USB power bank: How can both voltage and current be given

[greypilgrim]

Hi.

I'm still confused about basic electricity. I have a USB power bank with two 1 A and one 2.1 A outputs. They all are specified with 5 V (all numbers from the manual, not measured myself).

I just can't bring this into agreement with Ohm's law U=RI. Shouldn't the resistance R only depend on the load that is connected to the outputs? So how can both U and I be given beforehand? Or is one of them or both a maximum value?

 

[phinds]

Hi.

I'm still confused about basic electricity. I have a USB power bank with two 1 A and one 2.1 A outputs. They all are specified with 5 V (all numbers from the manual, not measured myself).

I just can't bring this into agreement with Ohm's law U=RI. Shouldn't the resistance R only depend on the load that is connected to the outputs? So how can both U and I be given beforehand? Or is one of them or both a maximum value?

You have a common misunderstanding. The RATED value is not the SUPPLIED value. The rated value is a maximum, above which the supplied value will start to cause problems w/ the source. The supplied value is whatever the load requires (up to the rated value, when, again, problems start).

 

[greypilgrim]

Thanks, that's what I figured.

I can only charge my iPad on the 2.1 A output. If I connect it to one of the 1 A outputs, it shows the charging symbol but the battery level doesn't increase. So I guess the resistance of the iPad is lower than 5 ohms, but this causes "problems", as you put it. What exactly happens then? Does the power bank increase some internal resistance until the current is 1 A? Or will the voltage drop from 5 V to a value such that the current becomes 1 A? Or does this depend on the power bank?

Another question: Since I can charge my smartphone on the 1 A output, but my iPad only on the 2.1 A output, this should mean that the iPad's resistance R=U/I is smaller than the smartphone's. Can this somehow be explained by the different battery capacities or are there different charge electronics or something like that?

 

[phinds]

Thanks, that's what I figured.

I can only charge my iPad on the 2.1 A output. If I connect it to one of the 1 A outputs, it shows the charging symbol but the battery level doesn't increase. So I guess the resistance of the iPad is lower than 5 ohms, but this causes "problems", as you put it. What exactly happens then? Does the power bank increase some internal resistance until the current is 1 A? Or will the voltage drop from 5 V to a value such that the current becomes 1 A? Or does this depend on the power bank?

The specific problem depends on the supply, but generally it is a lowering of the output voltage. It can also be overheating and damage to the supply circuitry.

Another question: Since I can charge my smartphone on the 1 A output, but my iPad only on the 2.1 A output, this should mean that the iPad's resistance R=U/I is smaller than the smartphone's. Can this somehow be explained by the different battery capacities or are there different charge electronics or something like that?

I think it has to be a difference in the charge electronics, not the battery capacity, but I'm not positive about that.

 

[CWatters]

I can only charge my iPad on the 2.1 A output. If I connect it to one of the 1 A outputs, it shows the charging symbol but the battery level doesn't increase.

If the iPad is ON then you are asking the Power Bank to provide current for two things... Charging the battery and running the iPad. If running the iPad takes more than 1A there won't be any left over to put into the battery.

I don't have an IPad so I can't check but... If you really get stuck it might be possible to charge the iPad from a 1A output if you switch the iPad OFF first rather than just allowing it to sleep. Many tablets will charge faster if you do this because virtually all of the available current can be used to charge the battery. I think on an iPad holding the power button down for 4 seconds will switch it off properly (requiring a reboot when you use it again).