Comparison of EMFs of Two Cells

The cell of e.m.f. E₁ is connected in the circuit through terminals 1 and 3 of key K₁. The balance point is obtained by moving the jockey on the potentiometer wire. E.m.f of cell E should be greater than the emfs of E₁ and E₂ separately (otherwise, balance point will not be obtained).

Let the balance point on potentiometer be at point Y₁ and length AY₁ = L₁.

The cell of e.m.f. E₂ is connected in the circuit through terminals 2 and 3 of the key K₂. Suppose balance is obtained at point Y₂ and length AY₂ = L₂.

Applying potentiometer principle,

E₁ = kL₁ and E₂ = kL₂

where k is the potential gradient along the wire AB. Hence,

E₁/E₂ = L₁/L₂