Comparison of two Capacitance using B.G.

Let C₁ and C₂ be the capacitance of the two given capacitors. These capacitors are connected to the end terminals of the DPDT key. A resistance of 1000Ω is introduced in P and 9000Ω in Q.

The capacitor C₁ is charged to the p.d. across P. The charge on C₁ is then discharged through the B.G. The throws in the B.G. are noted before. The mean throw θ₁ is found out.

With the same resistances in P and Q, the handle of the DPDT key is thrown on the side of C₂. C₂ is charged to the same potential across P.

The charge on C₂ is then discharged through the B.G. The mean throw θ2 is found out.

Let V be the p.d. across the terminals of P. Then,
In the first case,         q₁ = C₁ X V = K θ₁ (1 + ½ λ)

In he second case    q₂ = C₂ X V = K θ₂ (1 + ½ λ)

. :            C₁ / C₂ = θ₁ / θ₂

The experiment is repeated for different values of P keeping (P + Q) constant.



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