Question

A galvanometer has a resistance of $16\Omega$.It shows full scale deflection, when a current of $20mA$ is passed through it. The only shunt resistance available is $0.06\Omega$ which is not appropriate to convert a galvanometer into an ammeter. How much resistance should be connected in series with the coil of galvanometer so that the range of ammeter is $8A?$

Answer 1

$G=16\Omega$

$lg=20mA$

$l=8A$

$S=?$

$S=\left(\frac{lg}{l-lg}\right)G$

$=\frac{20}{8000-20}\times 16$

$=\frac{20\times 16}{7980}$

$=40.1\times {10}^{-3}\Omega$

$=0.04\Omega$