A full wave p-n junction diode rectifier uses a load resistance of `1300 Omega `. The internal resistance of each diode is 9`Omega `. Find the efficiency of this full wave rectifier.

Options

A. `72% `
B. ` 80.64%`
C. `75% `
D. `79%`

Answer

To find the efficiency of a full wave p-n junction diode rectifier, we can use the following formula:

\[ \text{Efficiency} (\eta) = 0.812 \times \frac{R_L}{R_L + R_F} \]

Where:
- \( R_L \) is the load resistance.
- \( R_F \) is the internal resistance of the diode.

### Step-by-Step Solution:

1. **Identify the values given in the problem:**
- Load resistance, \( R_L = 1300 \, \Omega \)
- Internal resistance of each diode, \( R_F = 9 \, \Omega \)

2. **Substitute the values into the efficiency formula:**
\[ \eta = 0.812 \times \frac{1300}{1300 + 9} \]

3. **Calculate the denominator:**
\[ R_L + R_F = 1300 + 9 = 1309 \, \Omega \]

4. **Now substitute back into the equation:**
\[ \eta = 0.812 \times \frac{1300}{1309} \]

5. **Calculate the fraction:**
\[ \frac{1300}{1309} \approx 0.993 \]

6. **Now calculate the efficiency:**
\[ \eta = 0.812 \times 0.993 \approx 0.806 \]

7. **Convert the efficiency to percentage:**
\[ \eta \times 100 = 0.806 \times 100 \approx 80.6\% \]

### Final Result:
The efficiency of the full wave rectifier is approximately **80.6%**.