If a galvanometer has full scale deflection current `I_(g)` and resistance G. A shunt resistance `R_(A)` is used to convert it into an ammeter of range `I_(o)` and a resistance `R_(V)` is connected in series to convert it into a voltmeter of range `V_(0)` such that `V_(0)=I_(0)G` then `R_(A)R_(V)` and `(R_(A))/(R_(V))` respectively are:

Options

A. `R_(A)R_(V)=G^(2)((I_(0)-I_(g))/(I_(g)))` and `(R_(A))/(R_(V))=((I_(g))/(I_(0)-I_(g)))^(2)`
B. `R_(A)R_(V)=G^(2)` and `(R_(A))/(R_(v))=(I_(g))/((I_(0)-I_(g)))`
C. `R_(A)R_(v)=G^(2)` and `(R_(A))/(R_(v))=((I_(g))/(I_(0)-I_(g)))^(2)`
D. `R_(A)R_(V)=G^(2)((I_(g))/(I_(0)-I_(g)))` and `(R_(A))/(R_(V))=((I_(0)-I_(g))/(I_(g)))^(2)`

Answer

To solve the problem, we need to derive the expressions for the shunt resistance \( R_A \) and the series resistance \( R_V \) used to convert a galvanometer into an ammeter and a voltmeter, respectively.

### Step 1: Understanding the Galvanometer to Ammeter Conversion

When converting a galvanometer to an ammeter, a shunt resistance \( R_A \) is connected in parallel with the galvanometer. The full-scale deflection current of the galvanometer is \( I_G \), and the range of the ammeter is \( I_0 \).

Using the current division rule, we can write:

\[ I_0 = I_G + I_S \]

Where \( I_S \) is the current through the shunt. The voltage across the galvanometer and the shunt is the same, so we can express the relationship as:

\[ I_G \cdot G = I_S \cdot R_A \]

Substituting \( I_S = I_0 - I_G \) into the equation gives:

\[ I_G \cdot G = (I_0 - I_G) \cdot R_A \]

Rearranging this equation, we can express \( R_A \):

\[ R_A = \frac{I_G \cdot G}{I_0 - I_G} \]

### Step 2: Understanding the Galvanometer to Voltmeter Conversion

For the conversion of a galvanometer to a voltmeter, a resistance \( R_V \) is connected in series with the galvanometer. The output voltage \( V_0 \) is given by:

\[ V_0 = I_G \cdot (G + R_V) \]

According to the problem, we have:

\[ V_0 = I_0 \cdot G \]

Substituting this into the voltmeter equation gives:

\[ I_0 \cdot G = I_G \cdot (G + R_V) \]

Rearranging this equation allows us to express \( R_V \):

\[ R_V = \frac{I_0 \cdot G}{I_G} - G \]

### Step 3: Finding the Product \( R_A R_V \)

Now, we can find the product \( R_A R_V \):

\[ R_A R_V = \left( \frac{I_G \cdot G}{I_0 - I_G} \right) \left( \frac{I_0 \cdot G}{I_G} - G \right) \]

Simplifying this expression:

1. Substitute \( R_V \):
\[ R_V = \frac{I_0 G - I_G G}{I_G} = \frac{(I_0 - I_G) G}{I_G} \]

2. Now substituting \( R_V \) into the product:
\[ R_A R_V = \frac{I_G \cdot G}{I_0 - I_G} \cdot \frac{(I_0 - I_G) G}{I_G} \]

3. The \( I_G \) terms cancel out:
\[ R_A R_V = G^2 \]

### Step 4: Finding the Ratio \( \frac{R_A}{R_V} \)

Now, we find the ratio \( \frac{R_A}{R_V} \):

\[ \frac{R_A}{R_V} = \frac{\frac{I_G \cdot G}{I_0 - I_G}}{\frac{(I_0 - I_G) G}{I_G}} = \frac{I_G^2}{(I_0 - I_G)^2} \]

### Final Results

Thus, we have:

1. \( R_A R_V = G^2 \)
2. \( \frac{R_A}{R_V} = \frac{I_G^2}{(I_0 - I_G)^2} \)