If the galvanometer current is 10 mA, resistance of the galvanometer is `40 Omega` and shunt of `2 Omega` is connected to the galvanometer, the maximum current which can be measured by this ammeter is

Options

A. 0.21A
B. 2.1A
C. 210A
D. 21A

Explanation

To solve the problem, we need to find the maximum current that can be measured by the ammeter when a shunt resistor is connected to the galvanometer. We will use the following information provided in the question:

- Current in the galvanometer (IG) = 10 mA = 10 × 10^-3 A
- Resistance of the galvanometer (RG) = 40 Ω
- Resistance of the shunt (S) = 2 Ω

### Step-by-Step Solution:

1. **Identify the formula for maximum current (I)**:
The formula for the maximum current that can be measured by the ammeter when a shunt is connected is:
\[ I = \frac{S + R_G}{S} \times I_G \]

2. **Substitute the known values into the formula**:
Substitute the values of S, RG, and IG into the formula:
\[ I = \frac{2 + 40}{2} \times (10 \times 10^{-3}) \]

3. **Calculate the numerator**:
Calculate the sum in the numerator:
\[ 2 + 40 = 42 \]

4. **Substitute back into the formula**:
Now substitute this back into the equation:
\[ I = \frac{42}{2} \times (10 \times 10^{-3}) \]

5. **Calculate the division**:
Calculate the division:
\[ \frac{42}{2} = 21 \]

6. **Multiply by the galvanometer current**:
Now multiply by the galvanometer current:
\[ I = 21 \times (10 \times 10^{-3}) = 21 \times 0.01 = 0.21 \text{ A} \]

7. **Final result**:
Therefore, the maximum current that can be measured by the ammeter is:
\[ I = 0.21 \text{ A} \]

### Summary:
The maximum current that can be measured by the ammeter is **0.21 A**.