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thought experiment to understand measuring equipment.

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dl09

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this is just a thought experiment, if i connect a 60,000 volt high voltage module to 1 end of a 1000 ohm resistor, and connect the other end to 1 prong of a multimeter and connect the other prong to the 60,000 volt high voltage module, completing the circuit, is there any reason to think the multimeter won't give accurate values? the 1000 ohm resistor won't get burnt out. the multimeter does not seem to be damaged.
 

ok but you dont have a circuit..and given an ideal multimeter it would say 0 volts
 

but in this thought experiment, the multimeter is suppose to measure current and the current is initially in the microamp range, then the current increase to the milliamp range. the current values change from 1 moment to the next, and the maximum current value is 10 milliamps and the minimum value is 3 milliamps. the current values you get are in that range. any reason to think these current values won't be accurate?
 

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
 

60,000 V through a 1000 ohm resistor to a current meter and back to the
60,000 V return is essentially 60 A

if the meter is a voltmeter, then the current will be about 60,000 V / 100Meg ohm (?)
or 0.6 mA (taking a guess at the impedance of a voltmeter)

so either i don't understand the original post,
or your thoughts are in need of a second thought

draw the circuit
label all the components
be sure it is readable
re-think
then re-post
 

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
 
Last edited:

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

2e6232ca-3552-4a5d-a647-fe20e10b887b.jpeg
here is a schematic of the circuit.

- - - Updated - - -

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?
 

Please stop with the crazy thought experiments and the wildly random assumptions....

Buy a 6 Volt battery and a few 6 Volt old-fashioned filament light bulbs and do some real experiments with a real multi-meter in circuit.

Wildly guessing about how electricity works without any understanding is just frustrating.
 
Last edited:

Hi,

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?

* but the voltage source has some resistance --> then show/draw the resistance in the circuit. How else can we know? And please mention those informations on the first post.
* we just don't know how much --> we don´t know either. If you don´t tell us, it is impossible to calculate even impossible to validate any thing of your ideas ...making the discussion useless.
* the total resistance of the circuit is much higher, which reduces the current to the milliamp range. --> you didn´t mention it before. Draw a circuit will all your informations in it.
* i just want to know if there is any reason to think the multimeter won't give accurate current values? --> the reason is in the information you hide...

Your circuit:
Why is there a voltmeter whan in post#3 you say:
the multimeter is suppose to measure current
Give clear informations.

Klaus
 

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......
 

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'm not so sure this is a thought experiment.
 

I have this picture in my mind of 40,000 'D' batteries in series and a poor little 0603 SMD resistor......

Brian.
 
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    d123

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Haven't read a confused thread like this since long.

Post #1 says "multimeter" whichout specifying a range.
Post #3 states it's operated in current measuring range
Post #7 however shows a voltmeter

Similar confusion about voltage source internal resistance
 

Hi, dl09,

Besides continuing dangerous lack of learning of elementary principles, even Ohm's Law, any DMM has specifications and a user manual - therein lies accuracy of voltmeter, ammeter, ohmmeter and whatever else it can measure. You don't understand, mate, most non-precision off-the-shelf consumer DMMs are +-5% accurate, at best, and cheaper ones won't auto-zero or have temperature compensation, etc. I sometimes wonder if your threads are real questions or tongue-in-cheek and sarcastically so bad intentionaly for teaching rather than learning purposes.
 

I think you guys have properly scared him off this thread/site now ...

I really hope not, too. Enthusiasm and curiosity are good things. Can only wish poster well and to continue with interest. From similar beginnings approach I think I had, meaning skipping learning before doing or considering xyz circuit properly or understanding need to understand components and theory well enough, and reaching for calculator being imperative, only mean I hope they do things safely and comprehend that a little research and basic rules go a long way.
 

Back to the question. That's my understanding of the discussed circuit:

60 kV voltage source, internal resistance limiting the current to maximal 10 mA, 1k resistor, current meter (multimeter in current range)

circuit.png

Alternatively, with a multimeter in voltage range in place of the current meter (according to post #7), the high voltage will cause a flashover inside the instrument and most likely destroy it.
 

@dl09
don't run away
we're trying to help

as you can read, everyone has a slightly different take on your question

but once you posted the drawing, everyone is on the same page -
the voltmeter will read 60 kV and there will be current that depends on the
impedance (resistance) of the voltmeter on the range that can read 60 kV.
the 1000 ohm resistor will be of no consequence, as the resistnce of
the voltmeter should be significantly higher.
and the internal resistance of the source may (or may not) be significant.

if FvM is correct in post 18, the internal resistance of your source seems quite high

basic lesson:
voltmeters are wired in parallel with the voltage you want to measure
in order to not disrupt the circuit, they have very high resistance, so very little current gets diverted through the meter.

ammeters are wired in series with the current you want to measure
in order to not disrupt the circuit, they have very low resistance, so very little voltage gets dropped across the meter.
 

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