Total Energy a 12V Battery Can Supply

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Homework Help Overview

The discussion revolves around calculating the total energy a 12 V battery rated at 200 Ah can supply. Participants are exploring the relationship between voltage, current, and energy, as well as the implications of using different equations in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the equation E = 1/2 C V^2 and question the appropriateness of using P = IV for energy calculations. There is confusion regarding the interpretation of amp-hours as a measure of current versus charge. Some participants also explore the calculation of capacitance and its relevance to the problem.

Discussion Status

The discussion is active with participants questioning assumptions and clarifying concepts related to electrical measurements. Some guidance has been offered regarding the correct interpretation of units and equations, but no consensus has been reached on the correct approach to the problem.

Contextual Notes

There is a noted confusion regarding the definitions of amp-hours and their application in energy calculations. Participants are also considering the necessity of a capacitor in the context of the equations being used.

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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
 
Last edited:
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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..
 
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.
 
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?
 
Last edited:
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?
 
Heh, whoops, wasn't thinking about actually needing a capacitor in it.

Anyway, now I got it, thanks all.
 

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