The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be

Options

A. `(E^(2))/(r)`
B. `(E^(2))/(2r)`
C. `(E^(2))/(3r)`
D. `(E^(2))/(4r)`

Answer

To find the maximum power dissipated in an external resistance \( R \) when connected to a cell of emf \( E \) and internal resistance \( r \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circuit Configuration**:
We have a circuit consisting of a battery with an emf \( E \) and an internal resistance \( r \). An external resistance \( R \) is connected in series with the battery.

2. **Condition for Maximum Power Dissipation**:
According to the maximum power transfer theorem, the power dissipated in the external resistance \( R \) is maximized when \( R \) is equal to the internal resistance \( r \). Therefore, we set:
\[ R = r \]

3. **Calculate the Current in the Circuit**:
The total resistance in the circuit is the sum of the internal resistance and the external resistance:
\[ R_{\text{total}} = R + r = r + r = 2r \]
The current \( I \) flowing through the circuit can be calculated using Ohm's law:
\[ I = \frac{E}{R_{\text{total}}} = \frac{E}{2r} \]

4. **Power Dissipated in the External Resistance**:
The power \( P \) dissipated in the external resistance \( R \) can be expressed as:
\[ P = I^2 R \]
Substituting the expression for \( I \):
\[ P = \left(\frac{E}{2r}\right)^2 R \]

5. **Substituting \( R \) with \( r \)**:
Since we have established that for maximum power, \( R = r \), we can substitute \( R \) in the power equation:
\[ P = \left(\frac{E}{2r}\right)^2 r \]

6. **Simplifying the Expression**:
Now, simplifying the expression:
\[ P = \frac{E^2}{4r^2} \cdot r = \frac{E^2}{4r} \]

7. **Final Result**:
Therefore, the maximum power dissipated in the external resistance \( R \) is:
\[ P_{\text{max}} = \frac{E^2}{4r} \]