Wednesday, Apr 07, 2010 at 10:48
okey dokey.
as you have comfirmed that you've used different wiring and it worked with a different battery we can discard any suggestion about sensitivity or the poly fuse issue.
what needs to be remembered is that a piece of wire and plugs and sockets might look like a good electric circuit they are just a resistance of different value.
a big piece of wire is a pretty small resistance and a smaller one is a bit bigger.
when you use a meter to measure the voltage on the end of a piece of wire ( with no other load on the wire ) it will indicate pretty much the same as if you measured it at the battery terminals. That's because a volt meter has a high input impedance ( impedance ~= resistance is near enough for DC ) and draws a tiny tiny amount of current. When we have a big load ( 4-5A by a running fridge ) the voltage drop across the cable is alot more ).
ie. the resistance of the cable /plugs/connectors might be 200 mOhms ( 0.2 Ohms ) pretty small in otherwords.
A digital meter will have an input impedance of ~10MOhms ( 10 million Ohms ).
And for the sake of the argument the battery is 12V.
NB changing any of the above does not alter the theory, just the final numbers a bit.
ok, have now got a cct with a battery, some wires and plugs and sockets and a digital volt meter.
12V ( battery ) -> 0.2 Ohms ( wire and stuff ) -> 10MOhms ( meter )
Current in the cct: V = I * R -> 12V = I * ( 10,000,000 + 0.2 )
I ~= 0.0000119999 and the voltage drop across the wire and plugs is 0.00001999 * 0.2 ~= 0.0000023999V or something extremely small, hence the meter will read close enough to the 12V of the battery.
Now suppose I connect a fridge to the end of the wires and plugs and it's running and it's drawing 5A.
My cable resistance won't change ( unless it changes temeperature and if it gets hotter it will increase in resistance ).
The voltage drop across the cables etc is now: V=I*R -> V = 5A * 0.2Ohms -> 1V, That's right 1V lost on the cable/plugs etc from just 0.2 Ohms of resistance.
The digital meter will read 11V and my fridge will see 11V
So hopefully this will show you why you must measure any voltages under the actual conditions you are trying to
test for and why it's easy to think you've tested the wiring when you measure at the end of a piece of wire.
In your system it still could be the 55Ah battery or the cabling ( or the plugs/sockets/battery terminals ) from the 55Ah to the fridge.
I'd suggest you get the 55Ah battery tested and/or use the new set of cable you made for the other battery to connect to the 55Ah battery.
I think the problem is somewhere in the wires/cable/plugs/55Ah battery.
Oh, just to make it more fun, a battery actually has some internal resistance as
well, if it increases as a battery starts to fail you get the same effect as having resistance in the wiring. under no load it reads ok but under load it has a lower voltage at the battery terminals. This is one of the things a battery load
test checks for.
gb
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