A current of 2A flows through a `2 Omega` resistor when connected across a battery. The same battery supplies a current of 0.5A when connected across a `9Omega` resistor The internal resistance of the battery is

Options

A. `1/3 Omega`
B. `1/4 Omega`
C. `5Omega`
D. `0.5Omega`

Answer

To find the internal resistance of the battery, we can use the information provided about the currents flowing through the resistors when connected to the battery. Let's break down the solution step by step.

### Step 1: Understand the given information
We have two scenarios:
1. A current of \( I_1 = 2 \, A \) flows through a resistor \( R_1 = 2 \, \Omega \).
2. A current of \( I_2 = 0.5 \, A \) flows through a resistor \( R_2 = 9 \, \Omega \).

### Step 2: Write the equations for the EMF of the battery
Using Ohm's law and the concept of internal resistance, we can express the EMF (\( E \)) of the battery in both cases.

For the first case:
\[ E = I_1 R_1 + I_1 r \]
Substituting the values:
\[ E = 2 \times 2 + 2r = 4 + 2r \quad \text{(1)} \]

For the second case:
\[ E = I_2 R_2 + I_2 r \]
Substituting the values:
\[ E = 0.5 \times 9 + 0.5r = 4.5 + 0.5r \quad \text{(2)} \]

### Step 3: Set the two equations equal to each other
Since both expressions represent the same EMF of the battery, we can set them equal to each other:
\[ 4 + 2r = 4.5 + 0.5r \]

### Step 4: Solve for the internal resistance \( r \)
Rearranging the equation:
\[ 4 + 2r - 0.5r = 4.5 \]
\[ 4 + 1.5r = 4.5 \]
Subtracting 4 from both sides:
\[ 1.5r = 0.5 \]
Dividing both sides by 1.5:
\[ r = \frac{0.5}{1.5} = \frac{1}{3} \, \Omega \]

### Conclusion
The internal resistance of the battery is:
\[ r = \frac{1}{3} \, \Omega \]

### Final Answer
The internal resistance of the battery is \( \frac{1}{3} \, \Omega \).

---