Next message: Beth Russell: "TWL: Steel boat construction - HELP"
No matter where the fault is, some of the batteries will have either the
positive or negative circuit disconnected, leaving them out of the circuit
depending on whether the bad link is in the positive or negative side.
As a result the drop in voltage will be greater under load than it would be
if all batteries were in the circuit.
It makes NO difference which end of the bank the positive and negative
supplies are taken from, it will always be the group of batteries on the far
side of the fault that are disconnected.
You say "If you take both positive and negative from the same end it will
never show - - -"
Of course it will show if they are taken from the same end. The open link
will remove some of the batteries and the voltage drop will be greater.
If you take the lead from the other end, those batteries that would have
been left out will now be the ones IN circuit BUT THOSE on the OTHER side of
the bad link will now be OUT of the circuit.
EITHER way some batteries will no longer be in the parallel circuit.
I don't know how else to explain it to you and this is my last attempt.
If you are still confused, contact me directly so we don't clutter the list.
Andina Foster,
> If there is a fault in the links between the batteries with the positive
> taken from one end and the negative from the opposite end of the bank it
> will show up as voltage drop on the output when you put the system
> under load.
> If you take both positive and negative from the same end it will never
> show as the batteries will be effectively disconnected from the system
> by the bad link.
> I am talking about batteries in parallel here in a series/parallel
> combination it wont be so obvious.
>
> Julian Tether
> Barge Parglena
> e-mail:
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