The diagram below shows a potentiometer set up. On touching the jockey near to the end X of the potentiometer wire, the galvanometer pointer deflects to left. On touching the jockey near to end Y of the potentiometer, the galvanometer pointer again deflects to left but now by a larger amount. Identify the fault in the circuit and explain, using appropriate equations or otherwise, how it leads to such a one-sided deflection.
OR
Following circuit was set up in a meter bridge experiment to determine the value X of an unknown resistance.
(a)Write the formula to be used for finding X from the observations.
(b) If the resistance R is increased, what will happen to balancing length?
As the potentiometer wire has uniform cross section throughout, the potential difference between X and any point at a distance l from A till point Y is,
V (l)= k l
where k is the potential drop per unit length, As the positive terminal of E1 is connected to X and on touching the jockey near to the end X of the potentiometer wire, the galvanometer pointer deflects to left and on touching the jockey near to end Y of the potentiometer, the galvanometer pointer again deflects to left but now by a larger amount, this indicates that the cell E1 is of larger emf than E which indicates that the galvanometer will never show zero deflection. For a galvanometer to show zero deflection somewhere on XY the emf of E > E1
OR
a) Formula: -
The meter bridge uses the concept of a balanced Wheatstone bridge to find the value of the unknown resistor so that the formula can be written as,
Where,
X is the unknown resistance,
R is the resistance of the resistance box,
LAB is the length of wire where till the null point is found.
It is written as to fulfill the condition for the balanced bridge.
b) The above formula can be re-written as
The above expression indicates that if the value of R is increased then the value of lAB will increase.
Conclusions: -
The value of x can be found as,
The value of R increases then the value of lAB also increases.