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Suppose I have a rechargeable 70 kWh battery , and I use 120 V rms from my house with a 10 A RMS circuit.

To find the max power of this circuit, is it just P = VI, since both the voltage and current appear in RMS?

So in this case P = 120 * 10 =1.2 kW?

Furthermore, suppose the battery was drained, in order to find out how long it takes to fully charge this battery, is the following correct?

70 kWh / 1.2 kW = 58.33 hours?

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  • \$\begingroup\$ You do realize 10A is just the fuse, right? You need something else to actually limit the current to this value. Unless specifically specified otherwise you cannot charge a battery with AC from your house. \$\endgroup\$ – jippie Feb 3 '16 at 19:40
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Your power calculation is correct. A 120 volt RMS sine wave actually goes from -170 to +170 volts, but use RMS for power calculations.

When charging a battery, you have the right idea, but you should be aware that charging a battery is not 100% efficient. That is, if you put 10 kW-hr into a charger/battery combination, the battery will not be able to put out 10 kW-hr. Charge efficiency is quite high during the early stages of charging, but gets less efficient toward the end, when the battery is nearly charged (the term for this is State of Charge, or SOC, which runs from 0% - fully discharged, to 100% - fully charged. Charge efficiency reduces as SOC increases). As a very rough number, for large lead-acid cells assume overall charge efficiency is about 50%, so your example would require 116.66 hours. Since the 50% number was not precise, call it 100 to 110 hours.

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  • \$\begingroup\$ The actual charge-discharge efficiency of a lead-acid battery is on the order of 70-80%. But when you also account for the efficiencies of the various external power converters, the overall system efficiency might be as low as 50-60%. \$\endgroup\$ – Dave Tweed Feb 3 '16 at 19:30
  • \$\begingroup\$ How did you go from the 58 hours to the 110? In general how would one calculate the time frame in theory? \$\endgroup\$ – learnmore Feb 3 '16 at 19:33
  • \$\begingroup\$ @learnmore - If the efficiency is 50% rather than 100%, it will take twice as long, right? So 2 times 58.33 is 116.66. And since the 50% is not precise, I rounded the number. I was aware (see Dave Tweed's comment) that I was being quite conservative in using 50%, so I rounded down (to 110) and then down a bit more (to 100). \$\endgroup\$ – WhatRoughBeast Feb 3 '16 at 19:38
  • \$\begingroup\$ 58.33/(50%) = 116.66. Really just translating to numbers, the original answer is very well written and answers your comment. \$\endgroup\$ – Vicente Cunha Feb 3 '16 at 19:38

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