1. Given data

• Galvanometer resistance $= G$
• Full scale deflection current $= I_g$

It is converted into:

• an ammeter of range $0$ to $I_0$ by connecting shunt resistance $R_A$
• a voltmeter of range $0$ to $V$, where $V = GI_0$, by connecting series resistance $R_V$

We must find $R_A R_V$ and $\dfrac{R_A}{R_V}$.

2. Find shunt resistance $R_A$ for ammeter conversion

For full scale ammeter reading $I_0$, galvanometer carries only $I_g$, and the remaining current goes through shunt: $I_s = I_0 - I_g$

Since galvanometer and shunt are in parallel, potential drop across both is same: $I_g G = (I_0 - I_g)R_A$

So, $R_A = \frac{I_g G}{I_0 - I_g}$

3. Find series resistance $R_V$ for voltmeter conversion

For voltmeter of range $V$, at full scale current through galvanometer is $I_g$.

Hence, $V = I_g(G + R_V)$

Given: $V = GI_0$

So, $GI_0 = I_g(G + R_V)$

$G + R_V = \frac{GI_0}{I_g}$

$R_V = \frac{GI_0}{I_g} - G$

$R_V = G\left(\frac{I_0}{I_g} - 1\right)$

$R_V = G\left(\frac{I_0 - I_g}{I_g}\right)$

4. Calculate product $R_A R_V$

$R_A R_V = \left(\frac{I_g G}{I_0 - I_g}\right)\left(G\frac{I_0 - I_g}{I_g}\right)$

Cancelling $I_g$ and $(I_0-I_g)$,

$R_A R_V = G^2$

5. Calculate ratio $\dfrac{R_A}{R_V}$

$\frac{R_A}{R_V} = \frac{\frac{I_g G}{I_0 - I_g}}{G\frac{I_0 - I_g}{I_g}}$

Cancelling $G$,

$\frac{R_A}{R_V} = \frac{I_g}{I_0 - I_g}\cdot\frac{I_g}{I_0 - I_g}$

$\frac{R_A}{R_V} = \left(\frac{I_g}{I_0 - I_g}\right)^2$

6. Compare with options

We obtained:

• $R_A R_V = G^2$
• $\frac{R_A}{R_V} = \left(\frac{I_g}{I_0 - I_g}\right)^2$

This matches Option D.

7. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So, the stored answer is correct.