Electronics > Beginners
Some noob questions
rstofer:
All of these complications with the analogy is why it is used for the first half hour of the first class and then put away, never to be discussed again.
All that is hoped for is an understanding that volts pushes current through resistance. The analogy should stop there because all the other electrical effects can't be described with the analogy. I had forgotten about mutual inductance, mentioned above. That clearly doesn't happen with water in a hose.
It is sufficient to understand that higher pressure in a hose will result in more water flow. Add to that the fact that more water will flow through a big pipe than a small pipe for the same pressure and you have just run out of the analogy. We're done, it's time to move to Ohm's Law and start playing with DC circuits.
OTOH, it isn't necessary to understand the physics behind electron flow. At least not until AC circuits and probably not until RF or semiconductor physics. Ohm's Law and Kirchhoff's Laws are sufficient for most anything a hobbyist is likely to run across unless they are into amateur radio.
The analogy is useful in that most people have played with a garden hose but beyond the trivial pressure, flow rate, pipe resistance, the analogy isn't helpful. Even flow resistance isn't a linear kind of thing. There are boundary conditions, laminar and turbulent flow and other issues that are not appropriate for consideration in the analogy. That's why there is the entire study of fluid dynamics. I wonder if they use the electrical analogy for the first five minutes of the discussion.
asmfan:
--- Quote from: rstofer on July 16, 2018, 09:28:54 pm ---It doesn't matter much because Ohm's Law relates all 3 variables and we can swap them back and forth at will.
--- End quote ---
For all practical purposes this statement is a useful simplification, and holds true when working with ohmic materials.
This simplification does not work when working with non-ohmic materials such as when experimenting with static electricity.
--- Quote ---Usually you see Ohm's law written in the following form:
V = I x R
However, in this form it is tempting to define voltage in terms of resistance and current. It is important to realize that R is the resistance of an ohmic material and is independent of V in Ohm's law. In fact, Ohm's law does not say anything about voltage; rather, it defines resistance in terms of it and cannot be applied to other areas of physics such as static electricity, because there is no current flow. In other words, you don't define voltage in terms of current and resistance; you define resistance in terms of voltage and current. That's not to say that you can't apply Ohm's law to predict what voltage must exist across the a known resistance, given a measured current. In fact, this is done all the time in circuit analysis.
--- End quote ---
- Paul Scherz and Simon Monk. "Practical Electronics for Inventors" Fourth Edition pg 24
rstofer:
--- Quote from: asmfan on July 17, 2018, 03:38:40 pm ---
--- Quote from: rstofer on July 16, 2018, 09:28:54 pm ---It doesn't matter much because Ohm's Law relates all 3 variables and we can swap them back and forth at will.
--- End quote ---
For all practical purposes this statement is a useful simplification, and holds true when working with ohmic materials.
This simplification does not work when working with non-ohmic materials such as when experimenting with static electricity.
--- End quote ---
And perfectly adequate for the purposes of a newbie thread. There is no point in expanding the discussion at this point in the learning process. Close enough...
Mr D:
Many thanks to all for being exceedingly patient with me, i'm gonna take my time and read through this thread a few times.
I want to ask one more favour. Could someone tell me the values of a few resistors i should buy that i can use to connect to my Fluke and power source so i can experiment with calculating ohms law? I guess the resistance should be in a range that it'll be clear on my multimeter, but not so low that i risk damaging my power supply or Fluke.
And does anyone have some links to a good lab power supply that will be useful for testing Ohm's law? Doesn't have to be very cheap, but i also don't want to spend more than necessary. There's so many available, i have no idea what to get.
Rick Law:
--- Quote from: Mr D on July 17, 2018, 07:59:36 am ---
--- Quote from: Jwillis on July 17, 2018, 03:08:29 am ---The flow or rate that the water moves is called "Amps" or Current.
--- End quote ---
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.
--- End quote ---
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.
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