Is it really as simple as W=V*A???
The issue is that I want to implement a "Phantom On" LED on a Pre. I understand the current limiting side to arrive at the value,
R=(V1-V2)/A1. [V1 being supply voltage, V2 being voltage drop across LED, and A1 being draw]
R =(48-2)/0.015
R = 3,066 (nearest size being 3.09k)
So once you plug the numbers into the watt equation....
W = 48 (or 46 to accommodate the voltage drop?)*0.015
W = 0.72
So do I need a 3.09k 3/4W resistor for this application?
The issue is that I want to implement a "Phantom On" LED on a Pre. I understand the current limiting side to arrive at the value,
R=(V1-V2)/A1. [V1 being supply voltage, V2 being voltage drop across LED, and A1 being draw]
R =(48-2)/0.015
R = 3,066 (nearest size being 3.09k)
So once you plug the numbers into the watt equation....
W = 48 (or 46 to accommodate the voltage drop?)*0.015
W = 0.72
So do I need a 3.09k 3/4W resistor for this application?