A galvanometer of resistance `25Omega` measures `10^(-3)A` . shunt required to increase range up tow 2 A is

Options

A. `12.5 mOmega`
B. `0.125 mOmega`
C. `0.125 Omega`
D. `1.25 mOmega`

Answer

To solve the problem of finding the shunt resistance required to increase the range of a galvanometer, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance of the galvanometer, \( G = 25 \, \Omega \)
- Current through the galvanometer, \( I_g = 10^{-3} \, A \)
- Total current for the new range, \( I = 2 \, A \)

2. **Understand the Circuit Configuration:**
- The galvanometer is connected in parallel with a shunt resistor \( S \).
- The current through the galvanometer is \( I_g \) and the current through the shunt is \( I - I_g \).

3. **Apply the Voltage Equality in Parallel:**
- Since the galvanometer and shunt are in parallel, the voltage across both must be the same:
\[ I_g \cdot G = (I - I_g) \cdot S \]

4. **Substitute the Known Values:**
- Substitute \( I_g = 10^{-3} \, A \), \( G = 25 \, \Omega \), and \( I = 2 \, A \):
\[ 10^{-3} \cdot 25 = (2 - 10^{-3}) \cdot S \]

5. **Simplify the Equation:**
- Calculate the left side:
\[ 25 \cdot 10^{-3} = 0.025 \]
- The right side becomes:
\[ (2 - 0.001) \cdot S = 1.999 \cdot S \]
- Therefore, we have:
\[ 0.025 = 1.999 \cdot S \]

6. **Solve for the Shunt Resistance \( S \):**
- Rearranging gives:
\[ S = \frac{0.025}{1.999} \]
- Approximating \( 1.999 \) as \( 2 \):
\[ S \approx \frac{0.025}{2} = 0.0125 \, \Omega \]

7. **Convert to Milliohms:**
- Since \( 0.0125 \, \Omega = 12.5 \, m\Omega \).

### Final Answer:
The shunt resistance required to increase the range up to 2 A is approximately \( 12.5 \, m\Omega \).