When the current flowing in a galvanometer is (1/n) of the total current, the resistance of the shunt will be

Options

A. G/n
B. ` (n-1)G`
C. `G//(n-1)`
D. `G//(n^(2) -1) `

Answer

To find the resistance of the shunt when the current flowing in the galvanometer is \( \frac{1}{n} \) of the total current, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Current Relationship**:
- Let \( I \) be the total current flowing through the circuit.
- The current through the galvanometer \( I_g \) is given as \( I_g = \frac{I}{n} \).
- Therefore, the current through the shunt \( I_s \) can be expressed as:
\[ I_s = I - I_g = I - \frac{I}{n} = I \left(1 - \frac{1}{n}\right) = I \left(\frac{n-1}{n}\right) \]

2. **Apply Ohm's Law**:
- The potential difference across the galvanometer \( V_g \) can be expressed using Ohm's law:
\[ V_g = I_g \cdot G \]
- Where \( G \) is the resistance of the galvanometer.

3. **Potential Difference Across the Shunt**:
- The potential difference across the shunt \( V_s \) is the same as that across the galvanometer because they are in parallel:
\[ V_s = I_s \cdot S \]
- Where \( S \) is the resistance of the shunt.

4. **Equate the Potentials**:
- Since \( V_g = V_s \), we can write:
\[ I_g \cdot G = I_s \cdot S \]

5. **Substitute the Values**:
- Substitute \( I_g \) and \( I_s \) into the equation:
\[ \left(\frac{I}{n}\right) \cdot G = \left(I \cdot \frac{n-1}{n}\right) \cdot S \]

6. **Simplify the Equation**:
- Cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ \frac{G}{n} = \frac{(n-1)}{n} \cdot S \]

7. **Solve for the Shunt Resistance \( S \)**:
- Rearranging gives:
\[ S = \frac{G}{n-1} \]

### Final Result:
Thus, the resistance of the shunt is:
\[ S = \frac{G}{n-1} \]