Correct Answer - Option 2 : 30.5 mA
Concept:
In a half-wave rectifier,
The average value of load current, \({I_{avg}} = \frac{{{V_m}}}{{\pi \left( {R + {R_L}} \right)}}\)
The RMS value of load current, \({I_{rms}} = \frac{{{V_m}}}{{2\left( {R + {R_L}} \right)}}\)
Vm is the peak value of supply voltage
R is the internal resistance
RL is the load resistance
Calculation:
Given that, supply voltage = 50 sin ωt
Peal voltage (Vm) = 50 V
Internal resistance (R) = 20 Ω
Load resistance (RL) = 800 Ω
The RMS value of load current,
\({I_{rms}} = \frac{{50}}{{2\left( {20 + 800} \right)}} = 30.5\;mA\)
