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Discussion in 'Electronic Basics' started by Average Shmo, Mar 14, 2006.

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  1. Average Shmo

    Average Shmo Guest


    Is there any way to know how many AMPs are comming out of a battery? Most
    batteries purchased at the shops dont tell you.

    I have a voltmeter here but that only (you guessed it) shows volts!

    Thanks in advance.
  2. Tom Biasi

    Tom Biasi Guest

    Amps is a rate of current flow. It depends on the load for a particular
    voltage. The battery will have a maximum rating which is determined by the
    physical makeup of the battery. You need the specs on your battery.
  3. Greg Neill

    Greg Neill Guest

    That's because it depends upon the load connected to the battery.
    If you know something about the load, like its power consumption or
    its resistance (assuming a resistive load), then you can work out
    the current being supplied at a given voltage. Otherwise, purchase
    an ammeter and measure the current!
  4. John Fields

    John Fields Guest

    The amount of current a battery can deliver into a load depends on
    the battery voltage, the internal resistance of the battery and the
    resistance of the load.

    If you had an ammeter you could connect it in series with the load
    and measure the current.

    If the load resistance was fixed and you knew what it was, or if you
    had an ohmmeter and could measure it, you could divide the load
    resistance into the battery voltage and get amps as the quotient.

    If you have neither an ammeter nor an ohmmeter you might be pretty
    much shit out of luck.


    You could get a very small valued resistor (a 'shunt') and put it in
    series with the load and the battery, then measure the voltage drop
    across the shunt.

    Your circuit would look like this: (View in Courier)

    | |
    | [LOAD]
    |+ |
    [BATTERY] +<--------+
    | | |+
    | | |

    and you could then determine the current in the shunt by using Ohm's

    I = ---

    Let's say that the resistor has a resistance of 0.1 ohm and your
    voltmeter reads 0.02 volts.

    Then, the current in the resistor will be:

    E 0.02V
    I = --- = ------- = 0.2 amperes
    R 0.1R

    since the current in a series circuit is everywhere the same, that
    200mA would also be flowing through the load and the battery, so
    that's how much current would be coming from the battery.

    There's a small error in there due to the resistance of the resistor
    causing less current to flow than if it weren't there, but we can
    get rid of it like this:

    First, measure the battery voltage when it's under load:

    | | |
    | [LOAD] |
    |+ | |
    [BATTERY] + |
    | | |+
    | | |

    Just for grins, let's say it's 11.8V. Record that voltage as E1.

    Second, measure the voltage across the shunt (as shown earlier) and
    record that as E2.

    Third, plug those values into this equation:

    R1 = ----------

    Where R1 will be the resistance of the load and R2 is the resistance
    of the shunt, like so:

    R2(E1-E2) 0.1R * (11.8V - 0.02V)
    R1 = ---------- = ----------------------- = 58.9 ohms
    E2 0.02V

    So, now that you know the resistance of the load, you can find the
    current the battery's pumping through it by removing the shunt,
    measuring the battery voltage with the load connected to it, and
    using Ohm's law:

    I = ---

    To find out what current is coming out of the battery.

    Or perhaps what you _really_ want to know is how to determine the
    capacity of a battery (that is, how many amps can you get out of it
    for how long)? ;)
  5. John Fields

    John Fields Guest

    charge :)
  6. I'm going to take a different take on the question. Perhaps he's asking
    how many amps the battery can supply at maximum load (lowest circuit

    Batteries are "current limited" by the size of the internal electrodes
    affected by the chemical process.

    A "AA" battery could source up to .5 Amps, "C" battery 1 Amp, and "D"
    cell up to 1.5 Amps. Two "D" cells in series is still 1.5 Amps. but two
    in parallel would be 3 Amps.

    A five pound lead-acid battery might source up to 20 Amps and a forty
    pound one perhaps 100 Amps as the larger battery has bigger lead
  7. John Fields

    John Fields Guest

  8. Tom Biasi

    Tom Biasi Guest

    OK John.
  9. Well, OK..I was primarily considering the internal components of the
    battery. But even then, you're referring to the internal resistance vs
    current sourcing capacity. A small nuance of difference, but its all
    part of the big picture.
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