Lead-acid battery - Current issues?

[elementcollector1]

I recently acquired a sealed lead-acid battery, 6V, 4.5 Ah, 20 h for use in electromagnet experiments. I assumed that this would output 4 amps, given the ratings previously mentioned. However, it only outputs about 0.06 or so. How can I get this up to 4 amps or more? I would like to use this, an F630 MOSFET and about 710 ft. of 32 AWG magnet wire for Arduino-controlled magnetic levitation, but it's a moot point if the current is so low. Is there a way to force it to get higher, even if it does shorten battery life?

Edited January 19, 2014 by elementcollector1

[John Cuthber]

According to this table the resistance of 32 gauge wire is about 164 ohms per thousand feet.

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/wirega.html

You have about 710 feet of it so that should give about 116 ohms.

Ohms law

I=V/R

tells us that the current will be

I=6/116 = 0.051 Amps

 

So your value of 0.06 Amps is pretty much what you would expect to get.

 

Why did you think you would get 4 Amps?

[Sensei]

I recently acquired a sealed lead-acid battery, 6V, 4.5 Ah, 20 h for use in electromagnet experiments. I assumed that this would output 4 amps, given the ratings previously mentioned.

 

1 A*h = 1 A * 3600 s

but 1 A*s = 1 C

so

4.5 Ah = 4.5*3600 = 16200 C

 

One electron has charge 1.6*10^-19 C

so 16200 C / 1.6*10^-19 = 1.0125*10^23 electrons

 

If you will have resistance 1.33 ohm, I=U/R=6/1.33 = 4.5 A, your battery will be working just 1 hour with such current.

With I=0.6 A (3.75*10^18 electrons per second) it'll be working ~7h 30m.

I don't know where you get 20h..

You would need 0.225 A to work it 20h.

Edited January 23, 2014 by Sensei

[elementcollector1]

Says 20 hours on the battery.

So, I presume that means that if it were to run at 4.5 A, it would be able to do so for 20 hours.

[John Cuthber]

You need to understand what it actually says on that battery.

" 6V, 4.5 Ah, 20 h"

Not 4.5A

It's a capacity, not acurrent.

The battery can deliver roughly 4.5 amps for 1 hour or 9 amps for half an hour or 0.25 amps for 18 hrs.

I say "roughly" because the efficiency of storage is a bit dependent on the current drawn.

So the manufacturer measures the capacity at a known current.

In this case they chose 0.225 Amps and the battery lasted 20 hrs.

20 hours * 0.225 amps =4.5 amp hours

[Sensei]

Says 20 hours on the battery.

So, I presume that means that if it were to run at 4.5 A, it would be able to do so for 20 hours.

 

I showed you how to calculate stuff in #3 post..

[elementcollector1]

And I read the post. I was reading off the battery.

Anyway, to get it up to the desired voltage (and therefore current), I probably need to build a boost converter.

[John Cuthber]

And I read the post. I was reading off the battery.

Anyway, to get it up to the desired voltage (and therefore current), I probably need to build a boost converter.

If you read the battery you would see that it said "4.5 AH" not" 4.5A"

 

What you probably need is thicker wire.

 

What are you actually trying to achieve?

[Sensei]

I found in shop such accumulator
http://sklep.avt.pl/akumulator-zelowy-6v-4-5ah.html

 

"pojemność (25 stopni C)

20 godzinna: 4.50Ah
10 godzinna: 4.28Ah
5 godzinna: 3.78Ah
1 godzinna: 2.70Ah"

 

It means that capacitance is variable, and depends on how fast or slow, you will be utilizing accumulator.

 

If you will be utilizing it for 20 hours, you will have capacitance 4.5 Ah = 16200 C. And you need I=0.225 A to last it that long.

The faster you will use accumulator, the smaller capacitance.

 

There is also showed dependency of capacitance on temperature in page that I linked.

 

There is also showed spontaneous discharge of accumulator in 3,6,9 months..