Accumulator Battery Internal Resistance & Charger Potential: A Doubt

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SUMMARY

The discussion centers on the calculation of current (I) in an accumulator battery with an emf of 6V and varying internal resistance. When the battery is fully discharged, its internal resistance is 10 Ohm, which drops to 1 Ohm when fully charged. With a charger maintaining a constant potential of 9V, the initial current upon connection is calculated to be 0.3A, while the current when fully charged is determined to be 3A. The analysis confirms that treating the emf as constant in this context is appropriate.

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vikcool812
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This is in continuation to https://www.physicsforums.com/showthread.php?t=330495
The question that initiated my doubt reads this -- " the internal resistance of an accumulator battery of emf 6V is 10 Ohm when fully discharged . as battery gets charged up , its internal resistance drops to 1 Ohm , . the battery when discharged is connected to a charger which maintains a constant potential of 9V . Find I when a)connections are just made b) when it is completely charged . "
ans for 1st part is 0.3V (OK), but ans for b is 3V , , here they are treating the emf const. isn't it wrong , Pls. help!
 
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That seems correct. The equivalent circuit of a battery is a voltage source in series with a resistance. They have given you that the battery is now fully charged with an applied voltage of 9 V across the terminals. Thus, the battery now has a voltage of 6 V and an internal resistance of 1 Ohm. This works out to 3 A.

By the way, they are asking for current, so the units are amps, not volts.
 

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