The galvanometer shown in the diagram has resistance `100 Omega` and current required for full-scale deflection is `10 mA`. Find the resistances `R_1, R_2` and `R_3` required to convert it into ammeter having ranges as indicated.
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Explanation

`G = 100 Omega, I_g = 10 mA = 10 xx 10^-3 = 0.01 A`
For the range `0.1 A, G` is in parallel to `(R_1 + R_2 + R_3)`

`I_g G = (I - I_g) (R_1 + R_2 + R_3)`
`R_1 + R_2 + R_3 = (I_g G)/(I - I_g) = (0.01 xx 100)/(0.1 - 0.01) = (100)/(9)` ...(i)
For the range `1 A, (G + R_3)` is in parallel to `(R_1 + R_2)`

`I_g (G + R_3) = (I - I_g) (R_1 + R_2)`
`I_g (G + R_1 + R_2 + R_3) = I(R_1 + R_2)`
`R_1 + R_2 = (0.01(100 + (100)/(9)))/(1) = (10)/(9)` ...(ii)
For the range `10 A,(G + R_2 + R_3)` is in parallel to `R_1`

`I_g (G + R_2 + R_3) = (I - I_g) R_1`
`I_g (G + R_1 + R_2 + R_3) = IR_1`
`R_1 = (0.01 xx(100 +(100)/(9)))/(10) = (1)/(9) Omega`...(iii)
`R_2 = (1)/(9) Omega, R_2 = (10)/(9) -(1)/(9) = 1 Omega`
`R_3 = (100)/(9) - (10)/(9) = 10 Omega`.