# Complete elementary beginner question

I know this forum is well advanced for the question I have so be kind :)
I have always saved DC adapters from all my electronics and often
reuse them in other devices. However today is the first time i've decided to try and learn what the significance of the volts and amps are. i usually just look for an exact match for instance 9V 1.2A + polarity etc.. but ive noticed that sometimes I can use an adapter with smaller voltage than the device specifies and it still works.
So based on my very rudimentary lesson on this today i'm now aware that the volts and amps combine to produce the power output the device needs. But can someone tell me if I have this understanding correct... lets say I have a device that requires an adaptor that is 12V .2A. Based on the formula of P = VI, wouldnt that mean that I could also use an adapter that was 6V .4A?? since both provide 2.4 watts of power??
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NewYawkah wrote:

Well, everyone has to start somewhere, and I see a surprising number of people who lack an understanding of this.
When speaking of a power source, the Amperage rating is the capacity, in other words it's the maximum amount of power the source is rated to supply. You could have a 12V 1,000A power supply and if the load only wants 0.2A, that's all it will draw. One caveat here is that many "wall wart" type power supplies are unregulated so if the load draws a lot less than the rated current, the voltage can drift up too high for the load. This is why if you measure the voltage without any load at all, it will often be 25-50% higher than what the label says, but will sag down at full rated load.
In a nutshell, pick a power supply that puts out the voltage the load wants, and make sure the amperage is equal or greater than the max the load will draw. Some loads are not at all picky about voltage, others are, knowing what you can get away with lacking an exact match is where you have to know something about the device in question.
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thanks so much for the explanation. So when you hear of "frying" a device for instance, would that mean then that its the voltage that is the more important variable not to overdo? I guess i'm not yet clear on the water hose analogy all the teaching materials use because they say voltage is the water pressure and current is the flow rate but in my mind, assuming a fixed diameter hose or resistance, i imagine if i put more pressure on the water coming through the hose, the flow rate would naturally increase as well wouldnt it?
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One additional rating must be taken into account. The power rating. As these devices are cheap and dirty units, the open circuit voltage can be much higher than the loaded voltage. Thus if you use a unit that is rated for 1 amp for a device that requires only 200 ma, you stand a chance of over voltageing it.
CP
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Excessive voltage can certainly fry something, but equally as common is connecting a power supply with the wrong polarity, that is the connector is wired such that + is where the device expects - and - is where it expects +. Some stuff is built with this in mind, plenty is not. Other devices expect an AC voltage, some just convert it to DC internally, others use the AC for some function and won't work properly on DC.
Water is not a perfect analogy for electricity, but it is one that generally works. Think of the amperage rating of a power supply as the diameter of a pipe carrying water and the voltage is the pressure. If you have a big fat pipe supplied by a constant pressure, you can attach many smaller pipes (loads) which will all see the same pressure (voltage) so long as the big pipe can flow sufficient volume to supply all the small ones.
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In article

You have the power formula OK but you do not understand it yet. Go read the Radio Amateur's Handbook. Wikipedia might have a good introductory section on electricity.
Learn about impedance matching. That will explain why voltage and current need to be properly chosen.
Meanwhile, thing of a lever or jack as a mechanical impedance matcher. To lift a car you need a certain amount of force. That is analogous to voltage. You also need to move the car some distance. Think of the rate of lifting as analogous to current. Both have to match the job.
Bill
--
An old man would be better off never having been born.