In a galvanometer 5% of the total current in the circuit passes through it. If the resistance of the galvanometer is G, the shunt resistance 'S' connected to the galvanometer is

Options

A. 19
B. G/19
C. 20 G
D. G/20

Answer

To find the shunt resistance \( S \) connected to a galvanometer, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Current Distribution**:
- Let the total current in the circuit be \( I \).
- The current passing through the galvanometer \( I_G \) is given as 5% of the total current \( I \).
- Therefore, we can express \( I_G \) as:
\[ I_G = \frac{5}{100} \times I = \frac{I}{20} \]

2. **Determine the Current through the Shunt**:
- The current flowing through the shunt resistance \( S \) is the remaining current after the galvanometer:
\[ I_S = I - I_G = I - \frac{I}{20} = \frac{20I}{20} - \frac{I}{20} = \frac{19I}{20} \]

3. **Apply Kirchhoff’s Law**:
- According to Kirchhoff's law, the potential difference across the galvanometer and the shunt must be equal. Therefore:
\[ I_G \cdot G = I_S \cdot S \]
- Substituting the values of \( I_G \) and \( I_S \):
\[ \left(\frac{I}{20}\right) \cdot G = \left(\frac{19I}{20}\right) \cdot S \]

4. **Rearranging the Equation**:
- We can rearrange the equation to solve for \( S \):
\[ S = \frac{I_G \cdot G}{I_S} = \frac{\left(\frac{I}{20}\right) \cdot G}{\left(\frac{19I}{20}\right)} \]
- The \( I \) terms cancel out:
\[ S = \frac{G}{19} \]

5. **Conclusion**:
- The shunt resistance \( S \) connected to the galvanometer is:
\[ S = \frac{G}{19} \]

### Final Answer:
The shunt resistance \( S \) is \( \frac{G}{19} \).

---