This question was previously asked in

PGCIL DT Electrical 13 Aug 2021 Official Paper (NR I)

Option 4 : 20 A and 12.7 A

Given, e = E_{m} sin ωt = 400 sin 314t

Hence, E_{m} = 400 volts,

The device offers an ohmic resistance (R) of 10 Ω

Hence, the maximum value of current (I_{m}) is, \(\dfrac{E_m}{R}=\dfrac{400}{10} = 40\ A\) .... (1)

Since, the flow of current in one direction, while preventing the flow of current in the opposite direction, hence resultant waveform from actual waveform can be drawn as,

Since, the resultant waveform looks like the output waveform of the Half-Wave Rectifier,

**For Half Wave Rectifier,**

RMS Value = \(\dfrac{A_m}{2}\) .... (2)

Average Value = \(\dfrac{A_m}{\pi}\) .... (3)

Where, A_{m} is the maximum value of an alternating quantity,

From equation (1),

A_{m} = I_{m} = 40 A

From equations (2) and, (3),

RMS Value of current = \(\dfrac{40}{2}=20\ A\)

Average Value of current = \(\dfrac{40}{\pi}=12.7\ A\)