Accumulator Battery Internal Resistance & Charger Potential: A Doubt

[vikcool812]

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!

 

[Born2bwire]

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.

 

[tomcue]

I would like to address your doubt and provide a response to the question at hand. Firstly, let us understand the concept of internal resistance in a battery. Internal resistance is the resistance that is present within the battery itself, which causes a voltage drop when current flows through it. This resistance can vary depending on the state of charge of the battery.

In the given scenario, we have an accumulator battery with an EMF of 6V and an internal resistance of 10 Ohms when fully discharged. As the battery gets charged, its internal resistance drops to 1 Ohm. Now, when the battery is connected to a charger with a constant potential of 9V, the current flowing through the circuit can be calculated using Ohm's law (I=V/R). For the first part of the question, when the connections are just made, the current would be 0.3A (9V/30 Ohms). This is because the internal resistance of the battery is still high, which results in a large voltage drop across it, leaving only 3V for the external circuit.

Now, let us move on to the second part of the question, where the battery is completely charged. In this case, the internal resistance of the battery has dropped to 1 Ohm. This means that the voltage drop across the internal resistance would be much lower, leaving a larger voltage (8V) for the external circuit. As the charger is maintaining a potential of 9V, the remaining 1V would be used to charge the battery. Therefore, the current in this case would be 1A (9V/9 Ohms).

It is not incorrect to treat the EMF as constant in this scenario. The EMF of a battery is a theoretical value that represents the maximum potential difference between the terminals of the battery. In reality, the EMF can vary depending on the state of charge and internal resistance of the battery. However, for the purpose of solving this problem, we can assume the EMF to be constant.

In conclusion, the answers provided for both parts of the question are correct. I hope this explanation has helped clear your doubts. If you have any further questions, please feel free to ask.