A cell having an emf E and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by

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Explanation

To solve the problem, we need to analyze the relationship between the potential difference \( V \) across the external resistance \( R \) and the resistance \( R \) itself when connected to a cell with an emf \( E \) and internal resistance \( r \).

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a cell with an emf \( E \) and internal resistance \( r \).
- The cell is connected to an external variable resistance \( R \).

2. **Applying Ohm's Law**:
- The total voltage provided by the cell is \( E \).
- The total resistance in the circuit is \( R + r \).
- The current \( I \) flowing through the circuit can be expressed as:
\[ I = \frac{E}{R + r} \]

3. **Finding the Potential Difference \( V \)**:
- The potential difference \( V \) across the external resistance \( R \) can be calculated using Ohm's Law:
\[ V = I \cdot R = \left( \frac{E}{R + r} \right) R \]
- Simplifying this gives:
\[ V = \frac{E R}{R + r} \]

4. **Analyzing the Behavior of \( V \) as \( R \) Increases**:
- As \( R \) increases from 0 to infinity:
- When \( R = 0 \):
\[ V = \frac{E \cdot 0}{0 + r} = 0 \]
- When \( R \) approaches infinity:
\[ V = \frac{E \cdot R}{R + r} \approx E \quad \text{(since } R \gg r\text{)} \]
- Therefore, \( V \) starts from 0 and approaches \( E \) as \( R \) increases.

5. **Graphical Representation**:
- The graph of \( V \) against \( R \) will start at the origin (0,0) and will asymptotically approach \( E \).
- This indicates that the graph is not a straight line but a curve that rises and flattens out as \( R \) increases.

6. **Conclusion**:
- The plot of potential difference \( V \) across the external resistance \( R \) is a curve that starts from the origin and approaches the emf \( E \) asymptotically.