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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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