The sensitivity of a galvanometer is 60 divisions/Amp. When a shunt is used, its sensitivity becomes 10 divisions/amp. If galvanometer is of resistance `20Omega`, · the value of shunt used is :

Options

A. `4Omega`
B. `5Omega`
C. `20Omega`
D. `2Omega`

Answer

To solve the problem step by step, we will use the given information about the galvanometer and the shunt resistance.

### Step 1: Understand the given data
- Sensitivity of the galvanometer without shunt (S₁) = 60 divisions/Amp
- Sensitivity of the galvanometer with shunt (S₂) = 10 divisions/Amp
- Resistance of the galvanometer (G) = 20 Ω

### Step 2: Relate the sensitivities to the currents
The sensitivity of a galvanometer can be expressed as:
\[ S = \frac{I_g}{I} \]
Where:
- \( I_g \) = current through the galvanometer
- \( I \) = total current through the circuit

From the sensitivities given:
\[ \frac{I_g}{I} = \frac{S_2}{S_1} = \frac{10}{60} = \frac{1}{6} \]

### Step 3: Set up the equation using the shunt formula
When a shunt (S) is used, the relationship can be expressed as:
\[ \frac{I_g}{I} = \frac{S}{S + G} \]
Substituting \( G = 20 \, \Omega \):
\[ \frac{1}{6} = \frac{S}{S + 20} \]

### Step 4: Cross-multiply to solve for S
Cross-multiplying gives:
\[ S + 20 = 6S \]

### Step 5: Rearrange the equation
Rearranging the equation:
\[ 20 = 6S - S \]
\[ 20 = 5S \]

### Step 6: Solve for S
Dividing both sides by 5:
\[ S = \frac{20}{5} = 4 \, \Omega \]

### Conclusion
The value of the shunt resistance used is \( 4 \, \Omega \).