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`

Explanation

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

### Step 1: Define the main current
Let the main current be \( I \).

### Step 2: Calculate the current through the galvanometer
Since 0.2% of the main current passes through the galvanometer, we can express this as:
\[ I_g = 0.002 \times I = \frac{2}{1000} \times I = \frac{I}{500} \]

### Step 3: Calculate the current through the shunt
The current that passes through the shunt (the rest of the current) can be calculated as:
\[ I_s = I - I_g = I - \frac{I}{500} = \frac{500I - I}{500} = \frac{499I}{500} \]

### Step 4: Set up the relationship between the galvanometer and shunt resistances
Let the resistance of the galvanometer be \( G \) and the resistance of the shunt be \( S \). The potential difference across the galvanometer and the shunt must be equal:
\[ I_g \cdot G = I_s \cdot S \]
Substituting the values of \( I_g \) and \( I_s \):
\[ \left(\frac{I}{500}\right) \cdot G = \left(\frac{499I}{500}\right) \cdot S \]

### Step 5: Simplify the equation
We can cancel \( \frac{I}{500} \) from both sides (assuming \( I \neq 0 \)):
\[ G = \frac{499}{500} S \]

### Step 6: Solve for the shunt resistance \( S \)
Rearranging the equation gives:
\[ S = \frac{500}{499} G \]

### Step 7: Calculate the total resistance of the ammeter
The total resistance \( R \) of the ammeter, which consists of the galvanometer and the shunt in parallel, is given by:
\[ R = \frac{G \cdot S}{G + S} \]
Substituting the value of \( S \):
\[ R = \frac{G \cdot \left(\frac{500}{499} G\right)}{G + \left(\frac{500}{499} G\right)} \]

### Step 8: Simplify the expression for \( R \)
Calculating the denominator:
\[ G + \frac{500}{499} G = G \left(1 + \frac{500}{499}\right) = G \left(\frac{499 + 500}{499}\right) = G \left(\frac{999}{499}\right) \]
Now substituting back into the equation for \( R \):
\[ R = \frac{G \cdot \frac{500}{499} G}{G \cdot \frac{999}{499}} = \frac{500 G^2 / 499}{999 G / 499} = \frac{500 G}{999} \]

### Final Result
Thus, the resistance of the ammeter is:
\[ R = \frac{G}{500} \]