The given figure represents a potentiometer with balance point at X.

What will happen to balance point if :

a) R is increased?

b) S is increased?

c) P is replaced by a cell with lower emf than Q?

Please give the entire explanation, especially for a) and b).

Thanks!:-)

Rates assured!

A-

if R is increased then the value of current flowing in potential metre shouid decrease which will result in lower value of potential gradient(drop of potential per unit length)

Hence to re-establish balance point u wouid require to shift the point X to right(i.e toward"s point B)

i m giving here some more explanation

initially when the resistance was R then balance point was at X suppose AX=L since potential metre is balanced at this position which mean potential difference b\w A and X is equal to EMF of cell Q(EMF coz no current is flowing through Q hence no potential drop when potential metre is balanced)

Next suppose u increased the resistance which will result decrease in value of current in potential metre so for producing same potential drop=EMF of cell Q u hav to shift the balance point toward"s B..

B-



As no current flows through galvanometre, cell Q , or resistance S when potential metre is at balanced position.



Hence there will be no effect of decreasing\increasing value of S..

if P is replaced by a cell with lower emf than Q then u can not achieve balance point.

now u can think y not............

Is it because the arm has very low pd across it, but the jockey is connected to high pd and so they can't balance each other?

Thanks a lot for the answers- the first one is now clear. But the answer to b) says that it will shift to the left.

Is the book answer wrong?

 if we increase the resistance S then the potential difference across point A and point X will decrease and so we have to shift the jockey towards left to make the potential difference of the potentiometer cell and the cell which we have attached thereby getting a null deflection point