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Earlier I asked for help about a high voltage power supply, and I was told not to worry too much about 20k volts because they have a very low current
I was left thinking about that
say whatever circuit that ends in an inductor and generates that voltage of 20k volts.
and I was told that the current runs at around 200 mA
how does this work?
I mean I was taught that V=iR
and AFAIK the resistance of a wire (whatever material) is dictated by its material properties, its density and whatnot, so even tho its not a "perfect wire" it exhibits certain resistance R that should be (should be?) fixed, right?
so If R is fixed, how can V and i be fixed too?
example:
I found a table that says that copper wire has 0.001588 Ohms per foot
lets say I have a 10 ft wire running those 20kV, so that means that the current
i= V/R
should be 20000 V / (.001588*10) Ohms= 1259445.84383 Amps
how come the current is 200 mA ?
is there some alteration to Ohms law or what's going on here?
I was left thinking about that
say whatever circuit that ends in an inductor and generates that voltage of 20k volts.
and I was told that the current runs at around 200 mA
how does this work?
I mean I was taught that V=iR
and AFAIK the resistance of a wire (whatever material) is dictated by its material properties, its density and whatnot, so even tho its not a "perfect wire" it exhibits certain resistance R that should be (should be?) fixed, right?
so If R is fixed, how can V and i be fixed too?
example:
I found a table that says that copper wire has 0.001588 Ohms per foot
lets say I have a 10 ft wire running those 20kV, so that means that the current
i= V/R
should be 20000 V / (.001588*10) Ohms= 1259445.84383 Amps
how come the current is 200 mA ?
is there some alteration to Ohms law or what's going on here?