In half wave rectifiers, a p-n diode with internal resistance 20 is used. If the load resistance of 2K is used in the circuit, then the efficiency of this half wave rectifier is(in percentage)
(A) 80.4%
(B) 40.2%
(C) 20%
(D) 50%

Hint: As we know, rectification efficiency is defined as the ratio between the output power to the ac input power. By using the efficiency formula we can find the efficiency of half wave rectifiers.

Complete step by step answer:
The efficiency \[\eta \] of half wave rectifier is given by,
\[\eta \]\[ = \dfrac{{0.406{R_L}}}{{{r_f} + {R_L}}}\]
\[\eta = \dfrac{{0.406}}{{1 + \dfrac{{{r_f}}}{{{R_L}}}}}\]
Here, ${R_L}$ is the resistance of the load resistor and ${r_f}$ is the resistance of the diode in forward biased condition.
So, efficiency is-\[\eta = \left( {\dfrac{{0.406}}{{1 + \dfrac{{{r_f}}}{{{R_L}}}}}} \right) \times 100\]
Here ${r_f}$ =\[20\Omega \]
${R_L}$ =\[2K\Omega \]
Now,
\[\Rightarrow \eta = \dfrac{{0.406}}{{\left( {1 + \dfrac{{20}}{{2000}}} \right)}} \times 100\]
\[\Rightarrow \eta = \dfrac{{0.406}}{{\left( {1 + 0.01} \right)}} \times 100\]
\[\Rightarrow \eta = \dfrac{{0.406}}{{\left( {1.01} \right)}} \times 100\]
\[\Rightarrow\eta = 0.4019 \times 100\]
\[\Rightarrow \eta = 40.2\% \]

So, the efficiency of the half wave rectifier is \[ = 40.2\% \], And the final answer is B.

Note: A rectifier is a device that converts alternating current (AC ) to direct current (DC). A half wave rectifier uses one diode whereas a full wave rectifier uses multiple diodes.
In a half wave rectifier, one half of each AC input cycle is rectified. It is defined as a type of rectifier that allows only one half-cycle of AC voltage waveform to pass and blocking the other half cycle. It used to convert AC voltage to DC voltage. The efficiency of half wave rectifiers is \[40.2\% \] whereas efficiency of the full wave rectifier is \[81.2\% \] .
Additionally there are different types of rectifiers- single phase, three phase rectifiers, half wave, full wave rectifiers, bridge rectifiers etc.