Determination of Parameter for Galvanometer Conversion

To find the value of $n$, we analyze the two conversion cases of the galvanometer:

1. Conversion into a Voltmeter: Let the galvanometer resistance be $G$ and its full-scale deflection current be $I_g$. Given: $I_g = 0.006\text{ A}$ When a resistance $R = 4990\ \Omega$ is connected in series with the galvanometer, it can measure a maximum voltage of $V = 30\text{ V}$.

The equation for the voltmeter is: $V = I_g(G + R)$

Substituting the given values: $30 = 0.006(G + 4990)$ $G + 4990 = \frac{30}{0.006} = 5000\ \Omega$ $G = 5000 - 4990 = 10\ \Omega$

---

2. Conversion into an Ammeter: When a shunt resistance $S$ is connected in parallel with the galvanometer, it can measure a maximum current of $I = 1.5\text{ A}$.

The potential difference across the galvanometer and the shunt resistance is the same: $I_g \cdot G = (I - I_g) \cdot S$

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

Substitute the values: $S = \frac{0.006 \times 10}{1.5 - 0.006} = \frac{0.06}{1.494}$

Multiplying the numerator and denominator by $1000$: $S = \frac{60}{1494} = \frac{10}{249}\ \Omega$

---

3. Finding the value of $n$: We are given that the shunt resistance is: $S = \frac{2n}{249}\ \Omega$

Equating the two expressions for $S$: $\frac{2n}{249} = \frac{10}{249}$ $2n = 10 \implies n = 5$