A cell which has an emf `1.5` V is connectedin 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 steps:

### Step 1: Identify the given values
- EMF of the cell (E) = 1.5 V
- Potential difference across the cell (V) = 1.25 V
- External resistance (R) = 10 Ω

### Step 2: Use the formula for internal resistance
The internal resistance (r) of the cell can be calculated using the formula:
\[ r = \frac{E - V}{I} \]
Where:
- \( I \) is the current flowing through the circuit.

### Step 3: Calculate the current (I)
Since the cell and the external resistance are in series, the current can be calculated using Ohm's law:
\[ I = \frac{V}{R} \]
However, we need to find the current using the EMF and the total resistance (which includes both the external resistance and the internal resistance).

The total voltage across the external resistance is given by:
\[ V = I \cdot R \]
Thus, we can rearrange this to find the current:
\[ I = \frac{V}{R} = \frac{1.25}{10} = 0.125 \, \text{A} \]

### Step 4: Substitute the values into the internal resistance formula
Now we can substitute the values into the internal resistance formula:
\[ r = \frac{E - V}{I} \]
Substituting the known values:
\[ r = \frac{1.5 - 1.25}{0.125} \]

### Step 5: Calculate the internal resistance
Now, calculate the numerator:
\[ 1.5 - 1.25 = 0.25 \]
Then divide by the current:
\[ r = \frac{0.25}{0.125} = 2 \, \Omega \]

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

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