The deflection in a galnometer falls from 50 divisions to 20 divisions, when a `12Omega` shunt is applied. The galvanometer resistance is

Options

A. `18Omega`
B. `24 Omega`
C. `30Omega`
D. `36 Omega`

Explanation

To find the resistance of the galvanometer (G), we can follow these steps:

### Step 1: Understand the relationship between deflection and current
The deflection in a galvanometer is directly proportional to the current passing through it. Let the initial current (when the deflection is 50 divisions) be \( I_0 \) and the current after applying the shunt (when the deflection is 20 divisions) be \( I' \).

### Step 2: Set up the equations for the currents
From the deflection values:
- Initially, \( I_0 = 50K \) (where \( K \) is a proportionality constant)
- After applying the shunt, \( I' = 20K \)

### Step 3: Relate the currents through the galvanometer and the shunt
When a shunt resistance \( R_s \) (12Ω) is connected in parallel with the galvanometer, the current through the galvanometer \( I' \) can be expressed as:
- \( I_0 = I' + I_s \)
Where \( I_s \) is the current through the shunt.

### Step 4: Use the ratio of currents
Since the deflection changes from 50 to 20, we can set up the ratio:
\[ \frac{I_0}{I'} = \frac{50}{20} = \frac{5}{2} \]
This means:
\[ I_0 = \frac{5}{2} I' \]

### Step 5: Express the current through the shunt
From the previous step, we can express \( I_s \) as:
\[ I_s = I_0 - I' = \frac{5}{2} I' - I' = \frac{3}{2} I' \]

### Step 6: Apply Ohm's Law
The potential difference across the galvanometer and the shunt is the same:
\[ I' \cdot G = I_s \cdot R_s \]
Substituting \( I_s \) and \( R_s \):
\[ I' \cdot G = \left(\frac{3}{2} I'\right) \cdot 12 \]

### Step 7: Simplify the equation
We can cancel \( I' \) from both sides (assuming \( I' \neq 0 \)):
\[ G = \frac{3}{2} \cdot 12 \]
\[ G = 18 \, \Omega \]

### Final Answer
The resistance of the galvanometer is \( G = 18 \, \Omega \).

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