A full wave p-n diode rectifier uses a load resistance of $1300\Omega $. No filter is used. If the internal resistance of each diode is $9\Omega $, then the efficiency of this full wave rectifier is
A. $80.64\% $
B. $40.6\% $
C. $13.9\% $
D. $100\% $

Hint: Efficiency of a full wave rectifier is defined as the ratio of the output DC power to the input DC power. So find the output DC power and input DC power. Output DC power is the product of the ripple factor and the load resistance. The ripple factor of a full wave rectifier is 0.812.

Complete step by step answer:
Rectifier is a device which is used for converting alternating current/voltage into direct current/voltage. In a full wave rectifier the upper p-n junction diode is forward biased and the lower p-n junction diode is reverse biased during the positive half of the input AC. During the negative half cycle, the upper p-n junction diode is reversed biased and the lower p-n junction diode is forward biased.

If ${R_L}$ is the load resistance and ${r_f}$ is the internal resistance of a diode in forward biased condition, then the efficiency of the full wave rectifier is given by the relation
$\eta = \dfrac{{0.812{R_L}}}{{{r_f} + {R_L}}}$,
where ${R_L}$ is the load resistance, ${r_f}$ is the internal resistance of a diode in forward biased condition and 0.812 is the ripple factor of the full wave rectifier.

$ \eta = \dfrac{{0.812}}{{1 + \dfrac{{{r_f}}}{{{R_L}}}}}$
[Dividing numerator and denominator by ${R_L}$]
Hence, the efficiency in percentage is given by,
$\eta = \dfrac{{0.812}}{{1 + \dfrac{{{r_f}}}{{{R_L}}}}} \times 100$
$ \Rightarrow \eta = \dfrac{{0.812}}{{1 + \dfrac{9}{{1300}}}} \times 100\\
\Rightarrow \eta = \dfrac{{0.812}}{{1 + \dfrac{9}{{1300}}}} \times 100 \\
\therefore \eta = \dfrac{{812 \times 130}}{{1309}} = 80.64\% $

Hence, the correct option is A.

Note: Use the value of the ripple factor correctly and identify the values of the load resistance and internal resistance appropriately. As the options are in percentage, so multiply by 100 to convert the ratio into percentage.