In a ammeter `0.2%` of main current passes through the galvanometer. If resistance of galvanometer is `G`, the resistance of ammeter will be

Options

A. `(1)/(499)G`
B. `(499)/(500)G`
C. `(1)/(500)G`
D. `(500)/(499)G`

Answer

To find the resistance of the ammeter when 0.2% of the main current passes through the galvanometer, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Current Distribution**:
- Let the total current in the circuit be \( I \).
- According to the problem, \( 0.2\% \) of the main current passes through the galvanometer.
- Therefore, the current through the galvanometer \( I_g = 0.002I \).
- The remaining current that passes through the shunt resistance \( I_s = I - I_g = I - 0.002I = 0.998I \).

2. **Applying Ohm's Law**:
- The potential drop across the galvanometer and the shunt resistance must be the same when they are connected in parallel.
- The potential drop across the galvanometer can be expressed as:
\[ V_g = I_g \times G = 0.002I \times G \]
- The potential drop across the shunt resistance can be expressed as:
\[ V_s = I_s \times S = 0.998I \times S \]

3. **Setting the Potential Drops Equal**:
- Since both potential drops are equal, we can set up the equation:
\[ 0.002I \times G = 0.998I \times S \]
- We can cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ 0.002G = 0.998S \]

4. **Solving for Shunt Resistance \( S \)**:
- Rearranging the equation gives:
\[ S = \frac{0.002G}{0.998} \]
- Simplifying this expression:
\[ S \approx \frac{G}{499} \]
- This approximation holds because \( 0.002/0.998 \) is approximately \( 1/499 \).

5. **Finding the Total Resistance of the Ammeter**:
- The total resistance of the ammeter \( R_a \) is the combination of the shunt resistance \( S \) and the galvanometer resistance \( G \) in parallel:
\[ R_a = \frac{G \cdot S}{G + S} \]
- Substituting \( S \):
\[ R_a = \frac{G \cdot \left(\frac{G}{499}\right)}{G + \frac{G}{499}} \]
- Simplifying this expression:
\[ R_a = \frac{G^2 / 499}{G(1 + \frac{1}{499})} = \frac{G^2 / 499}{G \cdot \frac{500}{499}} = \frac{G}{500} \]

### Final Result:
The resistance of the ammeter is:
\[ R_a \approx \frac{G}{499} \]