Question

Figure 2 shows a power supply circuit comprising a transformer coupling the mains voltage on the left to the full-wave rectifier circuit on the right. It is given that the mains voltage (Vmains) is 230 V RMS, the mains frequency (fin) is 50 Hz, the transformer T1 has a ratio of 9:1, the load resistance (RL) is 1 k. and the capacitance of the capacitor (C) is 470 uF. a. Calculate the following: 1. the transformer AC output RMS voltage for V2 ii. the transformer AC output peak-to-peak voltage for V2 iii. the ripple peak-to-peak voltage (Vpk-pk) across the 1 k load resistance. You can assume that the voltage across the resistor RL is: V1 = Vpk-pk 12. b. If the load resistance is reduced to 5002, what happens to the ripple peak-to-peak voltage (Vpk-pk)? fuse chassis RI] = Figure 2 Full-wave rectifier with transformer

Answer 1

niven Full wave ractifier Umine = 230 Volt fm = 50Hz Tuon ration (trom former ratio) = 9!1 . R2 = 1kr - 4704F PHVSE Vmains We know that Tun Ration of tooms former Vacams) = 930 2556 volt Va CRMs) - 25.56 Volt (a) ©

We know that Vm 52 Vyms of Full wowe rach fier is. where Im is peak voliage - Vm = Voms: 25.56 J2 Vm= 25, 56 52 = 36.14 Volt Peak to peak voltage of g = 2Vm = 72.28 Volt I . We know about the formula of sipple peak-to beak volterge (Vr) = 2 foc for peak to peak sipple (Vr) we have to calculate IDC Also we know that DC Olp Voltage VPK PK - Vr Vpc = Vm - Vr/ Voc or V₂ = (VPK - APK)/2 Given V/2 = Vm - Vy Vm= Vr = 36.14 Volt Upeak to beak sipple - 36.14 Volt

6 Peak to peak ripple (VV) or VPK: NOK VEL-PK = The Resistanie reduced to soor the Vpk pk chonge above assumed value will not follow here Voc = Vm= V/2 = Vm - 4 pc also know that VDC = IDCRL Joche = vm- Toe C R = soon Toc [ R + efica] - Vm for 50 ? C= 470010 Tee [ 500 + Axisów uro] um Vm = .07074 amp IDC= 510-63 APP TIL Inc VPK:No O Đ7074 10 6 2x 50 x 470 2 foc Vok-pk = 1.5 osvolt Upk- Voks 1.5 1 Volt RL reduced the ripple peak to peak to Hage also reduced.