A galvanometer of resistance `22.8Omega` measures 1 A. How much shunt should be used , so that it can be used to measure 20 A ?

Options

A. `1 Omega`
B. `2Omega`
C. `1.2 Omega`
D. `2.2 Omega`

Answer

To solve the problem of determining the shunt resistance required for a galvanometer to measure a higher current, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Resistance of galvanometer (G) = 22.8 Ω
- Current through galvanometer (Ig) = 1 A
- Total current to be measured (I) = 20 A

2. **Calculate Current through Shunt**:
- The current passing through the shunt (Is) can be calculated as:
\[ I_s = I - I_g = 20 \, \text{A} - 1 \, \text{A} = 19 \, \text{A} \]

3. **Use Ohm's Law**:
- Since the galvanometer and the shunt are in parallel, the potential difference across both is the same. According to Ohm's Law:
\[ V = I_g \cdot G = I_s \cdot S \]
- Where S is the shunt resistance.

4. **Set Up the Equation**:
- From the above relationship, we can write:
\[ I_g \cdot G = I_s \cdot S \]
- Substituting the known values:
\[ 1 \, \text{A} \cdot 22.8 \, \Omega = 19 \, \text{A} \cdot S \]

5. **Solve for Shunt Resistance (S)**:
- Rearranging the equation to solve for S:
\[ S = \frac{I_g \cdot G}{I_s} = \frac{1 \, \text{A} \cdot 22.8 \, \Omega}{19 \, \text{A}} \]
- Calculating the value:
\[ S = \frac{22.8}{19} \approx 1.2 \, \Omega \]

6. **Conclusion**:
- The required shunt resistance to allow the galvanometer to measure 20 A is approximately **1.2 Ω**.