A moving coil galvanometer of resistance `100Omega` is used as an ammeter using a resistance `0.1 Omega`. The maximum diflection current in the galvanometer is `100 muA`. Find the minimum current in the circuit so that the ammeter shows maximum deflection

Options

A. `100.1 mA`
B. `1000.1mA`
C. `10. 01mA`
D. `1.01mA`

Explanation

To solve the problem, we need to find the minimum current in the circuit (I) that will cause the galvanometer to show maximum deflection. We can use the relationship between the current through the galvanometer (Ig), the total current (I), and the shunt resistance (S).

### Step-by-Step Solution:

1. **Identify the given values:**
- Resistance of the galvanometer (Rg) = 100 Ω
- Shunt resistance (S) = 0.1 Ω
- Maximum deflection current in the galvanometer (Ig) = 100 μA = 100 × 10^(-6) A

2. **Use the formula for the current in the circuit:**
The relationship between the total current (I), the current through the galvanometer (Ig), and the shunt resistance (S) can be expressed as:
\[ \frac{I - I_g}{S} = \frac{I_g}{R_g} \]
This equation states that the potential difference across the shunt resistance (I - Ig) is equal to the potential difference across the galvanometer (Ig).

3. **Rearranging the equation:**
Rearranging the above equation gives:
\[ I - I_g = \frac{I_g}{R_g} \cdot S \]
Therefore,
\[ I = I_g + \frac{I_g \cdot S}{R_g} \]

4. **Substituting the values:**
Now substitute the known values into the equation:
\[ I = 100 \times 10^{-6} + \frac{100 \times 10^{-6} \cdot 0.1}{100} \]

5. **Calculating the second term:**
\[ I = 100 \times 10^{-6} + \frac{100 \times 10^{-6} \cdot 0.1}{100} = 100 \times 10^{-6} + 0.1 \times 10^{-6} \]
\[ I = 100 \times 10^{-6} + 0.1 \times 10^{-6} = 100.1 \times 10^{-6} \]

6. **Converting to milliampere:**
To convert the current into milliampere:
\[ I = 100.1 \times 10^{-6} \text{ A} = 100.1 \text{ mA} \]

### Final Answer:
The minimum current in the circuit so that the ammeter shows maximum deflection is **100.1 mA**.