Max current flow when connecting two batteries

simonfraser

simonfraser

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So I am about to set up a Victron energy cyrex
two identical 110Ah AGMs
one battery may be 13.6 V other 12.4V
now the ‘relay’ closes and current flows
how much current is going to flow ?
alternator can provide 20Ah one side, solar 3Ah other side
the wire is going to be 30cm to each, what gauge ?
tnx
 
The full alternator output could flow.
Not clear what that is . You say 20Ah. What's that mean ??? What is the max out put possible in Amps ? Do you mean that the alternator is only 20 A max output
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If you are going to fit the "start assist" feature the wire must be heavy enough to carry the full Cyrix 5 secs cranking rating current. (180 A ). 25mm2 will be adequate butI'd suggest using the same as your battery cables.

If you are not fitting the "start assist" you only need to allow for the max alternator output. Thats 1.5mm2 If the alternator really is only 20A.There is no need to allow for voltage drop with such short connections. Same as battery cables might be a good idea in case you or someone else adds the start assist inthe future

Fit a fuse in one (or both, if you prefer), as indicated in the Cyrix instruction manual sized to suit the wiring used but not greater than the Cyrix rating (120 A presumably)

Ps learn the difference between Amps (A) and ampere hours (Ah)

On 2nd thoughts don't go smaller than 2.5mm2
 
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simonfraser said:
So I am about to set up a Victron energy cyrex
two identical 110Ah AGMs
one battery may be 13.6 V other 12.4V
now the ‘relay’ closes and current flows
how much current is going to flow ?
alternator can provide 20Ah one side, solar 3Ah other side
the wire is going to be 30cm to each, what gauge ?
tnx
Click to expand...
If you are connecting directly the two batteries together, à very large current will flow, possibly hundreds of amps until something burns through.
 
simonfraser said:
the alternator can deliver a max of 20Ah, but that all depends on how full the battery is of course ;)
not going to fit the start assist

so the max current flow from one battery to the other is due to the alternator, not the voltage difference between the two batteries ?
Click to expand...
unless I'm mistaken, current is measured in Amps not Amphours. alternators the same, so your outboard alternator has at full whack 20A output?
 
westernman said:
If you are connecting directly the two batteries together, à very large current will flow, possibly hundreds of amps until something burns through.
Click to expand...
The Cyrix is rated at 120 amps
There should be a fuse to protect the wring .
The fuse must not be greater than 120A
Hundreds of amps will not flow !
 
VicS said:
The Cyrix is rated at 120 amps
There should be a fuse to protect the wring .
The fuse must not be greater than 120A
Hundreds of amps will not flow !
Click to expand...
If he connects the two batteries together, lots of amps will flow from the higher voltage one to the lower.
May be the VE Cyrix has separate connections for the different batteries with diodes to prevent one battery charging the other.

But what I said stands. If one battery is at 13.6V and the other at 12.4V and the terminals are directly connected together, then current will flow which will be only limited by the internal resistance of the of two batteries (which for Lithium is very low) and the wiring. Which if chunky will also be low resistance. If not so chunky, then it will heat up and melt.
 
westernman said:
If he connects the two batteries together, lots of amps will flow from the higher voltage one to the lower.
May be the VE Cyrix has separate connections for the different batteries with diodes to prevent one battery charging the other.

But what I said stands. If one battery is at 13.6V and the other at 12.4V and the terminals are directly connected together, then current will flow which will be only limited by the internal resistance of the of two batteries (which for Lithium is very low) and the wiring. Which if chunky will also be low resistance. If not so chunky, then it will heat up and melt.
Click to expand...

If it is all installed as the manual directs this cannot happen
AND, FYI
AGM batteries are not lithium so no need to introduce that as a red herring

To avoid unnecessary fuse blowing , and for that reason only , not to do with all this nonsense about things burning through, I will revise my suggestion of using 2.5mm2 cable in favour of 25mm2 I suggested earlier. Its rated at 170 A but the fuse(s) must not be greater than 120 A for a Cyrix ct
 
In connecting 2 batteries to parallel that are different voltages as said current depends on the voltage difference and the resistance of the wire. (pos and negative circuit)
I would not be too concerned about wire getting too hot. The resistance of light wire will limit current and high current will only be for a short period. Perhaps of more concern might be the relay contacts which make the initial contact and conduct the high current. A sophisticated system might have 2 relays the first switching in a connection of significant resistance like 1 ohm then later in time another relay for full low resistance connection. But then I am not sure that would be worth doing just make sure you have a chunky relay which of course you need if it is also to be used for engine start boost. ol'will
 
William_H said:
In connecting 2 batteries to parallel that are different voltages as said current depends on the voltage difference and the resistance of the wire. (pos and negative circuit)
I would not be too concerned about wire getting too hot. The resistance of light wire will limit current and high current will only be for a short period. Perhaps of more concern might be the relay contacts which make the initial contact and conduct the high current. A sophisticated system might have 2 relays the first switching in a connection of significant resistance like 1 ohm then later in time another relay for full low resistance connection. But then I am not sure that would be worth doing just make sure you have a chunky relay which of course you need if it is also to be used for engine start boost. ol'will
Click to expand...
The Cyrix is designed to support start assistance for a short time. ( 30 secs)
 
simonfraser said:
the alternator can deliver a max of 20Ah, but that all depends on how full the battery is of course ;)
not going to fit the start assist

so the max current flow from one battery to the other is due to the alternator, not the voltage difference between the two batteries ?
Click to expand...
Think you have your terminology wrong. The alternator will be rated in amps not ampere hours. 20ah could be 1amp for 20 hrs or 20 amp for 1 hr!
 
simonfraser said:
'If one battery is at 13.6V and the other at 12.4V and the terminals are directly connected together, then current will flow which will be only limited by the internal resistance of the of two AGM batteries'
yep, so how much current, what's the formulae to work that out ?
Click to expand...

Ohm's Law
 
simonfraser said:
'If one battery is at 13.6V and the other at 12.4V and the terminals are directly connected together, then current will flow which will be only limited by the internal resistance of the of two AGM batteries'
yep, so how much current, what's the formulae to work that out ?
Click to expand...

The take the difference between the two voltages and divide by the combined internal resistance of the two batteries
A car battery is said to have an internal resistance of 0,02ohms so take 0.04 ohms as the combined resistance, although the partially discharged battery will have a slightly higher resistance
Assuming you batteries are about car battery size
(13.6 - 12.4)/0.04 = 30. Therefore initially 30 amps will flow but the lower voltage will slowly rise and the higher one will quickly fall. The current will, therefore quickly fall to a much lower figure.

You can determine the acual internal resistance of a battery by measuring the open circuit volts once it has rested for at least 24 hours and any remaining surface charge has been discharged (V1).
Then connect a load ( eg a car headlamp bulb or even a flasher bulb) with an ammeter in series Wait until the the readings have stabilised then read the battery volts again ( V2 ) and the current (I)

The internal resistance = (V1 - V2)/I

.
 
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