Potentiometer Circuit 1: Balancing the Current - Solving for Ix and Rx

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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 .

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Homework Equations

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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conscience said:
E0−(I−Ix)R=0E0−(I−Ix)R=0E_0 - (I-I_x)R = 0
That's not correct.
Current through R is not I-Ix throughout.
What happens at point d?
 
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cnh1995 said:
That's not correct.
Current through R is not I-Ix throughout.
What happens at point d?

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 ?
 
conscience said:
E0−(I−Ix)Rx−I(R−Rx)=0E0−(I−Ix)Rx−I(R−Rx)=0E_0 - (I-I_x)R_x - I(R-R_x) = 0
Yes.
 
$$I_x = \frac{E_0R_x -E_x R}{RR_x -R_x^2+Rr}$$

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

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 .
Looks good to me.
 
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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 ?
 
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## .
 
conscience said:
I think resistance rrr has to be included
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.
 
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conscience said:
Applying KVL in right loop clockwise ,

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

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 ?
 
conscience said:
In other words , Ix=0Ix=0I_x =0 , as and when Ex=IRxEx=IRxE_x=IR_x condition is achieved .
Yes.
 
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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 .
As and 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 .

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 :smile:
 
conscience said:
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
It is correct. In fact, I said something much similar in this recent thread.
https://www.physicsforums.com/threads/find-the-value-of-resistance-in-the-circuit.921298/
See #22 and #29.
 
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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 ?
 
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@gneill , do you believe a constant current flowing in the potentiometer wires is a necessity ? Is a rheostat really required in the primary circuit ?
 
conscience said:
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)
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.
 
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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 .
 
conscience said:
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 .
Yes, that's how it is performed in laboratory.
But the rheostat's function is still the same.
cnh1995 said:
I believe it is used to adjust the potential gradient along the potentiometer wire.
 
cnh1995 said:
I believe it is used to adjust the potential gradient along the potentiometer wire.

But potential gradient is adjusted by altering current in primary circuit .
 
conscience said:
But potential gradient is adjusted by altering current in primary circuit .
...which is done by altering the rheostat position, isn't it?
 
Right .

Thanks !