amperage

[artv]

Hi,..All..........when you hook bat's in series,..voltage increases,...when you hook,..parallel...current increases??......How do you determine the value of the strenght of the current,.........or measure the available power,.......when you hook Par,..seems I lose ~2/3 the amps,...........they don't add like series voltage...........artv

[rossw]

Hi,..All..........when you hook bat's in series,..voltage increases,...when you hook,..parallel...current increases??......How do you determine the value of the strenght of the current,.........or measure the available power,.......when you hook Par,..seems I lose ~2/3 the amps,...........they don't add like series voltage...........artv

When you hook batteries in series, the voltage does indeed increase to the numerical sum of the individual batteries.
Ie, 4 x 12V in series makes 48V.

When you put them in parallel, the current doesn't automatically increase over say a single 12V battery, but the POTENTIAL to deliver current increases. That is, the "internal resistance" of the source drops so it can deliver a larger current. A 120 watt load will draw 10 amps from a 12V battery. Putting 2 or more batteries in parallel won't increase the current because the load only *needs* 10A at 12V.

The only time the current automatically does increase, is if you're comparing a given load (that can run from a variety of voltage sources and take the same POWER - so not just a light or heating element - it needs to be a switch-mode power supply) - at different voltages.

Lets take that 120W load. 120W @ 24V will take 5 amps.
Reconnect the batteries so they're all in parallel, but now at 12V, you will need 10 amps for the load.
Reconnect the batteries so they're all in series, but now at 48V and you will only need 2.5A for the same load.

All bets are off with a plain old load like a dump-load resistor. A 4.8 ohm resistor will draw 5A from a 24V supply, and use 120 watts.
The same 4.8 ohm resistor on 12V will take 2.5A but only dissipate 30 watts, while at 48V it'll take 10A and dissipate 480 watts.

Hope that helps rather than just confuses!

[ghurd]

It is all Ohm's Law.
Everything Ross said can be shown with Ohm's Law.

You do not specify the load.
Some loads are not obviously following Ohm's Law, but they do follow it.
Diodes are not a resistance, they are a nearly fixed voltage drop.
And regular light bulbs are a more confusing device because the resistance changes with heat caused by the amps.
G-

[artv]

Hi   GHurd....the load, is the meter itself??....artv

[ghurd]

Nope!

Measuring amps with the only load being the meter will result in blown fuses.
Or melted meter leads.
Or other bad things.

Measuring amps with the only load being the meter is very nearly a dead short.
There is practically nothing to limit the amps.
Might be a great idea to Google "how to use a multimeter" before it is toast.

Loads are things like light bulbs, power resistors, motors, TV, computer....
G-

[artv]

very confused been reading and have reread about ohms law over and over.......I have two identical batteries each reads .64volts with meter......hook them in series they read~1.28v......hook par.,~.64v.......hook a 1k resistor in series with one reads~.1ma,..with the two paralleled,....~.64v,...~.15-.13ma,..........why isn't the reading .2ma........E=I*R....,.64v divided by 1000 ohms =.64ma........none of these #'s make any sense..........even if my decimal places are wrong the meter dosen't = the math.....I even went and bought a new meter ,but both old and new give the same readings......the book I have says to break the circut and put the ammeter in series ...I hook the + side of the resisitor to, + of battery, hook the + side of meter to, - of resistor, -side of meter to - of battery..........I've reversed the resistor reading stays the same..........ohms law dosen't seem to work,....obviously I'm doing something wrong but what???............artv

[ghurd]

The problem there is the batteries.
They are so dead they can not supply that much current without the voltage dropping.

64v divided by 1000 ohms =.64ma
That is correct.  And that is exactly what it would be if the battery was able to supply it, and the meter was accurate enough to read it.

Find a new alkaline battery.  That will be about 1.6V or so.  It will send 1.6ma through the 1K resistor.
G-

[artv]

Ghurd ....thanx for the reply ............are you saying that ohms law is only good for minimal values,....below these limits the law dose'nt apply...........artv

[ghurd]

Nope!

I am saying the battery is so empty it can not supply the power it looks like it can.

You have 2 meters.

Set one to test the amps.
Set the other to test the voltage on the resistor.
Do the tests at the same time.

Remember, the battery numbers are very low.  It will have some error in the meter.

You measured 0.1ma in the resistor.
That means the battery voltage at that time was pulled down to 0.1V.
And they will be measued that way when done at the same time.

0.1V = 0.0001A x 1000 ohms

For numbers that small, I will trust my calculator over the meter every time.
G-

[artv]

Ghurd..............is this drop you refer to ,......what is required for the load ,...is it the resistance of the load.......or are they the same thing............I'm thinking that alot of these rules need some looking over.........artv 

[ghurd]

Not sure how to explain it, except you will see it with both meters connected.

Connect the volt meter to the battery if you wish.  The result is the same 0.1V.

The battery is so low it can not supply 0.6ma and maintain 0.6V.
You have a 15 liter pail (battery), with 100ml of water (amps) in it, and you are trying to pour out 10 liters (amps).
The amps are not there to get out.  When you try, the voltage drops to 0.1V.

Guessing these are some kind of flashlight (torch) batteries.
They are considered dead at between 1.0V and 0.8V, depending on the load.
0.64V is too far past dead to be of any use at all.

The problem is not Ohm's Law.  The problem is the batteries.
Buy a new pair of batteries.
G-

[artv]

Hey.........rossw ,....read your answer ,...more than a dozen times still confused,......I'm old and slow,lol......Ghurd why does a load pull down the voltage ?.....if I have 12 volts with a bunch of weak batteries hooked in series,.......hook a little motor out of a cd player to it the volts drop right off the scale.............does this mean volts are useless?...........its all about amperage?......btw thank you for your patience,...........I'm sure you feel like reaching through and strangeling me,....I know I would........artv

[birdhouse]

artv-
think of batteries like the fuel tank in your car.  batteries (12v) would read somewhere around 14 volts for a full tank.  12 volts would be a half tank, and below that, your running out of gas. 

if your looking to do small amperage tests with 12 volt.  start with a decent car battery or something.  a bunch of near dud AA's are going to be hard to get any info out of because they will go dead so quickly, and have a range of voltage as they are going dead.

does this make sense? 

adam

[rossw]

Hey.........rossw ,....read your answer ,...more than a dozen times still confused,......I'm old and slow,lol......Ghurd why does a load pull down the voltage ?.....if I have 12 volts with a bunch of weak batteries hooked in series,.......hook a little motor out of a cd player to it the volts drop right off the scale.............does this mean volts are useless?...........its all about amperage?......btw thank you for your patience,...........I'm sure you feel like reaching through and strangeling me,....I know I would........artv

Which part(s) didn't make any sense? Or perhaps which parts *did* make sense?

I'll jump in before Ghurd does, perhaps - but as to why does the load pull down the voltage is a somewhat more complex question to answer. "It just does, ok?" isn't an answer you're going to be happy with either I suspect

Lets take the torch batteries we assume you're using. They have some chemicals in them and dis-similar metals. Between them all, a chemical reaction happens that makes POWER. Typically, around 1.5 volts at some measurable current. Larger cells like D cells can supply a lot more current than smaller cells like AA or AAA. But it's enough to understand that a chemical process makes volts (that's like electrical "pressure" to push electrons around). The "flow" of electrons is the current - amps. So you need both volts (pressure) and flow (amps) to do any useful work.

When the battery is "flat", the chemicals are all used up and the reaction that makes the power is barely working. It's still creating a small voltage, but there just are not enough chemicals left to make a decent amount of power. When you try to draw amps out of it the voltage drops.

It's a bit like gasoline engine. When you have stale fuel the engine will run, it might even make nearly full revs - but as soon as you try to put any load on it, it coughs and splutters and dies. If you're quick and take the load off, it'll recover. "Stale" fuel is like the used-up chemicals in the battery - it just can't make the POWER you want.

[artv]

rossw.......thanx for the explanation ,.....It makes things a little more clear,..........from what I understand batteries work on chemical reaction,......excess of electrons at neg. post,....shortage of electrons at pos. post.............connect the two through a circut causes flow ,....- to +,...........to balance out ,........this is the part that does'nt make sense,......if the negative terminal is in excess of electrons,...the positive terminal is missing electrons ,...why don't they just balance within the battery itself...........sorry but it just does'nt make any sense...........artv

[rossw]

,......if the negative terminal is in excess of electrons,...the positive terminal is missing electrons ,...why don't they just balance within the battery itself...........sorry but it just does'nt make any sense...........artv

While not strictly accurate, and not necessarily applicable to all battery chemistry, think of it like this if it will help you visualise it:

The positive plate and the electrolyte react, the electrolyte "pulls" electrons out of the positive plate. The electrolyte hangs onto them harder than the positive plate can "pull" them back out, so it's a one-way battle.

The same thing at the negative plate. It pulls electrons out of the electrolyte and doesn't want to give them back. When the battery is open-circuit (nothing connected) this "stealing" of electrons stabilises at the "strength" of the reactions - the negative plate is as "full" as it can get, and the electrolyte is as "full" as it can get.

Connecting a load across + and - plates lets the electrons flow - the chemical reactions now just form a "pump" to keep pulling electrons from + plate to - plate and the load just keeps leaking them back!

This also explains to some extent the voltage drop you see with a flat battery. The "pump" is stuffed and can hardly pump anything, so when you do try to pull something out of it, the pressure drops (volts) and the flow is barely a trickle (amps).

[ghurd]

artv,
You need something like a "VVS" to get this completely understood...
Such as the "Basic and Simple Variable Voltage Supply Kit" available from ghurd.info
PM me your mailing address.
G-

[artv]

rossw.........thats the best discriptive explantion I"ve ever read.......Thank-You........Ghurd.....I already have low v supply........just need to learn how to use it.......artv

[rossw]

rossw.........thats the best discriptive explantion I"ve ever read.......Thank-You

Hooray!

* rossw chalks one up on the board!

[artv]

Hi...what is the blue and white build up of material on the outer suface of a battery??......I put ten in series and,....three in parallel,.....to make 11 in series,..will the 10 ,...being lower,...bring the 3 parallel,...down to the series #.......artv

[ghurd]

That gook is the chemicals that formerly made the battery work.
It is corrosive and will eat the test clips, battery holder springs, and cloth.
Throw them away.

If they are rechargeable nicd batteries, the stuff is very VERY bad.
Do NOT get any gook on you.
Put them in a plastic bag.
Turn them in for recycling as soon as possible.

Get some new batteries.
Alkaline D cells have the most power for the money.
Non alkaline batteries are a waste of money.

See
http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws
For what you are asking about the battery configuration.
The voltage will be the same as 11 in series.
The 10 will simply go dead faster than the 3.
G-

[artv]

if its' corrosive can't we use the heat as energy?...........artv

[JW]

Im assuming were talking about lead acid batterys here.

Im not so sure how common the use of felts are. In general, the felts are green and red. Im not sure what there soaked with, but they do seem to deter- corrosion at the battery terminals. You will seen them on automotive battery posts, under the terminal.

Most people I know will say, "pop a can of cola soda" then pour that on the terminals, and it will "eat the corrosion off" I have found that the "house hold cleaner" 409 will also cut the sulfide, and restore "bright copper" to battery terminals..

JW

[ghurd]

Use the heat?
I guess... if we were VERY patient!
The heat is about as fast flowing as a car fender rusting.  (Not really that slow, but very slow)

Like Ross said, "a chemical reaction happens that makes POWER"

The corrosion happening inside the battery is what happens when the electrons move from one terminal to the other.
Following the car fender idea, when the battery insides have rusted into dust, the battery is dead.

Following the heat idea, that is why batteries get hot when drained too fast.

You may be able to answer you next question by reading this,
http://en.wikipedia.org/wiki/Sacrificial_anode

G-

[artv]

first off I'd like to say thanks for your patience to everyone who responded ,...rossw you said that if the load requires a certain amount of power ,...paralleling the batteries won't increase the amps because the load dosen't require it.....so how do you check a source to see its' max available amps?......do you just put a big load on it like a 18 volt cordless drill or bigger?...........artv

[rossw]

first off I'd like to say thanks for your patience to everyone who responded ,...rossw you said that if the load requires a certain amount of power ,...paralleling the batteries won't increase the amps because the load dosen't require it.....so how do you check a source to see its' max available amps?......do you just put a big load on it like a 18 volt cordless drill or bigger?...........artv

There are some (quite expensive) instruments that will measure the internal resistance of batteries. This will tell you what they can "theoretically" deliver. Add in the cable, switch, fuse etc resistances and you'll be getting somewhere in the "realistic" area.

My own cells claim a "short circuit" current of something like 32,000 amps (from memory). That's a *SCARY* amount of current. Each cell is only 2V, but thats still 60KW of power. Think of it as being like 60 one-bar radiators on one spot. It's serious power.

I've got 24 cells in series, and 3 banks in parallell - so it could push some *truly* dangerous power. (Thats also why I added a 200 amp HRC fuse right at the battery post of each string).

For more modest batteries, you could use a decent voltmeter across the terminals of the battery. Apply a reasonably substantial load (the exact load would depend on your battery type, capacity and condition) and monitor how far the voltage sags and how quickly.

[artv]

Hi ....I hooked two double A batteries in series,...2.24v, added 1k resistor, 2.13ma. without the resistor(dead short), , 29ma, but dropping, ~.1 ma /sec the dropping of the amp readings slows around 21ma. The cells I have read 4v , add resistor,1.5ma,without (ds), 2.5ma but dosen't drop like the AA's. The voltage readings here don't suit ohms law,I guess basically I don't understand the dropping of values ?   "how far the voltage sags and how quickly"........confused......artv

[rossw]

Hi ....I hooked two double A batteries in series,...2.24v, added 1k resistor, 2.13ma. without the resistor(dead short), , 29ma, but dropping, ~.1 ma /sec the dropping of the amp readings slows around 21ma. The cells I have read 4v , add resistor,1.5ma,without (ds), 2.5ma but dosen't drop like the AA's. The voltage readings here don't suit ohms law,I guess basically I don't understand the dropping of values ?   "how far the voltage sags and how quickly"........confused......artv

Dejavu.

two AA batteries in series won't give you 2.24V unless they're dead flat.
If they're dead flat, you won't get any sensible test results from them.
I think this is where we came in....