Measuring resistivity of dielectric liquid
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Is it possible to reliably determine the resistivity of a highly resistive liquid above the breakdown voltage (i.e. after ionising the substance)?
electricity
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$begingroup$
Is it possible to reliably determine the resistivity of a highly resistive liquid above the breakdown voltage (i.e. after ionising the substance)?
electricity
$endgroup$
add a comment |
$begingroup$
Is it possible to reliably determine the resistivity of a highly resistive liquid above the breakdown voltage (i.e. after ionising the substance)?
electricity
$endgroup$
Is it possible to reliably determine the resistivity of a highly resistive liquid above the breakdown voltage (i.e. after ionising the substance)?
electricity
electricity
asked Mar 30 at 15:04
J. DoeJ. Doe
506
506
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You can determine the resistivity of anything, by measuring the voltage that causes a current to flow.
It's only useful to say that you've 'determined' the resistivity if it's reasonably linear (so stays the same as the voltage varies) and constant (is the same tomorrow as today).
The ionised fluid after a breakdown due to excessive voltage tends to have a conductivity that varies over many orders of magnitude, determined by the power that's being dissipated in it, the output impedance of the power source, and any impurities in it. It is very, very non-constant, and non-linear. Any breakdown may release material from the electrodes, changing the impurity level.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can determine the resistivity of anything, by measuring the voltage that causes a current to flow.
It's only useful to say that you've 'determined' the resistivity if it's reasonably linear (so stays the same as the voltage varies) and constant (is the same tomorrow as today).
The ionised fluid after a breakdown due to excessive voltage tends to have a conductivity that varies over many orders of magnitude, determined by the power that's being dissipated in it, the output impedance of the power source, and any impurities in it. It is very, very non-constant, and non-linear. Any breakdown may release material from the electrodes, changing the impurity level.
$endgroup$
add a comment |
$begingroup$
You can determine the resistivity of anything, by measuring the voltage that causes a current to flow.
It's only useful to say that you've 'determined' the resistivity if it's reasonably linear (so stays the same as the voltage varies) and constant (is the same tomorrow as today).
The ionised fluid after a breakdown due to excessive voltage tends to have a conductivity that varies over many orders of magnitude, determined by the power that's being dissipated in it, the output impedance of the power source, and any impurities in it. It is very, very non-constant, and non-linear. Any breakdown may release material from the electrodes, changing the impurity level.
$endgroup$
add a comment |
$begingroup$
You can determine the resistivity of anything, by measuring the voltage that causes a current to flow.
It's only useful to say that you've 'determined' the resistivity if it's reasonably linear (so stays the same as the voltage varies) and constant (is the same tomorrow as today).
The ionised fluid after a breakdown due to excessive voltage tends to have a conductivity that varies over many orders of magnitude, determined by the power that's being dissipated in it, the output impedance of the power source, and any impurities in it. It is very, very non-constant, and non-linear. Any breakdown may release material from the electrodes, changing the impurity level.
$endgroup$
You can determine the resistivity of anything, by measuring the voltage that causes a current to flow.
It's only useful to say that you've 'determined' the resistivity if it's reasonably linear (so stays the same as the voltage varies) and constant (is the same tomorrow as today).
The ionised fluid after a breakdown due to excessive voltage tends to have a conductivity that varies over many orders of magnitude, determined by the power that's being dissipated in it, the output impedance of the power source, and any impurities in it. It is very, very non-constant, and non-linear. Any breakdown may release material from the electrodes, changing the impurity level.
answered Mar 30 at 15:19
Neil_UKNeil_UK
78.7k285182
78.7k285182
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