When a galvanometer is shunted with a `4Omega` resistance the deflection is reduced to` 1//5`. If the galvanometer is further shunted with a `2Omega` wire the new deflection will be (assuming the main current remains the same)

Options

A. `5//13` of the deflection when shunted with `4 Omega` only
B. `8//13` of the deflection when shunted with `4 Omega` only
C. `3//4` of the deflection when shunted with `4Omega` only
D. `3//13` of the deflection when shunted with `4Omega` only

Answer

A

Explanation

When only `4 Omega` resistance is
shunted, `(i_(g))_(1) = i//5`
`Gxxi//5 = 4 xx (4//5) implies G = 16Omega`
`(4//3)/(16+(4//3))i=(1)/(13)`itherefore Hence, (A) is correct