The way I read it is that figure 5025 is milliamperes, not millivolts, i.e., the current drawn by the pump when running (5.025 amps).
What you are measuring is voltage drop E across the fuse. The fuse has a certain resistance R and, by Ohm’s Law you can calculate the current I drawn. The table assumes a specific (unstated) resistance for the fuses.
E=IR
Therefore I=E/R


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