A galvanometer of `100 Omega` resistance gives full scale deflection when 10 mA of current is passed. To convert it into 10 A range ammeter, the resistance of the shunt required will be

Options

A. `-10 Omega`
B. `1 Omega`
C. `0.1 Omega`
D. `0.01 Omega`

Answer

To solve the problem of converting a galvanometer into an ammeter, we need to determine the resistance of the shunt required. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Resistance of the galvanometer, \( R_g = 100 \, \Omega \)
- Full scale deflection current of the galvanometer, \( I_g = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A} \)
- Desired maximum current for the ammeter, \( I = 10 \, \text{A} \)

### Step 2: Calculate the current through the shunt
When the galvanometer is connected in parallel with the shunt resistor, the total current \( I \) will split between the galvanometer and the shunt. The current through the galvanometer is \( I_g \), and the current through the shunt \( I_s \) can be calculated as:
\[ I_s = I - I_g \]
Substituting the known values:
\[ I_s = 10 \, \text{A} - 10 \times 10^{-3} \, \text{A} = 10 \, \text{A} - 0.01 \, \text{A} = 9.99 \, \text{A} \]

### Step 3: Use the formula for shunt resistance
The relationship between the galvanometer current, the shunt current, and their resistances can be expressed as:
\[ \frac{I_g}{I_s} = \frac{R_s}{R_g} \]
Where \( R_s \) is the resistance of the shunt. Rearranging this gives:
\[ R_s = R_g \cdot \frac{I_s}{I_g} \]

### Step 4: Substitute the known values into the formula
Now substituting the values we have:
\[ R_s = 100 \, \Omega \cdot \frac{9.99 \, \text{A}}{10 \times 10^{-3} \, \text{A}} \]
Calculating the right side:
\[ R_s = 100 \, \Omega \cdot \frac{9.99}{0.01} = 100 \, \Omega \cdot 999 = 99900 \, \Omega \]

### Step 5: Conclusion
Thus, the resistance of the shunt required to convert the galvanometer into a 10 A ammeter is approximately:
\[ R_s \approx 99900 \, \Omega \]