Comparison of emf’s of Two cells using B.G.

Let EL of ED be the emfs of two given cells (Laclanche cell and Daniel cell).These Cells are connected to the end terminals of a DPDT key. Two resistance boxes P and Q are connected in series with the central terminals of the DPDT key. The p.d. across P is applies to the capacitor through the vibrator and charging terminal. The charge on the capacitor can be discharged through the B.G.

Put P = 1000Ω and Q = 9000Ω so that P + Q = 10000Ω. The cell of emf EL is included in the circuit. Capacitor C is charged and then discharged through the B.G. The first throw is found. After reversing commutator, the throw is found again. Let the mean be θ1.

P.d. across P = ___E_L P___
        __________(P + Q)
Charge on the capacitor C = q1 = ____E_LP____ X C
             ______________________    (P + Q)

. :            __ E_LP_ C ∞ θ₁
        __     (P + Q)

Now the handle of the DPDT key is thrown to the side of the Daniel cell of emf E_D. The experiment is repeated for the same values of P and Q.
The mean throw θ₂ is found. Then,

        __ ELP_ C ∞ θ₂
          (P + Q)

Dividing Eq. (1) by Eq.(2), EL / ED = θ₁ / θ₂
The experiment is repeated for different values of P, keeping (P + Q) = 10000Ω always.