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I have a Fluke clamp on reading a 11 amps on for an AC load. I bought two different AC-DC current probes for the scope. The DC readings are great and match dc loads. When reading the AC load for the 11amps on the scope, I get 160mv peak or 320 peak to peak. The probe claims 40amp scale use 1mv=100ma. This gives me 16,000ma peak or 32,000ma peak to peak. How does this calculate to 11amps?

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AC DMM's assume sine input and convert to RMS

so if they measure Vpp then conversion is 1/(2√2) to \$I_{RMS}= \frac{32A_{pp}}{(2*√2)}=11.3A\$

If your input is clipped it will read less than true RMS or spikes more than true RMS.

and if they measure precision FW rectified average it is converted again to RMS=Vavg x 2/π or Vavg= 0.637Vpk or Half Wave rectified RMS=1/π * Vavg

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  • \$\begingroup\$ Some DMM are true-RMS DMM and don't assume a sine input. It will still be the RMS value ultimately and not Vpp. \$\endgroup\$ – Orhym May 5 '18 at 16:45
  • \$\begingroup\$ True, If he gets what I computed can you assume I am correct. and if says TRUE RMS, they will understand "True" means independent of waveform but perhaps BW limited. Or did you think I was not aware of that? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 '18 at 17:11

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