Two cells of e.m.f.s E₁ and E₂ (E₁>E₂ ) are connected as shown in figure. When the potentiometer is connected between A and B , the balancing length of the potentiometer wire is 3.60 m . On connecting the potentiometer between A and C , the balancing length is 0.90 m . The ratio E₁ / E₂ is

Step-by-step solution

Let k be the potential gradient of the potentiometer wire. According to the potentiometer principle, the potential difference across a wire length is proportional to that length: V = kL . When connected between points A and B, the balanced length L₁ = 3.60 m measures the e.m.f. of cell E₁ , giving: E₁ = k 3.60 When connected between A and C, the configuration measures the net e.m.f. of E₁ and E₂ connected in opposition. Given E₁ > E₂ , the effective e.m.f. is E₁ - E₂ , balanced at length L₂ = 0.90 m , so: E₁ - E₂ =