9V Battery Current

[Chad Waldman]

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

 

[Robert Baer]

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

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.

 

[E. Rosten]

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.

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)

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for /s 15 d f pop 240 420 m 0 1 3 { 4 2 1 r sub -1 r show } for showpage

 

[Kim]

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

 

[Rodney Kelp]

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

 

[andy]

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

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.

 

[Michael A. Covington]

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.

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

 

[Derelict]

Go and buy an ammeter! They are not expensive.

<>snip<>

-Ed

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.

 

[John Larkin]

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

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

 

[Jonathan Kirwan]

RatShack 9v alkaline, I got 6 amps into a dead-short ammeter for the
first second or so, and then it dropped rapidly.

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.

After about a minute,
it was down to 1 amp and the battery was getting pretty hot,

The curve suggests that this should happen at about 80-90% discharge.

so I quit.

And scratch one ratshack battery.

Jon

 

[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

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

 

[Jonathan Kirwan]

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?)

An accumulation of opposing charge on those bubbles, perhaps?

Jon

 

[Robert Baer]

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

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

 

[Jonathan Kirwan]

In fact, it is *stupid* to make such a useless measurement.

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

 

[Qwerty Keyboard]

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

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.

 

[John Larkin]

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

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

John

 

[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.

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

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

;-)

 

[andy]

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

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.

 

[andy]

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

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

;-)

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

 

[Rubicon]

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.