Explain comparison of emf of two cell by using potentiometer with necessary diagiam.

Explanation

As shown in figure a battery having emf E and internal resistance r variable resistance R and switch `K_(1)` is connected between two end A and B of potentiometer.
`rArr` To compare emf of two cell `epsilon_(1) and epsilon_(2) `, positive terminal of both cell is connected with A and negative terminal of cell is connected with point 1 and 2 of three way switch with terminal 3 of three way switch galvanometer and jockey key is connected in series. Jockey key can be moved on wire.
`rArr` First point 1 and 3 of three way switch are connected hence `epsilon_(1) ` cell is in the cell By sliding jockey key on wire obtained point `N_(1)` such that galvanometer shows zero deflection.
Let `AN_(1) = l_(1)`
By using Krichhoff.s second law for loop,
`AN_(1) G31A`
`phi l_(1) + 0 - epsilon_(1) = 0`
`therefore phi l_(1) = epsilon_(1) " " ` .... (1)
`rArr` Now connect point 2 and 3 and obtain point `N_(1)` such that galvanometer show zero deflection,
`AN_(2) = l_(2)`
For `AN_(2) G 32 A `loop,
`phi l_(1) + 0 - epsilon_(2) = 0`
`therefore phi l_(2) = epsilon " " `..... (2)
By taking ratio of (1) and (2),
` (epsilon_(1))/(epsilon_(2)) = (l_(1))/(l_(2)) `....(2)
when emf of one of the cell is given then from value of `l_(1) and l_(2)` by using equation (3) value of unknown emf can be obtained.
`rArr` By using potentiometer potential difference of very small value can be measured.