What is the charging current for a 120 ampere-hour storage battery?

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To charge a 120 ampere-hour storage battery in 8 hours, the charging current can be calculated by dividing the ampere-hours by the charging time, resulting in a current of 15 amperes. For a lead-acid battery designed for a continuous discharge of 180 amperes over 5 hours, the time to discharge at a steady current of 220 amperes can be determined using a proportion, yielding approximately 4.5 hours. The discussion emphasizes the simplicity of these calculations, suggesting that they do not require complex RC equations. Participants also engage in light-hearted banter about the problem's complexity. Overall, the conversation highlights basic principles of battery charging and discharging calculations.
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Charging current

I am total stumped on the following to questions so if someone could show me how to get started that would be good thanks.

1) A storage battery rated at 120 ampere-hours is to be fully charged at constant current in 8 hours. Calculate the charging current.

2) A lead-acid battery is designed to give continuous discharge of 180 amperes for 5 hours. Calculate the time it will be discharged by a steady current of 220 amperes.
 
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Why don't you use the RC equations for charging and discharging? I would assume you can treat a rechargable battery like a capacity in series with a resistor.
 
Hint: ampere-hours is a product.
 
DDuardo, were you teasing?

1) A storage battery rated at 120 ampere-hours is to be fully charged at constant current in 8 hours. Calculate the charging current.
A current is measured in amperes. To go from ampere-hours to amperes, you divide by hours! Think about it.

2) A lead-acid battery is designed to give continuous discharge of 180 amperes for 5 hours. Calculate the time it will be discharged by a steady current of 220 amperes.
This is really a "proportion" problem. Let t be the time required by the 220 ampere current. Then x is to 220 as 5 is to 180.
Or, in more modern terms, x/220= 5/180.
 
Thank you, HallsofIvy. Yes, it was a joke. I was trying to make the poor guy complicate the problem even further seeing how you just have to multiple a couple of numbers to get the solution. :smile:
 
Gosh, you're evil!
 

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