A battery of EMF E produces currents 4 A and 3 A when connected to external resistance ` 1 Omega` and `2 Omega` respectively. The internal resistance of the battery is

Options

A. `0.5 Omega`
B. `2 Omega`
C. `1.5 Omega`
D. `1 Omega`

Answer

To find the internal resistance of the battery, we can use the information given about the currents produced when connected to different external resistances.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- When connected to an external resistance of \( R_1 = 1 \, \Omega \), the current \( I_1 = 4 \, A \).
- When connected to an external resistance of \( R_2 = 2 \, \Omega \), the current \( I_2 = 3 \, A \).

2. **Apply Ohm's Law**:
The relationship between the EMF \( E \), internal resistance \( r \), and the external resistance \( R \) is given by:
\[ I = \frac{E}{R + r} \]
where \( I \) is the current flowing through the circuit.

3. **Set Up the Equations**:
For the first case with \( R_1 = 1 \, \Omega \):
\[ 4 = \frac{E}{1 + r} \quad \text{(Equation 1)} \]
For the second case with \( R_2 = 2 \, \Omega \):
\[ 3 = \frac{E}{2 + r} \quad \text{(Equation 2)} \]

4. **Express EMF \( E \) in terms of \( r \)**:
From Equation 1:
\[ E = 4(1 + r) = 4 + 4r \quad \text{(1)} \]
From Equation 2:
\[ E = 3(2 + r) = 6 + 3r \quad \text{(2)} \]

5. **Set the Two Expressions for \( E \) Equal**:
Since both expressions equal \( E \):
\[ 4 + 4r = 6 + 3r \]

6. **Solve for \( r \)**:
Rearranging the equation:
\[ 4r - 3r = 6 - 4 \]
\[ r = 2 \, \Omega \]

7. **Conclusion**:
The internal resistance of the battery is \( r = 2 \, \Omega \).