Question

two primary cells of emf's E1 and E2 are connected to the potentiometer wire AB.If the balancing length for the two combinations of the cells are 250 cm and 400 cm .Find the ratio of E1 and E2

Answer 1

E1/E2 =( 250+400)/(400-250) =650÷150= 13/3
ratio is 13:3

Answer 2

Hey dear,

¶ Potentiometer-
It's device used -
- To compare emf of two cells
- To find internal resistance of cell
- To measure potential difference across a resistor.

◆ Answer-
E1:E2 = 13:3

◆ Explaination-
# Given-
l1 = 250 cm
l2 = 400 cm

# Solution-
According to potentiometer law,
E1 / E2 = (l2+l1) / (l2-l1)
E1 / E2 = (400+250) / (400-250)
E1 / E2 = 650 / 150
E1 / E2 = 13 / 3

Therefore, ratio of E1:E2 = 13:3 .

Hope this helps...