"James Hahn" wrote in message
news
"Meindert Sprang" wrote in message
...
snip

There are two possible scenarios: one battery wears more quickly than
the
other, eventually reaching a state where the cell voltages reach higher
values due to increasing internal resistance. Thus the charges shuts off
too
early, leaving the better battery not topped up.
The charger does not monitor cell voltages. It monitors the voltage
across
all the cells in series.

That is exactly what I meant and exactly the reason why it can go wrong. So
I would like to invite you to shoot holes in my following argument:

The charge of a battery is the product of current x time.

Both batteries are in series and one load is connected to the set, operating
at 24V. Another load is connected across only one battery, operating at 12V.
So it is evident that one battery is discharged more than the other.

I start to charge the set in series, so the current through both batteries
is exactly the same. Since one battery is discharged more than the other and
the current throug both is the same, one battery must be charged longer that
the other. Exactly how am I going to achive that with the same current
through both batteries?
One battery will reach the full state before the other but is still being
charged with full current because the other battery hasn't reached the
voltage that corresponds with full charge.

Now, if you can come up with an valid argument why this shouldn't damage one
battery, I bow and take my hat off.

Meindert

Hello Meindert,

I agree with your analysis and go a step beyond: even if
there is no differential load across one of the batteries,
uneven charging is possible (or likely) because of
differences in the batteries themselves.

Regards,

Chuck

Meindert Sprang wrote:
"James Hahn" wrote in message
news
"Meindert Sprang" wrote in message
...

snip

There are two possible scenarios: one battery wears more quickly than

the

other, eventually reaching a state where the cell voltages reach higher
values due to increasing internal resistance. Thus the charges shuts off
too
early, leaving the better battery not topped up.

The charger does not monitor cell voltages. It monitors the voltage

across

all the cells in series.


That is exactly what I meant and exactly the reason why it can go wrong. So
I would like to invite you to shoot holes in my following argument:

The charge of a battery is the product of current x time.

Both batteries are in series and one load is connected to the set, operating
at 24V. Another load is connected across only one battery, operating at 12V.
So it is evident that one battery is discharged more than the other.

I start to charge the set in series, so the current through both batteries
is exactly the same. Since one battery is discharged more than the other and
the current throug both is the same, one battery must be charged longer that
the other. Exactly how am I going to achive that with the same current
through both batteries?
One battery will reach the full state before the other but is still being
charged with full current because the other battery hasn't reached the
voltage that corresponds with full charge.

Now, if you can come up with an valid argument why this shouldn't damage one
battery, I bow and take my hat off.

Meindert

"Meindert Sprang" wrote in message ...
"James Hahn" wrote in message
news
"Meindert Sprang" wrote in message
...
snip

There are two possible scenarios: one battery wears more quickly than
the
other, eventually reaching a state where the cell voltages reach higher
values due to increasing internal resistance. Thus the charges shuts off
too
early, leaving the better battery not topped up.
The charger does not monitor cell voltages. It monitors the voltage
across
all the cells in series.

That is exactly what I meant and exactly the reason why it can go wrong. So
I would like to invite you to shoot holes in my following argument:

The charge of a battery is the product of current x time.

Both batteries are in series and one load is connected to the set, operating
at 24V. Another load is connected across only one battery, operating at 12V.
So it is evident that one battery is discharged more than the other.

I start to charge the set in series, so the current through both batteries
is exactly the same. Since one battery is discharged more than the other and
the current throug both is the same, one battery must be charged longer that
the other. Exactly how am I going to achive that with the same current
through both batteries?
One battery will reach the full state before the other but is still being
charged with full current because the other battery hasn't reached the
voltage that corresponds with full charge.

Now, if you can come up with an valid argument why this shouldn't damage one
battery, I bow and take my hat off.

Meindert

Duh! the two batteries are in series. The one with the 12v tap cannot become
more discharged than the one without because the battery without the tap
will charge the one with. Or the one with will discharge the one without
until they are both the same. This is the same thing that happens with
individual cells in a 12v battery. With two 12v batteries in a series one
simply has more cells in a series.

CN

"Capt. Neal®" wrote in message
...
Duh! the two batteries are in series. The one with the 12v tap cannot
become
more discharged than the one without because the battery without the tap
will charge the one with. Or the one with will discharge the one without
until they are both the same. This is the same thing that happens with
individual cells in a 12v battery. With two 12v batteries in a series one
simply has more cells in a series.

If you became familiar with the concept of batteries in series and batteries
in parallel, then come back and review your own reply.

Meindert

"Meindert Sprang" wrote in message ...
"Capt. Neal®" wrote in message
...
Duh! the two batteries are in series. The one with the 12v tap cannot
become
more discharged than the one without because the battery without the tap
will charge the one with. Or the one with will discharge the one without
until they are both the same. This is the same thing that happens with
individual cells in a 12v battery. With two 12v batteries in a series one
simply has more cells in a series.

If you became familiar with the concept of batteries in series and batteries
in parallel, then come back and review your own reply.


Wrong, wrong, wrong, wrong, wrong, wrong and incorrect!
How can some people's minds be so ineffective at thinking?

Just what the hell do you think a 12v lead acid battery is, anyway?
One giant 12v cell? NOT! A twelve-volt, flooded, lead/acid battery
consists of six cells of approx 2.2v each. These cells are in a series.
When the battery is charged, electrons move through all the cells or
the battery will not be charged. Open one cell and the whole battery
will show dead because no current can move though the collection
of cells in a series but the cells that are not open will show approx.
2.2 volts each. As long as the cells are all connected and none of
them are open (or shorted) the battery works as a unit.

Perhaps you ******s would understand it better using flashlight D cells
as an example. Take one D cell that is half charged (Ni-Cad) and put it into
a two-cell flashlight in the company of another NI-Cad) D cell that is fully
charged and turn on the switch. The bulb will light and current will
pass through the circuit. As well as working to light the bulb, the
fully-charged cell will discharge into the half charge cell until
the voltage in both cells equalizes.

There is no difference between two 12v batteries in a series. As long
as the two batteries are are part of a circuit, which is the case in any
yacht (unless an isloation switch is turned off so the circuit is no more)
then, as is the case in the flashlight example, the two batteries will
equalize and they will do so even if one battery has more of a load
on it than the other because of a 12v tap to run a VHF.

Anytime the two 12v batteries in series are not part of a circuit then
the battery with the 12v tap will have its voltage lowered and the
battery without the tap will remain unaffected. Two 12v batteries
connected in series but not part of a circuit remain, for all practical
purposes, two separate batteries.

You need to learn how circuits work before you spew your nonsense,
Mr. Sprang. It seems to me you're confused when it comes to knowing
the difference between volts (pressure) and amps (flow). Batteries
always have voltage unless discharged but they only have amperage if
they are part of a circuit. Since the batteries, connected as part
of a circuit in Nigel's yacht, are in series in the circuit they will
both have equal voltage as long as they remain part of said circuit.


I hope this helps.

CN

Capt. Neal® wrote:

You need to learn how circuits work before you spew your nonsense,
Mr. Sprang.

Hmm. Perhaps he should stop being a professional electronics engineer
too, eh?

Pete

"Pete Verdon" d wrote in message ...
Capt. Neal® wrote:

You need to learn how circuits work before you spew your nonsense,
Mr. Sprang.

Hmm. Perhaps he should stop being a professional electronics engineer too, eh?

Pete

He needs to go back to school!

CN

In article ,
Capt. Neal® wrote:

He needs to go back to school!

CN

You Sir, are a complete, and utter Moroooon.... (Bugs Bunny Definition)
Now please take you DC Electrical Theories over to one of the
alt.engineering Newsgroups and see if they fly over there...... I can
hear the "Rolling on the floor, Laughing" already.....

We here have been vary patient with you, but your entertainment value
is about run it's course, and your noninformative posts could actually
cost unlearned folks, money and time. I only hope your Navigation skills
aren't on a par with you engineering skills.


Me

"Capt. Neal®" wrote in message
...
Wrong, wrong, wrong, wrong, wrong, wrong and incorrect!
How can some people's minds be so ineffective at thinking?

I know it is pointless to argue with you but:

Perhaps you ******s would understand it better using flashlight D cells
as an example. Take one D cell that is half charged (Ni-Cad) and put it
into
a two-cell flashlight in the company of another NI-Cad) D cell that is
fully
charged and turn on the switch. The bulb will light and current will
pass through the circuit. As well as working to light the bulb, the
fully-charged cell will discharge into the half charge cell until
the voltage in both cells equalizes.

Perhaps you could try to envision in which direction current flows through
the empty cell in this example and next, try to envision in which direction
current flows when *charging* a cell. Or even better: try this example for
yourself.

I hope this helps.

Certainly not.

Meindert

"Meindert Sprang" wrote in message ...
"Capt. Neal®" wrote in message
...
Wrong, wrong, wrong, wrong, wrong, wrong and incorrect!
How can some people's minds be so ineffective at thinking?

I know it is pointless to argue with you but:

Perhaps you ******s would understand it better using flashlight D cells
as an example. Take one D cell that is half charged (Ni-Cad) and put it
into
a two-cell flashlight in the company of another NI-Cad) D cell that is
fully
charged and turn on the switch. The bulb will light and current will
pass through the circuit. As well as working to light the bulb, the
fully-charged cell will discharge into the half charge cell until
the voltage in both cells equalizes.

Perhaps you could try to envision in which direction current flows through
the empty cell in this example and next, try to envision in which direction
current flows when *charging* a cell. Or even better: try this example for
yourself.

I hope this helps.

Certainly not.

Meindert


So even an engineer might understand. . .

http://hyperphysics.phy-astr.gsu.edu.../leadacid.html

Engineers look at a battery as a physical object while an electrician looks
at it as a container for a chemical reactions that store and release electricity.

Higher voltage than a fully charge battery can supply, when applied to the
battery terminals drives the chemical reaction and changes it from releasing
electrons to storing electrons but does not reverse the current as most dumb
engineers claim.

Read the above link carefully and click on all the links and perhaps you
will understand the error of your thinking.

You're welcomd.

CN