Connect parallel batteries with wrong poles

[The Tortoise-Man]

I would like to do now (theoretically) a couple of rather naive things with two batteries with voltages $V_1$ and $V_2$ in parallel charged with an abstract load with resistance $R$.

Usually, one learns in elementary electronics that the only "right" configuration with batteries in parallel is given by (M1), at least in case $V_1=V_2$:

Now the story becomes weird. What is the precise physical explanation why following configuration causes damage where we swap the poles, or asking differently how to model the system's behavior in following case - call it (M2):

The "layman argument" which I often read but which not persuades me is that it causes short circuit in this area:

Well, but applying the same "logic" one could also say that there is also a short circuit
in (M1), or not?

Why this "short circuit argument" is valid for (M2), but not for (M1).

Therefore up to now I'm not happy with this way of reasoning, but maybe I'm missing the whole point. Can this argumentation refute my objection that seemingly the same reasoning may be naively thinking applied to (M1), too? What is the precise difference in the two cases? In other words, why connecting poles in "right" way - as in (M1) - not causes a short circuit - but (M2) does?

If this "Dr Google argumentation" is indeed too sloppy, is there a reasonable advanced argumentation explaining why configuration in (M2) damages the system? Is there a reasonable mathematical model predicting such behavior? Possibly, is it sufficient to reasoning with Kirchhoff's junction and voltage rules in naive way or does one need more advanced techniques from electrochemistry to answer the problem qualitatively satisfactory?

What about the story with $V_1 \neq V_2$? Which behaviour of the system is going to be predicted for configs (M1) and (M2)?

 

[berkeman]

I would like to do now (theoretically) a couple of rather naive things with two batteries with voltages $V_1$ and $V_2$ in parallel charged with an abstract load with resistance $R$.

Usually, one learns in elementary electronics that the only "right" configuration in parallel is given by (M1), at least in case $V_1=V_2$:

View attachment 334881

Now the story becomes weird. What is the precise physical explanation why following configuration causes damage where we swap the poles (M2):

First, please don't ever do this in real life. Depending on the size and capacity of the batteries, you can start a fire or have the batteries explode, spewing corrosive stuff everywhere (including all over you and your face). Please don't ask me how I know this...

You need to think about the accurate models of the batteries. Each is modeled by the zero-source-resistance voltage source in series with some $R_s$ source resistance. To the extent that the source voltages are different (worst case reversed like you mention, less worst case but still problematic if they are paralleled with slightly different battery voltages), you will get fault currents flowing that depend on the voltage difference and the respective $R_s$ values.

Those fault currents generate heating in those batteries, which can cause fires and the exploding batteries that I mentioned.

Have a nice day.

 

[The Tortoise-Man]

You need to think about the accurate models of the batteries.
Each is modeled by the zero-source-resistance voltage source in series with some $R_s$ source resistance. To the extent that the source voltages are different (worst case reversed like you mention, less worst case but still problematic if they are paralleled with slightly different battery voltages), you will get fault currents flowing that depend on the voltage difference and the respective $R_s$ values.

Yes, that's the issue. I'm wondering what kind of model one should looking for "profound" enough to predict such behavior, not "too complicated" in order to be computable at all.

Okay, say we add too each battery a small resistence $R_s$, so replace (M1) by

and apply Kirchhoff rules to voltages and currents. Respectively the same game for (M2) except that we change $V_1$ by $-V_1$, ie with opposite sign.

Does this model already suffice to model the batteries to predict the behavior, or do we need something more advanced?

 

[berkeman]

Okay, let's dismiss reversed batteries out of the gate please. That's a really bad case and where I almost got kilt (PTSD).

So think about 12V batteries of moderate size that you try to parallel when they do not have the same charging history. They may have $V_s$ different by 0.5V, and $R_s$ of about 0.1 Ohm. How much power does that imply being dissipated when they are connected in parallel, and what kind of heating does that imply?

 

[The Tortoise-Man]

So think about 12V batteries of moderate size that you try to parallel when they do not have the same charging history. They may have $V_s$ different by 0.5V, and $R_s$ of about 0.1 Ohm. How much power does that imply being dissipated when they are connected in parallel, and what kind of heating does that imply?

Wait, so you suggest to apply the model from #3 for case (M1) , right? But then I not understand the choice of parameters you are suggesting to analyse the circuit with. Firsly, you assume $V_1=V_2=12V$ and both $R_s=0,1$ Ohm. But then performing Kirchhoff's voltage rule $V_i+ V_s+V_R=0$ for $i=1,2$ going through two circles trough first and second battery leads by symmetry that that both source resistances have same voltage $V_{s,1}=V_{s,1}$. (see the updated picture in #3)

Therefore I not understand what you mean by that they may differ by 0.5V. (So far I understand you correctly, by $V_s$ you mean the voltage at a source resistance right? Or do I misunderstand the configuration you are suggesting?

 

[Vanadium 50]

. What is the precise physical explanation why following configuration causes damage where we swap the poles, or asking differently how to model the system's behavior in following case - call it (M2):

You don't.

The model is for idealized components, not components bursting into flames and spewing acid or other chemical nastiness everywhere.

 

[renormalize]

Firsly, you assume $V_1=V_2=12V$ and both $R_s=0,1$ Ohm.

Therefore I not understand what you mean by that they may differ by 0.5V.

No, @berkeman does not mean $V_1=V_2=12V$; he says they differ by $0.5V$. So for example, set $V_1=12V$ and $V_2=11.5V$, apply Kirchhoff's laws and see what you get.

 

[Averagesupernova]

@The Tortoise-Man
What makes you think one battery can have current flowing out the positive terminal and the other one out the negative?

 

[Borek]

Let's try in some more layman terms.

Ignore the R (load) for now, you connect just two batteries.

Current flows form + to - (let's not get nitpicky about the details).

If you connect the batteries in parallel (+ to +, - to -) each of them tries to put some current into the circuit, but as they are opposed, nothing happens.

If you connect batteries + to -, they each put highest current they can into the circuit, heating everything till the chaos starts.

 

[The Tortoise-Man]

You don't.

The model is for idealized components, not components bursting into flames and spewing acid or other chemical nastiness everywhere.

Of course not, but let's keep things a bit more solution oriented. A well choosen model (...even thought every model is an idealization, the question is how "much" real information this idealization captures) could for sure give at least hints - when performing calculations "inside the framework" of the choosen occure some "strange" results (eg resulting values "beyond" the scale for which the model is conceived) - that there may some anomalous behavior take place.

Of course that depends highly on the model one has choosen. In the concrete situituation above, for example, if the current strength throught one path is many scales bigger then through the others, then that may give a hint that the components at this path may be going to be destroyed.So the question becomes which model is appropriate to detect (...or at least give hints) that taking wrong poles (ie as in pic M2) something anomalous happens? What about that one I suggested in #3? Or is this one "not sensitive" for the anomalous behavior appearing in that case? Would you suggest another one?

No, @berkeman does not mean $V_1=V_2=12V$; he says they differ by $0.5V$. So for example, set $V_1=12V$ and $V_2=11.5V$, apply Kirchhoff's laws and see what you get.

Thanks for the correction. Ok, before I perform the standard calculations, is it correct that - except from my misinterpretation with $V_1# and$V_2## (...so let assume# #V_1=12V$and$V_2=11.5V##) - that else @berkman's suggested configurations are adressed to the model in the picture in #3, ie each of the two wires with a battery $V_i$ carries also a resistance $R_s=0.1$ Ohm as in the linked picture depicted? Or is there only one resistance $R_s=0.1$ "shared" by both batteries? (...sorry for penibility but before I'm starting to calculate I would like to assure with which model I should work)

 

[berkeman]

Model each battery as a voltage source $V_s$ in series with its internal resistance $R_s$.

 

[renormalize]

Thanks for the correction. Ok, before I perform the standard calculations, is it correct that - except from my misinterpretation with $V_1$ and $V_2$ (...so let assume$V_1=12V$ and $V_2=11.5V$) - that else @berkman's suggested configurations are adressed to the model in the picture in #3, ie each of the two wires with a battery $V_i$ carries also a resistance $R_s=0.1$ Ohm as in the linked picture depicted? Or is there only one resistance $R_s=0.1$ "shared" by both batteries? (...sorry for penibility but before I'm starting to calculate I would like to assure with which model I should work)

Yes, just as you show in your diagram in post #3, each battery has its own internal resistance of $0.1\Omega$. So: set $V_1=12V$, $V_2=11.5V$, $R_s=0.1\Omega$, choose some value for your "load" resistor $R$ like $1\Omega$, and use Kirchhoff's laws to find the currents in the two branches.

 

[berkeman]

Yes, just as you show in your diagram in post #3, each battery has its own internal resistance of $0.1\Omega$. So: set $V_1=12V$, $V_2=11.5V$, $R_s=0.1\Omega$, choose some value for your "load" resistor $R$ like $1\Omega$, and use Kirchhoff's laws to find the currents in the two branches.

Actually, I'd suggest that they leave off the load resistor for now, since the worst case will be when only the two batteries are connected in parallel...

 

[Baluncore]

There are two forms of Marx impulse generator.

The more common Marx "voltage pulse" generator involves charging capacitors slowly in parallel, then switching them into series, to produce a sudden high-voltage discharge.
https://en.wikipedia.org/wiki/Marx_generator

The less common Marx "current pulse" generator involves charging capacitors in parallel, then switching them into a series loop, to produce a high-current, causing a magnetic pulse.

When you physically connect two batteries with reverse polarity, you are discharging them in the same way as a Marx "current pulse" generator. You should expect a momentary broad-band magnetic pulse, across the radio spectrum.

 

[sophiecentaur]

Let's try in some more layman terms.

Agreed (for the benefit of the OP). Basically, if you connect a well charged battery in parallel with a partially discharged battery, their emfs will be slightly different and the well charged battery will charge the other one, losing a lot of energy.

In some rare situations, battery equipment can use batteries in parallel. Suppliers always say that the batteries should all be of the same type, make and start off freshly charged. That's the only way to avoid such a parallel system dying very quickly, rather than having a double battery life.

 

[Borek]

Basically, if you connect a well charged battery in parallel with a partially discharged battery, their emfs will be slightly different and the well charged battery will charge the other one, losing a lot of energy.

Yep. I believe it was with some Metz flashes: they had an external battery pack, connected in parallel with the internal one, and the manual was very clear about not mixing batteries with different charge level, stating connecting them can even end with a fire.

 

[sophiecentaur]

an external battery pack, connected in parallel

If you can afford to lose a few millivolts you can always connect two batteries by putting a diode in series with each. That will stop reverse charging and will only cost you a fraction of a volt delivered. I had an outdoor weather station with six AA's and it ran and ran.

 

[berkeman]

you can always connect two batteries by putting a Schottky diode in series with each.

Fixed that for you.