A galvanometer whose resistance is `120 Omega` gives full scale deflection with a curretn of `0.5 A` so that it can read a maximum current of `10 A`. A shunt resistance is added in parallel with it. The resistance of the ammeter so formed is

Options

A. `0.3 Omega`
B. `0.6 Omega`
C. `0.8 Omega`
D. `1.2 Omega`

Answer

B

Explanation

`i_g/I=S/(S+G)therefore(50 xx 10^(-3))/10=S/(S+120)therefore 1/200=S/(S+120)therefore 200S=S+120 therefore S=120/199Omegatherefore` The combined resistance of the shunted galvanometer is`1/R=1/G+1/S``therefore=(GS)/(G+S)=((120xx120)/199)/(120+(120)/(199))=((120 xx 120)/(199))/((120x200)/199)=(120xx120)/(120 xx 200)=0.6Omega` Note: Novel method: We can solve this without calculating the shunt resistance. Effective resistance of the ammeter `(R')=(GS)/((G+S)) and I_g/I=(s/(S+G))xx 1/G=(R')/G therefore R'=(Gi_g)/I=120xx(50 xx 10^(-3))/10=120 xx 1/(200)=0.6 Omega`.