In an ammeter, 4% of the total current is passing through the galvanomter. If the shunt resistance is `5 Omega`, then the resistance is` ` Omega`, then the resistance of the galvanometer will be

Options

A. `30 Omega`
B. `60 Omega`
C. `120 Omega`
D. `240 Omega`

Explanation

To find the resistance of the galvanometer in the ammeter, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Current Distribution**:
- Given that 4% of the total current (I) is passing through the galvanometer, we can denote the current through the galvanometer as \( I_G = 0.04I \).
- Consequently, the current through the shunt resistor (S) will be \( I_S = I - I_G = I - 0.04I = 0.96I \).

2. **Define the Ratio of Currents**:
- The ratio \( N \) of the total current to the galvanometer current is given by:
\[ N = \frac{I}{I_G} = \frac{I}{0.04I} = \frac{1}{0.04} = 25 \]

3. **Use the Formula for Shunt Resistance**:
- The formula relating the shunt resistance (S), galvanometer resistance (G), and the current ratio (N) is:
\[ S = \frac{G}{N - 1} \]
- Rearranging this to find the galvanometer resistance (G):
\[ G = S \times (N - 1) \]

4. **Substituting the Known Values**:
- We know \( S = 5 \, \Omega \) and \( N = 25 \):
\[ G = 5 \times (25 - 1) = 5 \times 24 = 120 \, \Omega \]

5. **Conclusion**:
- The resistance of the galvanometer is \( G = 120 \, \Omega \).

### Final Answer:
The resistance of the galvanometer is **120 ohms**.

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