- #1
tellmesomething
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- Homework Statement
- The potential difference between the terminals of a 6V battery is 7.2 V when it is being charged by a current of 2A.
- Relevant Equations
- None
First let us talk about a case where the battery gets discharged, I have a circuit with a battery and a load.
The emf of the battery is 6V and the potential difference is a little less because the battery has internal resistance.
When a charge moves from the anode and reaches the load it loses all its energy into a useful form of energy for the load. It comes back to the positive potential, and neutralises a charge on the cathode(reduction). Now inside the battery in the electrolyte there are two types of forces on the ions electric field force and non electrostatic force. They were in equilibrium before the circuit was closed. Now since the field becomes a little less the chemical/non electrostatic force forces the positive ions to go to the cathode to sort of "revive this charge which just got neutral" and it forces the negative ions to go to the anode to do the same. However not all of the charge gets revived as theres internal resistance. Hence this chemical force stays always a little more than the electrostatic force.
Now say we want to increase this chemical force more so that even after facing the same amount of resistance it can revive the total charge it loses. I charge the battery. I plug it to a direct current source. The negative terminal of the dc source is connected to the negative electrode of the battery and the positive terminal of the dc source is connected to the positive terminal. This time I have a potential difference of 7.2 volts across the battery and the battery has an emf of 6V. Excess electrons form the negative terminal goes to the negative electrode, reduction occurs. The little charge that was left to revive revives, and it possibly releases ions into the solution?(I am not sure) this increases the chemical force as the solution becomes more ionic which causes the negative ion to move to the anode and the positive ion to move to the cathode despite of the obstruction / internal resistance. The excess electrons in the positive cathode from before discharging goes to the positive terminal of the dc source?
Again I am not sure about ions and how they are neutralised during discharging and separated during charging... Please let me know if the above is any correct.
Also its tempting to think that 1.2 V is the potential drop due to the resistance but I dont concretely know why? If the dc supply was a 6V machine would this charging not have worked?
I understand this might be a very crude way to look at this if not completely wrong. Please let me know if I am thinking even in the right direction
The emf of the battery is 6V and the potential difference is a little less because the battery has internal resistance.
When a charge moves from the anode and reaches the load it loses all its energy into a useful form of energy for the load. It comes back to the positive potential, and neutralises a charge on the cathode(reduction). Now inside the battery in the electrolyte there are two types of forces on the ions electric field force and non electrostatic force. They were in equilibrium before the circuit was closed. Now since the field becomes a little less the chemical/non electrostatic force forces the positive ions to go to the cathode to sort of "revive this charge which just got neutral" and it forces the negative ions to go to the anode to do the same. However not all of the charge gets revived as theres internal resistance. Hence this chemical force stays always a little more than the electrostatic force.
Now say we want to increase this chemical force more so that even after facing the same amount of resistance it can revive the total charge it loses. I charge the battery. I plug it to a direct current source. The negative terminal of the dc source is connected to the negative electrode of the battery and the positive terminal of the dc source is connected to the positive terminal. This time I have a potential difference of 7.2 volts across the battery and the battery has an emf of 6V. Excess electrons form the negative terminal goes to the negative electrode, reduction occurs. The little charge that was left to revive revives, and it possibly releases ions into the solution?(I am not sure) this increases the chemical force as the solution becomes more ionic which causes the negative ion to move to the anode and the positive ion to move to the cathode despite of the obstruction / internal resistance. The excess electrons in the positive cathode from before discharging goes to the positive terminal of the dc source?
Again I am not sure about ions and how they are neutralised during discharging and separated during charging... Please let me know if the above is any correct.
Also its tempting to think that 1.2 V is the potential drop due to the resistance but I dont concretely know why? If the dc supply was a 6V machine would this charging not have worked?
I understand this might be a very crude way to look at this if not completely wrong. Please let me know if I am thinking even in the right direction