A cell which has an emf `1.5` V is connection series with an external resistance of `10Omega`. If the potential difference across the cell is `1.25` V, then the internal resistance of the cell is `("in" Omega)`

Options

A. 2
B. `0.25`
C. `1.5`
D. `0.3`

Answer

To find the internal resistance of the cell, we can use the following formula:

\[ r = \frac{E}{V} - 1 \times R \]

Where:
- \( r \) = internal resistance of the cell
- \( E \) = emf of the cell
- \( V \) = potential difference across the cell
- \( R \) = external resistance

### Step-by-Step Solution:

1. **Identify the given values:**
- Emf of the cell, \( E = 1.5 \, \text{V} \)
- Potential difference across the cell, \( V = 1.25 \, \text{V} \)
- External resistance, \( R = 10 \, \Omega \)

2. **Substitute the values into the formula:**
\[ r = \frac{E}{V} - 1 \times R \]
\[ r = \frac{1.5}{1.25} - 1 \times 10 \]

3. **Calculate \( \frac{E}{V} \):**
\[ \frac{1.5}{1.25} = 1.2 \]

4. **Subtract 1 from the result:**
\[ 1.2 - 1 = 0.2 \]

5. **Multiply by the external resistance \( R \):**
\[ r = 0.2 \times 10 = 2 \, \Omega \]

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

### Final Answer:
The internal resistance of the cell is \( 2 \, \Omega \).

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