A half wave rectifier is used to supply 100 volts dc to a load of 800 ohm.the diode resistance is 200 ohm.What is the ac voltage required?
(A) $392.5 V$
(B) $280 V$
(C) $\dfrac{392.5}{\sqrt{2}} V$
(D) $\dfrac{196.25}{\sqrt{2}} V$

Hint
The half wave rectifier only passes the one half of the input sine wave (either positive or negative) and rejects the other half. The output of the half wave rectifier is pulsating DC. The ripple in the output waveform can be reduced using the filter.

Complete step by step answer
Given that, output dc voltage , Vdc = 50 V
Load resistance RL = 800 ohm
Input resistance Ri = 200 ohm
According to the ohm’s law,
${V_{dc}} = {I_{dc}} \times {R_L}$
V= DC voltage
I = current flow in DC
R = load resistance
Now, we know ${I_{dc}} = \dfrac{{{I_0}}}{\pi }$
So, ${V_{dc}} = \dfrac{{{I_0}}}{\pi } \times {R_L}$
Or, ${V_{dc}} = \dfrac{{{R_L}}}{\pi } \times \left( {\dfrac{{{V_0}}}{{{R_i} + {R_L}}}} \right)$
Or, ${V_0} = \dfrac{{{V_{dc}} \times \pi ({R_i} + {R_L})}}{{{R_L}}}$
$\therefore {V_0} = \dfrac{{50 \times (22/7)(200 + 800)}}{{800}}$
So, $V_0 = 196.43 V$.
So, the ac voltage required is $196.43 V$.

Note
Ohm's Law states that the current flowing in a circuit is directly proportional to the applied potential difference and inversely proportional to the resistance in the circuit. In other words by doubling the voltage across a circuit the current will also double.