The flow or rate that the water moves is called "Amps" or Current.
Ok, but what is being measured here? The physical speed that the current flows? Because water can move at different speeds. Can charge also move a different speeds?
Or is it the amount of electric charge per second that passes a certain point?
Because in the water analogy, you could have the same volume of water moving past a fixed point, whether it's a wide, deep, slow moving stream, or a 1 meter diameter pipeline at extremely high pressure.
Mr. D,
We need to back up a bit here. Your wall-wart has the
potential to supply
energy to whatever you plug the power output into.
That potential to provide energy is measured in
Voltage and
Current it can supply to the load.
Voltage is measured in
V (volts), and
Current is measured in
A (Amps or Amperage).
Current is generally denoted as
I. When your output is merely your DMM, the DMM itself is a load. This is actually unsafe, but lets ignore that safety issue for the time being and just understand what went on.
* When the wall-wart is set to
3 Volt. The
2A reading means the DMM is taking up 2Amps when the DMM itself is the load. 2A*3V=6Watt, it is using up
6 watts of power. The DMM is a heater putting out 6 Watts of power somewhere inside the DMM.
* When the wall-wart is set to
12 Volt. The
2A reading means the DMM is again taking up 2Amps when the DMM itself is the load. 2A*12V=24Watt, it is using up
24 watts of power. This is a lot of power, when applied for enough time, it may melt the whole darn thing.
Even at 6 Watt, it is a lot of heat to dissipate. Your DMM can cook itself. Since your DMM so far survived, lets continue to see what is going on.
Case 1:(Your DMM is set to measure current)
Imagine you have a FAN connected to that output with the DMM in-line (in series) like this:
power+ to
DMM (red) DMM (black) to
FAN+ FAN- to
power-Now your DMM is measuring the
current flowing through the DMM itself. Since the DMM is in series with the FAN, the
current is the same as the
current flowing through the FAN. (Just as a garden hose, the amount of water going through the hose is the same as the amount of water going through the nozzle.)
Case 2:(First make sure your DMM is set to measure voltage)
On the other hand, if you have the FAN and the DMM connected in parallel and then to power like this:
power+ to
DMM(red) and
DMM(red) is also connected to
FAN+ power- to
DMM(black) and
DMM(black) is also connected to
FAN-Now, your DMM is measuring the
voltage being supplied by the power. Consider voltage (aka potential) is the pressure (how hard it is pushing). The higher the
voltage (the harder it pushes), the more
current would flow assuming the darn thing didn't burst. The amount of energy being delivered would be V*I (ie: Wattage =
voltage times
current, recall,
current is denoted as
I). With your DMM, it seems pushing at 3V or 12V results in the same
current (2Amps), so it was likely designed limit - just as the diameter of a garden hose limits the amount of water that can flow.
Why your initial test was unsafe:If you understand what is described thus far, I can explain why your initial connection (with DMM measuring current directly connected to the output) is unsafe. Without the FAN in between, nothing is there to use the power. So, the DMM is taking the full load. Good that your DMM limits that to just 2A - it is still not good for long, but it survives for the short duration of heating.
Case 1 and 2 wrap up:Now if you want to try that yourself, find a FAN that operates in 12V and 3V - or use an incandescence light bulb such as a bulb for your car. Make sure it is incandescence bulb. (LED bulbs designed to function at specific voltage. It would not do well in our voltage-changing experiment.) When you connect the bulb at 12V, you should see it lights up - you can measure the current (when in series as depicted in case1 above), or you can measure the voltage (when in parallel as depicted in case2 above), now you can evaluate the Wattage the bulb is putting out.
Incandescence auto bulbs (as resisters) is a good way for your experimentation until you can buy some real resisters. Remember, resisters (bulb) will heat up. A 24watt 12volt bulb will dissipate the heat properly as long as you don't over-volt the darn thing by applying more volt than it can take.
Depending on where you are, mail order may be the best way to get some real resisters. Tayda is good place to get electronic parts. Very few real resisters would take 24watts (as your initial connection).
https://www.taydaelectronics.com/If you understand the post thus far, you can dig into Ohms law where you can begin to figure our how V=I*R works.
Case 3:If you got this far, an interesting thing to see would be to remove the light bulb with your DMM (set to voltage) connected to the power alone. This is the NO-LOAD voltage put out by your wall-wart. Depending on design, this should be higher than when the FAN was also connected.
Footnote: when in series, the DMM actually caused a very small voltage drop, but that is too deep a lesson for now.