Electronic – Question about the placement of resistors on a Marx generator

No, you are not right. AC5 is a pin of the chip. You have to place a 27 Ohms resistor near to that pin and connect the actual external circuitry to that resistor (on the diagram, this is called the AC5-Load).

AC5 and AC5-Load are different things: AC5 is the pin, AC5-Load is the circuitry that should be connected to that pin (apart from the resistor, which is for Synchronous 245 mode).

Since you are making the coil yourself, you can wind it to various voltage versus current tradeoffs. It seems you used too many turns of too thin wire, which therefore has a high DC resistance and needs high voltage to generate the large magnetic field you want.

Use less turns of a thicker wire. That will require more current to generate the same magnetic field, but a lower voltage to cause the current.

You can maybe achieve this with the wire you have by breaking down the "several thousand turns" into several winding and connecting them in parallel. For example, let's say you have one winding of 3000 turns now, which a DC resistance of 300 Ω. 10 A would require 3 kV, and provide 30 kAmp-turns of magnetic field. If you make that into three windings of 1000 turns each, then each winding has 100 Ω DC resistance. Put in parallel, the combined coil has 33 Ω resistance. It would require 1 kV and draw 30 A to produce the same 30 kAmp-turns magnetic field. This is actually still unrealistic. You want to go lower resistance and higher current overall. I just used these numbers as example to show how to calculate equivalent magnetic field.

I would find the largest capacitors you can get your hands on, and design the coil for that. This means putting enough current thru the wire for a short time to heat it just below damage level. The time will be so short that there won't be opportunity to conduct the heat away. Therefore, figure all the energy goes into heating the copper. Figure the maximum copper temperature you are willing to tolerate, use the specific heat of copper to find the energy per unit length of wire, then figure how much of that wire you need to use to absorb all the energy of the capacitor.