That's true, depends on the situation...😅

@@thedoubleeguy I mean Laplace transforms are lowkey just shortcutted differential equations. Like if you didn’t have a transform table you’d just be doing differential equations anyway. But in most circuit analysis I’ve seen as a third year electrical student we prefer Laplace.

@@albertohart5334 You're correct, using the Laplace transform is another valid and often more common way to solve this type of problem especially in the electrical engineering field. I just decided to use variation of parameters for this specific problem in order to demonstrate the concept.

Hey brother, that's a fantastic question. First, we know the LED has a voltage drop of 1.8-2.2V so we can take the average to be around 2V. Two double AA batteries in series are going to add up to a total V supply of 3V so that means we're going to have about a 1V voltage drop across the resistor. Depending on your configuration, you probably have around 20 mA of current running through your LED so the ideal resistor value is going to be R = V/I. After doing the math, a 50-100 ohm resistor should do the trick. If you have other elements like an inductor in your circuit with a switch, that could also cause some serious problems. An inductor resists change in current, so when you switch off your circuit, the inductor will provide whatever voltage it needs to maintain the same current. This could give you upwards of 100 V in your small circuit which could fry diodes and transistors. I hope this made some sense. Worse comes to worse, you can use a heftier resistor but it really depends on your configuration.

@@thedoubleeguy Thank you, Brother! Makes perfect sense. 🤠