A single phase full bridge diode rectifier is required to be designed for AC-DC rectification without transformer and energized from AC input of 230V 50Hz supply. without a filter capacitor. The load resistor is approximately 100 Ohm. The average value of the load current would be:Select one:a. 5.4 Ab. 3.25 Ac. 2.0 Ad. 3.8 A
Question
A single phase full bridge diode rectifier is required to be designed for AC-DC rectification without transformer and energized from AC input of 230V 50Hz supply. without a filter capacitor. The load resistor is approximately 100 Ohm. The average value of the load current would be:Select one:a. 5.4 Ab. 3.25 Ac. 2.0 Ad. 3.8 A
Solution
The average value of the load current can be calculated using Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R).
In this case, the voltage is 230V and the resistance is 100 Ohms.
So, I = V/R = 230V/100 Ohms = 2.3A
However, this is the RMS value of the current. The average value of the current for a full wave rectifier without a filter capacitor is given by (2*sqrt(2)Vpeak)/(piR).
The peak voltage (Vpeak) for a sinusoidal waveform is given by Vrmssqrt(2), so Vpeak = 230Vsqrt(2) = 325V approximately.
Substituting these values into the formula gives:
Iavg = (2*sqrt(2)325V)/(pi100 Ohms) = 2.9A approximately.
So, the closest answer to this value is 3.25A. Therefore, the correct answer is b. 3.25 A.
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