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When a galvanometer of resistance G iws converted into an ammeter of range IA then the current passing through shunt (S) isA. `I_s=((G)/(G+S))I`B. `I_s=((S)/(S+G))I`C. `I_s=((SI=G)/(S))`D. `I_s=SxxG`

Answer» Correct Answer - A


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