The e.m.f. of a cell is E volts and internal resistance is r ohm. The resistance in exteral circuit is also r ohm. The p.d across the cell will be

Options

A. `E//2`
B. `2E`
C. `4E`
D. `E//4`

Answer

To find the potential difference (p.d.) across the cell, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
- The e.m.f. of the cell is \( E \) volts.
- The internal resistance of the cell is \( r \) ohms.
- The external resistance in the circuit is also \( r \) ohms.

2. **Understand the Circuit Configuration**:
- The internal resistance of the cell and the external resistance are in series. Therefore, the total resistance in the circuit is the sum of the internal and external resistances:
\[ R_{\text{total}} = r + r = 2r \]

3. **Apply Ohm's Law**:
- According to Ohm's Law, the current \( I \) flowing through the circuit can be calculated using the formula:
\[ I = \frac{E}{R_{\text{total}}} = \frac{E}{2r} \]

4. **Calculate the Potential Drop Across Each Resistance**:
- The potential drop across the internal resistance \( r \) is given by:
\[ V_{\text{internal}} = I \cdot r = \left(\frac{E}{2r}\right) \cdot r = \frac{E}{2} \]
- The potential drop across the external resistance \( r \) is the same:
\[ V_{\text{external}} = I \cdot r = \left(\frac{E}{2r}\right) \cdot r = \frac{E}{2} \]

5. **Determine the Potential Difference Across the Cell**:
- The potential difference across the cell is the e.m.f. minus the potential drop across the internal resistance:
\[ V_{\text{across cell}} = E - V_{\text{internal}} = E - \frac{E}{2} = \frac{E}{2} \]

### Final Answer:
The potential difference across the cell will be \( \frac{E}{2} \) volts.

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