Multi-Cell Theory Update
By Mike Johnston

Several months ago I wrote a little paper on the advantages of using multiple electrolysis cells, connected to form a Series Circuit, over a single cell or even cells connected in parallel. In looking back on this paper I still feel that I was correct in the general idea that I tried to present even though I now realize that I was wrong in some respects and, in addition, did not adequately explain the concept that I was trying to relay. This paper is intended to both clarify and improve upon
those original ideas in several ways and to introduce a new design for series cells. Speaking of series cells I would like to briefly review some facts about series circuits. In a series circuit the same current flows through every part. It makes no difference how many parts or devices you  have. As long as the identical current passes through each, it is a series circuit. The current that flows through a series circuit depends upon the source voltage and the total circuit resistance. When there is only one load (cell) in a circuit IT provides the total resistance (other than the resistance of the wires and the internal resistance of the power source). However, when there are a series of loads the total circuit resistance is the sum of the resistances of the individual loads (cells). So if a circuit has five loads of one ohm resistance each it has a total resistance of 5 x 1 = 5 ohms.

To find the current that a series circuit will draw you determine the total circuit resistance by adding the resistances of each load (cell). Then you use Ohm's law (I=E/R) to find the current. In the above example if we were using a 12 volt power source then our current through the circuit would be I= 12volts/5ohms and so I = 2.4 amps.

Once the current is known it is easy to figure the total power used. P = I2R = (2.4 x 2.4) x 5 = 28.8 watts of power used in our example. An easier way is to multiply the source voltage by the current; 12 x 2.4 = 28.8.

Next the voltage drop. I think that this is a much misunderstood concept as I see it applied to electrolysis cells. For example, in my earlier paper I mentioned a 2 volt drop per cell. I got that number quoted to me by someone and as my experimental data was close to that number I accepted it as such. Other people believe that a certain minimum voltage is required for electrolysis to take place and confuse that number with the voltage drop. In actuality Faraday's law says that
the amount of H2 released in an electrolysis cell depends entirely upon the amount of current which passes. If you can maintain a high enough current at 1 volt it should still have the same effect as the same current at 300 volts. The only difference being that at 300 volts you could get that current to be maintained over substantially more cells. Ideally though the minimum voltage is 1.4 volts.
Back to the voltage drop. In a series circuit the total voltage dropped across all of the loads is equal to the source voltage. Simple isn't it? If you run a current from a 12 volt power source through one cell your voltage drop will be 12 volts. If you run current from a 12 volt power source through 300 cells your voltage drop will be 12 volts. In the single cell all of the voltage will be dropped across that one cell. In 300 cells (of equal resistance) the voltage drop per cell is determined by multiplying current by resistance (E = IR). In this case our voltage drop per cell would depend upon finding our current and resistance. Say we have a total resistance of 2 ohms across all of our 300 cells. We can then find the current that this setup would draw by using the formula I = E/R = 12volts/2ohms = 6amps. So, knowing this we can say that E = IR = 6amps x 2ohms = 12 volts dropped across the 300 cells. So then, if we want to know the voltage drop in each cell we would have to find the resistance of each cell by dividing our total resistance by the total number of cells (since all have equal resistances) R individual = 2ohms/300 = .0066666 ohms per cell. Then we can find our voltage drop for each cell by using the formula E = IR = 6amps x .0066666 ohms = .0399996 volts per cell. If we then multiply our voltage drop per cell by the total number of cells (300) we find that our total voltage dropped across all of the cells is 11.99988 volts.

This is just an example. It is not intended to be necessarily workable because there indeed might be a minimum voltage per cell in order for electrolysis to take place. I have not disproven that to date in an experimental manner and so for all examples that follow I will allow for more voltage being "left" to drop over the last cell.  The first factor that I believe needs to be specifically addressed in relation to series cells is resistance. I have seen resistance losses in electrolysis cells commonly quoted as being from 40 - 60 % as though this were some kind of "given", unchangeable. From the most basic understanding of Ohm's law it soon beams apparent that this is not the case. No circuit has a certain "built in" resistance unless someone builds it in. To assume a "standard" resistance for all possible designs of electrolysis cells is like saying that no matter how you design an         electrical circuit, or which specific components you include, or exclude, it will always have exactly the same resistance in it!

Obviously this is not true or we wouldn't have computers would we? Or for a simpler analogy, what if you put a 1200cc motorcycle motor into a truck tractor? Would you complain then because that setup wouldn't work properly? From that evidence you could come to believe that an internal combustion motor could never be able to power a big truck. Someday
though someone would figure out that perhaps a larger or different kind of motor would work better.

In electrolysis cells resistance plays just as big a role as in any other electrical circuit. You CAN adjust this factor within a very wide set of parameters, I hope to illustrate this in this paper and to show how this ability to design a specific resistance into a series of cells can and will, I believe, allow the construction of a series cell which is efficient enough to separate a sufficient quantity of water to provide H2(g) to run a car for example. I have seen many people (myself included) discussing many and varied designs for electrolysis cells but I have yet to see anyone show WHY they believe that their design would work. People just seem to accept certain things like "larger electrodes liberate more H2" without really going into the "why" of it. That's too
bad because that is where the answer lies.

First, a look at Ohm's law. The following simple formulas establish the basis for most of the examples I will use in this paper. I am sure that they are already familiar to most readers but I include them anyway in an attempt to be thorough. I = E/R, R = E/I, E = IR (I =Current, E = Volts and R = Resistance) and finally P = I2R for power losses. Also important is the formula for finding how much H2 will be released from water in a given amount of time by the application of a specific electrical current. This formula is; m = .0000105 x I x t  where I = Current, M = mass of H2 produced (in grams) and t = time. The number (.0000105) is obtained by dividing the atomic mass of the substance to be separated (H2) by it's valence (Faraday's Second Law).

First I want to show how much current must pass through an electrolysis cell in order to liberate the H2 contained in one liter of water in one hour. That should be a sufficient amount to run many types of small internal combustion motors and for this example I would like to concentrate primarily on such motors. So if we take the mass of H2 to be produced; 2.016 gm/mole x 55 mole/liter = 110.88 gmH2/liter. Then taking that number and dividing it by .0000105 gives us 10,560,000 amps which
must pass through our cell in one hour or, dividing that number by 60 gives us 176,000 amps per minute or, dividing that by 60 gives us 2933.33 amps per second. Pretty big current isn't it? Not too many small generators can deliver that can they? Remember though, that is the amount that would have to pass through ONE cell (ignoring any catalysts for now). What if we were to divide that current up among a group of cells? A much smaller amount of current would have to pass through them
all as each one would produce an amount of H2 equal to the quantity of current passing through it. As long as the SAME current flowed through all the cells their cumulative output of H2 would be all of their individual outputs added together and would equal the amount produced by a much larger current passing through ONE cell.

If we divide the above total current (2933.33amps per cell) by 50 cells we find that we must maintain a current of 58.66 amps in each of those cells to separate the desired quantity of H2 from water in the desired amount of time. That seems to be a much easier amount of current to produce and work with doesn't it? So there we have the first of the necessary factors for our Ohm's law equations. We have established the amount of H2 gas that we need to produce to run a specific device and have then gone on to find the current that we need to maintain in a certain number of electrolysis cells to give us the desired quantity of H2 gas in the necessary amount of time.

Next we must determine either our desired voltage or resistance. I will use voltage as I want to get more deeply into the resistance factor later. Let's pretend we want to use an automotive alternator to produce our H2 gas, We can assume then that we have a 12 volt power source. Now that we have this information we can find out what our maximum total resistance must be across all the cells and in each cell for this device to work as intended. To do this we use Ohm's formula of R = E/I which is
expressed here as R= 12volts/58.66amps, and so R = .205 ohms total resistance across all of our cells or .205amps/50(cells) = .00409 ohms resistance per cell. It seems simple and straightforward enough doesn't it?

What about resistance though? How is it determined and how can it be changed? Many books have tables of the resistances of various metals. I won't bother to list them here. I would like to focus on copper. It is a good conductor with low resistance. Silver is a better conductor with lower resistance but it is significantly more expensive and so less easily available for most researchers. Copper is also actually fairly inert and the right combination of electrode metal and type or concentration of electrolyte needs to be found so that the electrodes do not dissolve but that is for later.

The resistance of any material depends upon the number of free electrons that the material has. One ampere is 6,280,000,000,000,000,000 free electrons passing any given point in a conductor (such as a wire) in one second. So a good conductor must have many free electrons available to allow several amperes of current to flow. Since current is a measurement of electrons passing a point in a conductor more free electrons can be made available by using a thicker piece of metal so
that more current can flow. In the case of a battery or an electrolysis cell this is already known. The larger the surface area of your electrodes the more free electrons are available to cause the separation of water. Also, don't forget that water (containing an electrolyte) is our other conductor and by increasing the surface area of our solid electrodes we are increasing the surface area of our water conductor (and thereby lower it's resistance)as well!

As an example; a piece of copper 2cm high and 1cm wide will have twice as many free electrons available at the point at which current is being measured as a piece only 1cm wide by 1cm high. The piece that is twice as high will conduct twice as much current. Since the amount of H2 that is released by the electrolytic separation of water depends totally on the amount of current that passes through the water then it seems obviously desirable to have control of the resistance of the cell as this is the factor which determines how much current flows and thereby how much H2 is released. So when you increase the width or height of your electrodes you are increasing their cross-sectional area. The greater the cross-sectional area of your electrodes, the lower their resistance. The lower their resistance, the more current flows. The more current that flows, the more H2 that is released.

On the other side of the coin, if you increase the length of your electrodes (in the direction of current flow) you increase your
resistance and lower the current flow. By doubling the length of a conductor you double it's resistance (or conversely by cutting said conductors length in half you also decrease it's resistance by one half). This causes more of your available current to be converted into heat energy and increases your power losses. Therefore, by choosing the proper metal for a conductor, and making it with a certain cross-section and length, you can produce any kind of resistance effect you want.

I thought that, at this point it would be good to figure out the resistance of a piece of copper of a certain size so that researchers might have a basis on which to figure the resistance of their cells. As I said I feel that copper is the best metal to use for experimental purposes. I decided to find the resistance of a piece of copper of 1 sq/in by 1/4 inch thickness. This way you can much more easily figure the resistance of any size electrode you might use.

If you look at a table of American Standard Wire Gauges you will see that it gives you the dimensions and typical resistances of commercial copper wire. From this table I selected B&S Gauge #2 wire. It has a cross-sectional area of .258 inch (a little over a quarter inch) and it is round rather than square so I am fudging just a bit. That aside I believe that despite this my results will be close to being accurate so here goes; the resistance of one thousand feet of this wire at 70 degrees Fahrenheit is .159 ohms. So then the resistance of 500 feet would be half of that or .159/2 = .0795 ohms. That is a piece of copper 1/4 inch square in cross-sectional area by 500 feet long. Remember we
only want it to be 1/4 inch long so we have to keep going. I will keep
dividing it in half until I reach my desired length;
250 feet of wire = .0795ohms/2 = .03975ohms per 250 feet
125 feet of wire = .03975ohms/2 = .019875ohms per 125 feet
62.5 feet of wire = .019875ohms/2 = .0099375ohms per 62.5 feet
31.25 feet of wire = .0099375ohms/2 = .0049687ohms per 31.25 feet
15.625 feet of wire = .0049687ohms/2 = .0024843ohms per 15.625 feet
7.8125 feet of wire = .0024843ohms/2 = .0012421ohms per 7.8125 feet
3.59125 feet of wire = .0012421ohms/2 = .000621ohms per 3.59125 feet
1.795625 feet of wire = .000621ohms/2 = .0003105ohms per 1.795625 feet
.8978125 feet of wire = .0003105ohms/2 = .0001552ohms per .8978125 feet
.4489062 feet of wire = .0001552ohms/2 = .0000776ohms per .4489062 feet
.2244531 feet of wire = .0000776ohms/2 = .0000388ohms per .2244531 feet
.1122265 feet of wire = .0000388ohms/2 = .0000194ohms per .1122265 feet
.0561132 feet of wire = .0000194ohms/2 = .0000097ohms per .0561132 feet
.0280566 feet of wire = .0000097ohms/2 = .0000048ohms per .0280566 feet

And there is our quarter inch in length (roughly). Now we have the resistance for a quarter inch cube of copper. Remember that if we increase the surface area of our conductor we lower the resistance again. By doubling the surface area we cut the resistance in half. We wanted the number for a square inch of copper so we have to adjust the numbers accordingly.

1/4" = .0000048 ohms so 1/2" = .0000048/2 = .0000024ohms
1/2" = .0000024ohms so 1" = .0000024/2 = .0000012ohms

There is the multiplier for figuring the resistance of your copper electrodes. Find the number of square inches and multiply by the number of ohms per square inch. Remember that they have to be 1/4 inch thick for this to be accurate.

Just for fun let's imagine an electrolysis cell in which we have two electrodes, each measuring 3" x 3". That would be 9 square inches each or 18 square inches total. So we would take the one square inch that we have the number for and double it to 2 square inches. That cuts the resistance in half, so;
2sq/in = .0000012ohms/2 = .0000006 ohms
4sq/in = .0000006ohms/2 = .0000003ohms
8sq/in = .0000003ohms/2 = .0000001ohms
16sq/in = .0000001 ohms/2 = .00000005ohms

At 16 square inches of copper 1/4 inch thick we have reached nearly zero resistance. The problem is that the conductor can only carry so much current and the power source can only supply so much current. For these reasons, even though near zero resistance would theoretically draw an huge amount of current and consequently supply a proportionally huge amount of H2, it isn't technically possible at this time. Therefore if we use a series of cells which maintain a current of realistic proportions across all of them we can engineer the setup to produce almost any amount of H2 from water while staying within the paramaters
of real world possibility. Just as an added note, the #2 gauge wire I started with has a rated current carrying capacity of 91-136 amps with rubber insulation or between 96-241amps with other insulation. These amounts of current are fairly realistic in terms of what can be produced by commonly available power sources.

I haven't factored in the resistance of the water and I don't have any actual experimental data on the resistivity of water specifically but for this paper I will assume that, if you place your electrodes as close together as the thickness of one of them the resulting small thickness of your water conductor will no more than double the total cell resistance. In batteries a sheet of porous wood or rubber is inserted between the electrode plates to prevent arcing between the electrodes.

Automotive batteries make use of these principles to achieve a total internal resistance of around 1.8ohms for the whole battery or 1.8/6 = .03ohms per cell. Remember that lead is 12.76 times as resistive as copper so if a copper plate were being used the resistance per cell would be around .03/12.76 = .0235109ohms or 1.8/12.76 = .1410658ohms for  the whole battery.

How much current would a cell like the one above (with copper electrodes) be able to draw from a 12volt power source?
I = E/R = 12/.1410658 = 85.066685 amps. Could it be made to draw more? Yes.  Increase the cross-sectional area of the electrodes and thereby reduce it's resistance. How much H2 could be produced in a cell such as the above ?  Since m = .0000105 x a x t then m= .0000105 x 85.066685 = .0008932001925 gm/sec x 6 cells = .005359201155 gm/sec for
the whole device or .3215520693 gm/min or 19.293124158 gm/hr.

Part 2

In part one I looked at resistance and how to modify it to suit the requirements of a specific setup. In this part I plan to show that there are two different ways to approach this with two very different results. I will also then post a design for what I consider to be the perfect experimental multi-cell. (above)

The two ways to do this are 1) find the necessary total resistance to draw a specific current at a specific voltage and then add identical cells on. In doing it this way you will always produce the same amount of H2 in your multiple cells as you would with one cell because you are multiplying/increasing the total circuit resistance with each cell that you  add on and thereby lessening the amperage across the circuit. The benefit from doing it this way is that you can produce the desired amount of H2 with less total amperage input. The total amount of H2 produced remains the same though no matter how many identical cells of
this type that you use.

As an example let's say that we have a power source which supplies 120 volts and 130 amps. To find the amount of H2 that such a current would yield in one cell you use the standard formula m=.0000105 x I x t = .0000105 x 130 x 1(sec)= .001365 gm/sec H2. This would work out to .001365gm/sec x 60=.0189gm/min x 60= 4.914 gm/hr.

The resistance of this cell (in order to let 130 amps pass) would have to be R=E/I=120/130=.9230769ohms.

The voltage drop across this cell would be E=IR=130 x .9230769ohms=119.99999volts

If you added on another cell, identical to the first, then your resistance would be .9230769ohms x 2= 1.8461538 ohms.

So then your amperage across the multi-cell circuit would be 1/2 of your amperage in one cell because your total resistance has doubled. I=E/R=120/1.8461538=65.000001amps across the system. Each cell would then produce m=.0000105 x 65.000001 amps x 1(sec)=.0006825gm/sec of H2 per cell or .001365gm/sec total, which is identical to the amount produced by running ALL of the available 130 amps through one cell. As I said the advantage is that you are using less of the available power to produce this same amount of H2 from water. The total amount of H2 produced would be .001365gm/sec x 60=.0819gm/min or .0819gm/min x 60=  4.914 gm/hr. Probably not enough to run an internal combustion motor of a size sufficient to turn your generator.

If it is done in the other way the results are different. I will walk through the steps required to do it this way and then give an example of the results over 1, 50 and 65 cells. The power source for this example will be the same as the first; 120volts 130amps.

STEP 1:  Find the maximum resistance that can be allowed for one cell so that the desired amperage can flow
through your cell at the voltage  supplied. To do this use R=E/I which in this case looks like this
R=120v/130a=.9230769ohms.
NOTE: Remember that resistance is the one variable that you have to work  with which will control the amount of H2 that is ultimately produced AND  that by making simple adjustments in such factors as the cross sectional area and thickness of your conductor(s), including both the electrodes and the water between them, You CAN achieve ANY RESISTANCE EFFECT THAT YOU WANT!

STEP 2:  Figure out the amount of H2 that your one cell (in the above example) would produce by using the formula
m=.0000105 x I x t, which in this case would be m=.0000105 x 130 x 1(sec) so m=.00365gm/sec,
.0819gm/min, 4.914gm/hr.
NOTE: As far as that formula goes, there is all of the H2 that you can produce at that amount of current. However the law says that you can produce that amount of H2 in EVERY cell through which it passes. This is great because in a series cell the SAME current passes through EVERY CELL IN THE SERIES. The problem then is simply to engineer a system in which the available current passes through all of your cells.

STEP 3: Figure out your voltage drop per cell by using the formula E=IR. In this case E=130 x .9230769=119.99999 volts
dropped across our one cell. Good. The law says that the entire voltage should be dropped in a series circuit whether
you have one cell or 50. In the case of multiple cells the voltage drops by an equal amount in each cell. In the case of
electrolysis we have decided that we need to maintain at least 1.8 volts in each cell so we cannot drop more than
that per cell.

STEP 4: Figure the watts of power consumed in your circuit by using the formula for "heating losses" W=I2R. Which in this
case is I=(130 x 130) x .9230769=15,599.999watts. Our available power is found by the formula w=E X I which is
w= 120 x 130= 15,600 watts or 15kw. We use up our entire available power here, everything balances.

This is the basic process to use in figuring multiple cells out as well. Remember that to achieve the results that I am
going to show in the following example I took the maximum resistance for ONE cell as found above and divided it by the total number of cells I wanted to create. In this way you can find the necessary resistance per individual cell.

One Cell
Power supply = 120 volts 130 amps
Current passing through system 130 amps
119.99999 volts dropped
15,600 total watts available
15,599.999 total watts used
.9230769 ohms maximum resistance
.001365 gm/sec H2 total produced

Fifty Cells
Power supply = 120 volts 130 amps
Current passing through system 130 amps
2.399995 volts dropped per cell
15,600 total watts available
15,599.999 total watts used
311.99935 watts used per cell
.9230769 ohms total resistance
.0184615 ohms maximum resistance per cell
.001365gm/sec H2 produced per cell
.06825gm/sec H2 produced total

Sixty Five Cells
Power supply = 120 volts 130 amps
Current passing through system 130 amps
1.846143 volts dropped per cell
15,600 total watts available
15,599.999 total watts used
233.99859 watts used per cell
.9230769 ohms total resistance
.0142011 ohms maximum resistance per cell
.001365gm/sec H2 produced per cell
.088725gm/sec H2 produced total

Beyond this number of cells the voltage drop is too great to maintain the minimum required voltage in each cell. As you can see the output of sixty five cells, designed in this way is 65 times as much as what is produced by one cell! Remember too that if you wanted to further increase the amount of H2 produced at this current all you would have to do is increase the potential difference of the power source (voltage). If you did that you could add on as many more cells as the increase
would allow and produce substantially MORE H2 at this SAME CURRENT.
END
MJ     