# Parallel batteries

Discussion in 'General Electronics Chat' started by Vorador, Jun 29, 2013.

Oct 5, 2012
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Hi,

I was wondering - purely from a theoretical standpoint (i.e no internal resistances involved) - that if two identical voltage sources and a resistor all are connected in parallel, what would be the behavior of such a circuit? Would the current through the resistor be larger than what a single source is capable of producing? If yes, then wouldn't that violate Ohm's law since I=V/R and V remains the same whether there's one, two or an infinite identical sources connected in parallel...

2. ### LDC3 Active Member

Apr 27, 2013
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Just because a voltage source can supply more (or infinite) current does not mean it will.

Oct 5, 2012
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Ooops. I posted this in the wrong section. Can the mods please move it to a relevant section? I'm really sorry for this.

LDC3, yeah sure, but what is exactly going to happen in a circuit like that? Will the value of current be determined simply by I=V/R, or is there some important detail that I'm missing?

4. ### studiot AAC Fanatic!

Nov 9, 2007
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Yes it's called internal resistance.
Every real voltage source has some.

The circuit analysis is then not difficult for parallel voltage sources, AC or DC.

I'm sure I have posted a graphical method here before.

5. ### LDC3 Active Member

Apr 27, 2013
920
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A good example is water in a pipe. The height of the pipe corresponds to voltage. The diameter of the pipe corresponds to resistance (the smaller the pipe, the more resistance it has). The flow of the water corresponds to current.
The higher the pipe reaches, the more pressure on the water and more water will flow through the pipe. I can have a large reservoir of water at the top, but the flow is limited by the resistance of the pipe. An electrical circuit is the same. The current is limited by the resistance of the load.

6. ### studiot AAC Fanatic!

Nov 9, 2007
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I'm not sure how this is relevant to parallel generators, where one may actually force current through the other.

7. ### strantor AAC Fanatic!

Oct 3, 2010
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The same amount of current would flow through the resistor. Only difference is each battery would supply half of that current.

8. ### LDC3 Active Member

Apr 27, 2013
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Okay, you want to know about 2 voltage sources connected in parallel with a load.
If the 2 sources have the same voltage output, each will supply half the current through the load.
If one source has a slightly higher voltage, it will supply more current (maybe all the current) through the load and the second source supplies less current.
If one source has a voltage above the second source, it will try to push current back through the second source. Depending on the source, there is the internal resistance and a maybe a voltage potential (like a diode) to overcome. For example, in the real world, a 12V battery needs at least 14V on the charge circuit to be charged fully.

Oct 5, 2012
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Oh thank you strantor for your succinct answer and understanding me correctly! That's exactly what I wanted to know.

Thanks to everyone who tried to help me here, but all I wanted to know was the current that would be produced in a resistor in a parallel circuit consisting of two parallel voltage sources.. I'm sorry for my poor communication skills, I should have made my question clear and concise.

Thanks again everyone and also the moderator who moved my thread to the appropriate section!

10. ### GopherT AAC Fanatic!

Nov 23, 2012
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Also, since each battery supplies half the current, the voltage drop will be 50% of a single battery's voltage drop (across the internal resistance).

11. ### MrChips Moderator

Oct 2, 2009
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Each battery would supply half of the current if:

1) the batteries are exactly the same voltage (which is never the case)

2) the internal resistance of each battery is the same (which is never the case).

12. ### nsaspook AAC Fanatic!

Aug 27, 2009
3,013
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Exactly, the slightest difference will unbalance the current and the effect gets worse as power levels increase. It's usually easier to balance parallel batteries by using very low external resistances like connecting the wiring in a staggered configuration.

http://www.smartgauge.co.uk/batt_con.html

Nov 9, 2007
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