Resistance of Moving Coil Galvanometer with Shunt and Series Resistor

Let $G$ be the resistance of the galvanometer coil and $I_g$ be the full-scale deflection current of the galvanometer.

Case 1: Galvanometer shunted to act as an ammeter When the galvanometer is connected in parallel (shunted) with a resistance $S = 2\ \Omega$, it gives a full-scale deflection for a total current $I = 500\ \text{mA} = 0.5\ \text{A}$.

The potential difference across the galvanometer is equal to the potential difference across the shunt resistance: $I_g G = (I - I_g) S$

Substitute the given values $S = 2\ \Omega$ and $I = 0.5\ \text{A}$: $I_g G = (0.5 - I_g) \times 2$ $I_g G = 1 - 2 I_g$ $I_g (G + 2) = 1 \implies I_g = \frac{1}{G + 2} \quad \text{--- (1)}$

Case 2: Galvanometer connected with a series resistance to act as a voltmeter When a resistance $R = 470\ \Omega$ is connected in series with the galvanometer, it gives full-scale deflection for an applied potential difference $V = 10\ \text{V}$.

The total voltage across the combination is given by: $V = I_g (G + R)$

Substitute the given values $V = 10\ \text{V}$ and $R = 470\ \Omega$: $10 = I_g (G + 470) \implies I_g = \frac{10}{G + 470} \quad \text{--- (2)}$

Solving for $G$: Equating the expressions for $I_g$ from equations (1) and (2): $\frac{1}{G + 2} = \frac{10}{G + 470}$

Cross-multiplying: $G + 470 = 10(G + 2)$ $G + 470 = 10G + 20$ $9G = 450$ $G = 50\ \Omega$

Thus, the value of resistance of the galvanometer coil is $50\ \Omega$.