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

Solution

$\begin{array}{cc}n=50 \mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}\mathrm{s}& A=2\times {10}^{-4} m^2\end{array}$
$\begin{array}{cc}B=0.02T& K={10}^{-4}\end{array}$
$\begin{array}{cc}{Q}_m=0.2 rad& R_g=50\mathrm{\Omega }\end{array}$
$\begin{array}{cc}{I}_A=0-1.0A& \tau =MB=C\theta , M=NIA \end{array}$
$B{I}_{g}NA=C\theta$
$0.02\times I_g\times 50\times 2\times {10}^{-4}={10}^{-4}\times 0.2$
$I_g=0.1A$
For galvanometer, resistance is to be connected to ammeter in shunt.

$I_g\times R_g=\left(I-I_g\right)S$
$0.1\times 50=\left(1-0.1\right)S$
$S=\frac{50}{9}=5.55$

Hence, the correct answer is 5.55.