Question

A galvanometer with resistance $100\mathrm{\Omega }$ 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 ___________.

Answer 1

Step 1. Given data

Total current passing through circuit is $10\mathrm{A}$

Total current passing through the galvanometer which corresponds to full-scale reading is $10\mathrm{mA}=0.010\mathrm{A}$

Resistance of the galvanometer is $100\mathrm{\Omega }$

Step 2. Formula to be used.

As we know that,

$\mathrm{I}={\mathrm{I}}_{\mathrm{g}}+{\mathrm{I}}_{\mathrm{s}}$

$\left(\mathrm{I}-{\mathrm{I}}_{\mathrm{g}}\right){\mathrm{R}}_{\mathrm{s}}={\mathrm{I}}_{\mathrm{g}}{\mathrm{R}}_{\mathrm{g}}$

Substitute the given values in the above equation.

So,

$\left(10-0.010\right){\mathrm{R}}_{\mathrm{s}}=0.010\times 100$

${\mathrm{R}}_{\mathrm{s}}=\frac{1}{9.99}$

${\mathrm{R}}_{\mathrm{s}}=0.1\mathrm{\Omega }$

Hence, A galvanometer with resistance $100\mathrm{\Omega }$ 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 $0.1\mathrm{\Omega }$.