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Neon discharge as high resistance value.

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  • harvey norris
    In the preceeding RCB neon discharge shown as http://groups.yahoo.com/group/teslafy/files/MARX/DSC00091.JPG there appears to be the discrepancy of the primary
    Message 1 of 1 , Mar 24, 2002
      In the preceeding RCB neon discharge shown as
      http://groups.yahoo.com/group/teslafy/files/MARX/DSC00091.JPG
      there appears to be the discrepancy of the primary
      input where apparently 336 ma is being inputed at 7
      volts, but a voltage rise occurs showing 8.35 volts
      inputing .29 Amps in the loop, thus with that voltage
      rise is the loss in amperage in the loop. These
      voltage and amperages as the reactive power input show
      approximately 2.35 watts. However at the primary only
      154 ma is measured, making the reactive input only
      8.35 * .154 = 1.28 watts. The reactive power
      measurement at secondary is also 1.92 watts.

      In that case then we have used 600 volts and 3.2 ma
      for the reactive input measurements to yeild 1.92
      VAR,(volt ampere reactive) It is known that VARs do
      not yeild true power input which is always less than
      power input derived using reactive component
      measurememnts. We might then assume that the true
      power input figure must be closer to that derived at
      the primary by 1.28 watts.
      Let us then compute the acting resistance of the
      discharge if 1.28 watts were expended. We already know
      that a measured 3.2 ma is occuring, but that the
      reactive power mesurements produce a wattage 50%
      greater that that inputed at primary. Because of the
      fact that dual channel scopings show little or no
      phase angle deviations between the applied and
      midpoint voltages, we might further suppose that those
      reactive power primary input measurments are in fact
      identical to true power ones.

      Thus we might initially consider the equation 1.28= I
      squared R, where R is to be determined by determining
      I^2= (.0032)^2= 1.024 *10^-5, where the R value then
      becomes that multiplier making the result 1.28, or the
      very large value of 125,000 ohms.

      Comparing these values to what actually occurs with
      the 600 volt secondary voltage, if the 125,000 ohms
      were purely resistive, we would expect a 4.8 ma
      conduction. It then appears as a higher ohmic value as
      a consequence of its reactance being higher than the
      resistive. In any case it seems to be easy to classify
      the neon discharge as having a very high resistance
      filament, and not a low one as formerly imagined. HDN



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      Tesla Research Group; Pioneering the Applications of Interphasal Resonances http://groups.yahoo.com/group/teslafy/

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