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

Options

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Explanation

To solve the problem, we need to analyze how the potential difference \( V \) across the external resistance \( R \) changes as \( R \) is increased. We will use the relationship between the electromotive force (emf) \( E \), the internal resistance \( r \), and the external 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 a variable external resistance \( R \).

2. **Current in the Circuit**:
- 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. **Potential Difference Across External Resistance**:
- The potential difference \( V \) across the external resistance \( R \) can be calculated using Ohm's law:
\[ V = I \cdot R \]
- Substituting the expression for \( I \):
\[ V = \left(\frac{E}{R + r}\right) R \]
- This simplifies to:
\[ V = \frac{E \cdot R}{R + r} \]

4. **Analyzing the Behavior of \( V \)**:
- As \( R \) increases, we need to analyze the behavior of \( V \):
- When \( R = 0 \):
\[ V = \frac{E \cdot 0}{0 + r} = 0 \]
- When \( R \) approaches infinity:
\[ V = \frac{E \cdot R}{R + r} \rightarrow E \quad \text{(as \( R \) becomes very large)} \]

5. **Graphical Representation**:
- The potential difference \( V \) starts at 0 when \( R = 0 \) and approaches \( E \) as \( R \) increases indefinitely.
- Therefore, the graph of \( V \) versus \( R \) will start at the origin (0,0) and asymptotically approach \( E \) as \( R \) increases.

6. **Conclusion**:
- The correct plot of potential difference \( V \) across \( R \) as \( R \) is increased is a curve that starts at the origin and approaches the value \( E \).