Can Two Batteries Simultaneously Act as Both the Source and Load in a Circuit?

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In the discussion, two scenarios involving battery connections are analyzed to understand circuit behavior. In Scenario 1, connecting the positive and negative terminals of two batteries results in no current flow, indicating an incomplete circuit. Conversely, Scenario 2, where the positive terminal of one battery connects to the negative terminal of another, allows current to flow, completing the circuit. The conversation highlights that while both scenarios involve loops, only Scenario 2 constitutes a functioning circuit due to the current flow. The calculations for current and power are derived using Ohm's law and Kirchhoff's laws, emphasizing the distinction between open and closed circuits.
Evil Bunny
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Probably a very simple question, but I'm not sure of the answer...

We have two batteries.

Scenario 1: The positive terminals of both batteries are connected together and the negative terminals are connected together. Little to no current flows, correct?

Scenario 2:
The positive terminal of battery A is connected to the negative terminal of battery B.
The positive terminal of battery B is connected to the negative terminal of battery A.
Now we have lots of current flowing, correct?

In scenario 1, apparently no circuit was completed, yet in scenario 2 we must have completed a circuit... Why? Did both batteries (of scenario 2) simultaneously become the sources and the loads? How would you calculate the current flow?
 
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Hi Evil Bunny! :smile:
Evil Bunny said:
In scenario 1, apparently no circuit was completed, yet in scenario 2 we must have completed a circuit... Why? Did both batteries (of scenario 2) simultaneously become the sources and the loads? How would you calculate the current flow?

This is exactly what happens if you connect two 1.5 V batteries together the "correct" way, and then complete the circuit with only a wire …

the total voltage will be 3 V, and if the total internal resistance (plus the negligible resistance of the wire) is R Ω, then the current will be 3/R A, and the power drain (although you can't see it) will be 9/R W.
 
Evil Bunny said:
In scenario 1, apparently no circuit was completed, yet in scenario 2 we must have completed a circuit... Why? Did both batteries (of scenario 2) simultaneously become the sources and the loads? How would you calculate the current flow?


This makes me wonder exactly what we mean by the word "circuit". We certainly have a loop of some kind in scenario 1, but no current. (A simple metal ring with no battery is a loop, but without a current, unless there's a time-varying magnetic flux through the loop, of course.)

Scenario 1 has two batteries in parallel, so that "loop" is in itself a voltage source, having the same voltage as a single battery, like in this picture. But, should we call that a "circuit"? An open circuit perhaps? (Image from http://itp.nyu.edu/physcomp/Notes/Batteries")

[PLAIN]http://itp.nyu.edu/physcomp/uploads/battery_parallel.jpg
 
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I guess not... it's only a circuit if the negative side makes it's way back to the positive side. And apparently the source becomes it's own (very low resistance) load when we short the two terminals. Interesting...
 
It is a circuit (loop) no matter which way you connect them. The resistance of the circuit is 2r where r is the internal resistance of the battery. The sum of the emfs is E-E in the first case and E+E in the second case. (see Kirchoff's second law, for example).
So in the first case
I*2r= 0 and I=0
and in the second case
I*2r= 2E so I= E/r.
These cases bot refer to a circuit with just batteries. No external load.
 
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