Okay Guy but forgive an old fella who isn't up to the latest, but is 9W the power of the entire MR-16 appliance or just the bulb itself
Incidentally I looked up the MR-16 and find it's not an LED appliance but halogen. A rewiring was suggested so I'd assume it's indeed now LED
Thus it would be interesting to learn whether the appliance contains a transformer or a resistor stepping down to the 2V or 3V required by a typical LED. If it seems heavier than you'd think it should, it's probably a transformer. To just the LED that's P = 3V x 3A = 9W, quite a potent LED I must say but a thorough Googling on the subject would take me a week
Yes at 84 I'm kinda slow
With internal resistor on the other hand you might note appliance shell getting pretty hot. Now 120V at almost 3 amp P=EI gives a power waste of approaching 400 w, hot as hell, so I'd guess transformer instead to give you P = EI at 3V x 3 A = 9W
But returning to the title I note a reference to 12 v, again suggesting that internal transformer. However the link above indicates that some such replacements use a switch-mode supply instead of transformer, suggesting DC might be preferable over AC
Let us mull it over a bit longer, and in the meantime just try different resistors.
But if the entire appliance
uses 9 watts for an effective current of roughly 0.1 A, then its effective resistance is (say) R = E/I = 120/ 0.1 = approx 1200 ohms so if you use a resistor of this value it should cut the current in half . Wattage? P = EI = about 120/2 x 0.1/2 = 3 W
.....maybe cutting brightness by two-thirds. We're getting into the relationship between voltage and wattage here, Guy, so let's put it off for the time being
If you feel I've been of no help, Guy, I can well understand. If anybody finds a typo or two or an outright miscalculation, or heaven forgive me, Guy, you attempt a reply to one of the earlier postings... it's because the a2k software had cut me off.....