Are Currents Summed in Parallel Battery Circuits?

[PainterGuy]

hi everyone,

please look at attach picture.

in the circuit total current through 20 ohm resistor = sum of all currents delivered by each cell.

is this correct?

if one of battery is reversed then current of such a battery will be subtracted from the sum. is this correct?

i am grateful for any help you can come up.

cheers

 

[Sciurus]

Hi, painterguy. That's sort of a strange thing to do. The current through the resistor is the sum of the current supplied to the other batteries if no current flows back through the batteries. I think you might partially recharge (or damage) the lower voltage batteries, depending on what type they are.

 

[Studiot]

No this is not true.

You are connecting 4 batteries in parallel here. Look at post#11 of this thread

https://www.physicsforums.com/showthread.php?t=446780&highlight=parallel+batteries

This shows how to handle two batteries connected in parallel. If you feel up to the maths you can extend the method to any number of parallel batteries.

go well

 

[Curl]

Putting batteries in parallel is not a good thing to do

 

[Dale]

You cannot do a problem like this without considering the internal resistance of the batteries.

 

[mrspeedybob]

You cannot do a problem like this without considering the internal resistance of the batteries.

If you reverse 1 battery you will have a series/parallel circuit. 3 batteries and the resistor will be in parallel, that section of the circuit will be in series with the 4'th battery which will then be short circuited back to the first 3. You will have twice the voltage pushing current through the internal resistance of the batteries and the resistance of the wire. Some may flow through the resistor also but for most combinations of batteries and wire it will be negligible.

 

[Dale]

If you reverse 1 battery you will have a series/parallel circuit. 3 batteries and the resistor will be in parallel, that section of the circuit will be in series with the 4'th battery which will then be short circuited back to the first 3. You will have twice the voltage pushing current through the internal resistance of the batteries and the resistance of the wire. Some may flow through the resistor also but for most combinations of batteries and wire it will be negligible.

None of that is relevant. The point is that if you consider ideal voltage sources you get a set of equations with V=2, V=10, V=15, etc. Which has no solution. If you want to get a system of equations with a solution then you need to consider the internal resistance.