Comparing emf of cell (Sum and Difference Method)

 To compare Emf of a cell (Sum and Difference Method)

Construction: 

Consider two cell of Emf  E₁ and E₂   respectively whose emf is to be compared. Let E be the Emf of a battery Both cell is connected to Galvanometer via key and galvanometer is connected to Jockey on another side as shown by green colour which is tapped on Potentiometer wire. Let battery of Emf E is connected to Point A and B which is length of the potentiometer wire . 

Working:

Case I : When Cell E₁  and  E₂ are connected to assist

Sum Method

Battery of Emf  E and Cell of Emf  E₁ and E₂ is introduced in the circuit. Cell of Emf  E₁ and E₂ is connected such that so as to assist each other. Now jockey is tapped on potentiometer wire to obtain a null point so that galvanometer shows zero deflection. Let point Q be the point where Null point is obtained, such that  Potential between point A and Q becomes equal to  E₁ + E_{2 .} Let point Q obtained is at a distance of  L₁ distance from point A.

From Potential Gradient Equation,

V_AQ  = σ L₁

E₁ + E₂ = σ L₁ -----(A)

Case II : When Cell E₁  and  E₂ are connected to oppose

Difference Method

Battery of Emf  E and Cell of Emf  E₁ and E₂ is introduced in the circuit. Cell of Emf  E₁ and E₂ is connected such that so as to oppose each other. Now jockey is tapped on potentiometer wire to obtain a null point so that galvanometer shows zero deflection. Let point P be the point where Null point is obtained, such that  Potential between point A and P becomes equal to  E₁ - E_{2 .} Let point Q obtained is at a distance of  L₂ distance from point A.

From Potential Gradient Equation,

V_AP  = σ L₂

E₁ - E₂ = σ L₂ -----(B)

From (A) and (B)

E₁ + E₂  /  E₁ - E₂ = σ L₁  /  σ L₂ 

By, componendo dividendo

E₁ + E₂  /  E₁ - E₂ = σ L₁  /  σ L₂ 

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