A galvanometer of resistance `40Omega` and current passing through it is `100muA` per divison. The full scale has 50 divisions. If it is converted into an ammeter of range 2A by using a shunt, then the resistance of ammeter is

Options

A. `(40)/(399)Omega`
B. `(4)/(399)Omega`
C. `0.01Omega`
D. `0.4Omega`

Answer

To solve the problem, we need to find the resistance of the ammeter when a galvanometer is converted into an ammeter using a shunt resistor. Here are the steps to derive the solution:

### Step 1: Identify the given values
- Resistance of the galvanometer, \( R_g = 40 \, \Omega \)
- Current per division of the galvanometer, \( I_g = 100 \, \mu A = 100 \times 10^{-6} A \)
- Total divisions on the galvanometer = 50
- Full scale current for the galvanometer, \( I_g = 50 \times 100 \, \mu A = 5000 \, \mu A = 5 \, mA = 0.005 \, A \)
- Maximum current for the ammeter, \( I = 2 \, A \)

### Step 2: Set up the equation for the shunt resistor
When the galvanometer is used as an ammeter, the total current \( I \) splits into two parts: the current through the galvanometer \( I_g \) and the current through the shunt resistor \( I_s \).
Thus, we have:
\[ I = I_g + I_s \]
Where:
\[ I_s = I - I_g \]
Substituting the values:
\[ I_s = 2 \, A - 0.005 \, A = 1.995 \, A \]

### Step 3: Apply the voltage equality across the galvanometer and shunt
Since the galvanometer and the shunt are in parallel, the voltage across both must be equal:
\[ I_g \cdot R_g = I_s \cdot R_s \]
Substituting the known values:
\[ 0.005 \, A \cdot 40 \, \Omega = 1.995 \, A \cdot R_s \]

### Step 4: Solve for the shunt resistance \( R_s \)
Calculating the left side:
\[ 0.005 \cdot 40 = 0.2 \, V \]
Now substituting this into the equation:
\[ 0.2 = 1.995 \cdot R_s \]
Now solving for \( R_s \):
\[ R_s = \frac{0.2}{1.995} \approx 0.10025 \, \Omega \]

### Step 5: Calculate the total resistance of the ammeter
The total resistance of the ammeter \( R_a \) is the resistance of the galvanometer in parallel with the shunt:
\[ \frac{1}{R_a} = \frac{1}{R_g} + \frac{1}{R_s} \]
Substituting the values:
\[ \frac{1}{R_a} = \frac{1}{40} + \frac{1}{0.10025} \]
Calculating the right side:
\[ \frac{1}{R_a} \approx 0.025 + 9.95 \approx 10.025 \]
Thus,
\[ R_a \approx \frac{1}{10.025} \approx 0.09975 \, \Omega \]

### Final Answer
The resistance of the ammeter is approximately \( 0.09975 \, \Omega \).