With a certain cell the balance point is obtained at `70 cm` from the zero and of the potentiometer wire. With another cell whose emf differs from the `60 cm` mark .Calculate the emf of the two cells

Answer

`1.4 V,1.2 V`

Explanation

The emf of the two cell are `epsilon` will and `(epsilon - 0.2) ` volt
Then`(epsilon)/(epsilon - 0.2) = (30)/(60) = (1)/(6)`
`or 6 epsilon = 7 epsilon - 1.4 or epsilon = 1.4 V`
emf of other cell `= epsilon - 0.2 = 1.4 - 0.2 = 1.2 V`