The maximum effeciency of full wave rectifier is

Options

A. `100%`
B. `25.20%`
C. `40.2%`
D. `81.2%`

Explanation

To find the maximum efficiency of a full wave rectifier, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Efficiency Formula**:
The efficiency (η) of a rectifier is defined as the ratio of the DC output power (P_dc) to the AC input power (P_ac):
\[ \eta = \frac{P_{dc}}{P_{ac}} \]

2. **Expressing DC Output Power**:
The DC output power can be expressed in terms of the DC voltage (V_dc) and the load resistance (R_L):
\[ P_{dc} = \frac{V_{dc}^2}{R_L} \]

3. **Expressing AC Input Power**:
The AC input power can be expressed in terms of the RMS voltage (V_rms) and the load resistance (R_L):
\[ P_{ac} = \frac{V_{rms}^2}{R_L} \]

4. **Substituting the Expressions**:
Now, substituting the expressions for P_dc and P_ac into the efficiency formula:
\[ \eta = \frac{\frac{V_{dc}^2}{R_L}}{\frac{V_{rms}^2}{R_L}} = \frac{V_{dc}^2}{V_{rms}^2} \]

5. **Finding V_dc and V_rms**:
For a full wave rectifier:
- The DC voltage (V_dc) is given by:
\[ V_{dc} = \frac{2V_m}{\pi} \]
- The RMS voltage (V_rms) is given by:
\[ V_{rms} = \frac{V_m}{\sqrt{2}} \]

6. **Substituting V_dc and V_rms into Efficiency**:
Now substitute these values into the efficiency formula:
\[ \eta = \frac{\left(\frac{2V_m}{\pi}\right)^2}{\left(\frac{V_m}{\sqrt{2}}\right)^2} \]

7. **Simplifying the Expression**:
Simplifying the expression:
\[ \eta = \frac{\frac{4V_m^2}{\pi^2}}{\frac{V_m^2}{2}} = \frac{4V_m^2}{\pi^2} \cdot \frac{2}{V_m^2} = \frac{8}{\pi^2} \]

8. **Calculating the Numerical Value**:
Now we can calculate the numerical value of the efficiency:
\[ \eta \approx \frac{8}{9.87} \approx 0.812 \]

9. **Converting to Percentage**:
To express the efficiency in percentage, multiply by 100:
\[ \eta \approx 81.2\% \]

10. **Conclusion**:
Therefore, the maximum efficiency of a full wave rectifier is approximately **81.2%**.