A potential difference of `0.75V` applied across a galvanometer causes a current of 15 mA to pass through it. If can be converted into ammeter of range of 25 A , the requried shunt should be

Options

A. `0.8 Omega`
B. `0.93 Omega`
C. `0.03 Omega`
D. `2.0 Omega`

Answer

To solve the problem of finding the required shunt resistance to convert a galvanometer into an ammeter, we can follow these steps:

### Step 1: Identify the given values
- Potential difference (V) across the galvanometer = 0.75 V
- Current through the galvanometer (Ig) = 15 mA = 15 × 10^(-3) A
- Desired current range of the ammeter (I) = 25 A

### Step 2: Calculate the resistance of the galvanometer (Rg)
The resistance of the galvanometer can be calculated using Ohm's law, which states that:
\[ R_g = \frac{V}{I_g} \]

Substituting the known values:
\[ R_g = \frac{0.75 \, \text{V}}{15 \times 10^{-3} \, \text{A}} \]

Calculating this gives:
\[ R_g = \frac{0.75}{0.015} = 50 \, \Omega \]

### Step 3: Use the formula for shunt resistance (S)
To convert the galvanometer into an ammeter, we use the formula:
\[ I_g = \frac{S}{S + R_g} \times I \]

Where:
- \( I_g \) = current through the galvanometer
- \( S \) = shunt resistance
- \( R_g \) = resistance of the galvanometer
- \( I \) = total current through the ammeter

Rearranging the formula to solve for S:
\[ S = \frac{I_g \cdot R_g}{I - I_g} \]

### Step 4: Substitute the known values into the formula
Substituting the values we have:
- \( I_g = 15 \times 10^{-3} \, \text{A} \)
- \( R_g = 50 \, \Omega \)
- \( I = 25 \, \text{A} \)

So:
\[ S = \frac{(15 \times 10^{-3}) \cdot 50}{25 - (15 \times 10^{-3})} \]

Calculating the denominator:
\[ 25 - 15 \times 10^{-3} \approx 25 \, \text{A} \] (since 15 mA is negligible compared to 25 A)

Now substituting back:
\[ S = \frac{(15 \times 10^{-3}) \cdot 50}{25} \]

Calculating this gives:
\[ S = \frac{0.75}{25} = 0.03 \, \Omega \]

### Final Answer
The required shunt resistance is \( S = 0.03 \, \Omega \).

---