How is the emf greater than 9 V in this scenario?

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SUMMARY

The discussion centers on the scenario where a battery with an emf of 9 V and an internal resistance of 2 Ω can exhibit a potential difference greater than 9 V. The correct answer to the posed question is option D, indicating that the current must be out of the negative terminal. This situation occurs when the battery is connected to a higher emf source, such as a battery charger, which reverses the typical discharge behavior of the battery.

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Homework Statement


[/B]
A battery has an emf of 9 V and an internal resistance of 2 Ω. If the potential difference across its terminals is greater than 9 V:

A. it must be connected across a large external resistance
B. it must be connected across a small external resistance
C. the current must be out of the positive terminal
D. the current must be out of the negative terminal
E. the current must be zero


2. Homework Equations

The Attempt at a Solution


[/B]
The answer is D but I'm confused on one basic thing:

When I connect a battery, what does it mean for the current to be coming out of the negative terminal? Under what circumstance would this ever happen? Any examples?
 
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eprparadox said:
Under what circumstance would this ever happen? Any examples?
If you hook it up to a source with a higher emf, e.g. a battery charger :rolleyes:
 
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If the current is coming out of the negative it's also going into the positive terminal. That's the reverse of what happens when a battery is being discharged by a load, so it must be being charged.
 

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