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Label on my phone charger says it's output is DC 5V and 2.1mA. I find this confusing, because at school I was thought that you can only talk about current when you know exact voltage AND RESISTANCE. So, using Ohms law, I thinks manufacturer assumes that resistance of my battery is exactly 5/2.1=2.38 Ohms.

But how can they assume that if people might use this charger for different types of batteries? Why don't they simply say it's 5V?

elecbegin
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    This question has been asked dozens of times before. (Usually from a more naive standpoint than yours) Short form: the current rating is a maximum value. – The Photon Oct 01 '17 at 17:47
  • @ThePhoton, there are many different shades of naivety, so referring to general "Power Supply Guide" may not meet the educational level of OP, and digesting the entire "guide" might not answer the immediate concern. – Ale..chenski Oct 01 '17 at 19:22
  • on older transformers (non-switching), the load matters such that you might not get 5v unless the load was 2.1A; ex: the output can be 9v open-circuit. _probably_ not relevant here. – dandavis Oct 02 '17 at 06:12

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That's not what that means.

A current rating of 2.1A means that the charger can support a max current of 2.1A. But it doesn't mean that it needs to output that at all times. It can just as easily output 1A or 0.5A.

Some chargers have a minimal output current required to stay active (like battery packs sometimes) or they shut down usually on the order of 50mA.

ratchet freak
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It is true, you can't say if a power supply/charger can "give" 2.1 A of current without loading it with a resistive load. What the manufacturer is telling you is that this charger won't be able to drive a load less than 2.38 Ohms (which corresponds to 2.1A as you calculated). If you load it more, either its voltage will drop down (as chargers do), or it will completely cut the output off.

As a matter of fact, your battery (or phone or whatever) doesn't have any definite resistance. There is an actual battery charger inside your phone/tablet that regulates the current intake. So the load might look like a 2.38 ohm from the Ohm's Law perspective, and the "effective" resistance will only go up (drawing less current) as the battery charge approaches the fully-charged state.

Ale..chenski
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  • Your last line needs to be de-garbled. Try "*... so it looks like a 2.38 ohm load in the worst case (flat) ...*". Note lowercase for SI units named after a person when spelled out. 'V' or 'volt', 'K' or 'kelvin' and just to be awkward, 'Ω' for 'ohm'. I wonder would Mr. Ωhm have liked this. – Transistor Oct 01 '17 at 19:04
  • @Transistor, I didn't fully get what you said, but I made some changes. – Ale..chenski Oct 01 '17 at 19:15
  • Sorry if I wasn't clear. "*So the load looks like a 2.38 Ohm or less ...*" should read so the load looks like a 2.38 ohm [lower case] **or more** ...*". (I know the load reduces as the resistance increases but you have "or less" associated with the resistance.) – Transistor Oct 01 '17 at 19:25
  • @Transistor, ok, ok, "more" duly noted. So you mean that "volts" and "amperes" should be spelled in low case, so should be "ohms" as well. ok. – Ale..chenski Oct 01 '17 at 19:35