A cell of emf 'E' and internal resistance 'r' is connected to a variable external resistor 'R'. The graph which gives the terminal voltage of cell 'V' with respect to R is :

Options

A.
B.
C.
D.

Explanation

To solve the problem of determining the relationship between the terminal voltage \( V \) of a cell and the external resistance \( R \), we can follow these steps:

### Step 1: Understand the relationship between terminal voltage, EMF, and internal resistance
The terminal voltage \( V \) of a cell can be expressed as:
\[ V = E - I \cdot r \]
where:
- \( E \) is the electromotive force (EMF) of the cell,
- \( I \) is the current flowing through the circuit,
- \( r \) is the internal resistance of the cell.

### Step 2: Express current \( I \) in terms of \( R \)
Using Ohm's law, the current \( I \) can be expressed as:
\[ I = \frac{V}{R + r} \]
Substituting this into the equation for \( V \):
\[ V = E - \left(\frac{V}{R + r}\right) \cdot r \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ V + \frac{Vr}{R + r} = E \]
Factoring out \( V \):
\[ V \left(1 + \frac{r}{R + r}\right) = E \]
This simplifies to:
\[ V \left(\frac{R + r + r}{R + r}\right) = E \]
\[ V \left(\frac{R + 2r}{R + r}\right) = E \]

### Step 4: Solve for terminal voltage \( V \)
Now, we can solve for \( V \):
\[ V = \frac{E(R + r)}{R + 2r} \]

### Step 5: Analyze the behavior of \( V \) as \( R \) changes
- When \( R = 0 \):
\[ V = \frac{E(0 + r)}{0 + 2r} = \frac{E}{2} \]
- When \( R \to \infty \):
\[ V \to E \]

### Step 6: Graph the relationship
1. When \( R = 0 \), \( V = 0 \).
2. When \( R \) increases, \( V \) increases and approaches \( E \) as \( R \) approaches infinity.
3. The graph of \( V \) versus \( R \) starts at \( 0 \) and asymptotically approaches \( E \).

### Final Result:
The graph of terminal voltage \( V \) against external resistance \( R \) is a curve that starts at \( 0 \) when \( R = 0 \) and approaches \( E \) as \( R \) becomes very large.

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