The deflection of galvanometer falls from 60 to 30, when 12 `Omega` shunt is connected across it. The galvanometer resistance is

Options

A. `12 Omega`
B. `36 Omega`
C. `24 Omega`
D. `30 Omega`

Answer

To find the resistance of the galvanometer (let's denote it as \( R_g \)), we can use the information given about the deflection of the galvanometer before and after connecting the shunt resistor.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The initial deflection of the galvanometer is 60 divisions.
- After connecting a 12 Ω shunt resistor, the deflection falls to 30 divisions.
- This indicates that the current through the galvanometer has decreased due to the shunt.

2. **Setting Up the Relationship**:
- Let \( I_g \) be the current through the galvanometer when the deflection is 60 divisions.
- Let \( I_s \) be the current through the shunt when the deflection is 30 divisions.
- The total current \( I \) flowing through the circuit is the sum of the currents through the galvanometer and the shunt:
\[ I = I_g + I_s \]

3. **Using the Deflection Ratios**:
- The deflection is directly proportional to the current through the galvanometer. Thus, we can write:
\[ \frac{I_g}{I_s} = \frac{60}{30} = 2 \]
- This implies that \( I_g = 2I_s \).

4. **Current through the Shunt**:
- Since the shunt resistor is 12 Ω, we can express the current through the shunt in terms of the voltage across it. The voltage across the galvanometer and the shunt is the same.
- The voltage across the shunt can be expressed as:
\[ V = I_s \cdot 12 \]

5. **Voltage across the Galvanometer**:
- The voltage across the galvanometer can also be expressed in terms of its resistance \( R_g \) and the current \( I_g \):
\[ V = I_g \cdot R_g \]

6. **Equating the Voltages**:
- Since both expressions represent the same voltage, we can set them equal to each other:
\[ I_g \cdot R_g = I_s \cdot 12 \]

7. **Substituting for \( I_g \)**:
- From step 3, we have \( I_g = 2I_s \). Substituting this into the voltage equation gives:
\[ 2I_s \cdot R_g = I_s \cdot 12 \]

8. **Solving for \( R_g \)**:
- We can divide both sides by \( I_s \) (assuming \( I_s \neq 0 \)):
\[ 2R_g = 12 \]
- Therefore, solving for \( R_g \):
\[ R_g = \frac{12}{2} = 6 \, \Omega \]

### Final Answer:
The resistance of the galvanometer is \( 6 \, \Omega \).