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Question:

A galvanometer has a current sensitivity of 1mA per division. A variable shunt is connected across the galvanometer and the combination is put in series with a resistance of 500Ω and a cell of internal resistance 1Ω. It gives a deflection of 5 divisions for a shunt of 5Ω and 20 divisions for a shunt of 25Ω. The emf of the cell is?

47.1V

77.1V

67.1V

57.1V

Solution:

RgIg = S(I − Ig) ⇒ I = Ig(Rg + S)/S
Case 1: S = 5 Ω, Ig = 5 mA ⇒ I1 = (Rg + 5)/5 (i)
Case 2: S = 25 Ω, Ig = 20 mA ⇒ I2 = 0.8(Rg + 25) = 0.8Rg + 20 (ii)
E = 501I + IgRg (iii)
Substituting (i) in (iii), E = 2505 + 506Rg (iv)
Substituting (ii) in (iii), E = 10020 + 420.8Rg (v)
Solving (iv) & (v), E = 47136 mV
E ≈ 47.1 V