How to make low current output for flyback transformer??

[claydioactive]

So Recently I have been wanting to make a Jacobs latter and I remembered I had a massive 26.5KV flyback transformer from a large crt projection television and figured It would be quite sufficient. I then had realized I had been told "If I catch you playing with lethal voltage, Ill take away all your equipment and components!". So I started thinking and remembered the overused term "its not voltage that kills its the current" . So is there a way I could get a output below (Hopefully) 5 milliamps and still be able to make a good sized Jacobs latter?

[MagicSmoker]

https://www.repairfaq.org/sam/jacobs.htm

[Zero999]

It's a little more complex than that. The output current is not mains frequency AC so is less deadly. In any case, there's no easy way to reliably limit the current to a safe level, especially with an unknown transformer. Follow safe working practises you won't get hurt.

[claydioactive]

Would It be possible to limit the input current extremely low using a current limiting resistor? And also, I have always wondered about power gain and loss from transformers. Lets say I input 10 volts 2 amps AC into a transformer and get 100 volts out. Does the amperage get converted into a higher voltage? I mean, Its impossible to gain energy in a transformer, Theres actually a loss right?

[Zero999]

Would It be possible to limit the input current extremely low using a current limiting resistor? And also, I have always wondered about power gain and loss from transformers. Lets say I input 10 volts 2 amps AC into a transformer and get 100 volts out. Does the amperage get converted into a higher voltage? I mean, Its impossible to gain energy in a transformer, Theres actually a loss right?

Of course it's impossible to get extra energy from a transformer. In a perfect transformer, if the voltage is multiplied by 100, then the current should be divided by 100.

In a real transformer, the voltage drops, as more current is drawn due to copper losses and leakage inductance. If you know the turns ratio, you could theoretically limit the current to the primary winding, therefore the current in the secondary by a factor of the turns ratio.