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# need help with this dc electricity problem (circuits) watch

1. This question was taken from edexcel physics unit 2 june 2012, question 18.

I have a confusion with part b. of it. I have posted the question, part b of the question and one of the mark scheme answers to this problem that confuses me. Please view the mark scheme answer before reading any further. The part b of the question states that the emf of the battery is now 12v but the mark scheme answer shows the terminal pd across the battery charger is 12v + pd across internal resistance of battery. How is this possible? If the emf of the battery is inclusive of the volts lost in the internal resistance, according to my theory the terminal pd should be just 12 v since it's the total volts lost between the terminals of the battery charger (or am I wrong) I ?am wrong and the mark scheme is right obviously but I don't understand why I am wrong. Please someone explain this to me.
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2. The simple answer is that when a battery is delivering current the terminal pd is indeed its emf minus pd lost on internal resistance.
When you are forcing current in the opposite direction through the battery the pd across the internal resistance is now in the opposite direction and adds to the emf to give the terminal pd.
The pd across the terminals must be greater than the emf in order to force current back through the battery to charge it.

A quick application of Kirchhoff's Rules will also show this.
3. (Original post by thebrahmabull)
This question was taken from edexcel physics unit 2 june 2012, question 18.

I have a confusion with part b. of it. I have posted the question, part b of the question and one of the mark scheme answers to this problem that confuses me. Please view the mark scheme answer before reading any further. The part b of the question states that the emf of the battery is now 12v but the mark scheme answer shows the terminal pd across the battery charger is 12v + pd across internal resistance of battery. How is this possible? If the emf of the battery is inclusive of the volts lost in the internal resistance, according to my theory the terminal pd should be just 12 v since it's the total volts lost between the terminals of the battery charger (or am I wrong) I ?am wrong and the mark scheme is right obviously but I don't understand why I am wrong. Please someone explain this to me.
The key phrases are:

E.M.F. = Electro Motive Force

Terminal Potential Difference.

The schematic diagram of the battery is a 'model' of how the battery behaves. In that context, there will be an internal chemical process (cells) producing a voltage 'pressure' (e.m.f.) with the 'potential' to do work. i.e. 12V worth of work potential = 12 joules per coulomb of charge.

But the key difference here is that the battery is being charged - current is flowing into the battery NOT out of it.

That means in order for 12V to be placed across the battery cells, there must also be a potential drop across the internal resistance.

The current demanded by the charging cells will cause work to be done by the battery internal resistance as the current flows in. Thus the difference in work potential between the e.m.f. of the battery and that available at the battery terminals is:

Vterm = Vemf + Vint res

Vterm = Vemf + (I x Rint res)

Vterm = 12 + (4.3 x 0.05) = 12 + 0.215 = 12.215V

i.e. the potential at the battery terminals is 12.2V with 0.215V of potential lost across the internal resistance as per the mark scheme.
4. (Original post by Stonebridge)
The simple answer is that when a battery is delivering current the terminal pd is indeed its emf minus pd lost on internal resistance.
When you are forcing current in the opposite direction through the battery the pd across the internal resistance is now in the opposite direction and adds to the emf to give the terminal pd.
The pd across the terminals must be greater than the emf in order to force current back through the battery to charge it.

A quick application of Kirchhoff's Rules will also show this.
(Original post by uberteknik)
The key phrases are:

E.M.F. = Electro Motive Force

Terminal Potential Difference.

The schematic diagram of the battery is a 'model' of how the battery behaves. In that context, there will be an internal chemical process (cells) producing a voltage 'pressure' (e.m.f.) with the 'potential' to do work. i.e. 12V worth of work potential = 12 joules per coulomb of charge.

But the key difference here is that the battery is being charged - current is flowing into the battery NOT out of it.

That means in order for 12V to be placed across the battery cells, there must also be a potential drop across the internal resistance.

The current demanded by the charging cells will cause work to be done by the battery internal resistance as the current flows in. Thus the difference in work potential between the e.m.f. of the battery and that available at the battery terminals is:

Vterm = Vemf + Vint res

Vterm = Vemf + (I x Rint res)

Vterm = 12 + (4.3 x 0.05) = 12 + 0.215 = 12.215V

i.e. the potential at the battery terminals is 12.2V with 0.215V of potential lost across the internal resistance as per the mark scheme.
thanks a lot! I actually asked these to two of my teacher's assistants before coming here and they weren't able to explain me. They should do some training with you lol

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