When cell of e.m.f. ' E₁ ^ is connected to potentiometer wire, the balancing length is ' ₁ ^ . Another cell of e.m.f. ' E₂ ^ (E₁>E₂ ) is connected so that two cells oppose each other, the balancing length is ' ₂ ^ . The ratio E ₁: E ₂ is

Step-by-step solution

A potentiometer's potential drop across any wire length is proportional to that length when current is constant and cross-section uniform. Let k represent the potential gradient. For cell E₁ with balancing length ₁ , the e.m.f. equals the potential drop: E₁ = k ₁ . When E₂ opposes E₁ , the effective e.m.f. becomes E₁ - E₂ and balances at length ₂ : E₁ - E₂ = k ₂ . Dividing these equations eliminates k : E₁ E₁ - E₂ = k ₁ k ₂ = ₁ ₂ Cross-multiplying yields E₁ ₂ = (E₁ - E₂) ₁ , which rearranges to E₂ ₁ = E₁ ( ₁ - ₂) .