A cell of emf E is connected across a resistance R. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell must be -

Options

A. `(2(E-V)V)/(R )`
B. `(2(E-V)R)/(E )`
C. `((E-V)R)/(V)`
D. `(E-V)R`

Explanation

To solve the problem, we need to find the internal resistance \( r \) of a cell given its electromotive force (emf) \( E \), the potential difference \( V \) across the terminals, and the external resistance \( R \).

### Step-by-Step Solution:

1. **Understand the Circuit**:
- We have a cell with emf \( E \).
- The cell has an internal resistance \( r \).
- The cell is connected to an external resistance \( R \).
- The potential difference across the terminals of the cell is \( V \).

2. **Use Ohm's Law**:
- The current \( I \) flowing through the circuit can be expressed in two ways:
- From the cell's perspective:
\[ I = \frac{E}{r + R} \]
- From the external resistance's perspective:
\[ I = \frac{V}{R} \]

3. **Set the Two Expressions for Current Equal**:
- Since both expressions represent the same current \( I \), we can set them equal to each other:
\[ \frac{E}{r + R} = \frac{V}{R} \]

4. **Cross-Multiply to Eliminate the Fractions**:
- Cross-multiplying gives:
\[ E \cdot R = V \cdot (r + R) \]

5. **Distribute and Rearrange the Equation**:
- Distributing \( V \) on the right side:
\[ E \cdot R = V \cdot r + V \cdot R \]
- Rearranging gives:
\[ E \cdot R - V \cdot R = V \cdot r \]

6. **Factor Out \( R \)**:
- Factoring \( R \) from the left side:
\[ R(E - V) = V \cdot r \]

7. **Solve for the Internal Resistance \( r \)**:
- Dividing both sides by \( V \):
\[ r = \frac{R(E - V)}{V} \]

### Final Answer:
The internal resistance \( r \) of the cell is given by:
\[ r = \frac{R(E - V)}{V} \]