Question

The maximum current in a galvanometer can be $10mA$ Its resistance is $10\Omega .$ To convert it into an ammeter of 1 A, what resistance should be connected in parallel with galvanometer $\left(\text{in}{10}^{-1}\Omega \right)?$

Answer 1

$I_G=10mA,R_G=10\Omega$

$R(I-I_G)=I_G\times R_G$

$R(1-0.001)=0.001\times 10$

$R=\frac{0.001}{0.999}$

$R=0.1\Omega$