A galvanometer with its coil resistance `25 Omega` requires a current of 1 mA for its full deflection. In order to construct an ammeter to read up to a current of 2A, the approximate value of the shunt resistance should be :

Answer

To solve the problem of finding the approximate value of the shunt resistance needed to convert a galvanometer into an ammeter, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of the galvanometer, \( R_g = 25 \, \Omega \)
- Full-scale deflection current of the galvanometer, \( I_g = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \)
- Desired maximum current for the ammeter, \( I = 2 \, \text{A} \)

2. **Understand the Circuit Configuration:**
- The galvanometer is connected in parallel with a shunt resistor \( R_s \).
- The total current \( I \) entering the parallel combination splits into two parts: the current through the galvanometer \( I_g \) and the current through the shunt resistor \( I_s \).
- Therefore, we can express the relationship as:
\[ I = I_g + I_s \]

3. **Voltage Across the Galvanometer:**
- The voltage across the galvanometer can be expressed using Ohm's law:
\[ V_{AB} = I_g \cdot R_g \]

4. **Voltage Across the Shunt Resistor:**
- The voltage across the shunt resistor is also equal to \( V_{AB} \):
\[ V_{AB} = I_s \cdot R_s \]

5. **Set the Voltages Equal:**
- Since both resistors are in parallel, we can set the voltages equal to each other:
\[ I_g \cdot R_g = I_s \cdot R_s \]

6. **Express \( I_s \) in terms of \( I \) and \( I_g \):**
- From the relationship \( I = I_g + I_s \), we can express \( I_s \) as:
\[ I_s = I - I_g \]

7. **Substitute \( I_s \) into the Voltage Equation:**
- Substitute \( I_s \) into the voltage equation:
\[ I_g \cdot R_g = (I - I_g) \cdot R_s \]

8. **Rearranging to Find \( R_s \):**
- Rearranging gives:
\[ R_s = \frac{I_g \cdot R_g}{I - I_g} \]

9. **Substituting Known Values:**
- Substitute \( I_g = 1 \times 10^{-3} \, \text{A} \), \( R_g = 25 \, \Omega \), and \( I = 2 \, \text{A} \):
\[ R_s = \frac{(1 \times 10^{-3}) \cdot 25}{2 - 1 \times 10^{-3}} \]

10. **Calculating \( R_s \):**
- Calculate the denominator:
\[ 2 - 1 \times 10^{-3} \approx 2 \, \text{A} \]
- Thus:
\[ R_s = \frac{25 \times 10^{-3}}{2} = \frac{25}{2000} = 0.0125 \, \Omega \]

11. **Final Answer:**
- Therefore, the approximate value of the shunt resistance \( R_s \) is:
\[ R_s \approx 0.0125 \, \Omega \]