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Uses of Ballistics Galvanometer

Absolute Capacitance of a capacitor

(i)    Two Resistance boxes P and Q are connected in series with an accumulator of emf E. A small resistance (=0.1 Ω) is taken in P and a large resistance box R are connected across P. With no resistance in R, the steady deflection d of the galvanometer in found. A suitable resistance is taken in R till the deflection becomes half. The resistance in R is the galvanometer resistance Rg. The experiment is repeated for various values of P keeping P + Q constant.

(ii)    The galvanometer coil is set oscillating freely in open circuit. The time for 10 oscillations is found and the period T is calculated.

(iii)    Resistances P1 (1000 Ω) and Q1 (9000 Ω) are included in the boxes P and Q respectively. Potential differences across P1 = V = EP1 /P1 + Q1
The drop of potential across P1 is used to charge the capacitor, by connecting the terminals Ch and V of the charge-discharge key.

Comparison of two Capacitances using B.G.

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

The capacitor C1 is charged on C1 is then discharged through the B.G. The throws in the B.G. are noted before and after reversing the commutator. The mean throw θ1 is found out.

With the same resistance in P and Q, the handle of the DPDT key is thrown on the side of C2. C2 is charge 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 + 1/2λ)
In the second case, q2 = C2 X V = K θ2 (1 + 1/2λ)
. : C1 /C2 = θ1 / θ2

The experiment is repaired for different values of P keeping (P + Q) constants.

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

Let EL and ED be the emfs of two given cells. These cells are connected to the end terminals of a DPDT key.  The p.d. across P is applied to the capacitor through the vibrator and charging terminal. The charge on the capacitor can be discharged through the B.G.

Put P = 1000 Ω and Q = 9000 Ω so that P + Q = 10000 Ω. 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 = ELP / (P + Q)
Now the handle of the DPDT key is thrown of the side of the Daniel cell of emf ED. The experiment is repeated for the same values of P and Q. The mean throw θ2 is found.

The experiment is repeated for different values of P, Keeping (P + Q) = 10000 Ω always.

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