Connect with us

9V Battery Current

Discussion in 'Electronic Basics' started by Chad Waldman, Aug 18, 2004.

Scroll to continue with content
  1. Chad Waldman

    Chad Waldman Guest

    I posted a message a week back but never received the answer I was
    looking for.

    I have an electromagnet hooked-up to a 9V battery. The magnet is
    strongest when the battery is at full charge. I want to replace the
    battery with an adapter to run the magnet from an outlet. What I am
    wondering is, what is the max current the battery is outputting when it
    is new? If you have an ammeter, can you please test the output of a new
    battery.

    Thank you,
    Chad
     
  2. Robert Baer

    Robert Baer Guest

    I do not have the "magnet" that you have for testing, and it is not
    possible for anyone else to do that testing without an accurate
    description (manufactureer and model number for starters).
    That *electro*magnet has a measureable resistance; one could calculate
    the current drain at 9 volts.
    Alternately, *you* could use a multimeter or a DVM to measure that
    current.

    Be advised, that most wall adaptors are unregulated, and so the output
    voltage is *always* higher than the rating, even at full rated load.
    That means, that your *electro*magnet will be stronger when you use a
    9V adaptor.
    An adaptor with even a 500mA rating would seem to be more than
    adequate.
     
  3. E. Rosten

    E. Rosten Guest


    Go and buy an ammeter! They are not expensive.

    I recently purchased a digital multimeter from Maplin for the sum of
    £2.99 (!!!!!!!!!). It includes voltmeter, ammeter (up to 10A unfused),
    transistor meter, diode meter and ohmmeter. It seems OK, this
    multimeter, and the only thing it is missing is a buzzer for some
    continuity tester.

    Anyway, I don't know how it is even remotely possible to make and sell a
    multimeter at a profit for that price (it probably isn't), and I don't
    know if they are still selling them that cheap. Anyway, a bloke in my
    local market is selling the same model for about £7, and an analogue
    model for about £6 (I might get one, since the response is somewhat faster).

    In short: multimeters are *very* *very* *very* *very* *very* *very*
    cheap, so buy one. They are invaluable for anything electrical.

    -Ed

    --
    (You can't go wrong with psycho-rats.) (er258)(@)(eng.cam)(.ac.uk)

    /d{def}def/f{/Times findfont s scalefont setfont}d/s{10}d/r{roll}d f 5/m
    {moveto}d -1 r 230 350 m 0 1 179{1 index show 88 rotate 4 mul 0 rmoveto}
    for /s 15 d f pop 240 420 m 0 1 3 { 4 2 1 r sub -1 r show } for showpage
     
  4. Kim

    Kim Guest

    Even the cheapest multimeter has a Ammeter function nowadays, and if you are
    planning to make this a hobby... then its a ABSOLUTE NECCESSITY!. As
    mentioned earlier, any old supply will work, anywhere between 250 and 500ma.
    Kim
     
  5. Rodney Kelp

    Rodney Kelp Guest

    No reason why it won't work with 12 volts. Use a car battery.
     
  6. andy

    andy Guest

    it doesn't just depend on the battery - it depends on the resistance of
    the magnet coils as well. The easiest way is to buy or borrow a meter and
    measure it.
     
  7. What you need is to use an ammeter to measure the current flowing through
    your electromagnet.
     
  8. Derelict

    Derelict Guest

    but an ammeter is easily destroyed. If you connect the meter across a
    battery, the battery will break its heart trying to provide infinite
    current. This could be bad for the battery, meter, test leads and
    nearby persons.

    Chad is asking the wrong question and, to optimize his device, should
    look into Ohm's Law. This will show him that the voltage for his
    electromagnet is not as important as the current it requests from the
    source.
     
  9. John Larkin

    John Larkin Guest

    RatShack 9v alkaline, I got 6 amps into a dead-short ammeter for the
    first second or so, and then it dropped rapidly. After about a minute,
    it was down to 1 amp and the battery was getting pretty hot, so I
    quit. A rechargable would likely behave differently.

    We'd have to know the magnet's resistance to estimate the actual
    current. How long does the battery last, by your standards?

    You should probably buy a cheap DVM and measure actual coil resistance
    and the voltage that makes it work OK.

    John
     
  10. I just checked the Energizer datasheet for it and they list a starting internal
    resistance of just under 2 Ohms, so that 6A seems about right.
    The curve suggests that this should happen at about 80-90% discharge.
    And scratch one ratshack battery. :)

    Jon
     
  11. John Larkin

    John Larkin Guest


    A 9v alkaline is good for maybe 1 a-h, so I probably didn't use it up
    much by shorting it for a minute. Now that it's cooled off and rested
    for a couple of hours, it's back up to 5.8 amps initial current. I
    think there's some polarization that goes on at high current (tiny
    bubbles or something) but it's temporary (some sort of catalytic
    reaction depolarizes it?)

    John
     
  12. An accumulation of opposing charge on those bubbles, perhaps?

    Jon
     
  13. Robert Baer

    Robert Baer Guest

    The short circuit current of the battery is not relevant, period.
    In fact, it is *stupid* to make such a useless measurement.
     
  14. I'm just a hobbyist and John can speak for himself, but I'd assumed that the
    concept John was thinking of here was that placing a (assumed for now) fixed
    voltage across a coil means the current rises at a rate proportional to V/L and
    that the limiting factor to that ramp would be the internal resistance of the
    battery (it is, after all, a weakling 9V) in series with the coil resistance.
    In other words, it would eventually (in short order, really) be limited that
    way. So measuring the dead-short current would certainly set an upper limit for
    discussion purposes and, if the coil's resistance is much less than the 2 Ohms
    for the 9V battery, not too far from the mark.

    Something like:


    | (V/L)
    | / .........> (V/R)
    | / .
    | / .
    | / .
    I | .
    | .
    | .
    | .
    | .
    |.
    \-----------------------------------------
    time

    In that case, testing the short circuit current seems to provide one of the few
    useful somethings that others can test and think about, given the lack of
    information about the coil, itself.

    Jon
     
  15. WTF? The only one who is *stoopid* here is you. Short circuit testing a
    battery provides valuable information about how the battery will behave
    under heavy loads and short circuit fault conditions. The original poster
    wasn't very specific about what kind of electromagnet was in use, but smart
    money bets it was a hobbyist's homebrew electromagnet with far too few turns
    with far to large wire and the 9V battery impedance provides most of the
    current limiting.

    You are clearly so stoopid to make such a comment you evidently belong in
    the lower two quartiles of this study.

    http://www.apa.org/journals/psp/psp7761121.html

    Even pond scum has more potential than you.

    http://www.acfnewsource.org/environment/pond_scum.html

    And your momma wears combat boots.

    Bitch.
     
  16. John Larkin

    John Larkin Guest


    How is learning something stupid? Oh, yeah, for some people, it is a
    waste of time.

    John
     
  17. John Fields

    John Fields Guest

    ---
    Q: "What's the maximum current a new 9V battery can supply?"

    A: "Here's how I wound my solenoid."

    ;-)
     
  18. andy

    andy Guest

    I've managed to get good results with a standard 250g coil of 24 swg wire
    (0.56 mm) unwound from the bobbin, split in 3 equal lengths, and then
    wound back on again, with the coils connected in parallel. The coil
    resistance is around 1 ohm. With a 1.2Ah 12V lead acid battery, this gives
    a peak current of about 10-11 amps. (The battery voltage drops to around
    10.5 volts from about 12.5)

    I'm using a piece of rebar (concrete reinforcing rod) as the core - this
    works the best out of the metal objects I've tried. Bolts are no good.

    With this setup, it's strong enough to push a neodymium magnet off the end
    of the core, and lift the core against gravity. Guessing about 50-100
    grams force when you try to pull the core out of the magnet with the
    current on. To make a proper push/pull solenoid, you'd need a permanently
    magnetised core, I think, which would be harder to find.
     
  19. andy

    andy Guest

    I thought it might be relevant. But I guess I was showing off a bit as
    well.
     
  20. Rubicon

    Rubicon Guest

    I read today that a 9V pp3 battery is only good for 50mA before the
    battery voltage rapidly drops.

    I am not sure that it's true but it does help to explain a few
    problems I'm having at the moment.

    Andrew.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-