A milliammeter of range `10 mA` has a coli of resistance `1 Omega`. To use it as an ammeter of range `1 A`, the required shunt must have a resistance of

Options

A. `(1)/(101) Omega`
B. `(1)/(100) Omega`
C. `(1)/(99) Omega`
D. `(1)/(9) Omega`

Explanation

To find the required shunt resistance for converting a milliammeter of range 10 mA into an ammeter of range 1 A, we can follow these steps:

### Step 1: Understand the Circuit Configuration
We need to connect a shunt resistor (S) in parallel with the galvanometer (G) to convert it into an ammeter. The galvanometer has a maximum current (IG) of 10 mA and a resistance (RG) of 1 Ω.

### Step 2: Define the Variables
- Maximum current through the galvanometer (IG) = 10 mA = 10 × 10^(-3) A
- Total current (I) = 1 A
- Resistance of the galvanometer (RG) = 1 Ω
- Current through the shunt resistor (IS) = I - IG

### Step 3: Calculate the Current through the Shunt Resistor
Using the values defined:
\[ IS = I - IG = 1 A - 10 \times 10^{-3} A = 1 A - 0.01 A = 0.99 A \]

### Step 4: Apply the Voltage Equality Condition
The potential difference across the galvanometer and the shunt resistor must be equal. Therefore, we can write:
\[ IG \cdot RG = IS \cdot S \]

### Step 5: Rearrange the Equation to Solve for S
Substituting the known values into the equation:
\[ 10 \times 10^{-3} \cdot 1 = 0.99 \cdot S \]
\[ 10 \times 10^{-3} = 0.99 S \]

Now, rearranging for S gives:
\[ S = \frac{10 \times 10^{-3}}{0.99} \]

### Step 6: Calculate the Value of S
Calculating the value:
\[ S = \frac{10^{-2}}{0.99} \approx 0.0101 \, \text{Ω} \]

### Step 7: Final Result
Thus, the required shunt resistance (S) is approximately:
\[ S \approx 0.0101 \, \text{Ω} \]