The deflection in a moving coil galvanometer is reduced to half, when it is shunted with a resistance `40 Omega`. The resistance of the galvanometer is

Options

A. `20 Omega`
B. `40 Omega`
C. `60 Omega`
D. `80 Omega`

Explanation

To solve the problem, we need to find the resistance of the galvanometer (R_g) when the deflection is reduced to half due to the shunt resistance (R_s) of 40 ohms. Here’s a step-by-step solution:

### Step 1: Understand the relationship between galvanometer current and shunt current
When a shunt resistance is added to a galvanometer, the total current (I) splits into two parts: the current through the galvanometer (I_g) and the current through the shunt (I_s). The relationship can be expressed as:
\[ I = I_g + I_s \]

### Step 2: Establish the condition of deflection
According to the problem, the deflection is reduced to half when the shunt is connected. This means:
\[ I_g' = \frac{I_g}{2} \]
Where \( I_g' \) is the new galvanometer current after the shunt is connected.

### Step 3: Relate the currents
From the relationship established in Step 1, we can write:
\[ I = I_g' + I_s \]
Substituting \( I_g' \):
\[ I = \frac{I_g}{2} + I_s \]

### Step 4: Express shunt current in terms of galvanometer current
The shunt current can also be expressed in terms of the shunt resistance and the voltage across the galvanometer:
\[ I_s = \frac{V}{R_s} \]
Where \( V \) is the voltage across the galvanometer and \( R_s \) is the shunt resistance. The voltage across the galvanometer can be expressed as:
\[ V = I_g \cdot R_g \]
Thus:
\[ I_s = \frac{I_g \cdot R_g}{R_s} \]

### Step 5: Substitute and simplify
Substituting \( I_s \) back into the equation from Step 3:
\[ I = \frac{I_g}{2} + \frac{I_g \cdot R_g}{R_s} \]

### Step 6: Solve for R_g
Now, we can rearrange the equation to isolate \( R_g \):
1. Multiply through by \( R_s \):
\[ I \cdot R_s = \frac{I_g \cdot R_s}{2} + I_g \cdot R_g \]
2. Rearranging gives:
\[ I_g \cdot R_g = I \cdot R_s - \frac{I_g \cdot R_s}{2} \]
3. Factor out \( I_g \):
\[ R_g = \frac{I \cdot R_s - \frac{I_g \cdot R_s}{2}}{I_g} \]

### Step 7: Substitute known values
We know \( R_s = 40 \Omega \) and the deflection is reduced to half, which implies:
\[ I_g' = \frac{I_g}{2} \]
This leads us to:
\[ R_g = R_s \]
Thus, substituting \( R_s = 40 \Omega \):
\[ R_g = 40 \Omega \]

### Conclusion
The resistance of the galvanometer is \( 40 \Omega \).