Total Energy a 12V Battery Can Supply

In summary, the conversation discusses the calculation of total energy that a 12 V battery with a rating of 200 Ah can supply. There is confusion about whether to use the formula P=IV or E=1/2 CV^2, and the conversation also touches on the units of Ah and how to convert to kWh. Ultimately, it is determined that P=IV is the correct approach and the correct calculation is 2.4 kWh. There is also a brief mention of capacitance and its relationship to charge.
  • #1
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Homework Statement


A 12 V battery is rated at 200 Ah (Amps hours). How much total energy can it supply?


Homework Equations


P=IV

The Attempt at a Solution


First I did E = 1/2 C V^2. (Got 4,320,000 J, although I'm not sure if that is correct). But now I need to convert it to kwh so my question is can I use P = IV in this manner:
P = 200 Ah * 12 V
P = 2400 Wh = 2.4 kWh
or do I need to manipulate my original energy finding?

4,320,000 J != 2.4 kWh so one must be wrong
 
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  • #2
Huh? amp hours? That's a new one.. You plugged in 200 Ah for I..meaning that Ah is a measurement of current? According to Wikipedia, Ah is a measurement of electrical charge..
 
  • #3
Where did 1/2 C V^2 come from? What did you use for C, and why?

P has units of power, not energy - to get energy you have to multiply by a time, which is what you did, though this might have been accidental since amp-hours are units of charge, not current.

Also, 2400 Wh is not equal to 24 kWh.
 
  • #4
Alright, using P=IV was the wrong approach

Heres how I calculated Capacitance:
First 200 Ah * 60 min/h * 60 s/min = 720,000 Coulombs.
CV = q
C*12 = 720,000 C
C = 60,000 F (which seems really high)

1/2 * 60,000 F * 12^2 = 4,320,000 J which is equal to 1.2 kWh ( i think )

Am I calculating the correct amount of charge?
 
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  • #5
CV = q ??
the C refers to capacitance. do you have a capacitor in the circuit?
Hint: what do you have to do to the equation P = IV to get units of energy on the left hand side?
 
  • #6
Heh, whoops, wasn't thinking about actually needing a capacitor in it.

Anyway, now I got it, thanks all.
 

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