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

• B
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
Gold Member
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).

davenn
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
Gold Member
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.

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