A galvanometer of resistance `22.8Omega` measures 1 A. How much shut 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 needed for a galvanometer to measure a higher current, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Resistance of the galvanometer, \( R_g = 22.8 \, \Omega \)
- Maximum current the galvanometer can measure, \( I_g = 1 \, A \)
- Total current to be measured, \( I_t = 20 \, A \)

2. **Calculate Current through the Shunt**:
- The current that will pass through the shunt, \( I_s \), can be calculated as:
\[ I_s = I_t - I_g = 20 \, A - 1 \, A = 19 \, A \]

3. **Apply the Voltage Equality for Parallel Components**:
- Since the galvanometer and the shunt are in parallel, the voltage across both must be the same:
\[ V_g = V_s \]
- The voltage across the galvanometer can be expressed as:
\[ V_g = I_g \times R_g = 1 \, A \times 22.8 \, \Omega = 22.8 \, V \]
- The voltage across the shunt can be expressed as:
\[ V_s = I_s \times R_s \]
- Where \( R_s \) is the shunt resistance we need to find.

4. **Set the Voltages Equal**:
- Since \( V_g = V_s \):
\[ 22.8 \, V = 19 \, A \times R_s \]

5. **Solve for Shunt Resistance**:
- Rearranging the equation to find \( R_s \):
\[ R_s = \frac{22.8 \, V}{19 \, A} \approx 1.2 \, \Omega \]

### Final Answer:
The required shunt resistance is approximately \( R_s \approx 1.2 \, \Omega \).

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