# AA batteries in parallel

Hey, so I tried to get a lot of Amps and primarly my tool are batteries, so I have 3 batteries that I have connected together in parallel (+ with the + and then - with the -) they are 1.6V Kirkland batteries that are all brand new. So when I took a machine I have to calculate the voltage and amps from a single batter i got 1.6 V which is what was supposed to be and I got 0.3 amps which is I guess normal for a AA battery? However when i connected them with a stripped down telephone wire (copper) and measured the same quantities I was expecting to get 1.6V and 0.9 amps... You know, pretty simple math there... However I got 1.6V and 0.3amps... My batteries are well connected but I keep getting a reading of 0.3 amps... What is happening? P.S the machine I was refering to is a multimeter.

anorlunda
Staff Emeritus
However when i connected them with a stripped down telephone wire (copper) and measured the same quantities

Connected them how? Please show a diagram of all the connections, and the load, and where you make measurements. You can use the UPLOAD button next to POST REPLY and PREVIEW to insert pictures into your post.

Ok so in the pic you can see (could be a bad pic but enough details tbh)

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sophiecentaur
Gold Member
2020 Award
I can't be certain of the circuit you are describing but it doesn't surprise me that the current that your battery or batteries were supplying would have been the battery volts divided by the resistance of the "machine" you were using. (possibly 1.6/0.3 = 5.33Ω) This would be the same for any number of batteries in parallel and it suggests that your batteries are each quite capable of delivering 0.3A without dropping any significant volts.
So I reckon that your Multimeter is the thing that's limiting the current. Its internal resistance makes it unsuitable for doing the measurement that way.
If you want to find how much current your batteries can supply, you need a range of known resistances lower than 5.5Ω and you need to measure the Voltage across the resistor. You can then do I = V/R to find the current that the battery is supplying. I'd imagine that an AA battery could supply up to 2A for a brief time.

What current can you get with two batteries in series? I would think you could expect higher than 0.3A; possibly 0.6A but the internal resistance of the batteries may limit the current to a bit below 0.6A.

PS You should make sure that you cave good contacts around your circuit it's no good just holding meter probes and juggling around with multiple batteries (sorry if that is obvious for you). Also you should take care that components don't get hot with high currents and burn your fingers.

Merlin3189
I can't be certain of the circuit you are describing but it doesn't surprise me that the current that your battery or batteries were supplying would have been the battery volts divided by the resistance of the "machine" you were using. (possibly 1.6/0.3 = 5.33Ω) This would be the same for any number of batteries in parallel and it suggests that your batteries are each quite capable of delivering 0.3A without dropping any significant volts.
So I reckon that your Multimeter is the thing that's limiting the current. Its internal resistance makes it unsuitable for doing the measurement that way.
If you want to find how much current your batteries can supply, you need a range of known resistances lower than 5.5Ω and you need to measure the Voltage across the resistor. You can then do I = V/R to find the current that the battery is supplying. I'd imagine that an AA battery could supply up to 2A for a brief time.

What current can you get with two batteries in series? I would think you could expect higher than 0.3A; possibly 0.6A but the internal resistance of the batteries may limit the current to a bit below 0.6A.

PS You should make sure that you cave good contacts around your circuit it's no good just holding meter probes and juggling around with multiple batteries (sorry if that is obvious for you). Also you should take care that components don't get hot with high currents and burn your fingers.
Ah that might be why to be honest... And yes lol I do know about the connections it's because I took it apart not too long ago and I hadn't put it back together so when he asked for the picture I just put them side by side, and in series you'll get the same current but different volts... However looking at the machine I do think there is a cap of current it will measure as there seems to be a 320mA logo but seems to be faded, so quite possibly it does have a cap. Thank you!

sophiecentaur
Gold Member
2020 Award
and in series you'll get the same current but different volts.
That's just an assertion which will only apply if the batteries are going into current limit (which I doubt, for AA cells). If the resistance of the meter is the main resistance in the circuit then the current should go up with the volts - unless there is something else going on that you haven't mentioned.

That's just an assertion which will only apply if the batteries are going into current limit (which I doubt, for AA cells). If the resistance of the meter is the main resistance in the circuit then the current should go up with the volts - unless there is something else going on that you haven't mentioned.

russ_watters
Mentor
Amperage is something you measure in a circuit. You don't have a circuit, you just have a bunch of batteries. Sitting there by themselves, the amperage is zero.

So what you really did is short them with the multi-meter. You are lucky you were only using batteries and weak ones at that, otherwise you would have burned-out or burned the fuse out of your multi-meter.

nasu and sophiecentaur
Amperage is something you measure in a circuit. You don't have a circuit, you just have a bunch of batteries. Sitting there by themselves, the amperage is zero.

So what you really did is short them with the multi-meter. You are lucky you were only using batteries and weak ones at that, otherwise you would have burned-out or burned the fuse out of your multi-meter.
No no, I do have a circuit, that picture was just the set up of how the batteries were connected...

russ_watters
Mentor
No no, I do have a circuit, that picture was just the set up of how the batteries were connected...
Right. So measuring amperage is meaningless. A bunch of batteries doesn't have an amperage.

sophiecentaur
Right. So measuring amperage is meaningless. A bunch of batteries doesn't have an amperage.
Well it's to build an electromagnet... But I need a lot of amps.

davenn
Gold Member
Well it's to build an electromagnet... But I need a lot of amps.

so have you built your coil ?
if so, how many turns on what size former and what wire gauge ?

russ_watters
sophiecentaur
Gold Member
2020 Award
so have you built your coil ?
if so, how many turns on what size former and what wire gauge ?
This can be a difficult exercise for someone not familiar with the business. @davenn wants that information so that you can work out the resistance as well as the magnetic field it can produce.
If you have a battery then the coil needs to have as many turns as possible, consistent with the battery volts and the available current. It's best to have some idea of a plan for coil winding or you can have too few turns and less Field than you could get - plus you could be taking more current than the battery can supply comfortably. Too many turns and the battery will not have the Volts to put sufficient through the coil.
This is not highly critical but it's worth while getting somewhere near an optimum.

so have you built your coil ?
if so, how many turns on what size former and what wire gauge ?
I've been testing more... So far I've got around a bunch of iron nails together around a cm in radius and I've wrapped around 600 times on it. I'm just trying to figure out the best option

sophiecentaur
sophiecentaur
Gold Member
2020 Award
I'm just trying to figure out the best option
You will certainly benefit from choosing things 'right'. That will involve having an idea about the resistance per metre of the wire you are using so that you can get the maximum sustainable current from your batteries.
If you can find just how much current the battery (or several in parallel) can supply (with the procedure I was describing above), you can use the best number of turns for the job.

I more dont have the skills/knowledge for that.. Im more doing a project to keep myself busy/hobby. I'm a young man who gets ideas of how to possibly help people and then try to pursue it...

sophiecentaur
Gold Member
2020 Award
I more dont have the skills/knowledge for that.. Im more doing a project to keep myself busy/hobby. I'm a young man who gets ideas of how to possibly help people and then try to pursue it...
It is worth while trying to get somewhere near the 'best' performance from anything you are making - for yourself or others. It's a simple matter of finding the diameter of the wire first and estimating the length that you used for the coil. I can help you to find what the resistance of that coil is. We can then decide, on the basis of that, whether it's better to use more or fewer turns of wire. If you could also measure the battery volts with and without the coil connected, that should be enough to work with.

russ_watters
Mentor
I more dont have the skills/knowledge for that.. Im more doing a project to keep myself busy/hobby. I'm a young man who gets ideas of how to possibly help people and then try to pursue it...
Fair enough. You can learn at least something by just randomly wrapping a bunch of wire around a nail and attaching a battery. Odds are good you'll run your batteries dead quickly, but that's a small price to pay.

davenn
It is worth while trying to get somewhere near the 'best' performance from anything you are making - for yourself or others. It's a simple matter of finding the diameter of the wire first and estimating the length that you used for the coil. I can help you to find what the resistance of that coil is. We can then decide, on the basis of that, whether it's better to use more or fewer turns of wire. If you could also measure the battery volts with and without the coil connected, that should be enough to work with.
I have 20 meteres of wire, 22 gauge and the volts is 1.6V

davenn
Gold Member
Well it's to build an electromagnet... But I need a lot of amps.

you have a little misunderstanding about current. Current isn't something you supply, it is something that is "drawn" from a PSU ( batteries etc)
and it will depend on the resistance of the load ( circuit) in this case your coil

So measure the resistance of your 20 metres of 22 gauge wire. it may be around 1 Ohm, 2 Ohms at most
After initial experiments ... you should consider increasing that to 2 or 3 times that length and possibly dropping to 26 or 28 gauge

Amps = Volts / Resistance
Amps = 1.6V / 2 Ohms
Amps = 0.8 A

several AA batteries in parallel will handle that current for a short while but as @russ_watters said, they wont last long

cheers
Dave

256bits and sophiecentaur
you have a little misunderstanding about current. Current isn't something you supply, it is something that is "drawn" from a PSU ( batteries etc)
and it will depend on the resistance of the load ( circuit) in this case your coil

So measure the resistance of your 20 metres of 22 gauge wire. it may be around 1 Ohm, 2 Ohms at most
After initial experiments ... you should consider increasing that to 2 or 3 times that length and possibly dropping to 26 or 28 gauge

Amps = Volts / Resistance
Amps = 1.6V / 2 Ohms
Amps = 0.8 A

several AA batteries in parallel will handle that current for a short while but as @russ_watters said, they wont last long

cheers
Dave
ohh ok that makes sense, i will try this thank you!

sophiecentaur and davenn
256bits
Gold Member
finding the diameter of the wire firs
22 gauge
I think it was mentioned that telephone wire was being used for the wrap.
That is a fairly thick wire, and.
That has a fairly thick plastic insulation making for a "loose" linking of the magnetic flux and the iron core, for the outer layers of wrap.
I would bet that after a few layers of wrap, a lot of the flux is not contributing to the magnetization of the core.
Better to use an wire that can be wrapped around the core in a dense bundle.
Look at some of the wrappings for motors, speakers, relays that use a thinner varnish or enamel as insulation.

possibly dropping to 26 or 28 gauge
Maybe even thinner wire and more wrap - 38 guage - after all, its NI ( ampere-turns ) that gives rise to the magnetomotive force.
https://www.britannica.com/science/magnetic-circuit#ref235761
More N--> More ohms --> less I --> batteries last longer, for the same NI.

davenn and sophiecentaur
sophiecentaur
Gold Member
2020 Award
That has a fairly thick plastic insulation making for a "loose" linking of the magnetic flux
Most electromagnets use multiple layers but the wire that's used is enamelled. That packs the turns in much more densely. With enamelled wire, it's much easier to wind neatly, which helps performance.

256bits
I think it was mentioned that telephone wire was being used for the wrap.
That is a fairly thick wire, and.
That has a fairly thick plastic insulation making for a "loose" linking of the magnetic flux and the iron core, for the outer layers of wrap.
I would bet that after a few layers of wrap, a lot of the flux is not contributing to the magnetization of the core.
Better to use an wire that can be wrapped around the core in a dense bundle.
Look at some of the wrappings for motors, speakers, relays that use a thinner varnish or enamel as insulation.

Maybe even thinner wire and more wrap - 38 guage - after all, its NI ( ampere-turns ) that gives rise to the magnetomotive force.
https://www.britannica.com/science/magnetic-circuit#ref235761
More N--> More ohms --> less I --> batteries last longer, for the same NI.
Oh wow ok, I shall look to find some then

As others have pointed out, you need only a set of nominal resistances to measure voltage, while you need a set of energy-depleting loads to measure amperage.

Resistors, in the sense of electronic components, are in general of negligible load once their resistance is overcome by adequate voltage, while ampere-measured loads, in the sense of objects that perform work, are ultimately heaters, and substantially deplete the power of a battery.

Please read the warnings and directions before verifying that your multimeter can display a/c 110v-125v when you insert the probes into a wall socket. To verify that the outlet can supply more than 15a, you need a 15a load -- for example, you can plug two or three 1500w heaters into a power strip that has a 15a breaker, turn them on, and watch the resultant current draw trip the breaker. Your multimeter alone can't tell you what that tells you.

On a car battery label, along with voltage, you can see ratings for such values as steady amps, peak or cold cranking amps, and total amp-hours. To run a 1000 watt microwave oven from a 12v battery, for a few minutes at a time, you'd want to see ratings in the hundreds of amps, and you'd want at least a few tens of amp hours to not run the battery down too soon. Cell phone batteries are similarly rated, but the results are usually reported primarily in terms of standby and in-use time.

To accurately determine the energy and power storage and transmission capacities of a battery set requires a more sophisticated testing apparatus set than the one you presented, which is fine for testing voltage, but insufficient for testing whether the battery can adequately supply your contemplated electromagnet.

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