A galvanometer has a current range o`f 15 mA `and a voltage range of` 0.75 V.` To convert this galvanometer into an ammeter of range `15 A`, the shunt resistance is

Options

A. `(100)/(999)Omega`
B. `(50)/(999)Omega`
C. `(200)/(999)Omega`
D. `(150)/(999)Omega`

Answer

B

Explanation

Maximum current=15mA, Maximum voltage=0.75 V `therefore`Resistance of the galvanometer`=(750x10^(-3))/(15x10^(-3))=50Omega I_g/I=S/(S+G) therefore(15 xx 10^(-3))/15=S/(S+50)therefore 1/(1000)=S/(S+50)therefore1000S=S+50 thereforeS=50/(999)Omega`