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)

Options

A. `0.60mA`
B. `0.08 mA`
C. `0.12 mA`
D. `0.05 mA`

Answer

D

Explanation

`I=4/(4+G) IRightarrow 4/(4+G)=1/5(given)where G=16 Omega`
`I''=(4//3)/(4/3+16) I`
`I''=1/13I`
or `I''=1/13(51')=5/13I'`
`i/5xxG=4xx4/5iRightarrow G = 16W`
In second case `(i-i')xx16=4/3i' R_(eq)=(2xx4)/(2+4)=4/3`
`4i-4i' = i/3 Rightarrow 4i= (13i')/3 Rightarrow i=12/13i`
`Rightarrow i-i'=1/13i=1/13xx0.65mA=0.05mA`