# Potentiometer circuit

1. Jul 31, 2017

### conscience

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

1) What is the relationship between Ix
and Rx ?
2) Prove that on varying resistance Rx , Ix reduces to zero .

2. Relevant equations

3. The attempt at a solution

Assume resistance of Galvanometer is $r$ .
Total resistance in middle branch is $R$ , Variable resistance is $R_x$

Applying KVL in right loop clockwise ,

$-E_x - I_xr + (I-I_x)R_x =0$

Applying KVL in the left loop clockwise ,

$E_0 - (I-I_x)R = 0$

Solving the above two equations ,

$$I_x = E_0 \frac{R_x}{rR} - \frac{E_x}{r}$$

Now when $R_x$ decreases , the first term decreases and consequently $I_x$ decreases .Finally at a particular value of $R_x$ , $I_x$ becomes zero .The current in the Galvanometer is zero and the circuit is said to be balanced .

Is this correct ?

Thanks

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2. Jul 31, 2017

### cnh1995

That's not correct.
Current through R is not I-Ix throughout.
What happens at point d?

3. Jul 31, 2017

### conscience

Sorry .

The total resistance R in the middle branch can be broken up in two parts such that $R_x + R' = R$ .

Applying KVL in right loop clockwise ,

$-E_x - I_xr + (I-I_x)R_x =0$

Applying KVL in the left loop clockwise ,

$E_0 - (I-I_x)R_x - I(R-R_x) = 0$

Is it OK now ?

4. Jul 31, 2017

### cnh1995

Yes.

5. Jul 31, 2017

### conscience

$$I_x = \frac{E_0R_x -E_x R}{RR_x -R_x^2+Rr}$$

Denominator will always be positive since $R >R_x$ .

As $R_x$ decreases ,first term in numerator decreases .As a consequence $I_x$ decreases . Finally for a particular value of $R_x$ , $I_x$ becomes 0 .

6. Jul 31, 2017

### cnh1995

Looks good to me.

7. Jul 31, 2017

### conscience

Thank you .

Just to keep things simple , I think there was no need to include resistance $r$ .The resistance of Galvanometer could have been neglected .

The final expression will be $$I_x = \frac{E_0R_x -E_x R}{RR_x -R_x^2}$$

Is there anything you would like to add to the analysis I have done ?

8. Jul 31, 2017

### conscience

No . I think resistance $r$ has to be included . If resistances of both battery $E_x$ and Galvanometer are neglected then potential difference across points 'ad' would be equal to emf $E_x$ at all times . Including $r$ would mean that only when $I_x = 0$ the potential difference across 'ad' would equal emf $E_x$ .

9. Jul 31, 2017

### cnh1995

As far as the potentiometer experiment is concerned, you need not use r as the null point does not depend on r.
But then, Rx=R becomes an invalid condition (two ideal unequal voltage sources in parallel).
So, I believe you should include r to make the circuit practical and free from contradictions such as infinite current and unequal voltage sources in parallel.

10. Jul 31, 2017

### conscience

Thank you

11. Aug 2, 2017

### conscience

From this equation I can also infer that just when the product of current from battery $E_0$ times resistance $R_x$ becomes equal to EMF $E_x$ , current $I_x$ becomes zero and null point is achieved .

In other words , $I_x =0$ , as and when $E_x=IR_x$ condition is achieved .

Can we put it this way ?

12. Aug 2, 2017

### cnh1995

Yes.

13. Aug 2, 2017

### conscience

I hope you are not reasoning it backwards in the sense , when $I_x =0$ , then $E_x=IR_x$ .

It's the other way round .
It may look like I am over thinking this , but actually I need to explain potentiometer to someone . So I am just trying to simplify the analysis as much as I can .Since you are an expert in circuit , I just want to make sure my thinking is correct

14. Aug 2, 2017

### cnh1995

15. Aug 3, 2017

### conscience

Thanks .

One last clarification .

In the figure given in OP , as we move the jockey along potentiometer wire , current $I$ in the primary circuit changes .

But in the textbook I have to refer states that , there is also a rheostat in the primary circuit maintaining a constant current flow in the potentiometer wire all the time (while jockey moves on the wire)

Is this the way ?

Last edited: Aug 3, 2017
16. Aug 3, 2017

### conscience

@gneill , do you believe a constant current flowing in the potentiometer wires is a necessity ? Is a rheostat really required in the primary circuit ?

17. Aug 3, 2017

### cnh1995

I don't see how a rheostat can maintain a constant current in the circuit. Constant current is not a requirement here. I believe it is used to adjust the potential gradient along the potentiometer wire.

18. Aug 3, 2017

### conscience

I think the rheostat comes in picture when we are required to take more than one reading .First time any suitable value of rheostat would work .While taking second reading , it's value could change resulting in different value of primary current .Potential gradient would change . If we are comparing EMF's , balance lenths would also change , but the ratio of EMF's would very nearly be same as the previous result . Then we could find the mean value of the results .

19. Aug 3, 2017

### cnh1995

Yes, that's how it is performed in laboratory.
But the rheostat's function is still the same.

20. Aug 3, 2017

### conscience

But potential gradient is adjusted by altering current in primary circuit .