The senstivity of a galvanometer that measures current is decreased by 1/40 times by using shunt resistance of `10Omega` . Then, the value of the resistance of the galvanometer is

Options

A. `400Omega`
B. `410Omega`
C. `30Omega`
D. `390Omega`

Answer

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

### Step-by-Step Solution:

1. **Understanding Sensitivity**:
The sensitivity of a galvanometer is defined as the deflection (θ) per unit current (i). We denote the initial sensitivity as \( S_1 = \frac{\theta}{i_1} \).

2. **Sensitivity After Shunt Connection**:
When a shunt resistance (R_s) is connected, the sensitivity decreases by a factor of \( \frac{1}{40} \). Therefore, the new sensitivity \( S_2 \) can be expressed as:
\[ S_2 = \frac{S_1}{40} \]

3. **Relating Currents**:
From the sensitivity definitions, we can relate the currents:
\[ S_2 = \frac{\theta}{i_2} \implies i_2 = 40 \cdot i_1 \]

4. **Current Distribution**:
When the shunt of resistance \( R_s = 10 \, \Omega \) is connected, the total current \( i_2 \) splits between the galvanometer and the shunt. The current through the shunt is:
\[ i_s = i_2 - i_1 = 40 \cdot i_1 - i_1 = 39 \cdot i_1 \]

5. **Voltage Across the Galvanometer**:
The voltage across the galvanometer (V_g) can be expressed as:
\[ V_g = i_1 \cdot G \]
where \( G \) is the resistance of the galvanometer.

6. **Voltage Across the Shunt**:
The voltage across the shunt can also be expressed as:
\[ V_s = i_s \cdot R_s = (39 \cdot i_1) \cdot 10 = 390 \cdot i_1 \]

7. **Equating Voltages**:
Since both the galvanometer and the shunt have the same voltage across them when connected in parallel, we can equate the two expressions:
\[ i_1 \cdot G = 390 \cdot i_1 \]

8. **Solving for G**:
Dividing both sides by \( i_1 \) (assuming \( i_1 \neq 0 \)):
\[ G = 390 \, \Omega \]

### Final Answer:
The resistance of the galvanometer is \( G = 390 \, \Omega \).

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