A galvanometer of resistance `R_(G)` is to be converted into an ammeter, with the help of a shunt of resistance R. If the ratio of the heat dissipated through galvanometer and shunt is 3:4, then

Options

A. `R = (3)/(4)R_(G)`
B. `(4)/(3)R_(G)=R`
C. `(9R)/(16) = R_(G)`
D. `(16R)/(9) = R_(G)`

Answer

To solve the problem, we need to find the relationship between the resistance of the galvanometer \( R_G \) and the shunt resistance \( R \) given that the ratio of heat dissipated through the galvanometer and the shunt is \( 3:4 \).

### Step-by-Step Solution:

1. **Understanding the Heat Dissipation Formula**:
The heat dissipated in a resistor can be expressed using the formula:
\[ H = \frac{V^2}{R} \cdot t \]
where \( H \) is the heat dissipated, \( V \) is the potential difference across the resistor, \( R \) is the resistance, and \( t \) is the time.

2. **Heat Dissipated in the Galvanometer**:
For the galvanometer, the heat dissipated \( H_G \) can be written as:
\[ H_G = \frac{V^2}{R_G} \cdot t \]

3. **Heat Dissipated in the Shunt**:
For the shunt, the heat dissipated \( H_S \) can be written as:
\[ H_S = \frac{V^2}{R} \cdot t \]

4. **Setting Up the Ratio**:
According to the problem, the ratio of the heat dissipated through the galvanometer to that through the shunt is given as:
\[ \frac{H_G}{H_S} = \frac{3}{4} \]

5. **Substituting the Heat Expressions**:
Substitute the expressions for \( H_G \) and \( H_S \) into the ratio:
\[ \frac{\frac{V^2}{R_G} \cdot t}{\frac{V^2}{R} \cdot t} = \frac{3}{4} \]

6. **Simplifying the Ratio**:
The \( V^2 \) and \( t \) terms cancel out:
\[ \frac{1}{R_G} \cdot R = \frac{3}{4} \]

7. **Rearranging the Equation**:
Rearranging gives us:
\[ R = \frac{3}{4} R_G \]

8. **Finding the Relationship**:
To express \( R_G \) in terms of \( R \):
\[ R_G = \frac{4}{3} R \]

### Conclusion:
Thus, the relationship between the shunt resistance \( R \) and the galvanometer resistance \( R_G \) is:
\[ R_G = \frac{4}{3} R \]