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

The use of a rectifier is to change alternating current and voltage into direct current and voltage. In a full wave rectifier, during the positive half of the input AC, the upper p-n junction diode is forward biased and the lower p-n junction diode is reverse biased. The lower p-n junction diode is forward biased and the higher p-n junction diode is reverse biased during the negative half cycle.
The relationship between the internal resistance of a diode under forward bias and the load resistance, R_L, determines the full wave rectifier's efficiency.

$$\begin{gathered} \eta =\frac{0.812R_L}{r_f+R_L} \\ \eta =\frac{0.812R_L}{1+{\displaystyle \frac{r_f}{R_L}}} \end{gathered}$$

Thus, the efficiency is represented as a percentage by,

$$\begin{gathered} \eta =\frac{0.812}{1+{\displaystyle \frac{r_f}{R_L}}}\times 100 \\ \eta =\frac{0.812}{1+{\displaystyle \frac{9}{1300}}}\times 100 \\ \eta =\frac{812\times 130}{1309} \\ \eta =80.64\% \end{gathered}$$

Hence the correct answer is 80.64%.