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#### EVDL List

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I've got two strings of ThunderSky cells, 35 cells each, one string of
200Ah cells and one string of 90Ah cells, all with shunt regulators on
them. The two strings are connected to each other through circuit
breakers on both the positive and negative terminals. I want to test
out charging the two strings in parallel, so I need to connect them
together. The 200Ah string is at currently 142V and the 90Ah string is
at 136V. I think I should be okay just flipping on the circuit
breakers, but I want to verify I'm thinking about it correctly. Using
an internal resistance of .005 ohms for the 200Ah cells and .015 Ohm
for the 90Ah cells (I think resistances both are actually higher, but I
want to be conservative in my estimates), then the current through the
cells when the circuit breakers are turned on and the cells start to
equalize would be I = V/R:

(142 - 136) / ((.005 * 35) + .(015 * 35)) = 9A

Have I thought about that correctly?

Thanks.

Bill Dennis

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#### EVDL List

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##### Registered
Joined
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72,624 Posts
Discussion Starter · ·
Bill Dennis wrote:
> I've got two strings of ThunderSky cells, 35 cells each, one string of
> 200Ah cells and one string of 90Ah cells, all with shunt regulators on
> them. The two strings are connected to each other through circuit
> breakers on both the positive and negative terminals. I want to test
> out charging the two strings in parallel, so I need to connect them
> together. The 200Ah string is at currently 142V and the 90Ah string is
> at 136V. I think I should be okay just flipping on the circuit
> breakers, but I want to verify I'm thinking about it correctly. Using
> an internal resistance of .005 ohms for the 200Ah cells and .015 Ohm
> for the 90Ah cells (I think resistances both are actually higher, but I
> want to be conservative in my estimates), then the current through the
> cells when the circuit breakers are turned on and the cells start to
> equalize would be I = V/R:
>
> (142 - 136) / ((.005 * 35) + .(015 * 35)) = 9A
>
> Have I thought about that correctly?

Yes, this calculation is correct. However, what matters most is the
actual resistance, rather than the theoretical resistance.

If it were me, I'd connect the two pack through an ammeter, so I can see
what the current actually is. Keep at least one fuse or circuit breaker
in series as well, in case the current gets too alarming.

--
Ring the bells that still can ring
Forget the perfect offering
There is a crack in everything
That's how the light gets in -- Leonard Cohen
--
Lee A. Hart, 814 8th Ave N, Sartell MN 56377, leeahart_at_earthlink.net

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