Hero999

All that means is that if the capacitor is too small, it won't be large enough to damp the oscillations but if it's a single, large value capacitor it won't be good either. Higher value capacitors tend to have poor high frequency characteristics and become inductive past the resonant frequency, the higher the value, the lower the resonant frequency.

Rather than using a single, high value capacitor, it's better to use a large capacitor and lower value capacitor in parallel with one another. The type of capacitor is also important; for example tantalum is better than electrolytic. I'd recommend a 1uF tantalum in parallel with a 100nF and 10nF ceramic capacitor on both the input and output. If you don't have a 1uf  tantalum, go with 22uF electrolytic.

The same basic blocking oscillator circuit, normally used to drive a flyback transformer, can also drive a centre tapped or dual primary mains transformer.

Here's an example of a capacitor charger using a mains transformer with a centre tapped primary. It should be able to drive a voltage multiplier circuit.

                    C1 1 uF                   D2 1N4948   R2                 +------||------+        T1    1.2kV PRV  1K 1W                 |              |           +-----|>|-----/\/\---+------o +                 | R1 4.7K, 1W  |   red  ||( blk                 |                 +-----/\/\-----+------+ ||(                     |                 |                  yel )||(                   +_|_ C2  + o----------------------------------+ ||(                    --- 300 uF                 |                  red )||(                   - |  450 V                 |      +--------------+ ||(                     |                 |  Q1  |                ||( blk                 | 6 to 12         |    |/ C                  +--------------------+------o - VDC, 2A         +----|    2N3055    Stancor P-6134             D1 _|_   |\ E           117 V Primary (blk-blk)         1N4007 /_\     |            6.3 VCT Secondary (red-yel-red)                 |      |  - o------------+------+


















http://www.repairfaq.org/sam/samschem.htm#schssi

No never remove the fuse from the meter.

The meter's insulation is only 1kVDC, if you try to measure 100kV, an arc will jump through the case which will both damage it and you'll get a nasty shock.

The current rating is not a problem. It's better to go higher than lower. Using a power relay to switch low current signals is only a problem at very low voltages.

The problem with using a lower voltage relay is the contacts can arc over and what's worse it it could be unpredictable.The relay may seem fine most of the time but it could suddenly arc over due to ambient ionising radiation.

The meter will have an impedance of 10M so to divide by 10, all you need to do is connect a 90M high voltage resistor in series with it. This can be made using nine 10M resistors in series, each resistor needs to be rated to 1/10th of voltage being measured.

The current rating is not a problem. It's better to go higher than lower. Using a power relay to switch low current signals is only a problem at very low voltages.

The problem with using a lower voltage relay is the contacts can arc over and what's worse it it could be unpredictable.The relay may seem fine most of the time but it could suddenly arc over due to ambient ionising radiation.

It could be used as a switch, amplifier or (as you've said) part of a current limiting circuit.

Bipolar transistors have so many applications that it's impossible to tell you what it's doing from what you've said. For example if I asked you "I've seen a flat head screw; what's it holding together?", how could you know the answer?

If you post a picture I might be able to tell you but even then the only way is to post a schematic.

A schematic would help us to answer the question.
I wouldn't recommend either. A MOSFET would probably be better.