
#1
Mar508, 04:10 PM

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A 12V car battery is rated at 120 Ah. Assuming that the terminal voltage remains 12 V, for how long can this battery supply power at 29 W? How much charge flows through this battery in this time? How much energy is initially stored in the battery?
I know that an Amp Hour is 3600 C I know that a Watt is a Joule / s I know that 1 J= 1C/V 3600 C * 120 So the initial energy stored in the battery is (3600*120) / 12V=36000 Joules 29 Watts=29 J/s So this battery is delivering 36000 Joules at a rate of 29 J/s 36000/29 = 1241 seconds. This is only about 20 minutes. This seems very short. I use 7Amp hour batteries at work. I've used them to power a light for a few hours. So my intuition tells me this answer can't be right. How much charge flows through the battery during this time? : I=Q/t = 432000 C / 1241 s = 348 Amps. But this assumes I got the seconds correct. Did I do this right? 



#2
Mar508, 04:22 PM

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#3
Mar508, 04:28 PM

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Thanks for the reply.
What's wrong with that part? It wants energy, so my answer should have units of Joules, right? A joule is a Coulomb / Volt, and the battery has 3600*120=432000 coulombs 43200 / 12 is 36000 Joules. At least my units work. What did I do wrong? Is energy in a battery not expressed in Joules. 



#4
Mar508, 04:34 PM

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battery (amp hours, total charge, total current)
1 watt hour is 3600 joules. Should I have expressed it as 10 watt hours?




#5
Mar508, 04:37 PM

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#6
Mar508, 04:49 PM

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oops. Thanks for that.
Now I get 2 days. I guess that's about right. A fresh car battery can burn a single light (I'm guessing about 22W) for about 2 days. So I imagine I answered the Charge part wrong too. ( I computed current). The charge should be 120Ah * 3600 C/Ah = 432000 C. Thanks for your help! 



#7
Jun2810, 10:29 PM

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Power[watt] = Voltage[Volt] * Current[Amp], so Current[Amp] = Power[watt] / Voltage[Volt], so Current[Amp] = 29 / 12 = 2.42 [Amp] and Time[h] = Capacity[Ah] / Current[Amp] Time[h] = 120 / 2.42 =~ 50[h] EURECA! The answer is 50 hours. Note: This calculation assumes the battery is an ideal source of current (which is not, since it has some internal resistance). Actual timing may be a few percent less than 50 hours. 


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