A galvanometer shows a reading of `0.65` mA. When a galvanometer is shunted with a `4 Omega` resistance, the deflection is reduced to `0.13` mA if the galvanometer is further shunted with a `2 Omega` wire, the new reading will be (the main current remains the same)

Answer

0.05

Explanation

`I^(') = 4/(4 + G) I`
`implies 4/(4 + G) = 1/5` (given) where `G = 16 Omega`
`I^(") = (4//3)/(4/3 + 16) I`

`I^(") = 1/13 I`
or `I^(") = 1/31(51^') = 5/13 I^(')`

`i/5 xx G = 4 xx 4/5 i implies G = 16 W`.
In secod case `(i - i^(')) xx 16 = 4/3 i^(')`
`R_(eq) = (2 xx 4)/(2 + 4) = 4/3`
`4i - 4i^(') = (i^')/3 implies 4i = (13 i)/(3)`
`implies i^(') = 12/13 i`
`implies i = i^(') = 1/13 i = 1/13 xx 0.65 mA = 0.05 mA`.