Measuring resistance using potentiometer

[AdityaDev]

Homework Statement

Homework Equations

For R, current I=E/R and for X, I=E/X... both are separate cases.

The Attempt at a Solution

[/B]
Why does the balancing change?
The potential difference across each resistor is E Itself right?

I tried by assuming an arbitary internal resistance r.
but in the book, it given as:
E2/E1 = iX/iR
but how is I same in this case when resistance is varied? They have given that I is the current in potentiometer wire... but how does the that current cause potential difference across R? It should the current E/R or E/X that passes through R and X.

 

[Faris Shajahan]

When I was also in 12th I had exactly the same problem...(NCERT)...

Your doubt is why the current is same and not different right?
Well, I asked my teacher and she simply got angry with me!

 

[AdityaDev]

Potential difference across balancing lengthfor case R Is ##V_1=\frac{V_0l_1}{L}##
And for X ##V_2=\frac{V_0l_2}{L}## where l1 and l2 are balancing lengths and V0 is the potential difference across entire length.
in the lower circuit, for R, ##i_1=E/R## and for X ##i_2=E/X## where E is the lower cell. So ##\frac{I_1}{I_2}=X/R##
V1/V2 = l1/l2
But ##\frac{V_1}{V_2}=\frac{I_1R}{I_2X}## if you substitute value of ratio of the currents which is equal to X/R, you get V1=V2.

 

[andrevdh]

The cell establishes a current through R (or X) irrespective if the balance point
have been found or not. All that the potentiometer does is it measures the potential
drop over the cell in this state, which is not e when it is balanced since the resistor
(R or X) is drawing current from it.

 

[Faris Shajahan]

Potential difference across balancing lengthfor case R Is ##V_1=\frac{V_0l_1}{L}##
And for X ##V_2=\frac{V_0l_2}{L}## where l1 and l2 are balancing lengths and V0 is the potential difference across entire length.
in the lower circuit, for R, ##i_1=E/R## and for X ##i_2=E/X## where E is the lower cell. So ##\frac{I_1}{I_2}=X/R##
V1/V2 = l1/l2
But ##\frac{V_1}{V_2}=\frac{I_1R}{I_2X}## if you substitute value of ratio of the currents which is equal to X/R, you get V1=V2.

Gosh, dude! I was fine but now you have confused me!

 

[AdityaDev]

The cell establishes a current through R (or X) irrespective if the balance point
have been found or not. All that the potentiometer does is it measures the potential
drop over the cell in this state, which is not e when it is balanced since the resistor
(R or X) is drawing current from it.

At balance point, the the current from upper cell will not pass through the lower cell because the potential difference across that part of wire Is same as that across the resistor.
also, since internal resistance is zero, V=iR=E from kirchhoff rule in lower loop.

 

[BvU]

At balance point, the the current from upper cell will not pass through the lower cell because the potential difference across that part of wire Is same as that across the resistor.
also, since internal resistance is zero, V=iR=E from kirchhoff rule in lower loop.

This last statement is in conflict with the text of post #1 : 58.3 cm vs 68.5 cm can't be with the same ##\epsilon##

 

[AdityaDev]

This last statement is in conflict with the text of post #1 : 58.3 cm vs 68.5 cm can't be with the same ##\epsilon##

Why should the balancing length change?

 

[andrevdh]

The cell establishes a current in the resistor.
This means that its terminal voltage would not
be e due to the voltage drop over its own internal resistor.
The potentiometer sees the terminal voltage over the cell ,
which is the same voltage as that over the R or X resistor.

 

[Raghav Gupta]

Why should the balancing length change?

The balancing length should change as we have a different potential difference whenever we connect different resistances.
Remember in potentiometer the secondary circuit draws no current from primary circuit when balance point is reached and so same current in each case.
But as we have different potential difference created because of resistance, we say
It's V₁ (not E₁) = iR for first case
and V₂ = iX for second case
I think you know the rest part of solving it.

 

[andrevdh]

The current is not the same in the two cases.

 

[Raghav Gupta]

The current is not the same in the two cases.

So you mean we have both different currents and different voltage at balance point in 2 cases?

 

[andrevdh]

Yes, due to the internal resistance of the cell the voltage drops
over the internal resistor will differ in the two cases and the currents
will differ due to the different resistances of R and X.

 

[Raghav Gupta]

The cell establishes a current in the resistor.
This means that its terminal voltage would not
be e due to the voltage drop over its own internal resistor.
The potentiometer sees the terminal voltage over the cell ,
which is the same voltage as that over the R or X resistor.
View attachment 79536

But Aditya is saying that current is same and voltage is different over the R or X resistor according to the answer he have seen.
He also have in mind that terminal voltage should be same.
So do you agree with him?

 

[andrevdh]

No, the voltage differs because the current differs,
and the current differs because the resistances R and X are not the same.

 

[Raghav Gupta]

No, the voltage differs because the current differs,
and the current differs because the resistances R and X are not the same.

So how that is true?

but in the book, it given as:
E2/E1 = iX/iR
but how is I same in this case when resistance is varied? They have given that I is the current in potentiometer wire... but how does the that current cause potential difference across R? It should the current E/R or E/X that passes through R and X.

 

[andrevdh]

E2/E1 = iX/iR

Maybe it should be V_R/V_X = length_R/length_X ?

 

[Raghav Gupta]

That's okay what next?

 

[Suraj M]

according to the answer he have seen.

NCERT answer? your going with that? It might be wrong !

 

[Raghav Gupta]

NCERT is right in most of cases.
I think Aditya you should refer the theory in NCERT for internal resistance of a cell by potentiometer.
There you will see E = I(r+R)
V=IR
Current is same as cell is producing one magnitude of current only.
Try applying that here.

 

[AdityaDev]

No, the voltage differs because the current and the current differs because the resistances R and X are not the same.

Voltage does not differ. If you put a voltmeter across Ror X, you will get the emf of cell.

 

[AdityaDev]

NCERT is right in most of cases.
I think Aditya you should refer the theory in NCERT for internal resistance of a cell by potentiometer.
There you will see E = I(r+R)
V=IR
Current is same as cell is producing one magnitude of current only.
Try applying that here.

I tried using internal resistance. But you will end up with too many unknowns.

 

[Raghav Gupta]

Voltage does not differ. If you put a voltmeter across Ror X, you will get the emf of cell.

You are wrong there.
We will not get emf but V= E-ir = iR

 

[AdityaDev]

You are wrong there.
We will not get emf but V= E-ir = iR

Neglect r.

 

[Raghav Gupta]

But we cannot neglect r, all cells have some r. Here in question r is not given doesn't mean you will neglect r,
that's the main difference between
Voltmeter and potentiometer.
When we connect voltmeter across a cell we get V= E-ir but by potentiometer E.

 

[AdityaDev]

But we cannot neglect r, all cells have some r. Here in question r is not given doesn't mean you will neglect r,
that's the main difference between
Voltmeter and potentiometer.
When we connect voltmeter across a cell we get V= E-ir but by potentiometer E.

For the sake of board exam, let's just learn the solutions.

 

[Raghav Gupta]

For the sake of board exam, let's just learn the solutions.

But it's better if you know the concepts as it makes learning easy.
Yeah, but where there is much difficulty like in this case ( I was also confused when I first saw your thread)
It's better to learn solutions.

 

[Faris Shajahan]

Wait a sec guys!
I asked my Physics Sir today and he gave me what I would call a sensible answer!

He said the circuit must be incomplete. This is because if we take the circuit shown in the figure, then as Aditya said, the balancing length would be the same (you see, the same emf is in parallel hence the voltage drop across the wire in both the cases must equal the emf)

But they say the answer must be different. Hence circuit diagram must be incomplete.
My sir says, there must be something such as a rheostat in series with the emf so as to keep the current constant in both the cases.

Right, don't you think?

 

[AdityaDev]

My sir says, there must be something such as a rheostat in series with the emf so as to keep the current constant in both the cases.

Right, don't you think?

The battery has internal resistance. They forgot to mention it

 

[Faris Shajahan]

The battery has internal resistance. They forgot to mention it

I don't think that would solve the problem. Because even then, the current in both would be different. The first current would be ##I_1=\frac{E}{R+r}## and the second, ##I_2=\frac{E}{X+r}##...

Hence there must be variable resistance whether or not there is internal resistance for the battery...the current must be kept constant.

 

[AdityaDev]

I don't think that would solve the problem. Because even then, the current in both would be different.

That is what we want. Now the potential difference across the resistors will be different.

 

[andrevdh]

Usually this experiment is done with the resistors connected in series to the cell, in which case the
current will be the same through both resistors.
For this particular setup I can only reduce it to two unknowns, the resistance X, and the internal resistance of the cell.

 

[Faris Shajahan]

That is what we want. Now the potential difference across the resistors will be different.

Yes, it will be different but still it would be impossible for us to calculate the value of ##X## as we don't have the value of ##r## given in the question.

Instead if they say there is a rheostat in series with ε in order to maintain constant current in both the cases, then we can calculate ##X## as
##X=R\frac{l_2}{l_1}##