The maximum power dissipated by external resistance R by a cell of an external omf E and internal resistance r is `E^(2)//4R` which is obtained for

Options

A. `Rltr`
B. `Rltr`
C. `R=r`
D. any value of R.

Answer

To solve the problem, we need to understand the concept of maximum power transfer theorem in electrical circuits. The theorem states that maximum power is transferred to the load (external resistance R) when the load resistance is equal to the source's internal resistance (r).

### Step-by-Step Solution:

1. **Understanding the Power Dissipation Formula**:
The power (P) dissipated across an external resistance (R) connected to a cell with an EMF (E) and internal resistance (r) can be expressed as:
\[ P = \frac{E^2}{(R + r)^2} \cdot R \]

2. **Finding Maximum Power Condition**:
To find the condition for maximum power, we need to differentiate the power equation with respect to R and set the derivative to zero. However, we can use the maximum power transfer theorem directly, which states that maximum power is transferred when:
\[ R = r \]

3. **Substituting the Condition into the Power Formula**:
If we substitute \( R = r \) into the power formula, we can find the maximum power:
\[ P_{max} = \frac{E^2}{(R + r)^2} \cdot R \]
Substituting \( R = r \):
\[ P_{max} = \frac{E^2}{(r + r)^2} \cdot r = \frac{E^2}{(2r)^2} \cdot r = \frac{E^2}{4r} \]

4. **Conclusion**:
Therefore, the maximum power dissipated by the external resistance R is given by:
\[ P_{max} = \frac{E^2}{4R} \]
This shows that the maximum power is achieved when the external resistance \( R \) is equal to the internal resistance \( r \).

### Final Answer:
The maximum power dissipated by external resistance R by a cell of external EMF E and internal resistance r is \( \frac{E^2}{4R} \), which is obtained when \( R = r \).