Hi,
In first post you did not mention "current", now you say you want a "current measurement".
So show
* a clear description
* a schematic - even hand drawn
* your calculation of current using Ohm´s law (How you come to 10mA or 3mA with the values of 60,000V and 1kOhm)
.. I´m close to consider the whole thread as useless ...
Klaus
Ohms law:
I = V/R
I = 60000/1000 = 60 Amps
W = VI = 60000 * 60 = 3600000 = 3.6 MegaWatts
Your resistor will probably vaporise (unless it is a 3.6 MegaWatt Resistor)
Your multi-meter will vaporise (or if you are lucky, will blow a "thought" fuse). EDIT: only a "thought" multi-meter will withstand 60000 volts. A real multi-meter will vaporise even with a blown fuse.
Treez seems to have a different understanding of your schematic than me, but we are both guessing. Agree with KlausST that clearer information is needed.
Also agree with KlausST about useless thread. Seems like nonsense to me, but maybe I just do not understand.
EDIT: wwfeldman beat me to it
but the voltage source has some resistance, we just don't know how much, the total resistance of the circuit is much higher, which reduces the current to the milliamp range. i just want to know if there is any reason to think the multimeter won't give accurate current values?
Give clear informations.the multimeter is suppose to measure current
from your diagram, the meter will measure 60000 volts...i think you know this?
There will be no current because an ideal voltmeter has no current flow....but a non-ideal one......
I think you guys have properly scared him off this thread/site now ...
I think you guys have properly scared him off this thread/site now ...
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