Question

The resistance of galvanometer is $99\omega$. TYhe necessary current for full scale deflection is 1 MA. If a shunt of $1\omega$ is connected with galvanometer

Answer 1

$I_g.G=(I-I_g).s$
$I-I_g=\frac{I_g.G}{s}$
$I-I_g=\frac{I_g.G}{s}$
$I=\frac{I_g.r}{S}+I_g$
$I=\frac{{10}^{-3}\times 99}{1}+{10}^{-3}$
$I=100\times {10}^{-3}=0.1Ampere$