A galvanometer having a coil resistance of `100 omega` gives a full scale deflection , when a current of `1 mA` is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of `10A` , is :

Options

A. `2Omega`
B. `0.1Omega`
C. `3Omega`
D. `0.01 Omega`

Answer

To solve the problem, we need to determine the value of the shunt resistance (Rs) required to convert a galvanometer into an ammeter that can measure up to 10 A while still allowing the galvanometer to show full-scale deflection at 1 mA.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Resistance of galvanometer (Rg) = 100 Ω
- Full-scale deflection current of galvanometer (Ig) = 1 mA = 0.001 A
- Total current to be measured (I) = 10 A

2. **Understanding the Circuit:**
- When the total current (I) of 10 A flows through the circuit, only 1 mA (Ig) will pass through the galvanometer. The remaining current will flow through the shunt resistance (Rs).

3. **Calculate the Current through the Shunt Resistance:**
- The current through the shunt resistance (Is) can be calculated as:
\[ Is = I - Ig = 10 A - 0.001 A = 9.999 A \approx 10 A \]

4. **Using Ohm's Law:**
- The potential difference across the galvanometer (Vg) when it shows full-scale deflection can be calculated using Ohm's Law:
\[ Vg = Ig \times Rg = 0.001 A \times 100 \, \Omega = 0.1 V \]

5. **Setting Up the Equation for the Shunt Resistance:**
- The potential difference across the shunt resistance (Vs) is the same as that across the galvanometer:
\[ Vs = Is \times Rs \]
- Since Vs = Vg, we can write:
\[ 0.1 V = 10 A \times Rs \]

6. **Solving for Rs:**
- Rearranging the equation gives:
\[ Rs = \frac{0.1 V}{10 A} = 0.01 \, \Omega \]

### Final Answer:
The value of the shunt resistance (Rs) required to convert the galvanometer into an ammeter is **0.01 Ω**.