Question

Two cells of emf’s E₁ and E₂ (E₁ > E₂ ) are connected in a potentiometer circuit so as to assist each other. The null point is obtained at 8.125 m from the high potential end of the potentiometer wire. When the cell with emf E₂ is connected so as to oppose the emf E₁ , the null point is obtained at 1.25 m from the same end. Compare the emf’s of the two cells.

Answer 1

Data : l₁ = 8.125 m (cells assisting), 

l₂ = 1.25 m (cells opposing)

E₁ + E₂ = Kl₁ and E – E₂ = Kl₂

where K is the potential gradient.

\(\therefore\) \(\cfrac{E_1+E_2}{E_1-E_2}\) = \(\cfrac{I_1}{I_2}\)

Hence, the ratio of the emf's of the two cells,