Conversion of Moving Coil Galvanometer into Ammeter with Shunt Resistance

To find the value of the required shunt resistance $S$, we first determine the full-scale deflection current $I_g$ of the moving coil galvanometer.

When a current $I_g$ flows through the galvanometer coil, the magnetic torque experienced by the coil is given by: $\tau_m = N I_g A B$

where:

• $N = 50$ is the number of turns,
• $A = 2 \times 10^{-4} \text{ m}^2$ is the area of each turn,
• $B = 0.02 \text{ T}$ is the magnetic field strength.

This magnetic torque is balanced by the restoring torque supplied by the suspension spring: $\tau_r = k \theta$

where:

• $k = 10^{-4} \text{ N m rad}^{-1}$ is the torsional constant of the spring,
• $\theta = 0.2 \text{ rad}$ is the angle of full-scale deflection.

At equilibrium, the magnetic torque equals the restoring torque: $N I_g A B = k \theta$

Solving for $I_g$: $I_g = \frac{k \theta}{N A B}$

Substitute the given values: $I_g = \frac{10^{-4} \times 0.2}{50 \times (2 \times 10^{-4}) \times 0.02}$

$I_g = \frac{2 \times 10^{-5}}{100 \times 10^{-4} \times 0.02} = \frac{2 \times 10^{-5}}{2 \times 10^{-4}} = 0.1 \text{ A}$

To convert the galvanometer into an ammeter measuring up to a maximum current of $I = 1.0 \text{ A}$, a shunt resistance $S$ must be connected in parallel with the galvanometer resistance $G = 50\ \Omega$.

Since the galvanometer and shunt are connected in parallel, the potential difference across both components is identical: $I_g G = (I - I_g) S$

Solving for $S$: $S = \frac{I_g G}{I - I_g}$

Substitute the values $I = 1.0 \text{ A}$, $I_g = 0.1 \text{ A}$, and $G = 50\ \Omega$: $S = \frac{0.1 \times 50}{1.0 - 0.1} = \frac{5}{0.9} = \frac{50}{9}\ \Omega \approx 5.56\ \Omega$

The required value of the shunt resistance is $\frac{50}{9}\ \Omega$ or approximately $5.56\ \Omega$ (or $5.55$).