The purpose of a rectifier to produce an output that is DC, or the purpose may...

 

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The purpose of a rectifier to produce an output that is DC, or the purpose may be to produce a voltage or current wave-form that has a specified DC component. In practice, the half-wave rectifier is used most often in low-power applications because the average current in the supply will not be zero, and nonzero average current may cause problems in transformer performance. While practical applications of this circuit are limited. a) The half wave rectifier of has vs (t) = 125sin(cot) V with supply frequency of 60 Hz and a load resistance R = 2092. Determine 1. The average load current (2 marks) ii. The rms load current (2 marks) iii. The power,P absorbed by the load (2 marks) iv. The apparent power, S supplied by the source. (2 marks) v. The power factor of the circuit. (2 marks) b) The half-wave rectifier has a 230 Vrms source at 50 Hz, R = 450 22 and C= 120uF with the a= 0.746 rad. i. Determine an expression for the output voltage (4 marks) ii. Find the peak to peak voltage variation on the output (2 marks) iii. Determine an expression for the capacitor current (4 marks) iv. Calculate the peak diode current (2 marks) v. Determine the value of capacitor, C such that AVo is 2 percent of Vm. (3 marks) The purpose of a rectifier to produce an output that is DC, or the purpose may be to produce a voltage or current wave-form that has a specified DC component. In practice, the half-wave rectifier is used most often in low-power applications because the average current in the supply will not be zero, and nonzero average current may cause problems in transformer performance. While practical applications of this circuit are limited. a) The half wave rectifier of has vs (t) = 125sin(cot) V with supply frequency of 60 Hz and a load resistance R = 2092. Determine 1. The average load current (2 marks) ii. The rms load current (2 marks) iii. The power,P absorbed by the load (2 marks) iv. The apparent power, S supplied by the source. (2 marks) v. The power factor of the circuit. (2 marks) b) The half-wave rectifier has a 230 Vrms source at 50 Hz, R = 450 22 and C= 120uF with the a= 0.746 rad. i. Determine an expression for the output voltage (4 marks) ii. Find the peak to peak voltage variation on the output (2 marks) iii. Determine an expression for the capacitor current (4 marks) iv. Calculate the peak diode current (2 marks) v. Determine the value of capacitor, C such that AVo is 2 percent of Vm. (3 marks)

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a Given Vst 125sint V frequency f 60 Hz load resistance R 2092 and rectifier circuit is a halfwave rectifier i The average load current is given by Iavg 1T0 to T Idt dt where T 1f 160 sec is the time ... View the full answer