Question

A moving coil galvanometer has $50$ turns and each turn has an area $2\times {10}^{-4}{m}^2$. The magnetic field produced by the magnet inside the galvanometer is $0.02T$. The torsional constant of the suspension wire is ${10}^{-4}Nm-ra{d}^{-1}$. When a current flows through the galvanometer, a full-scale deflection occurs if the coil rotates by $0.2rad$. The resistance of the coil of the galvanometer is $50\Omega$. This galvanometer is to be converted into an ammeter capable of measuring current in the range $0-1.0A$. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is

Answer 1

$n=50$ turns,
$A=2\times {10}^{-4}{m}^2$
$B=0.02T$
$C={10}^{-4}$
$\theta =0.2rad$
$R_g=50\Omega$

for a galvanometer total difllectio is proptional to total current.
$\theta \propto i_g$

$T=MB=C\theta ,M=nIA$
$BINA=C\theta$ .....(1)

using above values,
$\Rightarrow 0.02\times I_g\times 50\times 2\times {10}^{-4}={10}^{-4}\times 0.210$
$I_g=0.1A$
For galvanometer, resistance is to be connected to ammeter in shunt.

$\Rightarrow I_g\times R_g=(I–I_g)S$
$\Rightarrow 0.1\times 50=(1–0.1)SS=\frac{50}{9}=5.55\Omega$