Question

long uniform potentiometer wire AB is having a constant potential gradient

along its length.The null points for the two primary cells of emf’s E1 & E2 connected in

the manner shown are obtained at a distance of 120 cm & 300 cm from end A. Find

(i) E1/E2 (ii) position of null point for the cell E1. ​

Answer 1

Answer:

Let k be the potential gradient of the potentiometer ,

for null point =120cm ,

as the cells are opposite to each other and are in series ,

ε

1

−ε

2

=120k ..........eq1 ,

for null point =300cm ,

as the cells are supporting to each other and are in series ,

ε

1

+ε

2

=300k ..........eq2 ,

solving eq1 and eq2 ,

ε

1

=210k and ε

2

=90k ,

(i) therefore ε

1

/ε

2

=210k/90k=7/3 ,

(ii) position of null point for ε

1

=kl

1

=210×k=210cm.

The potentiometer is said to be sensitive when a small displacement of jockey from null point , produces a large deflection in galvanometer . For that potential gradient (potential drop per unit length of wire of poentiometer) should be small , as potential gradient is given by ,

k=V/l , where l is the length of wire of potentiometer and V is the potential drop across it ,

so larger the l, smaller the k . hence we should use a wire of large length to increase the sensitivity.

Explanation:

please follow me

Answer 2

Answer:

In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null point for the two primary cells of emfs. and. connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A.

Explanation:

long uniform potentiometer wire AB is having a constant potential gradient

along its length.The null points for the two primary cells of emf’s E1 & E2 connected in

the manner shown are obtained at a distance of 120 cm & 300 cm from end A. Find

(i) E1/E2 (ii) position of null point for the cell E1.