A galvanometer with resistance 100 \Omega gives full scale deflection with a current of 10 \mathrm{~mA}. The value of shunt, in order to convert it into an ammeter of 10 ampere range, will be
A -10 \Omega
B 1 \Omega
C 0.1 \Omega
D 0.01 \Omega
A galvanometer with resistance 100 \Omega gives full scale deflection with a current of 10 \mathrm{~mA}. The value of shunt, in order to convert it into an ammeter of 10 ampere range, will be
A -10 \Omega
B 1 \Omega
C 0.1 \Omega
D 0.01 \Omega

Correct option is
C 0.1 \Omega

    \[\begin{array}{l} \mathrm{I}=\mathrm{I}_{\mathrm{g}}+\mathrm{I}_{\mathrm{s}} \\ \left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{R}_{\mathrm{s}}=\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}} \end{array}\]

Given,

    \[\begin{array}{l} \mathrm{I}=10 \mathrm{~A} \\ \mathrm{Ig}=10 \mathrm{~mA}=0.010 \mathrm{~A} \\ \mathrm{R}_{g}=100 \Omega \end{array}\]

Substituting the values,

    \[\begin{array}{l} (10-0.010) \mathrm{R}_{\mathrm{s}}=0.010 \times 100 \\ \Rightarrow \mathrm{R}_{\mathrm{s}}=\frac{1}{9.99}=0.1 \end{array}\]