A multirange current meter can be constructed by using a galvanom… - Vidyadip

Question

A multirange current meter can be constructed by using a galvanometer circuit as shown in Fig. We want a current meter that can measure 10mA, 100mA and 1A using a galvanometer of resistance 10Ω and that prduces maximum deflection for current of 1mA. Find $S_1, S_2$ and $S_3$ that have to be used.

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Answer

Key concept: 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 $I_gG = (l - I_g) S$, where $I_g$ is the range of galvanometer, G is the resistance of galvanometer.

For measuring $I_1 = 10mA : IG.G = (I_1 - IG)(S_1 + S_2 + S_3)$
For measuring $I_2 = 100mA : IG(G + S_1) = (I_2 - IG)(S_2 - S_3)$
For measuring $I_3 = 1A : IG(G + S_1 + S_2) =(I_3 - IG)(S_3)$
Gives $S_1 = 1Ω, S_2 = 0.1Ω $ and $S_3 = 0.01Ω.$

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