To determine the internal resistance of a cell by using a potentiometer, the null point is at 1 ~m when the cell is shunted by 3 resistance and at a length 1.5 ~m when cell is shunted by 6 resistance. The internal resistance of the cell is.

Step-by-step solution

Consider the following diagram: The potential drop across the external shunt resistance R is equal to the potential drop across the battery of emf E and internal resistance r . aligned & V=E-i r=i R & i= E (R+r) & V=i R= E R (R+r) aligned V L , i.e., the balance length is directly proportional to the potential drop. Therefore, E R (R+r) L In case of internal resistance measurement by potentiometer, considering two balance lengths (L₁, L₂ ) corresponding to shunt resistors (R₁, R₂ ) respectively, V₁ V₂ = L₁ L₂ = E R