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The resistance of galvanometer is 999Ome...

The resistance of galvanometer is `999Omega`. A shunt of `1 Omega` is connected to it.If the main current current is `10^(-2)A`, what is the current flowing through the galvanometer .

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`G = 999 Omega , S = 1 Omega, i=10^(-2) A, i_g = ? `
`i_g = i( (S)/(G +S) ) = 10^(-2) xx ( (1)/(999 + 1) ) = 10^(-5) A`
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