# Batteries connected in parallel

What will happen if two batteries were connected in parallel, and how could the total voltage and current be distributed among the resistors?
Like in this attachment considering that the internal resistance of the 12V battery is 3 ohms while the resistance of the 5V battery is 2 ohms

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berkeman
Mentor
What will happen if two batteries were connected in parallel, and how could the total voltage and current be distributed among the resistors?
Like in this attachment considering that the internal resistance of the 12V battery is 3 ohms while the resistance of the 5V battery is 2 ohms
View attachment 66866
The 5V battery will likely catch fire.

Voltage sources are not connected in parallel, unless their output voltages are closely matched.

The 5V battery will likely catch fire.

Voltage sources are not connected in parallel, unless their output voltages are closely matched.

Why would it catch fire ??

berkeman
Mentor
Why would it catch fire ??
Because you are driving an overcurrent backward through the 5V battery (and drawing a large current out of the 12V battery). That's generally a bad thing to do with batteries, depending on what chemistry they are.

An ideal voltage source has a very low output resistance, so when you try to connect voltage sources of different output voltages in parallel, you will be drawing large currents and forcing large reverse currents. Not a good idea.

Voltage sources can be connected in series, and current sources can be connected in parallel.

Because you are driving an overcurrent backward through the 5V battery (and drawing a large current out of the 12V battery). That's generally a bad thing to do with batteries, depending on what chemistry they are.

An ideal voltage source has a very low output resistance, so when you try to connect voltage sources of different output voltages in parallel, you will be drawing large currents and forcing large reverse currents. Not a good idea.

Voltage sources can be connected in series, and current sources can be connected in parallel.

I want to overlook all the problems that might arise due to this connection, I will prevent the reverse current by readjusting the position of one of the sources and I will control the situation by perfect, ideal circumstances to prevent any damage, even If that's impossible, I'm just curios how would the voltage and current be distributed among the resistors (theoretically speaking).

davenn
Gold Member
2019 Award
I will prevent the reverse current by readjusting the position of one of the sources
and how do you propose to do that ?

Dave

and how do you propose to do that ?

Dave

I'll make the two sources pump voltage in the same direction

davenn
Gold Member
2019 Award
again how will you do that

show us how ( I know how, I want to see if you do )

Dave

again how will you do that

show us how ( I know how, I want to see if you do )

Dave

I'm not the one who did this circuit I just found it on one of the websites and I copied it here I don't know how can I show you, I'm just saying in this circuit I'll reverse the position of one of the sources

berkeman
Mentor
I'm not the one who did this circuit I just found it on one of the websites and I copied it here I don't know how can I show you, I'm just saying in this circuit I'll reverse the position of one of the sources
If you do that with batteries, you will most likely damage one or both of them.

If you do it with power supplies, the higher voltage power supply will supply all of the load resistor current, and the lower voltage power supply will not supply any current to the load (its output would be shut off by the higher voltage at its output.

phinds
Gold Member
2019 Award
I'm not the one who did this circuit I just found it on one of the websites and I copied it here I don't know how can I show you, I'm just saying in this circuit I'll reverse the position of one of the sources
Huh? What does that even MEAN ?

If you do that with batteries, you will most likely damage one or both of them.

If you do it with power supplies, the higher voltage power supply will supply all of the load resistor current, and the lower voltage power supply will not supply any current to the load (its output would be shut off by the higher voltage at its output.

Thanks a lot that's a clear answer, I just want to ask one more thing, what if the voltage of both sources were nearly equal as you mentioned earlier

berkeman
Mentor
Thanks a lot that's a clear answer, I just want to ask one more thing, what if the voltage of both sources were nearly equal as you mentioned earlier
If they are batteries, they will fight each other a little bit, but level off at some intermediate output voltage.

For power supplies, the higher voltage will still dominate and shut off the lower voltage power supply. Most power supplies can only source current, they cannot sink current. So the lower voltage power supply will basically have its output stage shut off by the higher voltage power supply.

There are some very sophisticated systems that have multiple power supplies that share sourcing current for loads, or act as redundant backup supplies to be sure that the failure of one power supply will not take down the system.

If they are batteries, they will fight each other a little bit, but level off at some intermediate output voltage.

.

How is the intermediate voltage calculated?

davenn
Gold Member
2019 Award
I'm not the one who did this circuit I just found it on one of the websites and I copied it here I don't know how can I show you, I'm just saying in this circuit I'll reverse the position of one of the source
but you are the one making a claim/statement. so why make the statement if you cant show how you will do it ? It makes the whole point of what you are talking about meaningless.

the whole idea here is to help further one's education. If you dont know how to do it or show how its done, dont claim that you can

Dave

but you are the one making a claim/statement. so why make the statement if you cant show how you will do it ? It makes the whole point of what you are talking about meaningless.

the whole idea here is to help further one's education. If you dont know how to do it or show how its done, dont claim that you can

Dave

You really misunderstand what I meant by " I can't show it ", I meant that I don't have any kind of software to redraw the circuit that visualizes what I mean. So I'm saying it in words instead of a diagram.

berkeman
Mentor
How is the intermediate voltage calculated?
It would depend on the relative sizes of the batteries and their state of charge. Given the same size/capacity batteries with equal charges but off by 0.2V in their output voltages, the sum would close to the average output voltage.

It would depend on the relative sizes of the batteries and their state of charge. Given the same size/capacity batteries with equal charges but off by 0.2V in their output voltages, the sum would close to the average output voltage.

Thumbs up, I'm really grateful

berkeman
Mentor
Thumbs up, I'm really grateful
Glad to help. Keep in mind that many vehicles have dual-battery systems for extra reliability. You might try a Google or Wikipedia search on those to see how they are connected in real life.

How is the intermediate voltage calculated?
Consider the following circuit:

http://www.ibiblio.org/kuphaldt/electricCircuits/DC/00207.png

The voltage at the node between the resistors is:

V = (V1/R1+V2/R3)/(1/R1+1/R3+1/R2)

Suppose R1 = R3 = R. Then:

V = (1/R)(V1+V2)/(2/R+1/R2)
V = (V1+V2)/(2+R/R2)

Now let R go to zero, and you end up with that illegal circuit you had in the first image. Of course, we've made some illegal assumptions, so it's probably not what happens in reality, but it sort of captures the idea. Assuming your batteries don't blow up, the average voltage will be close to the output voltage.

The idea of a battery is that it pumps out as much current as possible to sustain a given voltage. However, in your illegal circuit, neither battery pumps out the correct amount of current to produce the correct voltage. Both batteries are still going to pump out what current they believe are correct, but the excess charge is going to travel from the stronger battery (12 V) to the weaker one (5 V) and mess with whatever chemical reaction is going on inside (which is described by the Nernst equation).