180000 volts.