Quote:
Originally Posted by Findm-Keepm
He's 20V low, and the 39 ohms is dropping that. My original calculations called for a ~13 ohm resistor. (see post 27, page 2) Take three 39 ohm resistors, parallel them, and place them in the circuit. That'll give you the proper (lower) voltage drop, which should give you the proper B+, and plenty of overhead on the wattage. .33A X 13 = 4.29W, and you've got (3) 5 watters in parallel.
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Minor correction - it's current squared times resistance or (0.33 * 0.33) * 13 or 1.5 watts like you calculated earlier. So only 0.5 watts per 39 ohm resistor.