A galvanometer has a current range of 15mA and voltage range of 750mv. To convert this galvanometer into an ammeter of range 25A, the sunt resistance required is nearly

Options

A. `0.2Omega`
B. `0.02Omega`
C. `0.03Omega`
D. `0.5Omega`

Answer

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

### Step 1: Understand the given values
- Maximum current through the galvanometer, \( I_G = 15 \, \text{mA} = 15 \times 10^{-3} \, \text{A} \)
- Maximum voltage across the galvanometer, \( V_G = 750 \, \text{mV} = 750 \times 10^{-3} \, \text{V} \)
- Desired maximum current for the ammeter, \( I = 25 \, \text{A} \)

### Step 2: Calculate the resistance of the galvanometer
The resistance \( R_G \) of the galvanometer can be calculated using Ohm's law:
\[ R_G = \frac{V_G}{I_G} \]
Substituting the values:
\[ R_G = \frac{750 \times 10^{-3}}{15 \times 10^{-3}} = \frac{750}{15} = 50 \, \Omega \]

### Step 3: Use the shunt resistance formula
To find the shunt resistance \( R_S \), we use the relation between the galvanometer current, shunt current, and total current:
\[ I_G = \frac{R_S}{R_S + R_G} \times I \]
Rearranging gives:
\[ I_G (R_S + R_G) = R_S \times I \]
This can be rearranged to:
\[ I_G R_S + I_G R_G = I R_S \]
\[ I_G R_G = R_S (I - I_G) \]
Thus, we can express \( R_S \) as:
\[ R_S = \frac{I_G R_G}{I - I_G} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ R_S = \frac{(15 \times 10^{-3}) \times 50}{25 - 15 \, \text{A}} \]
Calculating the denominator:
\[ R_S = \frac{(15 \times 10^{-3}) \times 50}{10} \]
Calculating the numerator:
\[ R_S = \frac{750 \times 10^{-3}}{10} = 75 \times 10^{-3} = 0.075 \, \Omega \]

### Step 5: Finalize the answer
Thus, the required shunt resistance to convert the galvanometer into an ammeter of range 25A is:
\[ R_S \approx 0.075 \, \Omega \]