[When you see it...]

I just posted this in the thread for an exam I just sat. Any help? BTW ignore the references to 'the question', all you need to know is that the voltmeter was being used to measure the emf of a cell:


For the question on the voltmeter attatched to the cell, was it connected in series or parallel. I know that normally they are connected in parallel but for this purpose it would make sense if it was connected in series as that would still work given the high resistance of voltmeters, taking a large proportion of the emf, therefore measuring a value for V close to the value of the cell.
I think I answered the calculations as though it was connected in series (the first one about the resistance of the voltmeter would have been the same either way) but did the explanation as though it was connected in parallel.
Hold on, I'll ask this in the main physics forum.


Series or parallel? I'm guessing series.

[When you see it...]

bump
Anyone?

[Stonebridge]

(Original post by When you see it...)
I just posted this in the thread for an exam I just sat. Any help? BTW ignore the references to 'the question', all you need to know is that the voltmeter was being used to measure the emf of a cell:


For the question on the voltmeter attatched to the cell, was it connected in series or parallel. I know that normally they are connected in parallel but for this purpose it would make sense if it was connected in series as that would still work given the high resistance of voltmeters, taking a large proportion of the emf, therefore measuring a value for V close to the value of the cell.
I think I answered the calculations as though it was connected in series (the first one about the resistance of the voltmeter would have been the same either way) but did the explanation as though it was connected in parallel.
Hold on, I'll ask this in the main physics forum.


Series or parallel? I'm guessing series.

If you connect a single voltmeter across the terminals of a cell and there is nothing else in the circuit, it doesn't matter whether you call it series or parallel.
It's actually both in such a simple circuit.
If that's not what you mean you will need to provide a circuit diagram.

[When you see it...]

(Original post by Stonebridge)
If you connect a single voltmeter across the terminals of a cell and there is nothing else in the circuit, it doesn't matter whether you call it series or parallel.
It's actually both in such a simple circuit.
If that's not what you mean you will need to provide a circuit diagram.

Yes that is exactly what I mean.
You get different results for the calculations (I think) and different explanations for questions such as 'why does it need a high resistance?' depending on whether you treat it as series or parallel.
For example:
You are told the current through the voltmeter and asked to find its resistance, given the EMF of a power source. Although the question mentions internal resistance, you can assume it to be negligible because the resistance of the voltmeter is probably massive. For this particular calculation, it is R=V/I regardless of whether you treat it as series or parallell (although if you take into account the internal resistance I think you get slightly different values because in series the internal resistance would take a proportion of the total voltage).
For the next question though, it asks:
Find the voltage across the internal resistance.
For this question, I assumed the circuit to be series and therefore used V=IR with the current through the voltmeter (as in a series circuit this would be the same as the current through the cell). However, if I treated the circuit as parallel, the current through the cell would be given as follows:
I = V/(Rcell +Rvoltmeter) - V/Rvoltmeter
I think this would be a different value to the current given in the question. This would change the value of V as V=Ir.
Also, in parallel, surely the voltage across it would just be the same as the EMF? It is just one 'arm' after all.
Anyway, looking back I think the circuit was meant to be in series and the answers were:
First question:
R= (E.M.F. - Ir)/I
Second question:
V=Ir
Third question (this was 'explain why a high resistance is needed?'):
So it takes an extremely high proportion of the p.d., therefore the voltage reading is close to the emf.

[When you see it...]

(Original post by Stonebridge)
If you connect a single voltmeter across the terminals of a cell and there is nothing else in the circuit, it doesn't matter whether you call it series or parallel.
It's actually both in such a simple circuit.
If that's not what you mean you will need to provide a circuit diagram.

Sorry, I should have mentioned the internal resistance in the OP.

[When you see it...]

Series or parallel?
I think series, can someone confirm?

[Stonebridge]

A voltmeter is always placed "in parallel" with the component you want to measure the pd across.
Having said that, in a simple circuit where you place a meter across a cell and nothing else, you have a simple series circuit where the current in the meter is the same as the current in the cell.
The total current in the circuit is I = E/(r+R) where r is internal resistance and R is that of the meter.
If the meter resistance is large then the current, I, will be small.
If the current, I, is small then the pd dropped across the internal resistance is small as this is = Ir.
This then means that the pd across the terminals of the cell as measured by the meter will be very close to the emf of the cell. The terminal pd V = E - Ir
So you want Ir to be as small as possible to measure the emf accurately.
So you use a voltmeter with as large a resistance as possible.

[When you see it...]

(Original post by Stonebridge)
A voltmeter is always placed "in parallel" with the component you want to measure the pd across.
Having said that, in a simple circuit where you place a meter across a cell and nothing else, you have a simple series circuit where the current in the meter is the same as the current in the cell.
The total current in the circuit is I = E/(r+R) where r is internal resistance and R is that of the meter.
If the meter resistance is large then the current, I, will be small.
If the current, I, is small then the pd dropped across the internal resistance is small as this is = Ir.
This then means that the pd across the terminals of the cell as measured by the meter will be very close to the emf of the cell. The terminal pd V = E - Ir
So you want Ir to be as small as possible to measure the emf accurately.
So you use a voltmeter with as large a resistance as possible.

Thanks for your insight.
I think, given the fact that there is effectively an extra resistor next to the cell, it is best modelled as a series circuit (although the answers I put in the exam, assuming it to be parallel, will be practically the same because the voltmeter had a high resistance).
Unfortunately, my exam board for physics have really specific markschemes and I don't think I'll get away with it - not even the answer marks.