The problem asks:

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?

First part is simple. Low power USB port is given as ~100mAh. So 680mAh/100mA = 6.8hours.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?

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?

Thanks in advance!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.

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