9V Battery Current

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

Reply to
Chad Waldman
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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

faster).

Reply to
Kim

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

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Reply to
Rodney Kelp

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.
Reply to
Robert Baer

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

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Reply to
E. Rosten

What you need is to use an ammeter to measure the current flowing through your electromagnet.

Reply to
Michael A. Covington

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.

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Reply to
andy

it

new

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.

Reply to
Derelict

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

Reply to
John Larkin

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

Reply to
John Larkin

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

Reply to
Jonathan Kirwan

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.

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Reply to
Qwerty Keyboard

An accumulation of opposing charge on those bubbles, perhaps?

Jon

Reply to
Jonathan Kirwan

The short circuit current of the battery is not relevant, period. In fact, it is *stupid* to make such a useless measurement.

Reply to
Robert Baer

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

John

Reply to
John Larkin

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

Reply to
Jonathan Kirwan

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

A: "Here's how I wound my solenoid."
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Reply to
John Fields

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.

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Reply to
andy

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

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Reply to
andy

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.

Reply to
Anonymous

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