If E is the emf of a cell of internal resistance r and external resistance R, then potential difference across R is given as

Options

A. `V=E//(R+r)`
B. `V = E`
C. `V=E//(1+r//R)`
D. `V=E//(1+R//r)`

Answer

To solve the problem, we need to find the potential difference (V) across the external resistance (R) in a circuit that includes a cell with an electromotive force (E) and an internal resistance (r).

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a cell with an EMF (E) and an internal resistance (r).
- The external resistance is denoted as (R).
- The circuit can be represented as a series circuit where the internal resistance (r) and the external resistance (R) are connected in series.

2. **Applying Ohm's Law**:
- The total resistance (R_total) in the circuit is the sum of internal resistance and external resistance:
\[ R_{total} = r + R \]

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

4. **Finding the Potential Difference (V) across External Resistance (R)**:
- The potential difference (V) across the external resistance (R) can be calculated using Ohm's law:
\[ V = I \cdot R \]
- Substituting the expression for current (I) we found earlier:
\[ V = \left(\frac{E}{r + R}\right) \cdot R \]
- Therefore, we can express the potential difference across the external resistance (R) as:
\[ V = \frac{E \cdot R}{r + R} \]

### Final Answer:
The potential difference across the external resistance (R) is given by:
\[ V = \frac{E \cdot R}{r + R} \]

---