Converting between mAh and C

Thread Starter

OckhamsHammer

Joined Apr 25, 2014
3
The problem asks:

A 680 mAh Lithium-ion battery is standard in a Motorola RAZR®. If this battery
is completely discharged (i.e., 0 mAh), how long will it take to recharge the battery to its full capacity of 680 mAh from a low-power USB port? How much charge is stored in the battery at the end of the charging process?
First part is simple. Low power USB port is given as ~100mAh. So 680mAh/100mA = 6.8hours.

For the second part and also the reason I'm posting here today. Charge is units of Coulombs so I have (mili-Coulombs/sec)*hrs. So I figured if I just multiply by 3600 seconds (ans= 2.448kC) I would get the correct answer. But the answer given (for both is just below this) they multiplied by 60 seconds: can someone tell me why?

A low-power USB port operates at 100 mA. Assuming that the charging current from the USB port remains at 100 mA throughout the charging process, the time required to recharge the battery is 680 mAh/ 100 mA = 6.8 h. The charge stored in the battery when full y charged is 680mAh • 60 s/h = 40,800 mAs = 40.8 As = 40.8 C.
Thanks in advance!
 
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Thread Starter

OckhamsHammer

Joined Apr 25, 2014
3
I get that this is all theoretical but I don't get how they are getting their answer without multiplying by how many seconds there are in a hour...I must be reading it wrong I just don't know..
 

WBahn

Joined Mar 31, 2012
26,398
The problem asks:



First part is simple. Low power USB port is given as ~100mAh. So 680mAh/100mA = 6.8hours.
This might just be a typo, but low power USB is rated for 100mA, not 100mAh.

For the second part and also the reason I'm posting here today. Charge is units of Coulombs so I have (mili-Coulombs/sec)*hrs. So I figured if I just multiply by 3600 seconds (ans= 2.448kC) I would get the correct answer. But the answer given (for both is just below this) they multiplied by 60 seconds: can someone tell me why?
Because they messed up. But at least they are tracking their units and so the mistake is easy to spot and correct. The have (60 s/h) as one of the terms but, as you've already indicated, this is clearly wrong and it should be (3600 s/h).

On a side note, it is not truly correct to say that this much charge has been stored in the battery because that implies that somewhere in the battery there is a charge separation comparable to what you would find in a similarly charged capacitor. But a battery is not a capacitor and the charge that is driven into it is almost instantly converted to chemical potential energy with very little charge separation remaining.
 
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