A Centre-tap transformer rectifier is required to be designed for AC-DC rectification with a transformation ratio 6:1:1 and energized from AC input of 230V 50Hz supply. with no filter capacitor. The load resistor is approximately 5.4 Ohm. The peak value of the load current would be:Select one:a. 20 Ab. None of the other given choicesc. 7 Ad. 10 A
Question
A Centre-tap transformer rectifier is required to be designed for AC-DC rectification with a transformation ratio 6:1:1 and energized from AC input of 230V 50Hz supply. with no filter capacitor. The load resistor is approximately 5.4 Ohm. The peak value of the load current would be:Select one:a. 20 Ab. None of the other given choicesc. 7 Ad. 10 A
Solution
The peak value of the load current can be calculated using Ohm's law, which states that the current is equal to the voltage divided by the resistance.
First, we need to find the peak voltage across the load resistor. The transformation ratio of the transformer is 6:1:1, which means that the voltage across the load resistor is 1/6 of the input voltage.
The peak value of an AC voltage is √2 times the RMS value. So, the peak value of the input voltage is 230V * √2 = 325.27V.
Therefore, the peak voltage across the load resistor is 325.27V / 6 = 54.21V.
Finally, we can calculate the peak value of the load current by dividing the peak voltage across the load resistor by the resistance:
I_peak = V_peak / R = 54.21V / 5.4Ω = 10.04A.
So, the closest answer is:
d. 10 A
Similar Questions
A single phase Centre-tap transformer rectifier is required to be designed for AC-DC rectification with a transformation ratio 6:1:1 and energized from AC input of 230V 50Hz supply. with no filter capacitor. The load resistor is approximately 5.4 Ohm. The minimum PIV Rating of each of the diodes is:Select one:a. 54 Vb. 108 Vc. 38 Vd. 76 V
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
Transformer utilization factor of a center tapped full wave rectifier is ___________1. A. 0.6232. B. 0.6783. C. 0.6934. D. 0.625
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 minimum PIV rating of each of the diodes would be:Select one:a. 650 Vb. 325 Vc. 200 Vd. 400 V
A half-wave rectifier circuit operates from a supply of 230 V (r.m.s.) at 50 Hz. The load consists of a 100 resistor in parallel with a capacitor of 1000 F. Neglecting the resistance of the rectifier element calculatethe mean output voltage and the r.m.s. value of the ripple voltage.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.