Tested the fuses by hooking them in series with a 55W 12V bulb with a set of "alligator" clip cables, and powered it with a 13.7V 20-amp power supply. According to ohms law (I=W/V) a 55W bulb should draw 4.01amps if fed 13.7V, which is exactly what I measured. 55W/13.7V = 4.01A.

Next I connected one fuse at a time in series without altering any other aspect of the circuit, and measured the voltage drop across the fuse itself. Again using ohms law I determined the internal fuse resistance by using the R=V/I formula with the known 4.01A current in the circuit.

Fuse ---- Voltage ---- Resistance
----------------------------------
8A red --- 0.083 V -- 20.69 milliohms
16A blu -- 0.043 V -- 10.72 milliohms
25A wth -- 0.032 V -- 7.98 milliohms

These resistance values appear to be quite consistent according to the metal cross section of the fuse. The larger the fuse rating, the larger it's metal cross section will be, and consequently it should have a lower resistance.

Now by using the known estimated internal resistance of each fuse, one can calculate the estimated actual voltage drop across it when subjected to its rated current rating, which would be pretty much the worst case scenario. For this I used yet another version of the ohms law formula: V=IR.

8A x 20.69 milliohms = 0.165 V
16A x 10.72 milliohms = 0.172 V
25A x 7.98 milliohms = 0.199 V

These IR (voltage) losses seem to be very reasonable, and should be proof that in my case I seem to have gotten good fuses from the "fastvw99" ebay seller.

Due to the overly positive test results, I did not bother to compare them to my existing aluminum fuses, but might do so later on just for the fun of it.

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W163 ML230 EURO SPEC 1999
W124 300E 1986