The deflection in a moving coil galvanometer falls from 50 divisions when a shunt of `2 Omega ` is applied. What is the resistance ( in ohms ) of the galvanometer ?

Answer

To solve the problem, we need to determine the resistance of the galvanometer (G) when a shunt resistance (S) of 2 ohms is applied, and the deflection changes from 50 divisions to 10 divisions.

### Step-by-Step Solution:

1. **Identify the Initial and Final Conditions:**
- Initial deflection = 50 divisions
- Final deflection with shunt = 10 divisions
- This means the current through the galvanometer (Ig) when the deflection is at 10 divisions is less than the current when it was at 50 divisions.

2. **Define the Current Values:**
- Let the full-scale deflection current be represented as I.
- Therefore, the current through the galvanometer when it was at 50 divisions is 50I.
- The current through the galvanometer when it is at 10 divisions is 10I, which we denote as Ig.

3. **Using the Current Division Rule:**
- The total current (I) splits between the galvanometer and the shunt.
- The current through the shunt (Is) can be expressed as:
\[ Is = I - Ig = I - 10I = 40I \]

4. **Applying the Relationship for Parallel Resistors:**
- The voltage across the galvanometer (Vg) and the shunt (Vs) must be equal since they are in parallel:
\[ Vg = Ig \cdot G = Is \cdot S \]
- Substituting the expressions for Ig and Is:
\[ 10I \cdot G = 40I \cdot 2 \]

5. **Simplifying the Equation:**
- Cancel I from both sides (assuming I ≠ 0):
\[ 10G = 80 \]
- Solving for G gives:
\[ G = \frac{80}{10} = 8 \, \text{ohms} \]

6. **Conclusion:**
- The resistance of the galvanometer is **8 ohms**.