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Question
A multirange current meter can be constructed by using a galvanometer circuit as shown in figure. We want a current meter that can measure 10 mA, 100 mA and 1A using a galvanometer of resistance 10 Ω and that prduces maximum deflection for current of 1mA. Find S1, S2 and S3 that have to be used
Solution
A galvanometer can be converted into ammeter by connecting a very low resistance wire (shunt S) connected in parallel with galvanometer. The relationship is given by IgG = (I – Ig) S, where Ig is the range of galvanometer and G is the resistance of galvanometer.
For measuring `I_1 = 10 mA: I_G.G = (I_1 - I_G)(S_1 + S_2 + S_3)`
For measuring `I_2 = 100 mA: I_G(G + S_1) = (I_2 - I_G)(S_2 + S_3)`
For measuring `I_3 = 1 A: I_G(G + S_1 + S_2) = (I_3 - I_G)(S_3)`
Gives `S_1 = 1 Ω, S_2 = 0.1 Ω`
And `S_3 = 0.01 Ω`
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