Question

A leclanche cell is balanced on length 280 cm of potentiometer wire. When daniel cell is joined in series with the cell the balance point is obtained at 480 cm distance. Calculate emf of two cells

Answer 1

Thus the emf of two cells is E1 : E2 = 7 : 5

Explanation:

Given data:

Length of potentiometer wire = 280 cm

• E  1  = ρl  1
• E  1  + E  2   = ρl  2

​Solution:

E  1 / E  2 =  l 2  / l  1

E  1 +E  2 / E  1    =  ρl  2  / ρl  1

E  1 +E  2 / E  1  ​=​ l  2 /  l  1

1 + E 2 / E1 = = l 2 / l 1

E2 / E1 = l2 / l1 - 1 = l2 - l1 / l1

E1 / E2 = l1 / l2 - l1 = 280 / 480 - 280

E1 / E2 = 280 / 200

E1 : E2 = 7 : 5

Thus the emf of two cells is E1 : E2 = 7 : 5

​  

 

​