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Energy output of a battery?

  1. May 18, 2014 #1
    There is an amount of energy stored in the battery. However, the rate of output would depend on
    the system its powering. What is the best formula to calculate the output energy from a battery?

    Also, what formula would calculate the stored energy of the battery?
  2. jcsd
  3. May 18, 2014 #2


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    If you know the battery voltage V, and the battery capacity AH in amp-hours, the energy stored by the battery in Joules is just E = V*AH*3600, where the 3600 is the number of seconds in an hour.
    Last edited by a moderator: May 18, 2014
  4. May 18, 2014 #3


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    The only way I know how to do it is by experiment:


    In this example, a hypothetical 2 volt fully charged battery is hooked to a 1 ohm load.
    After the first hour, the average of the continuous power is taken as the energy expended.
    1.9 watts expended over an hour yields 1.9 watt-hours.

    After the second hour, this is done again, and added to the previous hour's energy consumption.
    This continues until the battery is either dead, or you've reached the minimum usable voltage.

    In practical applications, it is almost always the "minimum usable voltage" which determines when you should stop discharging a battery. So in the above example, if 1.5 volts were the minimum practical voltage, the battery would be considered to have an effective capacity of 5.5 watt hours, or 19,800 joules.

    1 watt-second(one watt of power delivered for one second) = 1 joule(energy)
    3600 watt-seconds = 1 watt-hour = 3600 joules
  5. May 18, 2014 #4
    I had a discussion with a fellow undergraduate, about the use of batteries in multiple applications and the rated capacity (Ah) and voltage, would be the maximum output for a short period of time. It would not be higher of course, but would drop with time. However, the time would highly increase from 3600 seconds to 3600x(hours).

    The lowest amount of energy used is what is drawn from the battery in a short period of time, thats why my concern to prove my argument :tongue:
  6. May 18, 2014 #5


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    It sounds like you are confused between energy and power. The power drawn from a battery is just the voltage times the current. The energy drawn from a battery in a short time t is just the power drawn times t. As the time t becomes very short, the energy drawn from the battery goes to zero.
  7. May 18, 2014 #6


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    The discharge current level affects the decay of the output voltage with respect to time. Depending on the discharge current, the battery can last for a short time or a long time. Keep in mind that using large discharge current levels is a bit wasteful, since you are pulling that large current through the series resistance of the battery, and heating up that resistance is wasteful of the battery's energy.

    Have a look at a typical set of discharge curves to help you understand:


    The dotted line is where pretty much all of the useful energy stored in the battery has been pulled out (time to recharge...).
  8. May 18, 2014 #7

    jim hardy

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    A battery stores its rated energy, which in watt-hours is (nominal volts) X (rated amp hours) watt-hours.

    Berkeman's post shows that the rate at which you withdraw that energy affects how much of it you actually get.

    Battery specifications tell you at what rate you should discharge the battery to get it all. Car batteries are typically rated in amp-hours for eight hour rate, that is connected to a load that'll pull the energy out over an eight hour period. Take its amp-hours and divide by eight to figure that load in amperes. At higher load you'll get less energy out, at lower load you'll get a little more.

    Smaller batteries like for flashlights or radios may be rated at other discharge rates.
    This link mentions twenty hours.

  9. May 18, 2014 #8


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    Search for Peukert factor: http://en.wikipedia.org/wiki/Peukert's_law
  10. May 19, 2014 #9
    A bit curious here...
    If energy stored in the battery = VI x 3600
    How can it last over 3600 seconds? I'm a bit confused at that point. I'd like to know the total energy stored in the battery, why would it be limited to only 3600 seconds? I think I myself started to indeed get confused...
  11. May 19, 2014 #10


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    The factor 3600 is just to convert from hours to seconds.
    The SI unit should be Ws (watt-second), but because of the timescales involved (you rarely use something for 1 second) it is often more convenient to use Watt-hours.
  12. May 19, 2014 #11
    But still, we can use the battery for shorter periods than 3600? And we change "t" to that value? While "P" stays the same.

    What I meant earlier, is that the formula makes sense to calculate the energy used within a certain time frame(which is usually 3600seconds). But it does not seems to give the "total" energy stored in the battery, because the battery would still have energy beyond 1 hour, not at the same power(it would be less)... but the energy stored might be higher than the value computed at 3600 seconds if that make sense?
  13. May 19, 2014 #12


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    There is no "t" in the equation you were given. A conversion factor is always equal to 1 and doesn't change anything in the equation. It exists only to fix the units: 3600s/hr =1

    So the answer was given in joules, which is a unit of energy with no time component.

    If it makes it easier, you can think of amp-hrs as how many amps it would take to drain the battery in 1 hr. Not exactly right, but could help with the understanding. You can always drain the battery in half the time, using twice the amps, which therefore leaves the answer unchanged.
  14. May 19, 2014 #13
    What about the formula best to calculate the output energy from the battery converted to heat?
  15. May 20, 2014 #14


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  16. May 20, 2014 #15


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    A battery has an ampere hour rating that says how long the battery will last if it is discharged at a certain current. That rating is for a certain discharge current, and the capacity varies with the size of current you draw.
    See http://media.digikey.com/pdf/Data Sheets/Energizer Battery PDFs/EN91.pdf for an example.

    The current being drawn drawn times the voltage is the power being supplied at any given instant. The summation of the power over time is the energy supplied.

    Some energy is dissipated inside the battery as heat, based on its internal resistance. All of the energy supplied by the battery is converted to kinetic energy, potential energy, heat, light, or whatever (not sure what you are asking, specifically).
  17. May 20, 2014 #16


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    The answer to this will depend upon the specific design of the battery - not just the basic type. To make a battery that delivers loads of current (starter motor) you need to make compromises about other aspects of performance - leakage and allowable depth of discharge - relates to the price you are prepared to pay, I think.

    Also, the term 'internal resistance' is often used more loosely than appropriate. The voltage drop under load is not due to a simple Ohmic resistance in series so the value that people use for internal resistance cannot necessarily be used over the whole range of currents delivered. It's a useful but not rigorous quantity.
  18. May 20, 2014 #17
    I was taught the equation for calculating energy lost due to resistance was: P= I^2 • R
  19. May 20, 2014 #18


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    You were probably taught that the Power is I2R.
    There is an important distinction between Energy and Power. (This is what the thread is about)
    Also, as I wrote earlier, the value of internal R is not necessarily the same for all currents.
  20. May 20, 2014 #19
    My bad. The equation I entered above for power (Energy/Time) lost due to resistance in electrical circuits is as I stated. I was commenting on the conversation as it was progressing. I did not realize I was not allowed to follow along since resistive losses were being discussed. I was trying to provide an formula to assist the originator of this thread as they had asked. I would like to suggest a site that they might find helpful. I found the following at Hyperphysics:

    ImageUploadedByPhysics Forums1400635099.996079.jpg ImageUploadedByPhysics Forums1400635118.967500.jpg
  21. May 20, 2014 #20
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