Does internal resistance affect balance point?

[jonny23]

Homework Statement

if a potentiometer has driver cell and wire AB ,then an unknown emf (emf<driver cell) is balanced at some length L. does the balance point change if internal resistance of driver cell is changed?

Homework Equations

V= kl K= potential gradient

The Attempt at a Solution

if we increase the internal resistance ,the potential drop across wire AB should decrease and so balance point must shift.. But the solution says it will not change.

 

[ehild]

Homework Statement

if a potentiometer has driver cell and wire AB ,then an unknown emf (emf<driver cell) is balanced at some length L. does the balance point change if internal resistance of driver cell is changed?

Homework Equations

V= kl K= potential gradient

The Attempt at a Solution

if we increase the internal resistance ,the potential drop across wire AB should decrease and so balance point must shift.. But the solution says it will not change.

You are right, the balance point will shift if you change the internal resistance of the driver cell. It will not depend on the internal resistance of the unknown cell
.

 

[jonny23]

the balance point will shift if you change the internal resistance of the driver cell.

But in solution( i have checked in 4 books) says it will not change since we have set constant potential gradient across the wire.. and i don't even understand it

 

[ehild]

It is wrong. The potential gradient across the wire is E/(r_i+R_wire), where r_i is the internal resistance of the driving source. Book solutions are wrong quite often. They certainly thought of the internal resistance of the measured cell.

 

[nipun_59]

Well the solution to it is simple i guess because potentiometer is null method device i.e it does not draw any current from the cell and thus there is no potential drop due to internal resistance of the cell

 

[ehild]

Well the solution to it is simple i guess because potentiometer is null method device i.e it does not draw any current from the cell and thus there is no potential drop due to internal resistance of the cell

Which cell do you talk about?

 

[rude man]

Well the solution to it is simple i guess because potentiometer is null method device i.e it does not draw any current from the cell and thus there is no potential drop due to internal resistance of the cell

Sounds confusing.
Disconnect the unknown emf for a minute. Then the voltage across L is variable by varying the driver cell resistance, obviously. So adjust the driver cell to equal the unknown emf. When you reconnect the unknown emf there will be zero G current since the two voltages are the same. Disconnect the unknown emf again and readjust the driver cell resistance, then reconnect the unknown emf. Is the G current still zero?

Think of connecting two batteries of exactly equal voltage, + to +, - to -. No current, right? But take two batteries of dissimilar voltages and what happens?