In a potentiometer arrangement for determining the emf of cell, the balance point of the cell in open circuit is 350 cm. When a resistance of `9 Omega` is used in the external circuit of cell , the balance point shifts to 300 cm. Determine the internal resistance of the cell.

Explanation

Given `l_(1) = 350`
`l_(2) = 300`
`R = 9Omega` ltbrlt The internal resistance can be be calculate as .
` r= (R(l_(1) - l_(2)))/(l_(2))`
` r= 9((350-300)/(300)) = (9xx50)/(300) = 1.5Omega`