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. `5/9` of the deflection when shunted with `4Omega` only
B. `5/13` of the deflection when shunted with `4Omega` only
C. `1/13` of the original deflection only
D. `1/9` of the original deflection only

Answer

B, C

Explanation

`I//5=(I(4))/(G+4)`
`G+4=20`
`G=16Omega`
`(I=(4//3))/(16+4/3)=I/13`