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There's a heap of these questions, so before you mark mine as duplicate I'll try and explain how mine is different.

I don't really know too much about voltage and current, wattage, resistance or ohms.

If I supply my Pi with a computer power source, versus a wall charger, how is its needs met? Will it continue to draw until the power source breaks, or will the Pi be under-provided? What if the power source is too great? I understand too high in voltage will destroy the device, but what about too high in current? Will it draw what it needs (which seems to be suggested everywhere (and also makes no sense to me)) or will it break?

I'm probably the most confused with LEDs and drawing energy, because multiple sources say a resistor is in place to protect the Pi, and not the LED. To my knowledge, a resistor is because the current coming out is too high, but posts I've read talk about too much demand on the GPIO pins. What does that mean? Does the Pi try to provide the energy and can't (breaking the Pi)? or is the LED provided with too much current (breaking the LED)?

closed as unclear what you're asking by Milliways, joan, goldilocks Jun 24 '17 at 11:50

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • There is no such thing as too much current See Raspberry Pi Power Limitations. What is your ACTUAL question (or is this just a rant)? – Milliways Jun 24 '17 at 9:27
  • This has nothing to do with the Pi. You need to find a basic electricity tutorial. – joan Jun 24 '17 at 10:45
  • Welcome -- but I do not see a coherent question here. Please take the tour to understand better how the site works, and read "What types of questions should I avoid asking?". Note that general questions about electricity belong on our larger sibling site, Electrical Engineering. – goldilocks Jun 24 '17 at 11:50
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    I will clarify one thing for you: "a resistor is because the current coming out is too high, but posts I've read talk about too much demand on the GPIO pins" -> Because in theory there is no limit on the amount of current that could come out if you, e.g., attached an output GPIO to ground driven high. However, in reality a short circuit will ruin the GPIO and/or Pi before infinity current is reached ;) The point is you must make sure the current is limited, based on resistance, to within the working limits of the pin (~20 mA). The pin will not do it all by itself. – goldilocks Jun 24 '17 at 11:54
  • It's not a rant, I'm just ill informed. I've never done anything in electronics. How do you expect that I know? @Milliways – yeeeeee Jun 24 '17 at 14:10
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What about too high in Current? Will it draw what it needs or will it break?

About the amperage, a circuit will draw what it needs. The rule is to keep the voltage matching with the source, and to have a bigger amperage on the power source than the need of the device.

If your first intuition was true (devices are breaking), a lot of electronical devices wouldn't work ! I.e. consider a speaker amplifier. It's consumption varies over time, depending 1) on the gain setting : more volume, more consumption 2) on the sound given as input : amplifying a silence consume less that amplifying a sound.

Obvisouly, sound engineers don't change their power supply when they raise or lower the volume button, nor at each drum kick ! Instead, they use a power supply big enough (in ampers) to work at high volume.

Does the Pi try to provide the energy and can't (breaking the Pi)? or is the LED provided with too much current (breaking the LED)?

Both may happen ! Take an exemple : Consider a LED on the GND/V+ GPIO Pins, without resistor.

The formula : I=U/r show that we could get an almost infinite intensity (Ampers) consumption when the resistance is negligeable. This would lead both the LED to break, and the polyfuse (an inboard over-current protection) to trigger. In this case, it's hard to predict the first element to break, it's a course between the fuse and the LED.

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