Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: scientizscht on 18/06/2020 11:03:18

Hello
If we have one Cu/Zn battery that has a voltage of 2V and current of 5A, it will produce power of 10W.
I assume for every Zn atom that leaves the Zn electrode, one Cu atom must join the Cue electrode in order for voltage and current to be produced.
How can I calculate the number of Zn atoms (and thus Cu atoms) that need to leave the Zn electrode per second so that the power of 10W is achieved?
If we join two of the above cells in series, then a 2+2=4V will be achieved and a current of 5A so that a 20W power is produced. In that case how many Zn atoms do they need to react (leave the Zn electrode)? Is it the same or double the number of Zn atoms of the single cell? It is not obvious because the current is the same 5A.
In other words, a 20W power can be achieved with 4V x 5A or 2V x 10A. Is the number of molecules that need to react the same in both of these cases or different?
Thanks!

If we have one Cu/Zn battery that has a voltage of 2V
It won't; however.
The number of zinc atoms that dissolve (in each cell) each second is equal to half the number of electrons transferred per second.
Zn > Zn^{++} + 2 e^{}

OK, that's fine but I am not asking about the ratio of Zn atoms to electrons.
I am asking about the total number of Zn atoms that react when two cells are in parallel or in series.
In series, you have 2+2V and 5A, while in parallel you have 2V and 5+5A. Is the number of atoms per second that react the same in both cases? Definitely the number of electrons flowing per second is different, but what about the number of Zn (or Cu) that react (or leave the electrode)?

I am not asking about the ratio of Zn atoms to electrons.
Yes you are, you just don't realise it..
One amp is about 6 *10^18 electrons per second.

I am not asking about the ratio of Zn atoms to electrons.
Yes you are, you just don't realise it..
One amp is about 6 *10^18 electrons per second.
OK, so you are saying that if we connect the two cells in series, 3*10^18 Zn atoms will react where when we connect the cells in parallel, 6*10^18 Zn atoms will react per second?

If we have one Cu/Zn battery that has a voltage of 2V
It won't; however.
The number of zinc atoms that dissolve (in each cell) each second is equal to half the number of electrons transferred per second.
Zn > Zn^{++} + 2 e^{}
Do you recognise that the conservation of energy means that producing twice as much power requires twice the rate of reaction of zinc?

If we have one Cu/Zn battery that has a voltage of 2V
It won't; however.
The number of zinc atoms that dissolve (in each cell) each second is equal to half the number of electrons transferred per second.
Zn > Zn^{++} + 2 e^{}
Do you recognise that the conservation of energy means that producing twice as much power requires twice the rate of reaction of zinc?
I appreciate that but I think you do not understand what I am asking.
I am not asking to compare one cell with two cells (in serial or parallel).
I am asking to compare two cells in serial with two cells in parallel.
They produce the same power, 20W. The current in the parallel cells is 10A and the current in the series cells is 5A. The rate of electron flow in the parallel cells is 10x 6x10^18 electrons per second.
The rate of electron flow in the series cells is 5x 6x10^18 electrons per second.
My question is how many atoms of Zn react in total per second in the parallel and in the series cells. Is it 5x 6x10^18 Zn atoms per second in the parallel cells and 2.5x 6x10^18 Zn atoms per second in the series cells?

It is important to recognise that the cells do not have eyes.

If we have one Cu/Zn battery that has a voltage of 2V
It won't; however.
The number of zinc atoms that dissolve (in each cell) each second is equal to half the number of electrons transferred per second.
Zn > Zn^{++} + 2 e^{}
Do you recognise that the conservation of energy means that producing twice as much power requires twice the rate of reaction of zinc?
I appreciate that but I think you do not understand what I am asking.
I am not asking to compare one cell with two cells (in serial or parallel).
I am asking to compare two cells in serial with two cells in parallel.
They produce the same power, 20W. The current in the parallel cells is 10A and the current in the series cells is 5A. The rate of electron flow in the parallel cells is 10x 6x10^18 electrons per second.
The rate of electron flow in the series cells is 5x 6x10^18 electrons per second.
My question is how many atoms of Zn react in total per second in the parallel and in the series cells. Is it 5x 6x10^18 Zn atoms per second in the parallel cells and 2.5x 6x10^18 Zn atoms per second in the series cells?
The current, and thus the power, is determined by the load. For two 2 volt batteries in series to supply 20 watts, they have to be connected to a 0.8 ohm load, and in parallel, they would need to be hooked up to a 0.2 ohm load.
Just putting the batteries in parallel doesn't by itself double the current. What it does is double the current limit for the supply. ( a load can draw more current before significantly drawing down the supply voltage.)
Thus if I attached a 0.2 ohm load to the series batteries, they would try to supply 20 amps and 80 watts of power, which could very well be beyond their limits.
And if I attached the 0.8 ohm load to the parallel batteries, they would only need to supply 2.5 amps and 5 watts.

I assumed he was using different loads and thus getting the currents and voltages he said he was.

I was not considering different loads. I am talking about the inherent voltage and current of cells.
If your cell has a current density of 2A/cm2 then with an electrode of 1cm2 you will be able to draw a maximum of 2A current.
This means that on your 1cm2 electrode, there are 2/2 x 6x10^18 atoms of Zn reacting each second.
If your cell works with Zn/Cu then this has a standard potential of 1.1V.
So the values 1.1V and 2A are inherent in the cell and they do not depend on the resistor. The maximum power you can draw from that cell is 2.2W.
I am not considering different resistors or anything like that.
The question is how many Zn atoms in total react per second when two of the above cells are in parallel and when in series.

If your cell current is limited at, for example, one amp then it will only deliver 1 amp into a short circuit and the voltage will be zero (as will the power delivered to the load).
So the values 1.1V and 2A are inherent in the cell and they do not depend on the resistor.
It seems that you don't understand how batteries work
https://en.wikipedia.org/wiki/Internal_resistance#Battery
However, it is important to consider the observation that batteries are unable to see.

Each atom transfers the same amount of charge across the cell.
Energy = charge x voltage. So if you double the voltage you only need to move half the charge around the circuit to do the same amount of work.
With the cells in series, the charge moved through cell 1 also moves through cell 2, so for each zinc atom that moves in cell 1, one zinc atom moves in cell 2. To do the same amount of work with the cells in parallel, for each atom that moves in cell 1, an atom must also move in cell 2. So the total number of atoms moved is the same in both configurations.
In practice, series connection is generally preferable: if a cell shortcircuits (which lithium cells in particular are prone to do) it will not damage other cells by drawing excess current. In critical applications you can make the system tolerant of opencircuit cells with bypass diodes. And of course the transmission losses are smaller with higher voltages.

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I am not considering different resistors or anything like that.
You've told us that you're drawing 10A from 2V, and 5A from 4V. You can't do that and keep the resistance the same as well. V=IR

If your cell current is limited at, for example, one amp then it will only deliver 1 amp into a short circuit and the voltage will be zero (as will the power delivered to the load).
So the values 1.1V and 2A are inherent in the cell and they do not depend on the resistor.
It seems that you don't understand how batteries work
https://en.wikipedia.org/wiki/Internal_resistance#Battery
However, it is important to consider the observation that batteries are unable to see.
One more bit of additional info: You will get the maximum power to an external load when that load's resistance is equal to the battery's. (impedance matching)

One more bit of additional info: You will get the maximum power to an external load when that load's resistance is equal to the battery's. (impedance matching)
You would never normally do that with a battery though, the efficiency's only 50%.

I am not considering different resistors or anything like that.
You've told us that you're drawing 10A from 2V, and 5A from 4V. You can't do that and keep the resistance the same as well. V=IR
Actually, I think you (sort of) can; if the load resistor is zero.
If we have one Cu/Zn battery that has a voltage of 2V and current of 5A, it will produce power of 10W.
A battery with an open circuit voltage of 2 V and short circuit current of 5A is plausible it would have an internal resistance of 2/5 =0.4 ohms
And, if you short circuited it, it would dissipate 10W
All as heat in the battery no electrical power would be delivered.
Now, if you put 2 in series you get an O/C voltage of 4V and an internal resistance of 0.8 ohms.
The current through the short circuit is still 5A
And the power "wasted" is 20 W of heat (and no electrical energy in the external circuit)
OK, Now connect them in parallel and, again, short circuit them
The total current flowing in the wire is 10A i.e. 5A through each cell
And the total power  still all as heat is 20 W.
Again, it's clear that batteries are unable to see.

You would never normally do that with a battery though, the efficiency's only 50%.
Which is the best you can get...
The explanation for a battery is given here
https://www.electronicstutorials.ws/dccircuits/dcp_9.html
But they then talk tosh about impedance matching of audio amplifiers.

I am not considering different resistors or anything like that.
You've told us that you're drawing 10A from 2V, and 5A from 4V. You can't do that and keep the resistance the same as well. V=IR
Actually, I think you (sort of) can; if the load resistor is zero.
If we have one Cu/Zn battery that has a voltage of 2V and current of 5A, it will produce power of 10W.
A battery with an open circuit voltage of 2 V and short circuit current of 5A is plausible it would have an internal resistance of 2/5 =0.4 ohms
And, if you short circuited it, it would dissipate 10W
All as heat in the battery no electrical power would be delivered.
Now, if you put 2 in series you get an O/C voltage of 4V and an internal resistance of 0.8 ohms.
The current through the short circuit is still 5A
And the power "wasted" is 20 W of heat (and no electrical energy in the external circuit)
OK, Now connect them in parallel and, again, short circuit them
The total current flowing in the wire is 10A i.e. 5A through each cell
And the total power  still all as heat is 20 W.
Again, it's clear that batteries are unable to see.
This is self contradictory. You start by arguing that there isn't any resistance, and then include the internal resistance in your calculation, at which point there is resistance.

You would never normally do that with a battery though, the efficiency's only 50%.
Which is the best you can get...
No it isn't. With a series loss only the efficiency is Rl/(Rl+Rs), which is an asymptote to 100% as Rl tends to infinity. With a source that has series and parallel losses (eg a transformer), the load for optimum efficiency is the geometric mean of Rs and Rp (provided that Rs<<Rp).

You will get the maximum power to an external load when that load's resistance is equal to the battery's.
This principle is very relevant in transmission lines (whether 50Hz or 5GHz), where the impedance is mainly capacitance and inductance (which don't absorb real power, like a resistor does).
However, when you have significant resistance, power dissipation in the source becomes a significant problem.
 If you dissipate 10W in your battery for very long, your battery might explode as the electrolyte vaporizes.
 In a nuclear power plant, if you dissipate 1 GW inside your alternator, it will quickly melt.
 In both cases, the design goal is to make the source resistance much less than the intended load resistance.
 That way, you transfer a greater % of the power to the load, and you avoid damaging the power source.

So the total number of Zn atoms that react per second are the same when the cells are in series (where the current is 5A) and when in parallel (where the current is 10A)?
Yes or no and what is that/these numbers?

You start by arguing that there isn't any resistance,
Did you miss this bit?
if the load resistor is zero.

So the total number of Zn atoms that react per second are the same when the cells are in series (where the current is 5A) and when in parallel (where the current is 10A)?
I have already pointed out a number of times that batteries can't see the outside world.
A consequence of that is that they can not, even in principle, act differently depending whether they are in series or parallel because they don't "know" it..
So why are you asking if they do?

So the total number of Zn atoms that react per second are the same when the cells are in series (where the current is 5A) and when in parallel (where the current is 10A)?
Yes or no and what is that/these numbers?
Yes. Given that each Zn atom carries 2 electrons, I'll leave it to you to work out how many.

I have already pointed out a number of times that batteries can't see the outside world.
I think you have confused the poor lad. The battery does quite different things if the outside world consists of a short circuit (lots of ionic activity) or an open circuit (no activity).

I have already pointed out a number of times that batteries can't see the outside world.
I think you have confused the poor lad. The battery does quite different things if the outside world consists of a short circuit (lots of ionic activity) or an open circuit (no activity).
True, but if it's delivering 10W then it's delivering 10W
The heat or reaction of zinc (and whatever the oxidant is) is a matter of thermodynamics, not electronics.
Also if it's delivering 5A then it's delivering 5A.
It doesn't know any better, so it eats up zinc at the same rate.

You start by arguing that there isn't any resistance,
Did you miss this bit?
if the load resistor is zero.
No I didn't miss it.
You can't claim the resistance isn't changing when you switch from series to parallel just because it's hidden inside the cells.

You start by arguing that there isn't any resistance,
Did you miss this bit?
if the load resistor is zero.
No I didn't miss it.
You can't claim the resistance isn't changing when you switch from series to parallel just because it's hidden inside the cells.
Yes I can.
It is 0.4 Ohms in each cell whether that cell is in series or parallel.
.
My point I thought I had laboured it hard enough but, here we go again is that the cell doesn't know what's happening outside. So each cell has a constant internal resistance. (OK in practice it's temperature dependent and slightly current + time dependent).
Cut to the chase.
Either tell us where I contradicted myself, or point out where I got the wrong numbers.

Yes I can.
It is 0.4 Ohms in each cell whether that cell is in series or parallel.
In which case you'd have to apply the same logic to the voltage and current and say they're not changing either. My comment was in response to the OP who was talking about double voltage with half current, but no change in resistance.
My point I thought I had laboured it hard enough but, here we go again is that the cell doesn't know what's happening outside. So each cell has a constant internal resistance.
Yes, I know, but that’s just repeating what I’d already said here:
Snip.JPG (181.46 kB . 1528x617  viewed 1516 times)
My original point to the OP was that you can't double the voltage in series without doubling the resistance, and you can't double the current in parallel without halving the resistance as he seemed to be arguing.

In which case you'd have to apply the same logic to the voltage and current and say they're not changing either.
You start by arguing that there isn't any resistance,
Did you miss this bit?
if the load resistor is zero.
No I didn't miss it.
It seems you missed it again.
There really is a difference between the current in a cell of a parallel array and the current in the load connected to them.
And there really is a difference between the open circuit voltages of a single cell and a pair of cells in series.
But 0.4 ohms is still 0.4 ohms.
Yes, I know, but that’s just repeating what I’d already said here:
...
Which was just you repeating what I had said here
It is important to recognise that the cells do not have eyes.
What you and the OP seem to have in common is a failure to distinguish clearly between the fixed internal resistance (which, he misunderstands that doesn't help) and the load resistance which one can vary at whim.

Vhf  I think your diagrams are adding to other folks' confusion! The cell symbol looks like a voltage source in parallel with a resistance, whereas a cell actually behaves as an ideal (zeroimpedance) voltage source in series with a resistance. It's OK for your simulator which knows what it is talking about, but not for those whose physics comes from textbooks!