# cells in mixed connection

#### flippineck

Sep 8, 2013
358
Can anyone explain the voltage arising from a mix of batteries. for example, see sketch

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#### davenn

Moderator
Sep 5, 2009
14,264
Paralleling uneven voltages is never a good plan
The higher voltage is going to discharge into the lower voltage source ie. the 3V into the 1.5V.
That isn't going to be good for the 1.5V battery

Dave

#### Gryd3

Jun 25, 2014
4,098
Paralleling uneven voltages is never a good plan
The higher voltage is going to discharge into the lower voltage source ie. the 3V into the 1.5V.
That isn't going to be good for the 1.5V battery

Dave
If resistors were in place to limit the amount of reverse current through the lower voltage source, would there be a mathematical formula that could be used to determine the voltage at any given point in a circuit given 2 or more different voltage sources?

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
If resistors were in place to limit the amount of reverse current through the lower voltage source, would there be a mathematical formula that could be used to determine the voltage at any given point in a circuit given 2 or more different voltage sources?

Yes. (this assumes certain things about the batteries, but for high enough resistances it is easy enough to do)

If you do the MITx 6.02x you'll learn many of the mathematical tricks needed to determine the exact formula you need (or system of equations you need to solve) for almost any configuration

#### Laplace

Apr 4, 2010
1,252
If resistors were in place to limit the amount of reverse current through the lower voltage source...
But these are not "voltage sources" (capable of infinite current), they are "dry cells" with internal resistance. What is not clear is whether the internal resistance is the same in both forward and reverse voltage directions. If this is a multiple choice question, I know which two answers I would not choose.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Sorry Flippineck, I didn't realise this was homework. I just read some of the last posts and responded to them.

This question is a bit like asking "Two identical cars are travelling in convoy north along a single lane road at 60 miles per hour. Another identical car is travelling south towards them at 60 miles per hour. After the inevitable head-on collision, what is the velocity of the cars?"

I think you can see that there are far more variables than you can easily cater for.

In this case, the new fresh batteries would practically have a voltage of more than 1.5V, but even ignoring this, Laplace has pointed out that any typical values you might assume will not necessarily apply to all of the batteries in this question.

Personally, I would take option 4, (V=?) because I am not certain it would be any of the other voltages listed. However, there are 2 of the remaining three answers which are more likely more incorrect than the other.

#### flippineck

Sep 8, 2013
358
Excellent analogy about the cars thanks. It's a multichoice question aimed at generating class discussion.. I understand the correct answer is theoretically ?

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Well, the voltage you measure at that point will not be either the open circuit voltage across 1 or 2 cells. It is likely to be somewhere between the two, but the exact value is hard to predict with any certainty, and it will probably change as the cells are affected by the inevitable current that will flow. There is a reasonable chance that the cells may get hot, and at least one will be in danger of rupturing.

Similar to the possible death of drivers in my analogy above, there may be more important considerations than the voltage you measure.

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