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Homework Help: Determine the maximum current discharged from a battery

  1. Jan 12, 2009 #1
    Hi, I'm not sure about how manufacturers calculate the maximum constant current that a battery can discharge. I understand that a battery capacity rating is given by 20hrs multiplied by the maximum current value.

    I learnt also that by shorting a 9V battery of internal resistance 0.01ohms, we will not get 900A of current running through a circuit, hence ohm's law does not apply.

    I'm quite sure this should be related to the energy stored within the battery, but due to the limits of my knowledge the only formula I know is energy= V x I x T. I cant put a value for time in this formula though!
  2. jcsd
  3. Jan 12, 2009 #2


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    A battery is usually thought of as an ideal voltage source (no resistance) in series with a resistance called the "internal resistance". Ohm's Law DOES apply. I don't know where you got the quote from, but I don't think it is quite right. Perhaps there is some picky little difference such as the voltage on the battery being a little more than 9 Volts when no current is drawn. I just measured one with a Voltmeter and it says 9.5 Volts (no load). So I would think of it as a 9.5 Volt ideal battery in series with some internal resistance and I would apply Ohm's Law. By hooking it up to a known resistance and measuring the voltage or current, I could calculate the internal resistance or anything else including the power supplied by the battery or energy released over a period of time.
  4. Jan 12, 2009 #3


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    That's really more of a guideline that you should slow charge a battery at 1/20 of it's capacity.

    You will very briefly until the battery is destroyed.
    Fortunately small alkaline batteries have a higher internal resistance. But lead acid do have internal resistances this low - you can get a current of 1000A from a car battery if you for example short the terminals together with a spanner/wrench !

    The maximum current, known as the short circuit current, is only limited by the internal resistance. The maximum practical current will depend on the design of the battery, the temperature, cooling etc and will be given by the maker.
  5. Jan 12, 2009 #4
    Sorry if I weren't making good sense here..I'm a freshman currently doing my 1st electrical engineering course.

    Because I learnt that ohm's law is only an approximation, in fact variation of current with voltage across an element is not always linear. Also the other day my prof was saying sth like " in fact a typical 9V battery can only deliver a maximum of 3-4A, and after a few seconds the current supply will decrease. "

    I could not understand how he calculate that. And I thought the max current perhaps vary with the time that we intend to use the battery? (again from energy = V x I x T)
  6. Jan 12, 2009 #5
    he was referring to this battery:

    " a typical small 9V battery (2010) "

    and telling us we would not get 900A if we were to short it
  7. Jan 12, 2009 #6


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    Ohm's law is correct - for an ohmic resistor!
    Batteries are normally modelled in a circuit as a perfect (unchanging) voltage in series with a resistor (the internal resistance) this gives you the correct answer at current where the battery is still working. ultimately if you take too much current you will be limited by the chemistry of the battery and it will either stop giving you any more current or be damaged (depending on the type).
    The internal resistance for very high currents or very short times might be different.

    I'm surprised that a PP3 has such a low internal resistance - they are normally made from 6x1.5V AAAA cells in series which should each have R=0.2 or so.

    See http://data.energizer.com/PDFs/BatteryIR.pdf
  8. Jan 12, 2009 #7
    You're right, it should be the chemistry of the battery that limits the current.
    Was trying to understand how to calculate such a current, trying too hard I guess..

    The main thing I've learnt is not to apply ohm's law blindly, as ohmic resistors can become non-ohmic subjected to high current/voltage conditions.
    And thanks a lot :)
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