Question

The necessary current for full scale deflection is $4mA$ for any galvanometer of resistance $99\Omega$. What would you do to convert it to an ammeter of range $0$ to $6A$?

Answer 1

Given $G=99\Omega$
Current at full deflection $=4mA$
Range of Ammeter $I=6A$
To connect a galvanometer into an ammeter, a low resistance is to added in parallel
$S=\frac{I_{g}G}{I-I_g}=\frac{4\times {10}^{-3}\times 99}{6-(4\times {10}^{-3})}$
$S=\frac{4\times 99\times {10}^{-3}}{(6000-4)\times {10}^{-3}}=\frac{396}{5996}=6.6\times {10}^{-2}\Omega$