I put the following, or something along these lines.
As the emf across the circuit is equal to sum of the emf of the individual components, then by replacing the battery with a with one of the same emf, but internal resistance, the voltmeter would read the same value.
(Cant actually remember what the question was, if you remind me I can write what I actually answered)
Question was 1.5v battery is replaced with one with significant internal resistance. explain why the value on the voltmemter at A and B doesnt change
What did people get for the last electricity where the battery had internal r
Do you mean the one where you had to work out of one of the resistors? Like 0.15 I think. Or the voltmeter one, where I said that the resistor is in parallel and the reciprocal of a big number is very small and so it would have little difference
I put the following, or something along these lines.
As the emf across the circuit is equal to sum of the emf of the individual components, then by replacing the battery with a with one of the same emf, but internal resistance, the voltmeter would read the same value.
I put that the internal resistance would reduce the terminal potential difference but the potential difference between the two points would be the same as the external arrangement of the resistances is the same therefore the voltage across each component will be less but the difference between A and B the same
Do you mean the one where you had to work out of one of the resistors? Like 0.15 I think. Or the voltmeter one, where I said that the resistor is in parallel and the reciprocal of a big number is very small and so it would have little difference
The one where thd voltmeter was placed between point a and b
I put that the internal resistance would reduce the terminal potential difference but the potential difference between the two points would be the same as the external arrangement of the resistances is the same therefore the voltage across each component will be less but the difference between A and B the same
I said something about the potential drops would be the same is that right
I put that the internal resistance would reduce the terminal potential difference but the potential difference between the two points would be the same as the external arrangement of the resistances is the same therefore the voltage across each component will be less but the difference between A and B the same
I put very similar to this but i had no idea it was a complete guess
I put that the internal resistance would reduce the terminal potential difference but the potential difference between the two points would be the same as the external arrangement of the resistances is the same therefore the voltage across each component will be less but the difference between A and B the same
I said that, the pd reading between A and B should be zero as there are no components in parallel to it
damn, i said the voltage would just be zero because it's in parallel with no components. But, wouldnt the voltmeter not record the emf as its not in paralel with the cell
The question asked why the volmeter would read the same.