By a cell a current of 0.9 A flows through 2 ohm resistor and `0.3 A` through 7 ohm resistor. The internal resistance of the cell is

Options

A. `0.5 Omega`
B. `1.0 Omega`
C. `1.2 Omega`
D. `2.0 Omega`

Answer

To find the internal resistance of the cell, we can use the information provided about the current flowing through two different resistors connected to the same cell.

### Step-by-Step Solution:

1. **Identify the given values:**
- Current through 2 ohm resistor, \( I_1 = 0.9 \, A \)
- Current through 7 ohm resistor, \( I_2 = 0.3 \, A \)
- Resistance of first resistor, \( R_1 = 2 \, \Omega \)
- Resistance of second resistor, \( R_2 = 7 \, \Omega \)

2. **Use Ohm's Law and the concept of internal resistance:**
The voltage across the cell can be expressed in terms of the current and resistance:
\[ V = I_1(R_1 + r) \quad \text{(for the first resistor)} \]
\[ V = I_2(R_2 + r) \quad \text{(for the second resistor)} \]
Here, \( r \) is the internal resistance of the cell.

3. **Set up the equations:**
From the first case:
\[ V = 0.9(2 + r) \quad \text{(1)} \]
From the second case:
\[ V = 0.3(7 + r) \quad \text{(2)} \]

4. **Equate the two expressions for V:**
Since both expressions equal \( V \), we can set them equal to each other:
\[ 0.9(2 + r) = 0.3(7 + r) \]

5. **Expand both sides:**
\[ 1.8 + 0.9r = 2.1 + 0.3r \]

6. **Rearrange the equation:**
Move all terms involving \( r \) to one side and constant terms to the other:
\[ 0.9r - 0.3r = 2.1 - 1.8 \]
\[ 0.6r = 0.3 \]

7. **Solve for \( r \):**
\[ r = \frac{0.3}{0.6} = 0.5 \, \Omega \]

### Final Answer:
The internal resistance of the cell is \( r = 0.5 \, \Omega \).

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