# Measuring resistance using potentiometer

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1. Feb 23, 2015

1. The problem statement, all variables and given/known data

2. Relevant equations

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

3. The attempt at a solution

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.

2. Feb 23, 2015

### 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!!!

3. Feb 23, 2015

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.

4. Feb 23, 2015

### 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.

5. Feb 23, 2015

### Faris Shajahan

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

6. Feb 23, 2015

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 kirchoff rule in lower loop.

7. Feb 23, 2015

### BvU

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 $\epsilon$

8. Feb 23, 2015

Why should the balancing length change?

9. Feb 24, 2015

### 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.

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10. Feb 24, 2015

### Raghav Gupta

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.

11. Feb 24, 2015

### andrevdh

The current is not the same in the two cases.

12. Feb 24, 2015

### Raghav Gupta

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

13. Feb 24, 2015

### 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.

14. Feb 24, 2015

### Raghav Gupta

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?

15. Feb 24, 2015

### andrevdh

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

16. Feb 24, 2015

### Raghav Gupta

So how that is true?

17. Feb 24, 2015

### andrevdh

Maybe it should be VR/VX = lengthR/lengthX ?

18. Feb 24, 2015

### Raghav Gupta

That's okay what next?

19. Feb 24, 2015

### Suraj M

20. Feb 24, 2015

### 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.