First up - I am an electronics noob, so I make any dumb mistakes, please correct me.
I have a LED strip. It takes an input voltage of 12V DC, and it's power is 24 watts. If I understand the calculation correctly, this is about 2 amps. If I have a 3 amp 12V DC power supply, how do I make sure that I don't overdrive the LEDs?
By Ohm's law, I calculate that a resister of (E/I, or Voltage/current) of 12/2 = 6 ohms is what I need. The thing that is confusing to me is the wattage of the resister. Do I need a 24 watt resister? Or do I need a (12V * 1amp) 12 watt resister? Or something else, as it seems like most resisters I have seen (and that I have) are under a watt.
The led strip most likely has built in current limiting resistors
and remember just because you have 3 amp supply doesn't mean your forcing 3 amps through it, you just have the ability to draw more, not that you have to
you could have a 12v 100amp supply and it'd be the same
Simply because it is the voltage that provides the "pressure" to force the current through the system. The LED strip is designed to operate at 12 volts (without any external current regulation) and it's internal design is such that it demands 2 amps. It matters not what your supply is capable of delivering, be it 2 amps or a 100 amps. As long as the drive voltage is 12, then 2 amps is demanded and that is what the power supply will deliver.
No, ohms law does not apply because a) ohms law only relates to linear circuits and b) the LED strip is a non-linear system. Altering the voltage feeding the LED strip by say plus or minus 10% will not necessarily produce a corresponding 10% change in current.
Linear circuits generally relate to passive components (for example resistors) whereas semiconductors are either active or non-linear devices.
No, ohms law does not apply because a) ohms law only relates to linear circuits and...
Just to clarify - Ohm's Law ALWAYS applies, but since in this (non-linear) set-up the resistance changes when we change the voltage, it might not be directly useful here.
Similarly, with capacitors and inductors, the voltage and current can be out-of phase so it can appear that Ohm's Law is not working correctly, but it's a physical law that always holds.
Power supply Amps rating is like the water tower down the road. Just because it can supply the 10 inch pipe coming out the bottom doesn't mean that my 1/2 water line going to the outside faucet can carry the same capacity.
Just to clarify - Ohm's Law ALWAYS applies, but since in this (non-linear) set-up the resistance changes when we change the voltage, it might not be directly useful here.
Similarly, with capacitors and inductors, the voltage and current can be out-of phase so it can appear that Ohm's Law is not working correctly, but it's a physical law that always holds.
If you believe this then perhaps you would be interested in a good deal on the Brooklyn Bridge.
Ohm's "law" is an empirical law based on observations of the behaviour of many kinds of conductor including metals and semiconductors - it generally holds for most kinds of conductor at normal temperatures and "low" current densities. It is an emergent property of electron (or hole) scattering.
It DOES NOT HOLD IN GENERAL. It doesn't work for superconductors, it doesn't work for thermionic valves, it doesn't work for gasses / plasmas, it doesn't work for semiconductor junctions nor for metal-semiconductor junctions, it breaks down at very high current densities as the assumption that drift velocity is much less than rms charge carrier velocity breaks down. It doesn't work for quantum-tunnelling devices.
Well as regard to power calculation then ohms law does apply in the respect that the voltage drop across a device is proportional to the current through a device. The power being the product of voltage and current.
In "normal" materials the constant of proportionality is unchanging with current or voltage. It can change with other things like temperature, light, magnetic field and so on.
However, in the situations mentioned above the constant of proportionality changes depending on the voltage and or current. So a device has an effective instantaneous resistance that is not in any way a constant. It can be used to calculate power dissipation but at that instance, and under those conditions only, and can not be extrapolated into other conditions like you can do with "normal" materiel.
So from the title of this post the simpler the question the more complex the answer can be.
