A galvanometer has a resistance of `100Omega.` potential difference of 100 mV between its terminals gives a full scale deflection. The shunt resistance required to convert it into an ammeter reading up to 5 A, is

Options

A. `0.01 Omega`
B. `0.02 Omega`
C. `0.03 Omega`
D. `0.04 Omega`

Explanation

To solve the problem of finding the shunt resistance required to convert a galvanometer into an ammeter, we can follow these steps:

### Step 1: Understand the given values
- Resistance of the galvanometer, \( R_G = 100 \, \Omega \)
- Full-scale deflection potential difference, \( V = 100 \, \text{mV} = 0.1 \, \text{V} \)
- Maximum current for the ammeter, \( I = 5 \, \text{A} \)

### Step 2: Calculate the current through the galvanometer at full-scale deflection
The current through the galvanometer when it shows full-scale deflection can be calculated using Ohm's Law:
\[ I_G = \frac{V}{R_G} \]
Substituting the known values:
\[ I_G = \frac{0.1 \, \text{V}}{100 \, \Omega} = 0.001 \, \text{A} = 1 \, \text{mA} \]

### Step 3: Use the formula for shunt resistance
The shunt resistance \( R_S \) can be calculated using the formula:
\[ R_S = \frac{I_G \cdot R_G}{I - I_G} \]
Where:
- \( I_G \) is the current through the galvanometer (1 mA),
- \( R_G \) is the resistance of the galvanometer (100 Ω),
- \( I \) is the total current (5 A).

### Step 4: Substitute the values into the formula
Substituting the values we found:
\[ R_S = \frac{0.001 \, \text{A} \cdot 100 \, \Omega}{5 \, \text{A} - 0.001 \, \text{A}} \]
Calculating the denominator:
\[ 5 \, \text{A} - 0.001 \, \text{A} = 4.999 \, \text{A} \]
Now substituting into the formula:
\[ R_S = \frac{0.1 \, \text{A} \cdot \Omega}{4.999 \, \text{A}} \approx \frac{0.1}{4.999} \, \Omega \approx 0.020004 \, \Omega \]

### Step 5: Round off the answer
The shunt resistance required is approximately:
\[ R_S \approx 0.02 \, \Omega \]

### Final Answer
The shunt resistance required to convert the galvanometer into an ammeter reading up to 5 A is approximately \( 0.02 \, \Omega \).

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