Resistors in Series and Parallel 

Thanks! Glad you found it helpful.

Thanks!

You need to calculate the power for each resistor individually to make sure none of them exceed their individual power rating.

Hi Keith! You would have to try and look up the specs of the board and see what the "output impedance" or "internal resistance" of the pin is. A mathematically ideal pin has zero impedance when used as an output (i.e. voltage does not drop at all when it's loaded) and infinite input impedance (does not draw any current from whatever it's measuring). In reality, the pin has some finite impedance in both cases. So when used as an output, and you add a pull-down resistor, you actually have two resistors in series forming a voltage divider, which will drop your output voltage. If you can find the internal resistance of your pin, then in theory you can calculate that voltage from this equation (where R1 is the internal resistance, R2 is your pull-down resistor, Vin is the ideal voltage from the board, e.g. 5V, and Vout is the actual output voltage): en.wikipedia.org/wiki/Voltage_divider#Resistive_divider (note that technically, this is assuming that whatever else you have your PWM pin connected to is not drawing any current, and all the current flows through R2). So, depending on what your minimum acceptable voltage is, you could calculate the minimum resistor value you need. However, that's all a very long-winded way of saying "just try a bigger pull-down resistor." I think 10k is a pretty common value. I don't think it *hurts* to oversize the resistor.

Hi - this is kind of a lot to answer in a single comment, you might want to check out my intro to circuits playlist: ru-vid.com/group/PLKL6KBeCnI3U6KNZEiitdtqvrxkBhpuOp.

It depends on the circuit, but in general for most beginner/hobby circuits it probably won't matter. Two resistors take up more physical space than one resistor, so it could matter if you were designing a printed circuit board (PCB). However, assuming the resistors all have the same power rating (1/4 watt is the most common), then the two resistors would be able to dissipate 1/2 watt total. So if for some reason you need to pass a lot of current through the resistor but don't have a 10K resistor rated at 1/2W, then you could use the two 5K resistors.

@@BenFinio i see, ty ty that was bugging me from the start lol

Hi - it's common to use current-limiting resistors with LEDs (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-EeCh68a1GEg.html) but they are not typically used in this manner with other types of loads. For example with a motor you would either decrease the voltage or use pulse-width modulation (PWM) to decrease the speed, not put it in series with a resistor. I'm not that familiar with sirens so I'm not sure of the best way to control the volume.

That's more complicated than what's going on in this video with just resistors, as it depends on the rest of the circuit connected to the blinker.

1/(1/100+1/47)=31.97

1÷(1÷100+1÷47)=31.97

Hm - is the "donation" mandatory, or can you just close that and download anyway? It's been a while since I downloaded it but it was free at the time. Tinkercad is a good alternative that runs in a browser and is free (for now at least).