Two cells , each of `emf E` and internal resistance `r`, are connected in parallel across a resistor `R`. The power delivered to the resistor is maximum if `R` is equal to

Options

A. ` r // 2`
B. `r`
C. ` 2r`
D. `0`

Explanation

To solve the problem, we need to find the value of the load resistance \( R \) that maximizes the power delivered to it when two cells, each with an emf \( E \) and internal resistance \( r \), are connected in parallel.

### Step-by-Step Solution:

1. **Understanding the Circuit Configuration**:
- We have two cells connected in parallel, each with an emf \( E \) and internal resistance \( r \).
- These cells are connected across a load resistor \( R \).

2. **Calculate the Equivalent Internal Resistance**:
- When two resistances are connected in parallel, the equivalent resistance \( R_{\text{internal}} \) can be calculated using the formula:
\[ R_{\text{internal}} = \frac{r \cdot r}{r + r} = \frac{r^2}{2r} = \frac{r}{2} \]
- Therefore, the equivalent internal resistance of the two cells in parallel is \( \frac{r}{2} \).

3. **Applying the Maximum Power Transfer Theorem**:
- According to the maximum power transfer theorem, maximum power is delivered to the load when the load resistance \( R \) is equal to the internal resistance of the source.
- Thus, we set:
\[ R = R_{\text{internal}} \]
- Substituting the value of \( R_{\text{internal}} \):
\[ R = \frac{r}{2} \]

4. **Conclusion**:
- The power delivered to the resistor \( R \) is maximum when:
\[ R = \frac{r}{2} \]

### Final Answer:
The power delivered to the resistor is maximum if \( R \) is equal to \( \frac{r}{2} \).

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