Two cells, each of e.m.f. E and internal resistance r are connected in parallel between the resistance R . The maximum energy given to the resistor will be, only when-

Options

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

Explanation

To solve the problem of finding the condition under which the maximum energy is given to the resistor when two cells are connected in parallel, we can follow these steps:

### Step-by-step Solution:

1. **Understanding the Circuit Configuration**:
- We have two cells, each with an electromotive force (e.m.f.) \( E \) and internal resistance \( r \), connected in parallel to an external resistor \( R \).

2. **Finding the Equivalent Internal Resistance**:
- When two identical resistances are connected in parallel, the equivalent resistance \( r_{eq} \) of the internal resistances is given by:
\[ r_{eq} = \frac{r}{2} \]

3. **Applying Kirchhoff's Law**:
- The total voltage across the circuit is equal to the e.m.f. of the cells, which is \( E \). The total resistance in the circuit is the sum of the external resistance \( R \) and the equivalent internal resistance \( r_{eq} \):
\[ R_{total} = R + r_{eq} = R + \frac{r}{2} \]

4. **Finding the Current in the Circuit**:
- Using Ohm's law, the current \( I \) flowing through the circuit can be expressed as:
\[ I = \frac{E}{R + \frac{r}{2}} \]

5. **Condition for Maximum Power Transfer**:
- The maximum power transfer theorem states that maximum power (and hence maximum energy) is transferred to the load when the load resistance \( R \) is equal to the internal resistance of the source. In this case, we set:
\[ R = r_{eq} = \frac{r}{2} \]

6. **Conclusion**:
- Therefore, for maximum energy to be given to the resistor \( R \), the condition must be:
\[ R = \frac{r}{2} \]

### Final Answer:
The maximum energy given to the resistor will be when \( R = \frac{r}{2} \).

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