Internal Resistance Affect the Balancing Length in a Potentiometer?

[tellmesomething]

Homework Statement

Theres a typical potentiometer setup I had of a dc current source connected in series with a resistor across a metre bridge with a wire mounted on it. I attached the ends of the metre bridge with the battery through connecting wires of very low resistance.
If end A is connected to positive terminal of the dc current source and the negative terminal is connected to the resistor Which is connected to B, and I connect another wire from A, I put a secondary cell with positive terminal towards the end A and a resistor of 600kohms in series with it and connected a galvanometer in series and then connect the galvanometers other terminal with a jockey.

Relevant Equations

None

I was asked if the balancing length at which the galvanometer shows no deflection be affected by the internal resistance of the dc source. I am not sure why my answers incorrect my reasoning was yes if I had put a higher resistance then the voltage drop across the metre bridge would decrease, meaning the gradient V/L would also decrease so the if the balancing point was being found at say 50cm it would be found at a length greater than that now...


Why is this wrong?

 

[gleem]

Please provide a diagram.

 

[tellmesomething]

Please provide a diagram.

 

[tellmesomething]

View attachment 351039

And technically aren't you measuring the voltage across the cell E instead of Emf since theres a resistance as Well @gleem

 

[cnh1995]

I was asked if the balancing length at which the galvanometer shows no deflection be affected by the internal resistance of the dc source

Are you sure the question mentions the internal resistance of the dc source and not of the secondary cell?
They both are dc sources after all.

 

[cnh1995]

And technically aren't you measuring the voltage across the cell E instead of Emf since theres a resistance as Well @gleem

That resistance will matter only if there is current through it. Is that the case at the null deflection point?

 

[tellmesomething]

That resistance will matter only if there is current through it. Is that the case at the null deflection point?

Fair enough. No. I got this one but previous question still remains. They did specifically mention dc source for this circuit and pointed at it during the practical exam.

 

[cnh1995]

my reasoning was yes if I had put a higher resistance then the voltage drop across the metre bridge would decrease, meaning the gradient V/L would also decrease so the if the balancing point was being found at say 50cm it would be found at a length greater than that now...

This looks correct.

 

[berkeman]

Sorry, that circuit drawing makes no sense to me. Why do you show the potentiometer wiper wiping across a straight wire segment?

 

[Tom.G]

Sorry, that circuit drawing makes no sense to me. Why do you show the potentiometer wiper wiping across a straight wire segment?

That is the way (very) old textbooks would draw, and occassionally show images of, "slidewire bridges." There would be a straight resistance wire stretched between two posts perhaps 2 or 3 feet apart.

IIRC, that configuration would often be used to find the approximate distance to a short circuit in telegraph lines or similiar.

Cheers,
Tom

 

[tellmesomething]

Sorry, that circuit drawing makes no sense to me. Why do you show the potentiometer wiper wiping across a straight wire segment?

I should have mentioned there's resistance throughout this wire.

 

[tellmesomething]

That is the way (very) old textbooks would draw, and occassionally show images of, "slidewire bridges." There would be a straight resistance wire stretched between two posts perhaps 2 or 3 feet apart.

IIRC, that configuration would often be used to find the approximate distance to a short circuit in telegraph lines or similiar.

Cheers,
Tom

I dont get you is this experiment not accurate?

 

[Tom.G]

I dont get you is this experiment not accurate?

Nothing wrong with the experiment that I can see.
I agree with @cnh1995 that your answer is correct.

My earlier post was aimed at @berkeman, trying to explain that the wire connected to the + end of "DC" (point A) was likely a high resistance wire.

Your HOMEWORK STATEMENT clearly states that source "DC" is a dc current source.

If it was a battery instead of a current source, then the value of the resistor on the Negative end would affect the operation. You would have to measure the voltage across the ends of the resistance wire instead of using the rated battery voltage.

Hope this helps!

Cheers,
Tom