MEASUREMENT OF LEAKAGE CURRENT

Farmelectrics · Mar 20, 2023

I WAS ONCE SHOWN HOW TO CALCULATE INSULATION RESISTANCE USING A CLAMP METER. SO I NEED HELP AS IVE FORGOTTEN. CAN ANYONE HELP WITH THIS FOLLOWING FORMULA 230 + (20X10-6) = 11.5Mohms how have the arrived at this. A clamp meter has been clamped around live and neutral tails at the consumer and 2mili amps leakage 0.02mA is apparently 20micro joules. HELP

Farmelectrics · Mar 20, 2023

Farmelectrics said:
I WAS ONCE SHOWN HOW TO CALCULATE INSULATION RESISTANCE USING A CLAMP METER. SO I NEED HELP AS IVE FORGOTTEN. CAN ANYONE HELP WITH THIS FOLLOWING FORMULA 230 + (20X10-6) = 11.5Mohms how have the arrived at this. A clamp meter has been clamped around live and neutral tails at the consumer and 2mili amps leakage 0.02mA is apparently 20micro joules. HELP

Avo Mk8 · Mar 20, 2023

You've copied the (admittedly fuzzy) formula from the document wrongly. It's division, not '+'
Resistance = voltage divided by current!
and of course 20 times 10 to the power of -6 for micro amps

0.02mA is 20 microamps! Not 2 milliamps
And keep away from Joules - you want ohms law and resistance, not energy!

PS you would normally want to measure insulation resistance at increased voltage, eg 500v or 1kV if in doubt about condition of the wiring, so the clamp meter method is not ideal!

sum is 230/(20x10^-6) ohms
Same as 230/20 x 10^6
= 11.5 x 10^6 ohms
= 11.5 Megohms (or Megaohms if you prefer!)

Avo Mk8 · Mar 20, 2023

Duplicate post for some reason....

Farmelectrics · Mar 20, 2023

Avo Mk8 said:
You've copied the (admittedly fuzzy) formula from the document wrongly. It's division, not '+'
Resistance = voltage divided by current!
and of course 20 times 10 to the power of -6 for micro amps
0.02mA is 20 microamps! Not 2 milliamps
And keep away from Joules - you want ohms law and resistance, not energy!

PS you would normally want to measure insulation resistance at increased voltage, eg 500v or 1kV if in doubt about condition of the wiring, so the clamp meter method is not ideal!

sum is 230/(20x10^-6) ohms
Same as 230/20 x 10^6
= 11.5 x 10^6 ohms
= 11.5 Megohms (or Megaohms if you prefer!)

THANKS FOR YOUR HELP AVO