If you have 4 x 1.5 volt identical cells (batteries) and you arrange them so you have 2 cells in series connected in parallel to the other 2 cells, so you have 3 volts powered by those 4 cells. If you place a load on one and only one of the cells and run that battery to zero volts, what wold we expect to be the state of charge of the other 3 cells?
Are you asking what voltage would be measured across your battery array when 1 cell is completely discharged? Do you want to know if it’s less than 3V somehow?
That would help answer it. Also would the cell in the series with the cell with the load be expected to stay fully charged? Would it attempt to overcharge? And yes the final voltage of the ‘battery bank’, which I would assume would be the sum of the fully charged cell + zero from the depleted cell, so 1.5 volts, but that would requite the other parallel cells to also add up to 1.5V, so perhaps 0.75V each as they would need to discharge?
I think OP is describing a pair of parallel batts in series with a second parallel pair.
The batteries in parallel can be thought of as one battery for charging or discharging. In other words, not when in storage and unloaded where parallel batteries tend to self-discharge.
The second pair of parallel cells are unaffected.
No the opposite, 2 sets of 2 batteries (cells) in series, those connected in parallel.
(-)------i|-i|------+
in parallel with another
(-)------i|-i|------+
So that the (-) are connected to each other and the + is also connected to the other (+)
The upper left cell is additionally hooked up to a load and drained.
It is going to depend on a few things. Model the circuit as its Thevenien equivalent. So four voltage sources each with a series resistor. Model a discharging cell by increasing its series resistance.
With no load on the composite 2x2 battery, the actual circuit is better viewed as two batteries back to back in series. Current flowing into the load across a single cell sees two sources, the single cell, and the series string of two forward facing cells and one backward facing cell. Open circuit voltage across the cell is obviously 1.5v.
Once you start to draw current things depend upon the physical realisation of the cell seeing reverse current. In reverse you will see a different effective series resistance than forward, and over time the disparity will get worse.
The final state of the system (when the single cell is discharged) will depend upon how the effectively reversed cell acts. It will be somewhere between fully charged and utterly cooked. The state of the other two cells is likely partially discharged.
Francis has it right. This scenario is quite common in solar arrays and must be accounted for. Solar arrays are often wired in series-parallel and have to have diodes placed at the right points to prevent back voltage.
So it really depend on the characteristics of the battery that will see a reverse charge. So the final voltage across the array can be anywhere from 1.5 volts and 3 volts I assume.
I also assume that the load across the single cell would have consumed more total energy with the cell as part of the array than it would have been on its own. And that extra energy was provided by the other parallel battery.
If they are perfectly the same, the two cells in parallel act as one cell of twice the volume.
Either both the cells in parallel are discharging, or the 2nd cell isn’t in the circuit as described.
The other cells in series are irrelevant when the circuit is only from ,they have no influence on the state of the discharging cell. Their state of charge or discharge is only due to the sum of the current through them, No current = no change… also, discharging the two cells as per the question, will stop the circuit working, the two discharged cells will stop conducting , so it would be better if they were taken out of the circuit…
For that reason, if you have a set of cells that were in storage, and they don’t operate the device, test them individually, there may be just one failed !
But as per the wisdom of doing what the OP said - do not do that, it causes cells to fail. Charge the cells equally and use them equally.
Please look at the diagram I attempted in post 5, this does not exist in my question. There are 2 batteries in parallel, each battery made up of 2 cells in series.
The answer in the real world is indeterminate as the cells will never be identical so, yeah, don’t do that. As a thought experiment though, what will happen is that the cells with reach some sort of equilibrium with each other, and the discharged cell will have a little voltage across it (was 0 V), its neighbour will have a voltage a bit above 1.5 V (was exactly 1.5 V), and the two fresh cells in series will both read a bit below 1.5 V each (were exactly 1.5 V each). A simple model is to treat each cell as an ideal voltage source in series with identical internal resistances, and if you do that then the top 2 cells will be 1.125 V each, the discharged cell will be 0.375 V, and its neighbour will be 1.875 V (assuming the load is disconnected, and noting that 1.125 + 1.125 = 1.875 + 0.375). In practice though, cell models don’t look like that: their internal resistances increase and the off-load voltages decrease as the cells becomes discharged, and even non-rechargeable cells will be charged up a bit if current is pumped into them, so once again the answer is indeterminate without knowing an accurate behavioural model of the cells.
Interestingly, cell polarities can become reversed under certain conditions, and it’s an experiment you can try at home on any cell that isn’t likely to burst into flames when abused (e.g., use zinc-carbon or alkaline AAs, not Li-ion). If you’ve got an old flashlight that uses an incandescent bulb (not LEDs) and uses 2 or more cells in series, turn it on and leave it overnight until the bulb no longer glows. The voltage across the bulb will now be close to zero, but the individual cell voltages will now be a mix of positive and negative voltages, so that their sum voltage is essentially zero. What has happened here is that the strongest cell has forced the weakest cell to reverse its polarity, e.g., if you have two 1.5 V cells in series and you load them until flat, the strong cell might wind up as 0.4 V (say), and the weak cell would then be -0.4 V.
It doesn’t happen too often, but a completely dead car battery can be connected to a charger wrong and re-charged “backwards”, and will cause some head scratching. The capacity is reduced considerably, but it will work.
Series string & parallel battery banks are often setup with multiple chargers to charge each cell in the string individually to get them all plussed up equally, or near enough. It is important that each cell is roughly the same voltage, temperature, resistance, age, etc. This is incidentally why flashlight manufacturers always cautioned not to mix and match new and old cells, different brands, because the discharge rate is different, causing flickering and usually electrolyte leaks.
The device wouldn’t work because the batteries are in a series and the one dead battery would lower the voltage below the necessary threshold.